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Figure 8-1 Voltage-divider

bias. (a) Circuit; (b) voltage
8-1 Voltage-Divider Bias
divider; (c) simplified circuit. Figure 8-1a shows the most widely used biasing circuit. Notice that the base
+VCC circuit contains a voltage divider (R1 and R2). Because of this, the circuit is
called voltage-divider bias (VDB).
+
+
R1
RC
Simplified Analysis
–
–
For troubleshooting and preliminary analysis, use the following method. In any
well-designed VDB circuit, the base current is much smaller than the current
+
+ through the voltage divider. Since the base current has a negligible effect on the
R2
RE
voltage divider, we can mentally open the connection between the voltage divider
–
–
and the base to get the equivalent circuit of Fig. 8-1b. In this circuit, the output of
the voltage divider is
(a ) R2
VBB VCC
+VCC R1 R2
Ideally, this is the base-supply voltage as shown in Fig. 8-1c.
As you can see, voltage-divider bias is really emitter bias in disguise. In
other words, Fig. 8-1c is an equivalent circuit for Fig. 8-1a. This is why VDB sets
R1
up a fixed value of emitter current, resulting in a solid Q point that is independent
of the current gain.
+VBB There is an error in this simplified approach, and we will discuss it in the
next section. The crucial point is this: In any well-designed circuit, the error in
R2 using Fig. 8-1c is very small. In other words, a designer deliberately chooses
circuit values so that Fig. 8-1a acts like Fig. 8-1c.

(b )

+VCC Conclusion
After you calculate VBB, the rest of the analysis is the same as discussed earlier
RC for emitter bias in Chap. 7. Here is a summary of the equations you can use to
analyze VDB:
R2
VBB VCC (8-1)
R1 R2
+
VBB VE VBB VBE (8-2)
RE
–
VE
IE (8-3)
RE
(c )
IC IE (8-4)
VC VCC ICRC (8-5)
VCE VC VE (8-6)
These equations are based on Ohm’s and Kirchhoff’s laws. Here are the steps in the analysis:
1. Calculate the base voltage VBB out of the voltage divider.
2. Subtract 0.7 V to get the emitter voltage (use 0.3 V for germanium).
3. Divide by the emitter resistance to get the emitter current.
4. Assume that the collector current is approximately equal to the emitter current.

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GOOD TO KNOW 5. Calculate the collector-to-ground voltage by subtracting the voltage

Since V E IC RE, Eq. (8-6) can also across the collector resistor from the collector supply voltage.
be shown as 6. Calculate the collector-emitter voltage by subtracting the emitter
VCE VCC IC RC IC RE voltage from the collector voltage.
or Since these six steps are logical, they should be easy to remember. After
VCE VCC IC (RC RE). you analyze a few VDB circuits, the process becomes automatic.

Example 8-1
What is the collector-emitter voltage in Fig. 8-2?

SOLUTION The voltage divider produces an unloaded output voltage of:

2.2 k
VBB 10 V 1.8 V
Figure 8-2 Example. 10 k 2.2 k
+10 V
Subtract 0.7 V from this to get:
VE 1.8 V 0.7 V 1.1 V
The emitter current is:
1.1 V
RC
IE 1.1 mA
R1 1k
3.6 kΩ
10 kΩ Since the collector current almost equals the emitter current, we can calculate
the collector-to-ground voltage like this:
2N3904
VC 10 V (1.1 mA)(3.6 k ) 6.04 V
R2 RE The collector-emitter voltage is:
2.2 kΩ 1 kΩ
VCE 6.04 1.1 V 4.94 V
Here is an important point: The calculations in this preliminary analy-
sis do not depend on changes in the transistor, the collector current, or the tem-
perature. This is why the Q point of this circuit is stable, almost rock-solid.

PRACTICE PROBLEM 8-1 Change the power supply voltage of Fig. 8-2 from 10 V to 15 V and solve
for VCE.

Example 8-2
Discuss the significance of Fig. 8-3, which shows a MultiSim analysis of the same circuit analyzed in the
preceding example.

SOLUTION This really drives the point home. Here we have an almost identical answer using a com-
puter to analyze the circuit. As you can see, the voltmeter reads 6.03 V (rounded to 2 places). Compare this
to 6.04 V in the preceding example, and you can see the point. A simplified analysis has produced essen-
tially the same result as a computer analysis.
You can expect this kind of close agreement whenever a VDB circuit has been well designed.
After all, the whole point of VDB is to act like emitter bias to virtually eliminate the effects of changing the
transistor, collector current, or temperature.

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Figure 8-3 MultiSim example.

PRACTICE PROBLEM 8-2 Using MultiSim, change the supply voltage of Fig. 8-3 to 15 V and
measure VCE. Compare your measured value to the answer of Practice Problem 8-1.

8-2 Accurate VDB Analysis

What is a well-designed VDB circuit? It is one in which the voltage divider appears stiff to the input resis-
tance of the base. The meaning of the last sentence needs to be discussed.

Source Resistance
Chapter 1 introduced the idea of a stiff voltage source:
Stiff voltage source: RS 0.01RL
When this condition is satisfied, the load voltage is within 1 percent of the ideal voltage. Now, let us extend
this idea to the voltage divider.

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Figure 8-4 (a) Thevenin resistance; (b) equivalent circuit; (c) input resistance of base.
+VCC

RC RC
R1

R1 R2 + +
RTH VCC RIN VCC
– –
+
R2 VBB
RE RE
–

(a) (b) (c)

What is the Thevenin resistance of the voltage divider in Fig. 8-4a? Looking back into the voltage
divider with VCC grounded, we see R1 in parallel with R2. As an equation:
RTH R1 R2
Because of this resistance, the output voltage of the voltage divider is not ideal. A more accurate
analysis includes the Thevenin resistance, as shown in Fig. 8-4b. The current through this Thevenin resis-
tance reduces the base voltage from the ideal value of VBB.

Load Resistance
How much less than ideal is the base voltage? The voltage divider has to supply the base current in Fig. 8-4b.
Put another way, the voltage divider sees a load resistance of RIN, as shown in Fig. 8-4c. For the voltage di-
vider to appear stiff to the base, the 100 1 rule:
RS 0.01RL
translates to:
R1 R2 0.01RIN (8-7)
A well-designed VDB circuit will satisfy this condition.

Stiff Voltage Divider

If the transistor of Fig. 8-4c has a current gain of 100, its collector current is 100 times greater than the base
current. This implies that the emitter current is also 100 times greater than the base current. When seen from
the base side of the transistor, the emitter resistance RE appears to be 100 times larger. As a derivation:
RIN dcRE (8-8)
Therefore, Eq. (8-7) may be written as:
Stiff voltage divider: R1 R2 0.01 dcRE (8-9)
Whenever possible, a designer selects circuit values to satisfy this 100 1 rule because it will produce an ul-
trastable Q point.

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Firm Voltage Divider

Sometimes a stiff design results in such small values of R1 and R2 that other problems arise (discussed later).
In this case, many designers compromise by using this rule:
Firm voltage divider: R1 R2 0.1 dcRE (8-10)
We call any voltage divider that satisfies this 10 1 rule a firm voltage divider. In the worst case, using a firm
voltage divider means that the collector current will be approximately 10 percent lower than the stiff value.
This is acceptable in many applications because the VDB circuit still has a reasonably stable Q point.

A Closer Approximation
If you want a more accurate value for the emitter current, you can use the following derivation:

VBB VBE
IE (8-11)
RE (R 1 R 2)/ dc

This differs from the stiff value because (R1 R2)/ dc is in the denominator. As this term approaches zero, the
equation simplifies to the stiff value.
Equation (8-11) will improve the analysis, but it is a fairly complicated formula. If you have a com-
puter and need a more accurate analysis obtained with the stiff analysis, you should use MultiSim or an
equivalent circuit simulator.

Example 8-3
Is the voltage divider of Fig. 8-5 stiff? Calculate the more accurate value of
Figure 8-5 Example.
emitter current using Eq. (8-11).
+10 V

SOLUTION Check to see whether the 100 :1 rule has been used:
Stiff voltage divider: R1 R2 0.01 dcRE

R1
RC The Thevenin resistance of the voltage divider is:
3.6 kΩ
10 kΩ
(10 k )(2.2 k )
2N3904 R1 R2 10 k 2.2 k 1.8 k
bdc = 200 10 k 2.2 k
R2
RE The input resistance of the base is:
2.2 kΩ
1 kΩ
dcRE (200)(1 k ) 200 k
and one-hundredth of this is:
0.01 dcRE 2k
Since 1.8 k is less than 2 k , the voltage divider is stiff.
With Eq. (8-11), the emitter current is
1.8 V 0.7 V 1.1 V
IE 1.09 mA
1k (1.8 k )/200 1k 9
This is extremely close to 1.1 mA, the value we get with the simplified analysis.
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The point is this: You don’t have to use Eq. (8-11) to calculate emitter current when the voltage di-
vider is stiff. Even when the voltage divider is firm, the use of Eq. (8-11) will improve the calculation for
emitter current only by at most 10 percent. Unless otherwise indicated, from now on all analysis of VDB
circuits will use the simplified method.

8-3 VDB Load Line and Q Point

Because of the stiff voltage divider in Fig. 8-6, the emitter voltage is held constant at 1.1 V in the following
discussion.

The Q Point
The Q point was calculated in Sec. 8-1. It has a collector current of 1.1 mA and a collector-emitter voltage of
4.94 V. These values are plotted to get the Q point shown in Fig. 8-6. Since voltage-divider bias is derived
from emitter bias, the Q point is virtually immune to changes in current gain. One way to move the Q point
in Fig. 8-6 is by varying the emitter resistor.
For instance, if the emitter resistance is changed to 2.2 k , the collector current decreases to:
1.1 V
IE 0.5 mA
2.2 k
The voltages change as follows:
VC 10 V (0.5 mA)(3.6 k ) 8.2 V
and
VCE 8.2 V 1.1 V 7.1 V
Therefore, the new Q point will be QL and will have coordinates of 0.5 mA and 7.1 V.
On the other hand, if we decrease the emitter resistance to 510 , the emitter current increases to:
1.1 V
IE 2.15 mA
510

Figure 8-6 Calculating the Q point.

+10 V
IC
R1 RC
10 kΩ IC(sat)
QH
3.6 kΩ
VB
Q
1.1 mA
RE
QL
R2
2.2 kΩ 1 kΩ
VCE
4.94 V VCE(cutoff)

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GOOD TO KNOW and the voltages change to:

Centering the Q point on a VC 10 V (2.15 mA)(3.6 k ) 2.26 V
transistor load line is and
important because it allows VCE 2.26 V 1.1 V 1.16 V
for the maximum ac output
voltage from the amplifier.
In this case, the Q point shifts to a new position at QH with coordinates of
Centering the Q point on the
2.15 mA and 1.16 V.
dc load line is sometimes
referred to as midpoint bias. Q Point in Middle of Load Line
VCC, R1, R2, and RC control the saturation current and the cutoff voltage. A change in any of these quantities
will change IC(sat) and/or VCE(cutoff ). Once the designer has established the values of the foregoing variables,
the emitter resistance is varied to set the Q point at any position along the load line. If RE is too large, the Q
point moves into the cutoff point. If RE is too small, the Q point moves into saturation. Some designers set the
Q point at the middle of the load line.

VDB Design Guideline

Figure 8-7 shows a VDB circuit. This circuit will be used to demonstrate a simplified design guideline to es-
tablish a stable Q point. This design technique is suitable for most circuits, but it is only a guideline. Other de-
sign techniques can be used.
Before starting the design, it is important to determine the circuit requirements or specifications. The
circuit is normally biased for VCE to be at a midpoint value with a specified collector current. You also need
to know the value of VCC and the range of dc for the transistor being used. Also, be sure the circuit will not
cause the transistor to exceed its power dissipation limits.
Start by making the emitter voltage approximately one-tenth of the supply voltage:
VE 0.1 VCC
Next, calculate the value of RE to set up the specified collector current:
VE
RE IE
Since the Q point needs to be at approximately the middle of the dc load line, about 0.5 VCC appears
across the collector-emitter terminals. The remaining 0.4 VCC appears across the collector resistor; therefore:
RC 4 RE
Next, design for a stiff voltage divider using the 100 :1 rule:
Figure 8-7 VDB design. RTH 0.01 RE
dc
+10 V
Usually, R2 is smaller than R1. Therefore, the stiff voltage divider equa-
tion can be simplified to:
+ R2 0.01 dc RE
RC
R1 V1
You may also choose to design for a firm voltage divider by using the
– 10 :1 rule:
R2 0.1 dc RE
+
In either case, use the minimum-rated dc value at the specified collector current.
R2 V2
RE Finally, calculate R1 by using proportion:
–
V1
R1 V2 R2

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Example 8-4
For the circuit shown in Fig. 8-7, design the resistor values to meet these specifications:
VCC 10 V VCE @ midpoint
IC 10 mA 2N3904’s dc 100–300
SOLUTION First, establish the emitter voltage by:
VE 0.1 VCC
VE (0.1) (10 V) = 1 V
The emitter resistor is found by:
VE
RE
IE
1V
RE 100
10 mA
The collector resistor is:
RC 4 RE
RC (4) (100 ) 400 (use 390 )
Next, choose either a stiff or firm voltage divider. A stiff value of R2 is found by:
R2 0.01 dc RE
R2 (0.01) (100) (100 ) 100
Now, the value of R1 is:
V1
R1 R
V2 2
V2 VE 0.7 V 1V 0.7 V 1.7 V
V1 VCC V2 10 V 1.7 V 8.3 V
8.3 V
R1 (100 ) 488 (use 490 )
1.7 V

PRACTICE PROBLEM 8-4 Using the given VDB design guidelines, design the VDB circuit of Fig. 8-7
to meet these specifications:
VCC 10 V VCE @ midpoint stiff voltage divider
IC 1 mA dc 70-200

8-4 Two-Supply Emitter Bias

Some electronic equipment has a power supply that produces both positive and negative supply voltages. For in-
stance, Fig. 8-8 shows a transistor circuit with two power supplies: 10 and 2 V. The negative supply forward
biases the emitter diode. The positive supply reverse biases the collector diode. This circuit is derived from emit-
ter bias. For this reason, we refer to it as two-supply emitter bias (TSEB).
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Figure 8-8 Two-supply emitter bias. Figure 8-9 Redrawn TSEB circuit.
+10 V
+
3.6 kΩ
3.6 kΩ
–
+
10 V
–
–
2.7 kΩ +
2.7 kΩ
+ 1 kΩ 1 kΩ

–
–
2V –2 V
+

Analysis
The first thing to do is to redraw the circuit as it usually appears on schematic diagrams. This means deleting
the battery symbols, as shown in Fig. 8-9. This is necessary on schematic diagrams because there usually is
no room for battery symbols on complicated diagrams. All the information is still on the diagram, except that
it is in condensed form. That is, a negative supply voltage of 2 V is applied to the bottom of the 1 k , and
a positive supply voltage of 10 V is applied to the top of the 3.6-k resistor.
GOOD TO KNOW When this type of circuit is correctly designed, the base current
will be small enough to ignore. This is equivalent to saying that the base
When transistors are biased using voltage is approximately 0 V, as shown in Fig. 8-10.
well-designed voltage-divider or The voltage across the emitter diode is 0.7 V, which is why
emitter-bias configurations, they 0.7 V is shown on the emitter node. If this is not clear, stop and think
are classified as beta-independent about it. There is a plus-to-minus drop of 0.7 V in going from the base to
circuits because the values of the emitter. If the base voltage is 0 V, the emitter voltage must be 0.7 V.
IC and V CE are unaffected by In Fig. 8-10, the emitter resistor again plays the key role in set-
changes in the transistor’s beta. ting up the emitter current. To find this current, apply Ohm’s law to the
emitter resistor as follows: The top of the emitter resistor has a voltage of 0.7 V, and the bottom has a volt-
age of 2 V. Therefore, the voltage across the emitter resistor equals the difference between the two voltages.
To get the right answer, subtract the more negative value from the more positive value. In this case, the more
negative value is 2 V, so:
VRE 0.7 V ( 2 V) 1.3 V
Once you have found the voltage across the emitter resistor,
Figure 8-10 Base voltage is ideally zero. calculate the emitter current with Ohm’s law:
+10 V 1.3 V
IE 1.3 mA
RC 1k
3.6 kΩ This current flows through the 3.6 k and produces a voltage drop that we
0V
subtract from 10 V as follows:
VC 10 V (1.3 mA)(3.6 k ) 5.32 V
–0.7 V
RB
2.7 kΩ
RE The collector-emitter voltage is the difference between the collector voltage
1 kΩ
and the emitter voltage:
–2 V VCE 5.32 V ( 0.7 V) 6.02 V

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When two-supply emitter bias is well designed, it is similar to voltage-divider bias and satisfies this
100 1 rule:
RB 0.01 dcRE (8-12)
In this case, the simplified equations for analysis are:
VB 0 (8-13)
VEE 0.7 V
IE (8-14)
RE
VC VCC ICRC (8-15)
VCE VC 0.7 V (8-16)

Base Voltage
One source of error in the simplified method is the small voltage across the base resistor of Fig. 8-10. Since
a small base current flows through this resistance, a negative voltage exists between the base and ground. In
a well-designed circuit, this base voltage is less than 0.1 V. If a designer has to compromise by using a
larger base resistance, the voltage may be more negative than 0.1 V. If you are troubleshooting a circuit like
this, the voltage between the base and ground should produce a low reading; otherwise, something is wrong
with the circuit.

Example 8-5
What is the collector voltage in Fig. 8-10 if the emitter resistor is increased to 1.8 k ?

SOLUTION The voltage across the emitter resistor is still 1.3 V. The emitter current is:
1.3 V
IE 0.722 mA
1.8 k
The collector voltage is:
VC 10 V (0.722 mA)(3.6 k ) 7.4 V
PRACTICE PROBLEM 8-5 Change the emitter resistor of Fig. 8-10 to 2 k and solve for VCE.

Example 8-6
A stage is a transistor and the passive components connected to it. Figure 8-11 shows a three-stage circuit
using two-supply emitter bias. What are the collector-to-ground voltages for each stage in Fig. 8-11?

SOLUTION To begin with, ignore the capacitors because they appear as open circuits to dc voltage and
currents. Then, we are left with three isolated transistors, each using two-supply emitter bias.
The first stage has an emitter current of:
15 V 0 .7 V 14.3 V
IE 0.715 mA
20 k 20 k
and a collector voltage of:
VC 15 V (0.715 mA)(10 k ) 7.85 V
Since the other stages have the same circuit values, each has a collector-to-ground voltage of ap-
proximately 7.85 V.
Summary Table 8-1 illustrates the four main types of bias circuits.

PRACTICE PROBLEM 8-6 Change the supply voltages of Fig. 8-11 to 12 V and 12 V. Then,
calculate VCE for each transistor.
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Figure 8-11 Three-stage circuit.

+15 V

10 kΩ 10 kΩ 10 kΩ

vout

vin

33 kΩ 33 kΩ 33 kΩ
20 kΩ 20 kΩ 20 kΩ

–15 V

Summary Table 8–1 Main Bias Circuits

Type Circuit Calculations Characteristics Where used

Base bias Few parts; Switch; digital

VBB 0.7 V
IB dependent;
+VCC RB
RC
fixed base
RB current
+VBB
IC IB

VCE VCC ICRC

Fixed emitter IC driver;

Emitter bias VE VBB 0.7 V
current; amplifier
+VCC VE independent
IE
RC RE
+VBB VC VC ICRC
RE

VCE VC VE

Voltage divider bias R2 Needs more

VB VCC
+VCC
R1 R2 resistors; Amplifier
independent;
VE VB 0.7 V
RC
needs only one
R1
power supply
VE
IE
RE
R2
RE
VC VCC ICRC

VCE VC VE

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Summary Table 8–1 (continued)

Type Circuit Calculations Characteristics Where used

Two-supply emitter bias VB 0V Needs positive Amplifier

+VCC
and negative
VE VB 0.7 V power
RC
supplies;
VRE VEE 0.7 V independent

RB VRE
RE IE
RE
–VEE
VC VCC ICRC

VCE VC VE

Figure 8-12 (a)

Base bias;
8-5 Other Types of Bias
(b) emitter- In this section, we will discuss some other types of bias. A detailed analysis of these types
feedback bias. of bias is not necessary because they are rarely used in new designs. But you should at
+VCC
least be aware of their existence in case you see them on a schematic diagram.

RC
Emitter-Feedback Bias
RB
Recall our discussion of base bias (Fig. 8-12a). This circuit is the worst when it comes to
setting up a fixed Q point. Why? Since the base current is fixed, the collector current
varies when the current gain varies. In a circuit like this, the Q point moves all over the
load line with transistor replacement and temperature change.
Historically, the first attempt at stabilizing the Q point was emitter-feedback
(a)
bias, shown in Fig. 8-12b. Notice that an emitter resistor has been added to the circuit.
The basic idea is this: If IC increases, VE increases, causing VB to increase. More VB
+VCC
means less voltage across RB . This results in less IB, which opposes the original increase
in IC . It’s called feedback because the change in emitter voltage is being fed back to the
base circuit. Also, the feedback is called negative because it opposes the original change
RC
RB in collector current.
Emitter-feedback bias never became popular. The movement of the Q point is
still too large for most applications that have to be mass-produced. Here are the equations
for analyzing the emitter-feedback bias:

RE VC C V BE
IE (8-17)
R E RB / dc
VE IERE (8-18)
(b)
VB VE 0.7 V (8-19)
VC VCC ICRC (8-20)

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Figure 8-13 (a) Example of emitter-feedback bias; (b) Q point is sensitive to changes in
current gain.
+15 V IC

430 kΩ 910 Ω
14.9 mA
9.33 mA bdc = 300

3.25 mA bdc = 100

VCE
100 Ω 15 V

(a) (b)

The intent of emitter-feedback bias is to swamp out the variations in dc; that is, RE should be much
greater than RB/ dc. If this condition is satisfied, Eq. (8-17) will be insensitive to changes in dc. In practical
circuits, however, a designer cannot select RE large enough to swamp out the effects of dc without cutting off
the transistor.
Figure 8-13a shows an example of an emitter-feedback bias circuit. Figure 8-13b shows the load
line and the Q points for two different current gains. As you can see, a 3 1 variation in current gain produces
a large variation in collector current. The circuit is not much better than base bias.

Collector-Feedback Bias
Figure 8-14a shows collector-feedback bias (also called self-bias). Historically, this was another attempt at
stabilizing the Q point. Again, the basic idea is to feed back a voltage to the base in an attempt to neutralize
any change in collector current. For instance, suppose the collector current increases. This decreases the col-

Figure 8-14 (a) Collector-feedback bias; (b) example; (c) Q point is less sensitive to
changes in current gain.
+VCC

RC
RB

(a )

IC
+15 V
15 mA
1 kΩ
200 kΩ
bdc = 300
8.58 mA

4.77 mA bdc = 100

VCE
15 V
(b ) (c)

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lector voltage, which decreases the voltage across the base resistor. In turn, this decreases the base current,
which opposes the original increase in collector current.
Like emitter-feedback bias, collector-feedback bias uses negative feedback in an attempt to reduce
the original change in collector current. Here are the equations for analyzing collector-feedback bias:

VCC V BE
IE (8-21)
R C RB / dc
VB 0.7 V (8-22)
VC VCC ICRC (8-23)

The Q point is usually set near the middle of the load line by using a base resistance of:

RB dcRC (8-24)

Figure 8-14b shows an example of collector-feedback bias. Figure 8-14c shows the load line and the
Q points for two different current gains. As you can see, a 3 1 variation in current gain produces less varia-
tion in collector current than emitter-feedback (see Fig. 8-13b).
Collector-feedback bias is more effective than emitter-feedback bias in stabilizing the Q point. Al-
though the circuit is still sensitive to changes in current gain, it is used in practice because of its simplicity.

Collector- and Emitter-Feedback Bias

Figure 8-15 Collector-
emitter feedback bias. Emitter-feedback bias and collector-feedback bias were the first steps toward a
+VCC
more stable bias for transistor circuits. Even though the idea of negative feedback
is sound, these circuits fall short because there is not enough negative feedback to
RC do the job. This is why the next step in biasing was the circuit shown in Fig. 8-15.
RB The basic idea is to use both emitter and collector feedback to try to improve the
operation.
As it turns out, more is not always better. Combining both types of
feedback in one circuit helps but still falls short of the performance needed for mass
production. If you come across this circuit, here are the equations for analyzing it:
RE
VC C V BE
IE (8-25)
RC RE RB/ dc

VE IERE (8-26)
VB VE 0.7 V (8-27)
VC VCC ICRC (8-28)

8-6 Troubleshooting
Let us discuss troubleshooting voltage-divider bias because this biasing method is the most widely used.
Figure 8-16 shows the VDB circuit analyzed earlier. Table 8-1 lists the voltages for the circuit when it is sim-
ulated with MultiSim. The voltmeter used to make the measurements has an input impedance of 10 M .
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9-1 Base-Biased Amplifier

In this section, we will discuss a base-biased amplifier. Although a base-biased amplifier is not useful in mass
production, it has instructional value because its basic ideas can be used to build more complicated amplifiers.

Coupling Capacitor
Figure 9-1a shows an ac voltage source connected to a capacitor and a resistor. Since the impedance of the
capacitor is inversely proportional to frequency, the capacitor effectively blocks dc voltage and transmits ac
voltage. When the frequency is high enough, the capacitive reactance is much smaller than the resistance. In
this case, almost all the ac source voltage appears across the resistor. When used in this way, the capacitor is
called a coupling capacitor because it couples or transmits the ac signal to the resistor. Coupling capacitors
are important because they allow us to couple an ac signal into an amplifier without disturbing its Q point.
For a coupling capacitor to work properly, its reactance must be much smaller than the resistance at
the lowest frequency of the ac source. For instance, if the frequency of the ac source varies from 20 Hz to
20 kHz, the worst case occurs at 20 Hz. A circuit designer will select a capacitor whose reactance at 20 Hz is
much smaller than the resistance.
How small is small? As a definition:
Good coupling: XC 0.1R (9-1)
In words: The reactance should be at least 10 times smaller than the resistance at the lowest frequency of op-
eration.
When the 10 1 rule is satisfied, Fig. 9-1a can be replaced by the equivalent circuit of Fig. 9-1b. Why?
The magnitude of impedance in Fig. 9-1a is given by:
Z R 2 XC2
When you substitute the worst-case into this, you get:
Z R2 (0.1R)2 R2 0.01R 2 1.01R 2 1.005R
Since the impedance is within half of a percent of R at the lowest frequency, the current in Fig. 9-1a is only
half a percent less than the current in Fig. 9-1b. Since any well-designed circuit satisfies the 10 1 rule, we can
approximate all coupling capacitors as an ac short (Fig. 9-1b).
A final point about coupling capacitors: Since dc voltage has a frequency of zero, the reactance of a
coupling capacitor is infinite at zero frequency. Therefore, we will use these two approximations for a capac-
itor:
1. For dc analysis, the capacitor is open.
2. For ac analysis, the capacitor is shorted.

Figure 9-1 (a) Coupling capacitor; (b) capacitor is an ac short; (c) dc open and ac short.
C
SHORT

V R V R DC

AC
( a) (b) (c)

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Figure 9-1c summarizes these two important ideas. Unless otherwise stated, all the circuits we analyze from
now on will satisfy the 10 1 rule, so that we can visualize a coupling capacitor as shown in Fig. 9-1c.

Example 9-1
Using Fig. 9-1a, if R 2 k and the frequency range is from 20 Hz to 20 kHz, find the value of C needed
to act as a good coupling capacitor.

SOLUTION Following the 10 :1 rule, XC should be ten times smaller than R at the lowest frequency.
Therefore,
XC 0.1 R at 20 Hz
XC 200 at 20 Hz

1
Since XC
2 fC

1 1
by rearrangement, C
2 fXC (2 )(20 Hz)(200 )

C 39.8 F

PRACTICE PROBLEM 9-1 Using Example 9-1, find the value of C when the lowest frequency is
1 kHz and R is 1.6 k .

DC Circuit
Figure 9-2a shows a base-biased circuit. The dc base voltage is 0.7 V. Because 30 V is much greater than 0.7 V,
the base current is approximately 30 V divided by 1 M , or:
IB 30 A
With a current gain of 100, the collector current is:
IC 3 mA
and the collector voltage is:
VC 30 V (3 mA)(5 k ) 15 V
So, the Q point is located at 3 mA and 15 V.

Amplifying Circuit
Figure 9-2b shows how to add components to build an amplifier. First, a coupling capacitor is used between
an ac source and the base. Since the coupling capacitor is open to direct current, the same dc base current ex-
ists with or without the capacitor and ac source. Similarly, a coupling capacitor is used between the collector
and the load resistor of 100 k . Since this capacitor is open to direct current, the dc collector voltage is the

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Figure 9-2 (a) Base bias; (b) base-biased amplifier.

+30 V

5 kΩ
1 MΩ
+15 V

bdc = 100
+0.7 V

(a)

+30 V

5 kΩ
1 MΩ

100 kΩ
100 mV

(b)

same with or without the capacitor and load resistor. The key idea is that the coupling capacitors prevent the
ac source and load resistance from changing the Q point.
In Fig. 9-2b, the ac source voltage is 100 V. Since the coupling capacitor is an ac short, all the ac
source voltage appears between the base and the ground. This ac voltage produces an ac base current that is
added to the existing dc base current. In other words, the total base current will have a dc component and an
ac component.
Figure 9-3a illustrates the idea. An ac component is superimposed on the dc component. On the pos-
itive half cycle, the ac base current adds to the 30 A of dc base current, and on the negative half cycle it sub-
tracts from it.
The ac base current produces an amplified variation in collector current because of the current gain.
In Fig. 9-3b, the collector current has a dc component of 3 mA. Superimposed on this is an ac collector cur-
rent. Since this amplified collector current flows through the collector resistor, it produces a varying voltage
across the collector resistor. When this voltage is subtracted from the supply voltage, we get the collector
voltage shown in Fig. 9-3c.
Again, an ac component is superimposed on a dc component. The collector voltage is swinging si-
nusoidally above and below the dc level of 15 V. Also, the ac collector voltage is inverted, 180° out of phase
with the input voltage. Why? On the positive half cycle of ac base current, the collector current increases, pro-
ducing more voltage across the collector resistor. This means that there is less voltage between the collector

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Figure 9-3 DC and ac components. (a) Base current; (b) collector current; (c) collector
voltage.
IB

30 mA

t
(a)

IC

3 mA

t
(b)

VC

15 V

t
(c)

and ground. Similarly, on the negative half cycle the collector current decreases. Since there is less voltage
across the collector resistor, the collector voltage increases.

Voltage Waveforms
Figure 9-4 shows the waveforms for a base-biased amplifier. The ac source voltage is a small sinusoidal volt-
age. This is coupled into the base, where it is superimposed on the dc component of 0.7 V. The variation in
base voltage produces sinusoidal variations in base current, collector current, and collector voltage. The total
collector voltage is an inverted sine wave superimposed on the dc collector voltage of 15 V.
Notice the action of the output coupling capacitor. Since it is open to direct current, it blocks the
dc component of collector voltage. Since it is shorted to alternating current, it couples the ac collector volt-
age to the load resistor. This is why the load voltage is a pure ac signal with an average value of zero.

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Figure 9-4 Base-biased amplifier with waveforms.

+30 V

5 kΩ
1 MΩ +15 V 0

0 +0.7 V

100 kΩ

Voltage Gain
The voltage gain of an amplifier is defined as the ac output voltage divided by the ac input voltage. As a de-
finition:
vout
AV = (9-2)
vin
For instance, if we measure an ac load voltage of 50 mV with an ac input voltage of 100 V, the voltage gain is:
50 mV
AV 500
100 V
This says that the ac output voltage is 500 times larger than the ac input voltage.

Calculating Output Voltage

We can multiply both sides of Eq. (9-2) by vin to get this derivation:
vout AV vin (9-3)
This is useful when you want to calculate the value of vout, given the values of AV and vin.
For instance, the triangular symbol shown in Fig. 9-5a is used to indicate an amplifier of any design.
Since we are given an input voltage of 2 mV and a voltage gain of 200, we can calculate an output voltage of:
vout (200)(2 mV) 400 mV
Figure 9-5 (a) Calculating output
voltage; (b) calculating input voltage.
Calculating Input Voltage
vin vout We can divide both sides of Eq. (9-3) by AV to get this derivation:
2 mV AV = 200
vout
vin (9-4)
AV
( a)
This is useful when you want to calculate the value of vin, given the val-
ues vout and AV. For instance, the output voltage is 2.5 V in Fig. 9-5b.
vin vout With a voltage gain of 350, the input voltage is:
AV = 350 2.5 V
2.5 V
vin 7.14 mV
350
(b)

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Figure 9-6 (a) Bypass

capacitor; (b) point E is an
9-2 Emitter-Biased Amplifier
ac ground. The base-biased amplifier has an unstable Q point. For this reason, it is not used
R E much as an amplifier. Instead, an emitter-biased amplifier (either VDB or TSEB)
with its stable Q point is preferred.
V C

Bypass Capacitor
(a)
A bypass capacitor is similar to a coupling capacitor because it appears open to di-
R E rect current and shorted to alternating current. But it is not used to couple a signal
AC
GROUND between two points. Instead, it is used to create an ac ground.
V Figure 9-6a shows an ac voltage source connected to a resistor and a
capacitor. The resistance R represents the Thevenin resistance as seen by the
capacitor. When the frequency is high enough, the capacitive reactance is much
(b)
smaller than the resistance. In this case, almost all the ac source voltage appears
across the resistor. Stated another way, point E is effectively shorted to ground.
When used in this way, the capacitor is called a bypass capacitor because it bypasses or shorts point
E to ground. A bypass capacitor is important because it allows us to create an ac ground in an amplifier with-
out disturbing its Q point.
For a bypass capacitor to work properly, its reactance must be much smaller than the resistance at
the lowest frequency of the ac source. The definition for good bypassing is identical to that for good coupling:
Good bypassing: XC 0.1R (9-5)
When this rule is satisfied, Fig. 9-6a can be replaced by the equivalent circuit of Fig. 9-6b.

Example 9-2
In Fig. 9-7, the input frequency of V is 1 kHz. What value of C is needed to
Figure 9-7 effectively short point E to ground?
R1 E

600 Ω SOLUTION First, find the Thevenin resistance as seen by the

+
V
R2
C
capacitor C.
1 kΩ
–
RTH R1 R2
RTH 600 1k 375
Next, XC should be ten times smaller than RTH. Therefore, XC 37.5 at
1 kHz. Now solve for C by
1 1
C
2 f XC (2 )(1 kHz)(37.5 )

C 4.2 F

PRACTICE PROBLEM 9-2 In Fig. 9-7, find the value of C needed if

R is 50 .

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Figure 9-8 VDB amplifier with waveforms.

+10 V

3.6 kΩ
10 kΩ +6.04 V 0

0 +1.8 V

+1.1 V

100 mV 2.2 kΩ 100 kΩ

1 kΩ

VDB Amplifier
Figure 9-8 shows a voltage-divider-biased (VDB) amplifier. To calculate the dc voltages and currents, men-
tally open all capacitors. Then, the transistor circuit simplifies to the VDB circuit analyzed in Chap. 8. The
quiescent or dc values for this circuit are:
VB 1.8 V
VE 1.1 V
VC 6.04 V
IC 1.1 mA
GOOD TO KNOW As before, we use a coupling capacitor between the source and
base, and another coupling capacitor between the collector and the load
In Fig. 9-8, the emitter voltage is resistance. We also need to use a bypass capacitor between the emitter
rock-solid at 1.1 V because of the and ground. Without this capacitor, the ac base current would be much
emitter bypass capacitor. Therefore, smaller. But with the bypass capacitor, we get a much larger voltage
any variations in the base gain. The mathematical details of why this happens are discussed in the
voltage appear directly across next chapter.
the BE junction of the transistor. In Fig. 9-8, the ac source voltage is 100 V. This is coupled
For example, assume that into the base. Because of the bypass capacitor, all of this ac voltage ap-
vin 10 mV pp. At the positive pears across the base-emitter diode. The ac base current then produces
peak of vin, the ac base voltage an amplified ac collector voltage, as previously described.
equals 1.805 V and VBE equals
1.805 V 1.1 V 0.705 V. At the
negative peak of vin, the ac base VDB Waveforms
voltage decreases to 1.795 V, and
then V BE equals 1.795 V 1.1 V Notice the voltage waveforms in Fig. 9-8. The ac source voltage is a
0.695 V. The ac variations in small sinusoidal voltage with an average value of zero. The base voltage
VBE (0.705 to 0.695 V) are what is an ac voltage superimposed on a dc voltage of 1.8 V. The collector
produces the ac variations in I C voltage is an amplified and inverted ac voltage superimposed on the dc
and V CE . collector voltage of 6.04 V. The load voltage is the same as the collec-
tor voltage, except that it has an average value of zero.
Notice also the voltage on the emitter. It is a pure dc voltage of 1.1 V. There is no ac emitter volt-
age because emitter is at ac ground, a direct result of using a bypass capacitor. This is important to remember
because it is useful in troubleshooting. If the bypass capacitor were to open, an ac voltage would appear be-
tween the emitter and ground. This symptom would immediately point to the open bypass capacitor as the
unique trouble.

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Discrete versus Integrated Circuits

The VDB amplifier of Fig. 9-8 is the standard way to build a discrete transistor amplifier. Discrete means that
all components like resistors, capacitors, and transistors are separately inserted and connected to get the final
circuit. A discrete circuit differs from an integrated circuit (IC), in which all the components are
simultaneously created and connected on a chip, a piece of semiconductor material. Later chapters will
discuss the op amp, an IC amplifier that produces voltage gains of more than 100,000.

TSEB Circuit
Figure 9-9 shows a two-supply emitter bias (TSEB) amplifier. We analyzed the dc part of the circuit in Chap. 8
and calculate these quiescent voltages:
VB 0V
VE 0.7 V
VC 5.32 V
IC 1.3 mA
Figure 9-9 shows two coupling capacitors and an emitter bypass capacitor. The ac operation of the
circuit is similar to that of a VDB amplifier. We couple a signal into the base. The signal is amplified to get
the collector voltage. The amplified signal is then coupled to the load.
Notice the waveforms. The ac source voltage is a small sinusoidal voltage. The base voltage has a
small ac component riding on a dc component of approximately 0 V. The total collector voltage is an inverted
sine wave riding on the dc collector voltage of 5.32 V. The load voltage is the same amplified signal with
no dc component.
Again, notice the pure dc voltage on the emitter, a direct result of using the bypass capacitor. If the
bypass capacitor were to open, an ac voltage would appear at the emitter. This would greatly reduce the volt-
age gain. Therefore, when troubleshooting an amplifier with bypass capacitors, remember that all ac grounds
should have zero ac voltage.

Figure 9-9 TSEB amplifier with waveforms.

+10 V

3.6 kΩ

+5.32 V 0

0 ≈0
100 kΩ

–0.7 V
100 mV 2.7 kΩ
1 kΩ

–2 V

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9-3 Small-Signal Operation

Figure 9-10 shows the graph of current versus voltage for the base-emitter diode. When an ac voltage is cou-
pled into the base of a transistor, an ac voltage appears across the base-emitter diode. This produces the si-
nusoidal variation in VBE shown in Fig. 9-10.

Instantaneous Operating Point

When the voltage increases to its positive peak, the instantaneous operating point moves from Q to the upper
point shown in Fig. 9-10. On the other hand, when the sine wave decreases to its negative peak, the instanta-
neous operating point moves from Q to the lower point.
The total base-emitter voltage of Fig. 9-10 is an ac voltage centered on a dc voltage. The size of the
ac voltage determines how far the instantaneous point moves away from the Q point. Large ac base voltages
produce large variations, whereas small ac base voltages produce small variations.

Distortion
The ac voltage on the base produces the ac emitter current shown in Fig. 9-10. This ac emitter current has the
same frequency as the ac base voltage. For instance, if the ac generator driving the base has a frequency of
1 kHz, the ac emitter current has a frequency of 1 kHz. The ac emitter current also has approximately the
same shape as the ac base voltage. If the ac base voltage is sinusoidal, the ac emitter current is approximately
sinusoidal.
The ac emitter current is not a perfect replica of the ac base voltage because of the curvature of the
graph. Since the graph is curved upward, the positive half cycle of ac emitter current is elongated (stretched)
and the negative half cycle is compressed. This stretching and compressing of alternate half cycles is called
distortion. It is undesirable in high-fidelity amplifiers because it changes the sound of voice and music.

Reducing Distortion
One way to reduce distortion in Fig. 9-10 is by keeping the ac base voltage small. When you reduce the peak
value of the base voltage, you reduce the movement of the instantaneous operating point. The smaller this
swing or variation, the less the curvature in the graph. If the signal is small enough, the graph appears to be
linear.

Figure 9-10 Distortion when signal is too large.

IE

VBE

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Figure 9-11 Definition of small-signal operation.

IE

LESS
10 mA THAN
1 mA p-p

VBE

Why is this important? Because there is negligible distortion for a small signal. When the signal is
small, the changes in ac emitter current are almost directly proportional to the changes in ac base voltage be-
cause the graph is almost linear. In other words, if the ac base voltage is a small enough sine wave, the ac
emitter current will also be a small sine wave with no noticeable stretching or compression of half cycles.

The 10 Percent Rule

The total emitter current shown in Fig. 9-10 consists of a dc component and an ac component, which can be
written as:
IE IEQ ie

where IE the total emitter current

IEQ the dc emitter current
ie the ac emitter current
To minimize distortion, the peak-to-peak value of ie must be small compared to IEQ. Our definition
of small-signal operation is:
Small signal: ie( pp) 0.1 IEQ (9-6)
This says that the ac signal is small when the peak-to-peak ac emitter current is less than 10 percent of the dc
emitter current. For instance, if the dc emitter current is 10 mA, as shown in Fig. 9-11, the peak-to-peak emit-
ter current should be less than 1 mA in order to have small-signal operation.
From now on, we will refer to amplifiers that satisfy the 10 percent rule as small-signal amplifiers.
This type of amplifier is used at the front end of radio and television receivers because the signal coming in
from the antenna is very weak. When coupled into a transistor amplifier, a weak signal produces very small
variations in emitter current, much less than the 10 percent rule requires.

Example 9-3
Using Fig. 9-9, find the maximum small signal emitter current.

SOLUTION: First, find the Q point emitter current, IEQ.

VEE VBE 2V 0.7 V
IEQ IEQ IEQ 1.3 mA
RE 1k

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Then solve for the small signal emitter current ie(pp)

ie(pp) 0.1 IEQ

ie(pp) (0.1)(1.3 mA)

ie(pp) 130 App

PRACTICE PROBLEM 9-3 Using Fig. 9-9, change RE to 1.5 k and calculate the maximum small
signal emitter current.

9-4 AC Beta
The current gain in all discussions up to this point has been dc current gain. This was defined as:
IC
dc (9-7)
IB
The currents in this formula are the currents at the Q point in Fig. 9-12. Because of the curvature in the graph
of IC versus IB, the dc current gain depends on the location of the Q point.

Definition
The ac current gain is different. It is defined as:
ic
(9-8)
ib
In words, the ac current gain equals the ac collector current divided by the ac base current. In Fig. 9-12, the
ac signal uses only a small part of the graph on both sides of the Q point. Because of this, the value of the ac
current gain is different from the dc current gain, which uses almost all of the graph.

Figure 9-12 AC current gain equals ratio of changes.

IC

IB

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Graphically, equals the slope of the curve at the Q point in Fig. 9-12. If we were to bias the tran-
sistor to a different Q point, the slope of the curve would change, which means that would change. In other
words, the value of depends on the amount of dc collector current.
On data sheets, dc is listed as hFE and is shown as hfe. Notice that capital subscripts are used with
dc current gain, and lowercase subscripts with ac current gain. The two current gains are comparable in value,
not differing by a large amount. For this reason, if you have the value of one, you can use the same value for
the other in preliminary analysis.

Notation
To keep dc quantities distinct from ac quantities, it is standard practice to use capital letters and subscripts for
dc quantities. For instance, we have been using:
IE, IC, and IB for the dc currents
VE, VC, and VB for the dc voltages
VBE, VCE, and VCB for the dc voltages between terminals
For ac quantities, we will use lowercase letters and subscripts as follows:
ie, ic, and ib for the ac currents
ve, vc, and vb for the ac voltages
vbe, vce, and vcb for the ac voltages between terminals
Also worth mentioning is the use of capital R for dc resistances and lowercase r for ac resistances. The next
section will discuss ac resistance.

9-5 AC Resistance of the Emitter Diode

Figure 9-13 shows a graph of current versus voltage for the emitter diode. When a small ac voltage is across
the emitter diode, it produces the ac emitter current shown. The size of this ac emitter current depends on the
location of the Q point. Because of the curvature, we get more peak-to-peak ac emitter current when the Q
point is higher up the graph.

Figure 9-13 AC resistance of emitter diode.

IE

VBE

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Definition
As discussed in Sec. 9-3, the total emitter current has a dc component and an ac component. In symbols:
IE IEQ ie
where IEQ is the dc emitter current and ie is the ac emitter current.
In a similar way, the total base-emitter voltage of Fig. 9-13 has a dc component and an ac compo-
nent. Its equation can be written as:
VBE VBEQ vbe
where VBEQ is the dc base-emitter voltage and vbe is the ac base-emitter voltage.
In Fig. 9-13, the sinusoidal variation in VBE produces a sinusoidal variation in IE. The peak-to-peak
value of ie depends on the location of the Q point. Because of the curvature in the graph, a fixed vbe produces
more ie as the Q point is biased higher up the curve. Stated another way, the ac resistance of the emitter diode
decreases when the dc emitter current increases.
The ac emitter resistance of the emitter diode is defined as:
vbe
re (9-9)
ie
This says that the ac resistance of the emitter diode equals the ac base-emitter voltage divided by the ac emit-
ter current. The prime ( ) in re is a standard way to indicate that the resistance is inside the transistor.
For instance, Fig. 9-14 shows an ac base-emitter voltage of 5 mV pp. At the given Q point, this sets
up an ac emitter current of 100 A pp. The ac resistance of the emitter diode is:
5 mV
re 50
100 A
As another example, assume that a higher Q point in Fig. 9-14 has vbe 5 mV and ie 200 A. Then, the
ac resistance decreases to:
5 mV
re 25
200 A
The point is this: The ac emitter resistance always decreases when the dc emitter current increases, because
vbe is essentially a constant value.

Figure 9-14 Calculating re .

IE

100 mA

VBE

5 mV

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Formula for AC Emitter Resistance

With solid-state physics and calculus, it is possible to derive the following remarkable formula for the ac
emitter resistance:
25 mV
re (9-10)
IE
This says that the ac resistance of the emitter diode equals 25 mV divided by the dc emitter current.
This formula is remarkable because of its simplicity and the fact that it applies to all transistor types.
It is widely used in industry to calculate a preliminary value for the ac resistance of the emitter diode. The de-
rivation assumes small-signal operation, room temperature, and an abrupt rectangular base-emitter junction.
Since commercial transistors have gradual and nonrectangular junctions, there will be some deviations from
Eq. (9-10). In practice, almost all commercial transistors have an ac emitter resistance between 25 mV/IE and
50 mV/IE.
The reason re is important is because it determines the voltage gain. The smaller it is, the higher the
voltage gain. Chapter 10 will show you how to use re to calculate the voltage gain of a transistor amplifier.

Example 9-4
What does re equal in the base-biased amplifier of Fig. 9-15a?

Figure 9-15 (a) Base-biased amplifier; (b) VDB amplifier; (c) TSEB amplifier.
+30 V

5 kΩ
1 MΩ

bdc = 100
100 kΩ
100 mV

(a)

+10 V

3.6 kΩ
10 kΩ

100 kΩ

100 mV 2.2 kΩ
1 kΩ

(b)

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Figure 9-15 (continued)

+10 V

3.6 kΩ

100 kΩ

100 m V 2.7 kΩ
1 kΩ

–2 V
(c)

SOLUTION Earlier, we calculated a dc emitter current of approximately 3 mA for this circuit. With
Eq. (9-10), the ac resistance of the emitter diode is:

25 mV
re 8.33
3 mA

Example 9-5
In Fig. 9-15b, what does re equal?

SOLUTION We analyzed this VDB amplifier earlier and calculated a dc emitter current of 1.1 mA. The
ac resistance of the emitter diode is:

25 mV
re 22.7
1.1 mA

Example 9-6
What is the ac resistance of the emitter diode for the two-supply emitter-bias amplifier of Fig. 9-15c?

SOLUTION From an earlier calculation, we got a dc emitter current of 1.3 mA. Now, we can calculate
the ac resistance of the emitter diode:

25 mV
re 19.2
1.3 mA

PRACTICE PROBLEM 9-6 Using Fig. 9-15c, change the VEE supply to 3 V and calculate r e.

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9-6 Two Transistor Models

To analyze the ac operation of a transistor amplifier, we need an ac equivalent circuit for a transistor. In other
words, we need a model for the transistor that simulates how it behaves when an ac signal is present.

The T Model
One of the earliest ac models was the Ebers-Moll model shown in Fig. 9-16. As far as a small ac signal is
concerned, the emitter diode of a transistor acts like an ac resistance re and the collector diode acts like a cur-
rent source ic. Since the Ebers-Moll model looks like a T on its side, the equivalent circuit is also called the
T model.
When analyzing a transistor amplifier, we can replace each transistor by a T model. Then, we can
calculate value of re and other ac quantities like voltage gain. The details are discussed in the next chapter.
When an ac input signal drives a transistor amplifier, an ac base-emitter voltage vbe is across the
emitter diode, as shown in Fig. 9-17a. This produces an ac base current ib. The ac voltage source has to sup-
ply this ac base current, so that the transistor amplifier will work properly. Stated another way, the ac voltage
source is loaded by the input impedance of the base.
Figure 9-17b illustrates the idea. Looking into the base of the transistor, the ac voltage source sees
an input impedance zin(base). At low frequencies, this impedance is purely resistive and defined as:
vbe
zin(base) (9-11)
ib
Applying Ohm’s law to the emitter diode of Fig. 9-17a, we can write:
vbe iere

Figure 9-16 T model of a transistor.

ic ic

ib n ib

n re

ie ie

Figure 9-17 Defining the input impedance of the base.

ic ic
ib

+ zin(base)
vbe re re

–
ie ie

(a) (b)

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Figure 9-18 model of a transistor.

ib

bre ic ic
ib

re

ie ie

(a) (b)

Substitute this equation into the preceding one to get:

vbe iere
zin(base)
ib ib
Since ie ic, the foregoing equation simplifies to:
zin(base) re (9-12)
This equation tells us that the input impedance of the base is equal to the ac current gain times the ac resis-
tance of the emitter diode.

The Model
Figure 9-18a shows the model of a transistor. It’s a visual representation of Eq. (9-12). The model is eas-
ier to use than the T model (Fig. 9-18b) because the input impedance is not obvious when you look at the T
model. On the other hand, the model clearly shows that an input impedance of re will load the ac voltage
source driving the base.
Since the and T models are ac equivalent circuits for a transistor, we can use either one when ana-
lyzing an amplifier. Most of the time, we will use the model. With some circuits like the differential amps in
Chap. 17, the T model gives a better insight into the circuit action. Both models are widely used in industry.

9-7 Analyzing an Amplifier

Amplifier analysis is complicated because both dc and ac sources are in the same circuit. To analyze ampli-
fiers, we can calculate the effect of the dc sources, and then the effect of the ac sources. When using the su-
perposition theorem in this analysis, the effect of each source acting alone is
GOOD TO KNOW added to get the total effect of all sources acting simultaneously.
There are other, more accurate
transistor equivalent circuits
(models) in addition to those shown
The DC Equivalent Circuit
in Figs. 9-16, 9-17, and 9-18. The simplest way to analyze an amplifier is to split the analysis into two
A highly accurate equivalent parts: a dc analysis and an ac analysis. In the dc analysis, we calculate the
circuit will include something dc voltages and currents. To do this, we mentally open all capacitors. The
called the base spreading circuit that remains is the dc equivalent circuit.
resistance r b and the internal With the dc equivalent circuit, you can calculate the transistor
resistance r c of the collector currents and voltages as needed. If you are troubleshooting, approximate
current source. This model is used answers are adequate. The most important current in the dc analysis is the
if exact answers are desired. dc emitter current. This is needed to calculate re for the ac analysis.

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Figure 9-19 DC voltage source is an ac short.

R R AC GROUND

VP VP
+
VCC
–

( a) ( b)

AC Effect of a DC Voltage Source

Figure 9-19a shows a circuit with ac and dc sources. What is the ac current in a circuit like this? As far as the
ac current is concerned, the dc voltage source acts like an ac short, as shown in Fig. 9-19b. Why? Because a
dc voltage source has a constant voltage across it. Therefore, any ac current flowing through it cannot pro-
duce an ac voltage across it. If no ac voltage can exist, the dc voltage source is equivalent to an ac short.
Another way to understand the idea is to recall the superposition theorem discussed in basic elec-
tronics courses. In applying superposition to Fig. 9-19a, we can calculate the effect of each source acting sep-
arately while the other is reduced to zero. Reducing the dc voltage source to zero is equivalent to shorting it.
Therefore, to calculate the effect of the ac source in Fig. 9-19a, we can short the dc voltage source.
From now on, we will short all dc voltage sources when analyzing the ac operation of an amplifier.
As shown in Fig. 9-19b, this means that each dc voltage supply point acts like an ac ground.

AC Equivalent Circuit
After analyzing the dc equivalent circuit, the next step is to analyze the ac equivalent circuit. This is the cir-
cuit that remains after you have mentally shorted all capacitors and dc voltage sources. The transistor can be
replaced by either the model or the T model. In the next chapter, we will show the mathematical details of
ac analysis. For the remainder of this chapter, let us focus on how to get the ac equivalent circuit for the three
amplifiers discussed up to now: base-biased, VDB, and TSEB.

Base-Biased Amplifier
Figure 9-20a is a base-biased amplifier. After mentally opening all capacitors and analyzing the dc equivalent
circuit, we are ready for the ac analysis. To get the ac equivalent circuit, we short all capacitors and dc volt-
age sources. Then, the point labeled VCC is an ac ground.
Figure 9-20b shows the ac equivalent circuit. As you can see, the transistor has been replaced by its
model. In the base circuit, the ac input voltage appears across RB in parallel with re . In the collector cir-
cuit, the current source pumps an ac current of ic through RC in parallel with RL.

VDB Amplifier
Figure 9-21a is a VDB amplifier, and Fig. 9-21b is its ac equivalent circuit. As you can see, all capacitors
have been shorted, the dc supply point has become an ac ground, and the transistor has been replaced by its
model. In the base circuit, the ac input voltage appears across R1 in parallel with R2 in parallel with re . In
the collector circuit, the current source pumps an ac current of ic through RC in parallel with RL.

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Figure 9-20 (a) Base-biased amplifier; (b) ac equivalent circuit.

+VCC

RC
RB

RL
vin

(a )

B C

vin RB b re ic RC RL

(b)

Figure 9-21 (a) VDB amplifier; (b) ac equivalent circuit.

+VCC

RC
R1

RL

vin R2
RE

(a )

B C

vin R1 R2 bre ic RC RL

( b)

TSEB Amplifier
Our last example is the two-supply emitter-bias circuit of Fig. 9-22a. After analyzing the dc equivalent
circuit, we can draw the ac equivalent circuit of Fig. 9-22b. Again, all capacitors are shorted, the dc source
voltage becomes an ac ground, and the transistor is replaced by its model. In the base circuit, the ac input

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Figure 9-22 (a) TSEB amplifier; (b) ac equivalent circuit.

+VCC

RC

RL

vin RB
RE

–VEE

( a)

B C

vin RB bre ic RC RL

( b)

voltage appears across RB in parallel with re . In the collector circuit, the current source pumps an ac current
of ic through RC in parallel with RL.

CE Amplifiers
The three different amplifiers of Figs. 9-20, 9-21, and 9-22 are examples of a common-emitter (CE) ampli-
fier. You can recognize a CE amplifier immediately because its emitter is at ac ground. With a CE amplifier,
the ac signal is coupled into the base, and the amplified signal appears at the collector.
Two other basic transistor amplifiers are possible. The common-base (CB) amplifier and the com-
mon-collector (CC) amplifier. The CB amplifier has its base at ac ground, and the CC amplifier has its col-
lector at ac ground. They are useful in some applications, but are not as popular as the CE amplifier. Later
chapters discuss the CB and CC amplifiers.

Main Ideas
The foregoing method of analysis works for all amplifiers. You start with the dc equivalent circuit. After cal-
culating the dc voltages and currents, you analyze the ac equivalent circuit. The crucial ideas in getting the ac
equivalent circuit are:
1. Short all coupling and bypass capacitors.
2. Visualize all dc supply voltages as ac grounds.
3. Replace the transistor by its or T model.
4. Draw the ac equivalent circuit.
Subsequent chapters will use this method to calculate the voltage gain, input impedance, and other charac-
teristics of amplifiers.
The process of using superposition to analyze a VDB circuit is shown in Summary Table 9-1.

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Summary Table 9–1 VDB DC and AC Equivalents

VCC = 10 V

RC
R1 3.6 kΩ
10 kΩ

RL
100 kΩ
+
R2
Original V
2.2 kΩ RE
circuit –
1 kΩ

VCC = 10 V • Open all coupling and

bypass capacitors.
R1 RC • Redraw the circuit.
10 kΩ 3.6 kΩ
• Solve the dc circuit’s Q
point:
DC
VB 1.8 V
circuit
R2 RE
VE 1.1 V
2.2 kΩ 1 kΩ IE 1.1 mA
VCE 4.94 V

+
R1 R2 RC RL
AC V
10 kΩ 2.2 kΩ
br´e
3.6 kΩ 100 kΩ
–
model

• Short all coupling and

bypass capacitors.
RC RL • Visualize all dc supply
3.6 kΩ 100 kΩ
voltages as ac grounds.
• Replace the transistor
by its or T model.
+
R1 R2 • Draw the ac equivalent
AC T V
10 kΩ 2.2 kΩ
br´e
circuit.
–
model
25 mV
• re 22.7 .
IEQ

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9-8 AC Quantities on the Data Sheet

Refer to the partial data sheet of a 2N3904 in Fig. 9-23 during the following discussion. The ac quantities ap-
pear in the section labeled “Small-Signal Characteristics.” In this section, you will find four new quantities
labeled hfe, hie, hre, and hoe. These are called h parameters. What are they?

H Parameters
When the transistor was first invented, an approach known as the h parameters was used to analyze and de-
sign transistor circuits. This mathematical approach models the transistor on what is happening at its termi-
nals without regard for the physical processes taking place inside the transistor.
A more practical approach is the one we are using. It is called the r parameter method, and it uses
quantities like and re . With this approach, you can use Ohm’s law and other basic ideas in the analysis and
design of transistor circuits. This is why the r parameters are better suited to most people.
This does not mean that the h parameters are useless. They have survived on data sheets because
they are easier to measure than r parameters. When you read data sheets, therefore, don’t look for , re , and
other r parameters. You won’t find them. Instead, you will find hfe, hie, hre, and hoe. These four h parameters
give useful information when translated into r parameters.

Relations Between R and H Parameters

For instance, hfe given in the “Small-Signal Characteristics” section of the data sheet is identical to the ac cur-
rent gain. In symbols this is represented as:
hfe
The data sheet lists a minimum hfe of 100 and a maximum of 400. Therefore, may be as low as 100 or as
high as 400. These values are for a collector current of 1 mA and a collector-emitter voltage of 10 V.
Another h parameter is the quantity hie, equivalent to the input impedance. The data sheets give a
minimum hie of 1 k and a maximum of 10 k . The quantity hie is related to r parameters like this:
h ie
re (9-13)
hfe
For instance, the maximum values of hie and hfe are 10 k and 400. Therefore:
10 k
re 25
400
The last two h parameters, hre and hoe, are not needed for troubleshooting and basic design.

Other Quantities
Other quantities listed under “Small-Signal Characteristics” include fT, Cibo, Cobo, and NF. The first, fT, gives
information about the high-frequency limitations on a 2N3904. The second and third quantities, Cibo and
Cobo, are the input and output capacitances of the device. The final quantity, NF, is the noise figure; it indi-
cates how much noise the 2N3904 produces.
The data sheet of a 2N3904 includes a lot of graphs, which are worth looking at. For instance, the
graph on the data sheet labelled current gain shows that hfe increases from approximately 70 to 160 when the
collector current increases from 0.1 mA to 10 mA. Notice that hfe is approximately 125 when the collector

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Figure 9-23 The 2N3904 partial data sheet. (Copyright Semiconductor Components Industries, LLC; used by permission)
2N3903, 2N3904

ELECTRICAL CHARACTERISTICS (TA 25°C unless otherwise noted)

Characteristic Symbol Min Max Unit

SMALL–SIGNAL CHARACTERISTICS

Current–Gain–Bandwidth Product (IC 10 mAdc, VCE 20 Vdc, f 100 MHz) 2N3903 fT 250 – MHz
2N3904 300 –

Output Capacitance (VCB 0.5 Vdc, IE 0, f 1.0 MHz) Cobo – 4.0 pF

Input Capacitance (VEB 0.5 Vdc, IC 0, f 1.0 MHz) Cibo – 8.0 pF

Input Impedance (IC 1.0 mAdc, VCE 10 Vdc, f 1.0 kHz) 2N3903 hie 1.0 8.0 k
2N3904 1.0 10

Voltage Feedback Ratio (IC 1.0 mAdc, VCE 10 Vdc, f 1.0 kHz) 2N3903 hre 0.1 5.0 10–4
2N3904 0.5 8.0

Small–Signal Current Gain (IC 1.0 mAdc, VCE 10 Vdc, f 1.0 kHz) 2N3903 hfe 50 200 –
2N3904 100 400

Output Admittance (IC 1.0 mAdc, VCE 10 Vdc, f 1.0 kHz) hoe 1.0 40 mhos

Noise Figure (IC 100 Adc, VCE 5.0 Vdc, RS 1.0 k , f 1.0 kHz) 2N3903 NF – 6.0 dB
2N3904 – 5.0

H PARAMETERS
VCE 10 Vdc, f 1.0 kHz, TA = 25°C
hoe, OUTPUT ADMITTANCE (m mhos)

100
300
50
hfe, CURRENT GAIN

200 20

10
100
5
70

50 2

1
30 0.1 0.2 0.3 0.5 1.0 2.0 3.0 5.0 10
0.1 0.2 0.3 0.5 1.0 2.0 3.0 5.0 10
IC, COLLECTOR CURRENT (mA)
IC, COLLECTOR CURRENT (mA)
Output Admittance
Current Gain
hre, VOLTAGE FEEDBACK RATIO (X 10–4)
hie, INPUT IMPEDANCE (k OHMS)

20 10

10 7.0
5.0
5.0
3.0
2.0
2.0
1.0

0.5 1.0
0.7
0.2 0.5
0.1 0.2 0.3 0.5 1.0 2.0 3.0 5.0 10 0.1 0.2 0.3 0.5 1.0 2.0 3.0 5.0 10
IC, COLLECTOR CURRENT (mA) IC, COLLECTOR CURRENT (mA)
Input Impedance Voltage Feedback Ratio

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10-1 Voltage Gain

Figure 10-1a shows a voltage-divider-biased (VDB) amplifier. Voltage gain was defined as the ac output
voltage divided by the ac input voltage. With this definition, we can derive another equation for voltage gain
that is useful in troubleshooting.

Derived from the Model

Figure 10-1b shows the ac equivalent circuit using the model of the transistor. The ac base current ib flows
through the input impedance of the base ( re ). With Ohm’s law, we can write:

vin ib re

In the collector circuit, the current source pumps an ac current ic through the parallel connection of RC and
RL. Therefore, the ac output voltage equals:

vout ic(RC RL) ib(RC RL)

Figure 10-1 (a) CE amplifier; (b) ac equivalent circuit with model; (c) ac equivalent circuit with T model.
+VCC

RC
R1

RL vout

vin R2
RE

(a)

ib

vin R1 R2 bre ic RC RL vout

(b)

ic
RC RL vout

vin R1 R2 re
ie

(c)

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Now, we can divide vout by vin to get:

vout ib(RC RL)
AV
vin ib re
which simplifies to:
(RC RL)
AV (10-1)
re

AC Collector Resistance
In Fig. 10-1b, the total ac load resistance seen by the collector is the parallel combination of RC and RL. This
total resistance is called the ac collector resistance, symbolized rc. As a definition:

rc RC RL (10-2)

Now, we can rewrite Eq. (10-1) as:

rc
AV (10-3)
re
In words: The voltage gain equals the ac collector resistance divided by the ac resistance of the emitter diode.

Derived from the T Model

Either transistor model gives the same results. Later, we will use the T model when analyzing differential am-
plifiers. For practice, let us derive the equation for voltage gain using the T model.
Figure 10-1c shows the ac equivalent circuit using the T model of the transistor. The input voltage
vin appears across re . With Ohm’s law, we can write:

vin iere

GOOD TO KNOW In the collector circuit, the current source pumps an ac current ic through
the ac collector resistance. Therefore, the ac output voltage equals:
The current gain Ai of a common-
emitter amplifier equals the ratio vout icrc
of the output current iout to the
input current i in. The output Now, we can divide vout by vin to get:
current, however, is not ic, as you
might think. The output current i out vout icrc
is the current flowing in the load AV
vin iere
RL. The equation for Ai is derived as
follows:
Since ic ie, we can simplify the equation to get:
V out /RL
Ai
Vin /Zin rc
AV
or re
Ai Vout /Vin Zin/RL
This is the same equation derived with the model. It applies to all CE
Since A v Vout /Vin, then A i can be
amplifiers because all have an ac collector resistance of rc and an emit-
stated as A i Av Zin/RL.
ter diode with an ac resistance of re .

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Example 10-1
What is the voltage gain in Fig. 10-2a? The output voltage across the load resistor?

Figure 10-2 (a) Example of VDB amplifier; (b) example of TSEB amplifier.
+10 V

3.6 kΩ
10 kΩ

10 kΩ vout

2 mV 2.2 kΩ
1 kΩ

(a)

+9 V

3.6 kΩ

2.2 kΩ vout

5 mV 10 kΩ
10 kΩ

–9 V
(b)

SOLUTION The ac collector resistance is:

rc RC RL (3.6 k 10 k ) 2.65 k
In Example 9-2, we calculated an re of 22.7 . So, the voltage gain is:
rc 2.65 k
AV 117
re 22.7
The output voltage is:
vout AVvin (117)(2 mV) 234 mV

PRACTICE PROBLEM 10-1 Using Fig. 10-2a, change RL to 6.8 k and find AV.

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Example 10-2
What is the voltage gain in Fig. 10-2b? The output voltage across the load resistor?

SOLUTION The ac collector resistance is:

rc RC RL (3.6 k 2.2 k ) 1.37 k

The dc emitter current is approximately:

9 V 0.7 V
IE 0.83 mA
10 k

The ac resistance of the emitter diode is:

25 mV
re 30
0.83 mA

The voltage gain is:

rc 1.37 k
AV 45.7
re 30

The output voltage is:

vout AV vin (45.7)(5 mV) 228 mV

PRACTICE PROBLEM 10-2 In Fig. 10-2b, change the emitter resistor RE from 10 k to 8.2 k and
calculate the new output voltage, vout.

10-2 The Loading Effect of

Input Impedance
Up to now, we have assumed an ideal ac voltage source, one with zero source resistance. In this section, we
will discuss how the input impedance of an amplifier can load down the ac source, that is, reduce the ac volt-
age appearing across the emitter diode.

Input Impedance
In Fig. 10-3a, an ac voltage source vg has an internal resistance of RG. (The subscript g stands for “genera-
tor,” a synonym for source.) When the ac generator is not stiff, some of the ac source voltage is dropped
across its internal resistance. As a result, the ac voltage between the base and ground is less than ideal.
The ac generator has to drive the input impedance of the stage zin(stage). This input impedance in-
cludes the effects of the biasing resistors R1 and R2, in parallel with the input impedance of the base zin(base).
Figure 10-3b illustrates the idea. The input impedance of the stage equals:
zin(stage) R1 R2 re

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Figure 10-3 CE amplifier. (a) Circuit; (b) ac equivalent circuit; (c) effect of input impedance.
+VCC

RC
R1

RG

RL
zin(stage) zin(base)
vg R2
RE

(a)

RG

zin(stage)
vg R1 R2 bre ic RC RL

(b)

RG

vg zin(stage) vin

(c)

Equation for Input Voltage

When the generator is not stiff, the ac input voltage vin of Fig. 10-3c is less than vg. With the voltage-divider
theorem, we can write:

zin(stage)
vin vg (10-4)
RG zin(stage)

This equation is valid for any amplifier. After you calculate or estimate the input impedance of the stage, you
can determine what the input voltage is. Note: The generator is stiff when RG is less than 0.01z in(stage).

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Example 10-3
In Fig. 10-4, the ac generator has an internal resistance of 600 . What is the output voltage in Fig. 10-4 if
300?

Figure 10-4 Example.

+10 V

3.6 kΩ
10 kΩ

600 Ω

10 kΩ vout

2 mV 2.2 kΩ
1 kΩ

SOLUTION Here are two quantities calculated in earlier examples: re 22.7 and AV 117. We will
use these values in solving the problem.
When 300, the input impedance of the base is:
zin(base) (300)(22.7 ) 6.8 k
The input impedance of the stage is:
zin(stage) 10 k 2.2 k 6.8 k 1.42 k
With Eq. (10-4), we can calculate the input voltage:
1.42 k
vin 2 mV 1.41 mV
600 1.42 k
This is the ac voltage that appears at the base of the transistor, equivalent to the ac voltage across the emit-
ter diode. The amplified output voltage equals:
vout AVvin (117)(1.41 mV) 165 mV
PRACTICE PROBLEM 10-3 Change the RG value of Fig. 10-4 to 50 and solve for the new amplified
output voltage.

Example 10-4
Repeat the preceding example for 50.

SOLUTION When 50, the input impedance of the base decreases to:
zin(base) (50)(22.7 ) 1.14 k
The input impedance of the stage decreases to:
zin(stage) 10 k 2.2 k 1.14 k 698
With Eq. (10-4), we can calculate the input voltage:
698
vin 2 mV 1.08 mV
600 698

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11-1 CC Amplifier
The emitter follower is also called a common-collector (CC) amplifier. The input signal is coupled to the
base, and the output signal is taken from the emitter.

Basic Idea
Figure 11-1a shows an emitter follower. Because the collector is at ac ground, the circuit is a CC amplifier.
The input voltage is coupled to the base. This sets up an ac emitter current and produces an ac voltage across
the emitter resistor. This ac voltage is then coupled to the load resistor.
Figure 11-1b shows the total voltage between the base and ground. It has a dc component and an ac
component. As you can see, the ac input voltage rides on the quiescent base voltage VBQ. Similarly, Fig. 11-
1c shows the total voltage between the emitter and ground. This time, the ac input voltage is centered on a
quiescent emitter voltage VEQ.
The ac emitter voltage is coupled to the load resistor. This output voltage is shown in Fig. 11-1d, a
pure ac voltage. This output voltage is in phase and is approximately equal to the input voltage. The reason

Figure 11-1 Emitter follower and waveforms.

+VCC

R1

vin R2 RE RL

(a )

VB VE

VBQ
VEQ

t t
(b ) (c )

vout VC

VCC

t
t

(d ) (e )

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GOOD TO KNOW the circuit is called an emitter follower is because the output voltage fol-
In some emitter-follower circuits, a
lows the input voltage.
small collector resistance is used to
Since there is no collector resistor, the total voltage between
limit the dc collector current in the
the collector and ground equals the supply voltage. If you look at the
transistor in case a short occurs
collector voltage with an oscilloscope, you will see a constant dc voltage
between the emitter and ground. If
like Fig. 11-1e. There is no ac signal on the collector because it is an ac
a small RC is used, the collector will
ground point.
have a bypass capacitor going to
ground. The small value of RC will Negative Feedback
have only a slight bearing on the Like a swamped amplifier, the emitter follower uses negative feedback.
dc operation of the circuit and no But with the emitter follower, the negative feedback is massive because
bearing at all on the circuit’s ac the feedback resistance equals all of the emitter resistance. As a result, the
operation. voltage gain is ultrastable, the distortion is almost nonexistent, and the
input impedance of the base is very high. The trade-off is the voltage gain, which has a maximum value of 1.

AC Emitter Resistance
In Fig. 11-1a, the ac signal coming out of the emitter sees RE in parallel with RL. Let us define the ac emitter
resistance as follows:
re RE RL (11-1)
This is the external ac emitter resistance, which is different from the internal ac emitter resistance re .

Voltage Gain
Figure 11-2a shows the ac equivalent with the T model. Using Ohm’s law, we can write these two equations:
vout ie re
vin ie(re re )
Divide the first equation by the second, and you get the voltage gain of the emitter follower:
re
AV (11-2)
re re
Usually, a designer makes re much greater than re , so that the voltage gain equals 1 (approximately). This is
the value to use for all preliminary analysis and troubleshooting.

Figure 11-2 AC equivalent circuits for emitter follower.

re⬘

vin R1 R2
re vin R1 R2 b (re + re⬘) re
vout vout

(a) (b)

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Why is an emitter follower called an amplifier if its voltage gain is only 1? Because it has a current
gain of . The stages near the end of a system need to produce more current because the final load is usually
a low impedance. The emitter follower can produce the large output currents needed by low-impedance
loads. In short, although it is not a voltage amplifier, the emitter follower is a current or power amplifier.

Input Impedance of the Base

Figure 11-2b shows the ac equivalent circuit with the model of the transistor. As far as the input impedance
of the base is concerned, the action is the same as that of a swamped amplifier. The current gain transforms the
total emitter resistance up by a factor of . The derivation is therefore
GOOD TO KNOW identical to that of a swamped amplifier:
In Fig. 11-3, the biasing resistors R 1 z in(base) (re re ) (11-3)
and R 2 lower z in to a value that is
For troubleshooting, you can assume that re is much greater than re ,
not much different from that of a
which means that the input impedance is approximately re.
swamped CE amplifier. This
The step-up in impedance is the major advantage of an emitter
disadvantage is overcome in most
follower. Small load resistances that would overload a CE amplifier can
emitter follower designs by simply
be used with an emitter follower because it steps up the impedance and
not using the biasing resistors R 1
prevents overloading.
and R 2. Instead, the emitter
follower is dc-biased by the stage
driving the emitter follower. Input Impedance of the Stage
When the ac source is not stiff, some of the ac signal will be lost across the internal resistance. If you want to
calculate the effect of the internal resistance, you will need to use the input impedance of the stage, given by:
z in(stage) R1 R2 (re re ) (11-4)
With the input impedance and the source resistance, you can use the voltage divider to calculate the input
voltage reaching the base. The calculations are the same as shown in earlier chapters.

Example 11-1
What is the input impedance of the base in Fig. 11-3 if 200? What is the input impedance of the stage?

Figure 11-3 Example.

+10 V

R1
RG 10 kΩ
600 Ω

VG R2
1V 10 kΩ
RE RL
4.3 kΩ 10 kΩ

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SOLUTION Because each resistance in the voltage divider is 10 k , the dc base voltage is half the supply
voltage, or 5 V. The dc emitter voltage is 0.7 V less, or 4.3 V. The dc emitter voltage is 4.3 V divided by 4.3 k ,
or 1 mA. Therefore, the ac resistance of the emitter diode is:
25 mV
re 25
1 mA
The external ac emitter resistance is the parallel equivalent of RE and RL, which is:
re 4.3 k 10 k 3k
Since the transistor has an ac current gain of 200, the input impedance of the base is:
zin(base) 200(3 k 25 ) 605 k
The input impedance of the base appears in parallel with the two biasing resistors. The input im-
pedance of the stage is:
zin(stage) 10 k 10 k 605 k 4.96 k
Because the 605 k is much larger than 5 k , troubleshooters usually approximate the input impedance of
the stage as the parallel of the biasing resistors only:
zin(stage) 10 k 10 k 5k

PRACTICE PROBLEM 11-1 Find the input impedance of the base and the stage, using Fig. 11-3, if
changes to 100.

Example 11-2
Assuming a of 200, what is the ac input voltage to the emitter follower of Fig. 11-3?

SOLUTION Figure 11-4 shows the ac equivalent circuit. The ac base voltage appears across zin. Because
the input impedance of the stage is large compared to the generator resistance, most of the generator voltage
appears at the base. With the voltage-divider theorem:
5k
vin 1V 0.893 V
5k 600

Figure 11-4 Example.

RG

600 Ω
VG zin
1V 5 kΩ re
3.03 kΩ

PRACTICE PROBLEM 11-2 If the value is 100, find the ac input voltage of Fig. 11-3.

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Example 11-3
What is the voltage gain of the emitter follower in Fig. 11-5? If 150, what is the ac load voltage?

Figure 11-5 Example.

+15 V

R1
RG 4.7 kΩ
600 Ω

R2
4.7 kΩ
VG RE RL
1V 2.2 kΩ 6.8 kΩ

SOLUTION The dc base voltage is half the supply voltage:

VB 7.5 V
The dc emitter current is:
6.8 V
IE 3.09 mA
2.2 k
and the ac resistance of the emitter diode is:
2 5 mV
re 8.09
3. 09 mA
The external ac emitter resistance is:
re 2.2 k 6.8 k 1.66 k
The voltage gain equals:
1.66 k
AV 0.995
1.66 k + 8.09
The input impedance of the base is:
zin(base) 150(1.66 k 8.09 ) 250 k
This is much larger than the biasing resistors. Therefore, to a close approximation, the input impedance of
the emitter follower is:
zin(stage) 4.7 k 4.7 k 2.35 k
The ac input voltage is:
2.35 k
vin 1V 0.797 V
600 + 2.35 k
The ac output voltage is:
vout 0.995(0.797 V) 0.793 V

PRACTICE PROBLEM 11-3 Repeat Example 11-3 using an RG value of 50 .

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11-2 Output Impedance

The output impedance of an amplifier is the same as its Thevenin impedance. One of the advantages of an
emitter follower is its low output impedance.
As discussed in earlier electronics courses, maximum power transfer occurs when the load imped-
ance is matched (made equal) to the source (Thevenin) impedance. Sometimes, when maximum load power
is wanted, a designer can match the load impedance to the output impedance of an emitter follower. For in-
stance, the low impedance of a speaker can be matched to the output impedance of an emitter follower to de-
liver maximum power to the speaker.

Basic Idea
Figure 11-6a shows an ac generator driving an amplifier. If the source is not stiff, some of the ac voltage is
dropped across the internal resistance RG. In this case, we need to analyze the voltage divider shown in
Fig. 11-6b to get the input voltage vin.
A similar idea can be used with the output side of the amplifier. In Fig. 11-6c, we can apply the
Thevenin theorem at the load terminals. Looking back into the amplifier, we see an output impedance z out. In
the Thevenin equivalent circuit, this output impedance forms a voltage divider with the load resistance, as
shown in Fig. 11-6d. If z out is much smaller than RL, the output source is stiff and vout equals vth.

CE Amplifiers
Figure 11-7a shows the ac equivalent circuit for the output side of a CE amplifier. When we apply Thevenin’s
theorem, we get Fig. 11-7b. In other words, the output impedance facing the load resistance is RC. Since the

Figure 11-6 Input and output impedances.

RG RG

vg AMPLIFIER RL vg zin vin

zin
(a ) (b)

RG zout

vg AMPLIFIER RL vth RL vout

zout
(c) (d )

Figure 11-7 Output impedance of CE stage.

RC

ic RC RL vth RL vout

(a) (b)

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Figure 11-8 Output impedance of emitter follower.

re⬘
RG R1 R2 A

RE RL

(a)

zout A

vth RL vout

(b)

voltage gain of a CE amplifier depends on RC, a designer cannot make RC too small without losing voltage
gain. Stated another way, it is very difficult to get a small output impedance with a CE amplifier. Because of
this, CE amplifiers are not suited to driving small load resistances.

Emitter Follower
Figure 11-8a shows the ac equivalent circuit for an emitter follower. When we apply Thevenin’s theorem to
point A, we get Fig. 11-8b. The output impedance z out is much smaller than you can get with a CE amplifier.
It equals:
RG R1 R2
z out RE re (11-5)

The impedance of the base circuit is RG R1 R2. The current gain of the transistor steps this im-
pedance down by a factor of . The effect is similar to what we get with a swamped amplifier, except that we
are moving from the base back to the emitter. Therefore, we get a reduction of impedance rather than an in-
crease. The stepped-down impedance of (RG R1 R2)/ is in series with re as indicated by Eq. (11-5).

GOOD TO KNOW Ideal Action

Transformers can also be used to
In some designs the biasing resistances and the ac resistance of the emit-
match impedances between source
ter diode become negligible. In this case, the output impedance of an
and load. Looking into the
emitter follower can be approximated by:
transformer,
RG
Np 2 z out (11-6)
z in RL
Ns
This brings out the key idea of an emitter follower: It steps the impedance
of the ac source down by a factor of . As a result, the emitter follower al-

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lows us to build stiff ac sources. Instead of using a stiff ac source that maximizes the load voltage, a designer
may prefer to maximize the load power. In this case, instead of designing for:
z out RL (Stiff voltage source)
the designer will select values to get:
z out RL (Maximum power transfer)
In this way, the emitter follower can deliver maximum power to a low-impedance load such as a stereo
speaker. By basically removing the effect of RL on the output voltage, the circuit is acting as a buffer between
the input and output.
Equation (11-6) is an ideal formula. You can use it to get an approximate value for the output im-
pedance of an emitter follower. With discrete circuits, the equation usually gives only an estimate of the out-
put impedance. Nevertheless, it is adequate for troubleshooting and preliminary analysis. When necessary,
you can use Eq. (11-5) to get an accurate value for the output impedance.

Example 11-4
Estimate the output impedance of the emitter follower of Fig. 11-9a.

SOLUTION Ideally, the output impedance equals the generator resistance divided by the current gain of
the transistor:
600
zout 2
300
Figure 11-9b shows the equivalent output circuit. The output impedance is much smaller than the load
resistance, so that most of the signal appears across the load resistor. As you can see, the output source of
Fig. 11-9b is almost stiff because the ratio of load to source resistance is 50.
PRACTICE PROBLEM 11-4 Using Fig. 11-9, change the source resistance to 1 k and solve for the
approximate zout value.

Example 11-5
Calculate the output impedance in Fig. 11-9a using Eq. (11-5).

SOLUTION The quiescent base voltage is approximately:

VBQ 15 V
Ignoring VBE, the quiescent emitter current is approximately:
15 V
IEQ 150 mA
100
The ac resistance of the emitter diode is:
25 mV
re 0.167
150 mA
The impedance seen looking back from the base is:
RG R1 R2 600 10 k 10 k 536
The current gain steps this down to:
RG R1 R2 536
1.78
300
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Figure 11-9 Example.

+30 V

10 kΩ

b = 300
600 Ω

1 VPP 10 kΩ
100 Ω 100 Ω

(a)

2Ω

vth 100 Ω

(b)

This is in series with re , so that the impedance looking back into the emitter is:
RG R1 R2
re 0.167 1.78 1.95

This is in parallel with the dc emitter resistance, so that the output impedance is:
RG R1 R2
zout RE re 100 1.95 1.91

This accurate answer is extremely close to the ideal answer of 2 . This result is typical of many
designs. For all troubleshooting and preliminary analysis, you can use the ideal method to estimate the out-
put impedance.

PRACTICE PROBLEM 11-5 Repeat Example 11-5 using an RG value of 1 k .

11-3 Cascading CE and CC

To illustrate the buffering action of a CC amplifier, suppose we have a load resistance of 270 . If we try to
couple the output of a CE amplifier directly into this load resistance, we may overload the amplifier. One way
to avoid this overload is by using an emitter follower between the CE amplifier and the load resistance. The
signal can be coupled capacitively (this means through coupling capacitors), or it may be direct coupled as
shown in Fig. 11-10.
As you can see, the base of the second transistor is connected directly to the collector of the first
transistor. Because of this, the dc collector voltage of the first transistor is used to bias the second transistor.
If the dc current gain of the second transistor is 100, the dc resistance looking into the base of the second tran-
sistor is Rin 100 (270 ) 27 k .

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