L20

5.2.3. Voltage Transfer Characteristics

5.2.4 Commonly Used Bipolar Circuits: DC
Analysis
Chapter 5
The Bipolar Junction Transistor
Donald A. Neamen (2009). Microelectronics: Circuit Analysis and Design,
4th Edition, Mc-Graw-Hill
Prepared by: Dr. Hani Jamleh, School of Engineering, The University of Jordan

2018-11-25 Electronics I - Dr. Hani Jamleh - JU 1

5.2.3 Voltage Transfer Characteristics
• A plot of the voltage transfer characteristics (output voltage versus
input voltage) can also be used to visualize:
1. The operation of a circuit or
2. The state of a transistor.
• The following example considers both an npn and a pnp transistor
circuit.

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EXAMPLE 5.6
• Objective: Develop the voltage transfer curves for
the circuits shown in Figures 5.27(a) and 5.27(b).
• Assume npn transistor parameters of:
𝑉𝐵𝐸 (𝑜𝑛) = 0.7𝑉, 𝛽 = 120, 𝑉𝐶𝐸 (𝑠𝑎𝑡) = 0.2𝑉, and 𝑉𝐴 = ∞.
• Assume pnp transistor parameters of:
𝑉𝐸𝐵 (𝑜𝑛) = 0.7𝑉, 𝛽 = 80, 𝑉𝐸𝐶 (𝑠𝑎𝑡) = 0.2𝑉, and 𝑉𝐴 = ∞.

2018-11-25 JUEE – Electronics I – Dr. Hani Jamleh Figures 5.27 3

EXAMPLE 5.6
• Solution (npn Transistor Circuit):
1. For 𝑉𝐼 ≤ 0.7𝑉, the transistor 𝑄𝑛 is cut off, so
that 𝐼𝐵 = 𝐼𝐶 = 0𝐴.
𝑉𝑂 = 𝑉 + − 𝐼𝐶 ∙ 𝑅𝐶 = 5𝑉.

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Plot direction
EXAMPLE 5.6
2. For 𝑉𝐼 > 0.7𝑉, the transistor 𝑄𝑛 turns on and
is initially biased in the forward active mode.
𝑉𝐼 −0.7 𝛽 𝑉𝐼 −0.7
𝐼𝐵 = and 𝐼𝐶 = 𝛽𝐼𝐵 =
𝑅𝐵 𝑅𝐵
𝛽 𝑉𝐼 − 0.7 𝑅𝐶
𝑉𝑂 = 5 − 𝐼𝐶 𝑅𝐶 = 5 −
𝑅𝐵
−𝛽𝑅𝐶 𝛽𝑅𝐶
= 𝑉𝐼 + 5 + ∗ 0.7
𝑅𝐵 𝑅𝐵
−120 ∗ 5𝑘 120 ∗ 5𝑘
𝑉𝑂 = 𝑉𝐼 + 5 + ∗ 0.7
150𝑘 150𝑘
= −4𝑉𝐼 + 7.8
NOTE: This equation is valid for 0.2 ≤ 𝑉𝑂 ≤ 5𝑉.
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Plot direction
EXAMPLE 5.6
3. When 𝑉𝑂 = 0.2𝑉, the transistor 𝑄𝑛 goes into
saturation. The input voltage is found from:
𝛽 𝑉𝐼 − 0.7 𝑅𝐶
𝑉𝑂 = 5 −
𝑅𝐵
120 𝑉𝐼 − 0.7 5𝑘
0.2 = 5 −
150𝑘
• which yields 𝑉𝐼 = 1.9𝑉.
• For 5 ≥ 𝑉𝐼 ≥ 1.9𝑉 , the transistor 𝑄𝑛 remains
biased in the saturation region.

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Plot direction
EXAMPLE 5.6
• Solution (pnp Transistor Circuit):
1. For 4.3 ≤ 𝑉𝐼 ≤ 5𝑉, the transistor 𝑄𝑝 is cut off, so
that 𝐼𝐵 = 𝐼𝐶 = 0𝐴.
𝑉𝑂 = 𝐼𝐶 ∙ 𝑅𝐶 = 0𝑉

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EXAMPLE 5.6
2. For 𝑉𝐼 < 4.3𝑉, the transistor 𝑄𝑝 turns on and is
biased in the forward-active mode.
(5−0.7)−𝑉𝐼 5−0.7 −𝑉𝐼
• 𝐼𝐵 = and 𝐼𝐶 = 𝛽𝐼𝐵 = 𝛽
𝑅𝐵 𝑅𝐵
5 − 0.7 − 𝑉𝐼
𝑉𝑂 = 𝐼𝐶 𝑅𝐶 = 𝛽𝑅𝐶
𝑅𝐵
𝛽𝑅𝐶 𝛽𝑅𝐶
=− 𝑉𝐼 + 4.3
𝑅𝐵 𝑅𝐵
80 ∗ 8𝑘 80 ∗ 8𝑘
𝑉𝑜 = − 𝑉𝐼 + ∗ 4.3 = −3.2𝑉𝐼 + 13.76
200𝑘 200𝑘
• NOTE: This equation is valid for 0 ≤ 𝑉𝑂 ≤ 4.8𝑉.

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EXAMPLE 5.6
3. When 𝑉𝑂 = 4.8𝑉, the transistor 𝑄𝑝 goes into
saturation, the input voltage is found from:
5 − 0.7 − 𝑉𝐼
4.8 = 80 8𝑘
200𝑘
• which yields 𝑉𝐼 = 2.8𝑉.
• For 0 ≤ 𝑉𝐼 ≤ 2.8𝑉 , the transistor 𝑄𝑝 remains
biased in the saturation mode.

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EXAMPLE 5.6 npn
• Comment: As shown in this example, the
voltage transfer characteristics are
determined by:
• Finding the range of input voltage values → that
biases the transistor in:
1. The cutoff mode,
2. The forward-active mode, or
3. The saturation mode. pnp

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5.2.4 Commonly Used Bipolar Circuits: DC Analysis
• There are a number of other BJT circuit configurations, in addition to
the common-emitter circuits that are commonly used.
• Several examples of such circuits are presented in this section. BJT
circuits tend to be very similar in terms of DC analysis procedures.
• The same basic analysis approach will work regardless of the appearance of
the circuit.
• We continue our DC analysis and design of bipolar circuits to increase
our proficiency and to become more comfortable with these types of
circuits.

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EXAMPLE 5.7: Common Emitter Containing an
Emitter Resistor 𝑅𝐸
• Objective: Calculate the characteristics of a circuit
containing an emitter resistor.
• For the circuit shown in Figure 5.30(a), let
𝑉𝐵𝐸 (𝑜𝑛) = 0.7𝑉 and 𝛽 = 75.
• Note that the circuit has both positive 𝑉 + and negative
𝑉 − power supply voltages.

Figure 5.30

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EXAMPLE 5.7: Common Emitter Containing an
Emitter Resistor 𝑅𝐸
• Solution 1 (𝑄-point values):
• Writing KVL equation around the B–E loop, we
have:
𝑉𝐵𝐵 = 𝐼𝐵 𝑅𝐵 + 𝑉𝐵𝐸 (𝑜𝑛) + 𝐼𝐸 𝑅𝐸 + 𝑉 − 𝑉𝐵𝐵
• Assuming the transistor is biased in the forward-
active mode, we can write: 𝐼𝐸 = 1 + 𝛽 𝐼𝐵
• We can then solve the Equation above for the base
current:
𝑉𝐵𝐵 − 𝑉𝐵𝐸 (𝑜𝑛) − 𝑉 − 1 − 0.7 − (−1.8)
𝐼𝐵 = = ⇒ 2.665 𝜇𝐴
𝑅𝐵 + 1 + 𝛽 𝑅𝐸 560𝑘 + (76)(3𝑘) Figure 5.30

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EXAMPLE 5.7: Common Emitter Containing an
Emitter Resistor
• The collector and emitter currents are:
𝐼𝐶 = 𝛽𝐼𝐵 = (75)(2.665𝜇𝐴) ⇒ 0.20 𝑚𝐴
𝐼𝐸 = 1 + 𝛽 𝐼𝐵 = (76)(2.665𝜇𝐴) ⇒ 0.203 𝑚𝐴
• From Figure 5.30(b), by applying KVL, the
collector–emitter voltage is:
𝑉𝐶𝐸 = 𝑉 + − 𝐼𝐶 𝑅𝐶 − 𝐼𝐸 𝑅𝐸 − 𝑉 −
= 1.8 − (0.20𝑚)(7𝑘) − (0.203𝑚)(3𝑘)
− (−1.8)=1.59V
• Since 𝑉𝐶𝐸 > 𝑉𝐵𝐸 𝑜𝑛 > 𝑉𝐶𝐸 𝑠𝑎𝑡 → the transistor is
biased in the forward-active mode, as initially assumed.
Figure 5.30

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EXAMPLE 5.7: Common Emitter Containing an
Emitter Resistor
• Solution 2 (load line): We again use Kirchhoff’s
voltage law around the C–E loop.
• From the relationship between the collector and
emitter currents, we find:
+ −
1+𝛽
𝑉𝐶𝐸 = 𝑉 − 𝑉 − 𝐼𝐶 𝑅𝐶 + 𝑅𝐸
𝛽
Voltage Sources Voltage Drops
76
= 1.8 − −1.8 − 𝐼𝐶 7𝑘 + 3𝑘
75
= 3.6 − 𝐼𝐶 (10.04𝑘) Figure 5.30

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EXAMPLE 5.7: Common Emitter Containing an
Emitter Resistor
𝑉𝐶𝐸 = 3.6 − 𝐼𝐶 (10.04𝑘)

Figure 5.30

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DESIGN EXAMPLE 5.8: common-base circuit
• Objective: Design the common-base circuit shown
in Figure 5.32 such that:
1. 𝐼𝐸𝑄 = 0.50 𝑚𝐴 and
2. 𝑉𝐸𝐶𝑄 = 4.0𝑉.
• Assume transistor parameters of:
Figure 5.32
• 𝛽 = 120 and 𝑉𝐸𝐵 (𝑜𝑛) = 0.7𝑉.

• Find 𝑅𝐸 , and 𝑅𝐶 .

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DESIGN EXAMPLE 5.8: common-base circuit
• Solution: Writing Kirchhoff’s voltage law equation
around the E-B loop (assuming the transistor is
biased in the forward-active mode), we have:
+
𝐼𝐸𝑄
𝑉 = 𝐼𝐸𝑄 𝑅𝐸 + 𝑉𝐸𝐵 (𝑜𝑛) + 𝑅𝐵
1+𝛽
0.5𝑚 Figure 5.32
5 = 0.5𝑚 𝑅𝐸 + 0.7 + (10𝑘)
121
• which yields:
𝑅𝐸 = 8.52 𝑘Ω

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DESIGN EXAMPLE 5.8: common-base circuit
• We can find:
𝛽 120
𝐼𝐶𝑄 = ∙ 𝐼𝐸𝑄 = (0.5) = 0.496 𝑚𝐴
1+𝛽 121
• Now, writing Kirchhoff’s voltage law equation around
the E-C loop, we have:
𝑉 + = 𝐼𝐸𝑄 𝑅𝐸 + 𝑉𝐸𝐶𝑄 + 𝐼𝐶𝑄 𝑅𝐶 + 𝑉 −
Figure 5.32
5 = (0.5𝑚)(8.52𝑘) + 4 + 0.496𝑚 𝑅𝐶 + (−5)
• which yields:
𝑅𝐶 = 3.51 𝑘Ω
• Comment: The circuit analysis of the common-base
circuit proceeds in the same way as all previous
circuits.
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 Transistor Configurations ... Common Base ,
Collector , Emitter ……
• There are three basic circuit configurations that can be used with
transistors.
• Known as:
1. Common Emitter,
2. Common Base, and
3. Common Collector,
• These three circuit configurations have different attributes.
• When designing a transistor circuit it is necessary to adopt the
transistor circuit configuration that will provide the required
attributes.
Source : http://www.radio-electronics.com/info/circuits/transistor/circuit-configurations.php
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 Origin of the Terminology
• The terminology used for denoting the three basic transistor
configurations indicates:
• The transistor terminal that is common to both input and output circuits.
• The term grounded, i.e. grounded base, grounded collector and
grounded emitter may also be used on occasions because the
common element signal is normally grounded.

Source : http://www.radio-electronics.com/info/circuits/transistor/circuit-configurations.php
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 Different Configuration Attributes
[for your reference only]

Source : http://onebyzeroelectronics.blogspot.com/2015/08/transistor-configurations-common-
base.html
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5.2.4 Commonly Used Bipolar Circuits: DC
Analysis

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DESIGN EXAMPLE 5.9
• Objective: Design a pnp bipolar transistor circuit
to meet a set of specifications.
• Specifications:
1. The circuit configuration to be designed is shown in
Figure 5.36(a).
2. The quiescent emitter-collector voltage is to be:
𝑉𝐸𝐶𝑄 = 2.5 𝑉
• Choices:
1. Discrete resistors with tolerances of ±10 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 are
to be used,
2. An emitter resistor with a nominal value of 𝑅𝐸
= 2𝑘Ω is to be used, and
3. A transistor with 𝛽 = 60 and 𝑉𝐸𝐵 (𝑜𝑛) = 0.7𝑉 is
available. Figure 5.36

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DESIGN EXAMPLE 5.9
• Solution (ideal 𝑸-point value):
• We have 𝑅𝐸 → Writing the KVL equation around the C–
E loop:
𝑉 + = 𝐼𝐸𝑄 ∙ 𝑅𝐸 + 𝑉𝐸𝐶𝑄 → 5 = 𝐼𝐸𝑄 (2𝑘) + 2.5
• Which yields 𝐼𝐸𝑄 from:
𝐼𝐸𝑄 = 1.25𝑚𝐴.
• The collector current is:
𝛽 60
𝐼𝐶𝑄 = · 𝐼𝐸𝑄 = (1.25𝑚) = 1.23𝑚𝐴
1+𝛽 61
• The base current is:
𝐼𝐸𝑄 1.25𝑚
𝐼𝐵𝑄 = = = 0.0205 𝑚𝐴 = 20.5𝜇𝐴
1+𝛽 61 Figure 5.36

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DESIGN EXAMPLE 5.9
• To find 𝑅𝐵 → Writing the KVL equation around
the E–B loop:
𝑉 + = 𝐼𝐸𝑄 ∙ 𝑅𝐸 + 𝑉𝐸𝐵 𝑜𝑛 + 𝐼𝐵𝑄 ∙ 𝑅𝐵 + 𝑉𝐵𝐵
5 = (1.25𝑚)(2𝑘) + 0.7 + 0.0205𝑚 𝑅𝐵 + (−2)
• Which yields:
𝑅𝐵 = 185𝑘Ω

Figure 5.36

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DESIGN EXAMPLE 5.9
• Solution (ideal load line): The load line
equation is:
+ +
1+𝛽
𝑉𝐸𝐶 = 𝑉 − 𝐼𝐸 ∙ 𝑅𝐸 = 𝑉 − 𝐼𝐶 𝑅𝐸
𝛽
61
𝑉𝐸𝐶 = 5 − 𝐼𝐶 (2𝑘) = 5 − 𝐼𝐶 (2.03𝑘)
60
• The load line, using the nominal value of 𝑅𝐸 ,
and the calculated 𝑄-point are shown in Figure
5.37(a).
Figure 5.37

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DESIGN EXAMPLE 5.9
• Trade-offs: As shown in Appendix C in the textbook, a standard
resistor value of 185 𝑘Ω is not available.
• We will pick a value of 180 𝑘Ω.
• We will consider 𝑅𝐵 and 𝑅𝐸 resistor tolerances of ±10 𝑝𝑒𝑟𝑐𝑒𝑛𝑡.
• The quiescent collector current is given by:
𝑉 + − 𝑉𝐸𝐵 (𝑜𝑛) − 𝑉𝐵𝐵 6.3
𝐼𝐶𝑄 = 𝛽 = (60)
𝑅𝐵 + 1 + 𝛽 𝑅𝐸 𝑅𝐵 + 61 𝑅𝐸
• Then, the load line equation is given by:
+
1 +𝛽 61
𝑉𝐸𝐶 = 𝑉 − 𝐼𝐶 𝑅𝐸 = 5 − 𝐼𝐶 𝑅𝐸
𝛽 60

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DESIGN EXAMPLE 5.9
61
• The extreme values (worst cases) of 𝑉𝐸𝐶 = 5− 𝐼 𝑅
𝑅𝐵 are: 60 𝐶 𝐸

180𝑘Ω − 10% = 162𝑘Ω

180𝑘Ω + 10% = 198𝑘Ω
• The Q-point values for the extreme
values of 𝑅𝐵 and 𝑅𝐸 are given in the
following table:

Figure 5.37
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DESIGN EXAMPLE 5.9
61
• Figure 5.37(b) shows the 𝑄-points for 𝑉𝐸𝐶 = 5− 𝐼 𝑅
60 𝐶 𝐸
the various possible extreme values of
𝑅𝐸 and 𝑅𝐵 .
• The shaded area shows the region in
which the 𝑄 − 𝑝𝑜𝑖𝑛𝑡 will occur over
the range of resistor values.
• Comment: This example shows that
an ideal 𝑄-point can be determined
based on a set of specifications, but,
because of resistor tolerance:
• The actual 𝑄-point will vary over a range
of values.
Figure 5.37
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Example 5.10 Transistor Circuit With a Load
Resistance 𝑅𝐿
• Objective: Calculate the characteristics of an
npn bipolar transistor circuit with a load
resistance 𝑅𝐿 .
• The load resistance can represent a second
transistor stage connected to the output of a
transistor circuit.
• For the circuit shown in Figure 5.38(a), the
transistor parameters are:
• 𝑉𝐵𝐸 (𝑜𝑛) = 0.7𝑉, and 𝛽 = 100.

Figure 5.38

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Example 5.10 Transistor Circuit With a Load
Resistance 𝑅𝐿
• Explanation: The load resistance can represent a
second transistor stage connected to the output of
a transistor circuit.

𝑅𝐿
Figure 5.38

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Example 5.10 Transistor Circuit With a Load
Resistance 𝑅𝐿
• Solution (𝑸-Point Values): Kirchhoff’s voltage
law equation around the B–E loop yields
𝐼𝐵 𝑅𝐵 + 𝑉𝐵𝐸 (𝑜𝑛) + 𝐼𝐸 𝑅𝐸 + 𝑉 − = 0
• Again assuming 𝐼𝐸 = 1 + 𝛽 𝐼𝐵 , we find
− 𝑉 − + 𝑉𝐵𝐸 𝑜𝑛 − −5 + 0.7
𝐼𝐵 = =
𝑅𝐵 + 1 + 𝛽 𝑅𝐸 10𝑘 + (101)(5𝑘)
⇒ 8.35 𝜇𝐴

Figure 5.38

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Example 5.10 Transistor Circuit With a Load
Resistance 𝑅𝐿
• The collector and emitter currents are:
𝐼𝐶 = 𝛽𝐼𝐵 = (100)(8.35𝜇𝐴) ⇒ 0.835𝑚𝐴
• and
𝐼𝐸 = 1 + 𝛽 𝐼𝐵 = (101)(8.35𝜇𝐴) ⇒ 0.843𝑚𝐴
• At the collector node, we can write:
𝑉 + − 𝑉𝑂 𝑉𝑂
𝐼𝐶 = 𝐼1 − 𝐼𝐿 = −
𝑅𝐶 𝑅𝐿
12 − 𝑉𝑂 𝑉𝑂
0.835𝑚 = −
5𝑘 5𝑘
• Solving for 𝑉𝑂 , we find:
𝑉𝑂 = 3.91𝑉
• The currents are then:
• 𝐼1 = 1.620 𝑚𝐴 and
• 𝐼𝐿 = 0.782 𝑚𝐴.
Figure 5.38

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Example 5.10 Transistor Circuit With a Load
Resistance 𝑅𝐿
• Referring to Figure 5.38(b), the collector–
emitter voltage is:
𝑉𝐶𝐸 = 𝑉𝑂 − 𝐼𝐸 𝑅𝐸 − −5
= 3.91 − (0.843𝑚)(5𝑘) − (−5) = 4.70𝑉

Figure 5.38

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Example 5.10 Transistor Circuit With a Load
Resistance 𝑅𝐿
• Solution (Load Line): The load line equation for
this circuit is not as straightforward as for previous
circuits.
• The easiest approach to finding the load line is to
make a “Thevenin equivalent circuit” of
𝑅𝐿 , 𝑅𝐶 , 𝑎𝑛𝑑 𝑉 + , as indicated in Figure 5.38(a).
• The Thevenin equivalent resistance is:
𝑅𝑇𝐻 = 𝑅𝐿 ∥ 𝑅𝐶 = 5𝑘 ∥ 5𝑘 = 2.5𝑘
• and the Thevenin equivalent voltage is
𝑅𝐿 +
5𝑘
𝑉𝑇𝐻 = ·𝑉 = · (12) = 6𝑉
𝑅𝐿 + 𝑅𝐶 5𝑘 + 5𝑘
Figure 5.38

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Example 5.10 Transistor Circuit With a Load
Resistance 𝑅𝐿
• The equivalent circuit is shown in Figure
5.38(c).
• The KVL equation around the C–E loop is
𝑉𝐶𝐸 = 6 − −5 − 𝐼𝐶 𝑅𝑇𝐻 − 𝐼𝐸 𝑅𝐸
101
= 11 − 𝐼𝐶 2.5𝑘 − 𝐼𝐶 ∙ (5𝑘)
100
• or
𝑉𝐶𝐸 = 11 − 𝐼𝐶 (7.55𝑘)

2018-11-25 JUEE – Electronics I – Dr. Hani Jamleh Figure 5.38 37

Example 5.10 Transistor Circuit With a Load
Resistance 𝑅𝐿
𝑉𝐶𝐸 = 11 − 𝐼𝐶 (7.55𝑘)
• 𝑉𝐶𝐸 𝐶𝑢𝑡𝑜𝑓𝑓 = 11𝑉,
11
• 𝐼𝐶 𝑆𝑎𝑡 = = 1.46𝑚𝐴
7.55

• The load line and the calculated

𝑄 -point values are shown in
Figure 5.39.

• Question: What is the effect of

Figure 5.39
attaching a load 𝑅𝐿 directly to the
output of a BJT circuit on the
load line?
2018-11-25 JUEE – Electronics I – Dr. Hani Jamleh 38
Example 5.10 Transistor Circuit With a Load
Resistance 𝑅𝐿
• Comment: Remember that the
collector current, determined from
𝐼𝐶 = 𝛽𝐼𝐵 , is the current into the
collector terminal of the transistor.
• It is not necessarily the current in the
collector resistor 𝑅𝐶 .

2018-11-25 JUEE – Electronics I – Dr. Hani Jamleh 39