EE212 Electronics I Lecture # 4

EE212
- Electronics I -
Lecture # 4

Instructor: M. Zübeyir Ünlü, PhD

zubeyirunlu@iyte.edu.tr

Lecture # 1 1
EE212 Electronics I Lecture # 4 The Book

Chapters of EE212 - Electronics I

Lecture # 1 2
EE212 Electronics I Lecture # 4 Outline
Chapter 4: Basic FET Amplifiers
Preview
4.1 The MOSFET Amplifier 206
4.2 Basic Transistor Amplifier Configurations 216
4.3 The Common-Source Amplifier 216
4.4 The Common-Drain (Source-Follower) Amplifier 227
4.5 The Common-Gate Configuration 234
4.6 The Three Basic Amplifier Configurations: Summary and Comparison 237
4.7 Single-Stage Integrated Circuit MOSFET Amplifiers 238
4.8 Multistage Amplifiers 254
4.9 Basic JFET Amplifiers 258 (Excluded!)
4.10 Design Application: A Two-Stage Amplifier 264
4.11 Summary
Problems

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EE212 Electronics I Lecture # 4 Preview

In this chapter, we emphasize the use of the FETs in linear amplifier applications. Linear amplifiers
imply that, for the most part, we are dealing with analog signals. The magnitude of an analog signal
may have any value, within limits, and may vary continuously with respect to time. Although a
major use of MOSFETs is in digital applications, they are also used in linear amplifier circuits.

We will begin to see how all-transistor circuits, that is, circuits with no resistors, can be designed.
Since MOS transistors are small devices, high-density all-transistor circuits can be fabricated as
integrated circuits.

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EE212 Electronics I Lecture # 4 Preview

In this chapter, we will:

• Investigate the process by which a single-MOS transistor circuit can amplify a small, time-varying
input signal.

• Develop the small-signal models of the transistor that are used in the analysis of linear amplifiers.

• Discuss the three basic transistor amplifier configurations.

• Analyze the common-source, source-follower, and common-gate amplifiers, and become familiar
with the general characteristics of these circuits.

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EE212 Electronics I Lecture # 4 Preview

• Compare the general characteristics of the three basic amplifier configurations.

• Analyze all-MOS transistor circuits that become the foundation of integrated circuits.

• Analyze multitransistor or multistage amplifiers and understand the advantages of these circuits
over single-transistor amplifiers.

• Develop the small-signal model of JFET devices and analyze basic JFET amplifiers. (Excluded!)

• As an application, incorporate MOS transistors in a design of a two-stage amplifier.

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EE212 Electronics I Lecture # 4 4.1 THE MOSFET AMPLIFIER

Objective: Investigate the process by which a single-transistor circuit can amplify a small, time-
varying input signal and develop the small-signal models of the transistor that are used in the
analysis of linear amplifiers.

In this chapter, we will be considering signals, analog circuits, and amplifiers. A signal contains
some type of information. For example, sound waves produced by a speaking human contain the
information the person is conveying to another person. A sound wave is an analog signal. The
magnitude of an analog signal can take on any value, within limits, and may vary continuously with
time. Electronic circuits that process analog signals are called analog circuits. One example of an
analog circuit is a linear amplifier. A linear amplifier magnifies an input signal and produces an
output signal whose magnitude is larger and directly proportional to the input signal.

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EE212 Electronics I Lecture # 4 4.1 THE MOSFET AMPLIFIER

In this chapter, we analyze and design linear amplifiers that use field-effect transistors as the
amplifying device. The term small signal means that we can linearize the ac equivalent circuit. We
will define what is meant by small signal in the case of MOSFET circuits. The term linear amplifiers
means that we can use superposition so that the dc analysis and ac analysis of the circuits can be
performed separately and the total response is the sum of the two individual responses.

The mechanism with which MOSFET circuits amplify small time-varying signals was introduced in
the last chapter. In this section, we will expand that discussion using the graphical technique, dc
load line, and ac load line. In the process, we will develop the various small-signal parameters of
linear circuits and the corresponding equivalent circuits.

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EE212 Electronics I Lecture # 4 4.1 THE MOSFET AMPLIFIER

4.1.1 Graphical Analysis, Load Lines, and Small-Signal Parameters

Figure 4.1 shows an NMOS common-source circuit with a time-varying

voltage source in series with the dc source. We assume the time-varying
input signal is sinusoidal. Figure 4.2 shows the transistor characteristics,
dc load line, and Q-point, where the dc load line and Q-point are
functions of vGS, VDD, RD, and the transistor parameters. For the output
voltage to be a linear function of the input voltage, the transistor must
be biased in the saturation region. (Note that, although we primarily use
n-channel, enhancement-mode MOSFETs in our discussions, the same
results apply to the other MOSFETs.)

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EE212 Electronics I Lecture # 4 4.1 THE MOSFET AMPLIFIER

Also shown in Figure 4.2 are the sinusoidal variations in the gate-to-source voltage, drain
current, and drain-to-source voltage, as a result of the sinusoidal source vi.

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EE212 Electronics I Lecture # 4 4.1 THE MOSFET AMPLIFIER

The total gate-to-source voltage is the sum of VGSQ and vi .

As vi increases, the instantaneous value of vGS increases, and the bias point moves up the
load line.

A larger value of vGS means a larger drain current and a smaller value of vDS.

For a negative vi (the negative portion of the sine wave), the instantaneous value of vGS
decreases below the quiescent value, and the bias point moves down the load line.

A smaller vGS value means a smaller drain current and increased value of vDS.

Once the Q-point is established, we can develop a mathematical model for the sinusoidal, or
small-signal, variations in gate-to-source voltage, drain-to-source voltage, and drain current.

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EE212 Electronics I Lecture # 4 4.1 THE MOSFET AMPLIFIER

The time-varying signal source vi in Figure 4.1 generates a time-varying component of the gate-
to-source voltage. In this case, vgs = vi , where vgs is the time-varying component of the gate-to-
source voltage.

For the FET to operate as a linear amplifier, the transistor must be biased in the saturation
region, and the instantaneous drain current and drain-to-source voltage must also be confined
to the saturation region.

When symmetrical sinusoidal signals are applied to the input of an amplifier, symmetrical
sinusoidal signals are generated at the output, as long as the amplifier operation remains
linear. We can use the load line to determine the maximum output symmetrical swing. If the
output exceeds this limit, a portion of the output signal will be clipped and signal distortion will
occur. Lecture # 1 12
EE212 Electronics I Lecture # 4 4.1 THE MOSFET AMPLIFIER

In the case of FET amplifiers, the output signal must avoid cutoff (iD = 0) and must stay in the
saturation region (vDS > vDS(sat)). This maximum range of output signal can be determined from
the load line in Figure 4.2.

Transistor Parameters

We will be dealing with time-varying as well as dc currents and voltages

in this chapter. Table 4.1 gives a summary of notation that will be used.

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From Figure 4.1, we see that the instantaneous gate-to-source voltage is

vGS = VGSQ + vi = VGSQ + vgs (4.1)

where VGSQ is the dc component and vgs is the ac component. The instantaneous drain current
is

iD = Kn (vGS − VT N)2 (4.2)

Substituting Equation (4.1) into (4.2) produces

iD = Kn [VGSQ + vgs − VT N]2 = Kn [(VGSQ − VT N) + vgs ]2 (4.3(a))

or

iD = Kn (VGSQ − VT N)2 + 2Kn(VGSQ − VT N) vgs + Kn v2gs (4.3(b))

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The first term in Equation (4.3(b)) is the dc or quiescent drain current IDQ, the second term is
the time-varying drain current component that is linearly related to the signal vgs, and the third
term is proportional to the square of the signal voltage. For a sinusoidal input signal, the
squared term produces undesirable harmonics, or nonlinear distortion, in the output voltage.
To minimize these harmonics, we require

vgs << 2(VGSQ − VT N) (4.4)

which means that the third term in Equation (4.3(b)) will be much smaller than the second
term. Equation (4.4) represents the small-signal condition that must be satisfied for linear
amplifiers. Neglecting the v2gs term, we can write Equation (4.3(b)) as

iD = IDQ + id (4.5)
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Again, small-signal implies linearity so that the total current can be separated into a dc
component and an ac component. The ac component of the drain current is given by

id = 2Kn(VGSQ − VT N) vgs (4.6)

The small-signal drain current is related to the small-signal gate-to-source voltage by the
transconductance gm. The relationship is (4.7)

The transconductance is a transfer coefficient relating output current to input voltage and
can be thought of as representing the gain of the transistor.

The transconductance can also be obtained from the derivative

(4.8(a))

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which can be written as (4.8(b))

The drain current versus gate-to-source voltage for the transistor biased in the saturation
region is given in Equation (4.2) and is shown in Figure 4.3. The transconductance gm is the
slope of the curve. If the time-varying signal vgs is sufficiently small, the transconductance gm is
a constant. With the Q-point in the saturation region, the transistor operates as a current
source that is linearly controlled by vgs. If the Q-point moves into the nonsaturation region, the
transistor no longer operates as a linearly controlled current source.

As shown in Equation (4.8(a)), the transconductance is directly proportional to the conduction

parameter Kn, which in turn is a function of the width-to-length ratio. Therefore, increasing the
width of the transistor increases the transconductance, or gain, of the transistor.

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EXAMPLE 4.1

Objective: Calculate the transconductance of an n-channel MOSFET.

Consider an n-channel MOSFET with parameters VT N = 0.4 V, k'n = 100μA/V2, and W/L = 25.
Assume the drain current is ID = 0.40 mA.

Solution: The conduction parameter is

Assuming the transistor is biased in the saturation region, the transconductance is determined
from Equation (4.8(b)) as

Comment: The value of the transconductance can be increased by increasing the transistor
W/L ratio and also by increasing the quiescent drain current.

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AC Equivalent Circuit

From Figure 4.1, we see that the output voltage is

vDS = vO = VDD − iD RD (4.9)

Using Equation (4.5), we obtain

vO = VDD − (IDQ + id ) RD = (VDD − IDQ RD) − id RD (4.10)

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The output voltage is also a combination of dc and ac values. The time-varying output signal is
the time-varying drain-to-source voltage, or

vo = vds = −id RD (4.11)

Also, from Equations (4.6) and (4.7), we have

id = gm vgs (4.12)

In summary, the following relationships exist between the time-varying signals for the circuit in
Figure 4.1. The equations are given in terms of the instantaneous ac values, as well as the
phasors. We have

vgs = vi or Vgs = Vi (4.13)

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and id = gm vgs or Id = gm Vgs (4.14)

and vds = −id RD or Vds = −Id RD (4.15)

The ac equivalent circuit in Figure 4.4 is developed by setting the dc sources in

Figure 4.l equal to zero. The small-signal relationships are given in Equations
(4.13), (4.14), and (4.15). As shown in Figure 4.l, the drain current, which is
composed of ac signals superimposed on the quiescent value, flows through the
voltage source VDD.
Since the voltage across this source is assumed to be constant, the sinusoidal current produces
no sinusoidal voltage component across this element. The equivalent ac impedance is
therefore zero, or a short circuit. Consequently, in the ac equivalent circuit, the dc voltage
sources are equal to zero. We say that the node connecting RD and VDD is at signal ground.
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EE212 Electronics I Lecture # 4 4.1 THE MOSFET AMPLIFIER

4.1.2 Small-Signal Equivalent Circuit

Now that we have the ac equivalent circuit for the NMOS amplifier circuit,
(Figure 4.4), we must develop a small-signal equivalent circuit for the
transistor.

Initially, we assume that the signal frequency is sufficiently low so that any
capacitance at the gate terminal can be neglected. The input to the gate thus
appears as an open circuit, or an infinite resistance. Equation (4.14) relates the
small-signal drain current to the small-signal input voltage, and Equation (4.7)
shows that the transconductance gm is a function of the Q-point. The resulting Figure 4.5 (a) Common-source
NMOS transistor with small-signal
simplified small-signal equivalent circuit for the NMOS device is shown in parameters and (b) simplified
small-signal equivalent circuit for
Figure 4.5. (The phasor components are in parentheses.) NMOS transistor
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EE212 Electronics I Lecture # 4 4.1 THE MOSFET AMPLIFIER

This small-signal equivalent circuit can also be expanded to take into account the finite output
resistance of a MOSFET biased in the saturation region. This effect, discussed in the last
chapter, is a result of the nonzero slope in the iD versus vDS curve.

We know that iD = Kn [(vGS − VT N)2 (1 + λvDS)] (4.16) where λ is the channel-length

modulation parameter and is a positive quantity. The small-signal output resistance, as
previously defined, is (4.17)

or ro = [λ Kn(VGSQ − VT N )2]−1 ≈ [λ IDQ]−1 (4.18)

This small-signal output resistance is also a function of the Q-point parameters.

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EE212 Electronics I Lecture # 4 4.1 THE MOSFET AMPLIFIER

The expanded small-signal equivalent circuit of the n-channel MOSFET is shown in Figure 4.6
in phasor notation. Note that this equivalent circuit is a transconductance amplifier in that the
input signal is a voltage and the output signal is a current. This equivalent circuit can now be
inserted into the amplifier ac equivalent circuit in Figure 4.4 to produce the circuit in Figure 4.7.

Figure 4.7 Small-signal equivalent circuit of

common-source circuit with NMOS transistor
model

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EE212 Electronics I Lecture # 4 4.1 THE MOSFET AMPLIFIER

EXAMPLE 4.2

Objective: Determine the small-signal voltage gain of a MOSFET circuit.

For the circuit in Figure 4.l, assume parameters are: VGSQ = 2.12 V, VDD = 5 V, and RD = 2.5 kΩ.
Assume transistor parameters are: VT N = 1 V. Kn = 0.80 mA/V2, and λ = 0.02V−1. Assume the
transistor is biased in the saturation region.

Solution: The quiescent values are IDQ ≈ Kn(VGSQ − VT N)2 = (0.8)(2.12 − 1)2 = 1.0 mA and

VDSQ = VDD − IDQ RD = 5 − (1)(2.5) = 2.5 V

Therefore,

VDSQ = 2.5 V > VDS(sat) = VGS − VT N = 1.82 − 1 = 0.82 V

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EE212 Electronics I Lecture # 4 4.1 THE MOSFET AMPLIFIER

which means that the transistor is biased in the saturation region, as initially assumed, and as
required for a linear amplifier. The transconductance is

gm = 2Kn(VGSQ − VT N) = 2(0.8)(2.12 − 1) = 1.79 mA/V

and the output resistance is

ro = [λ IDQ]−1 = [(0.02)(1)]−1 = 50 kΩ

From Figure 4.7, the output voltage is

Vo = −gmVgs(roǁRD)

Since Vgs = Vi , the small-signal voltage gain is

Av = Vo / Vi = −gm(roǁRd ) = −(1.79)(50ǁ2.5) = −4.26

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EE212 Electronics I Lecture # 4 4.1 THE MOSFET AMPLIFIER

Comment: The magnitude of the ac output voltage is 4.26 times larger than the magnitude of
the input voltage. Hence, we have an amplifier. Note that the small-signal voltage gain contains
a minus sign, which means that the sinusoidal output voltage is 180 degrees out of phase with
respect to the input sinusoidal signal.

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EE212 Electronics I Lecture # 4 4.1 THE MOSFET AMPLIFIER

Problem-Solving Technique: MOSFET AC Analysis

Since we are dealing with linear amplifiers, superposition applies, which means that we can
perform the dc and ac analyses separately. The analysis of the MOSFET amplifier proceeds as
follows:

1. Analyze the circuit with only the dc sources present. This solution is the dc or quiescent solution.
The transistor must be biased in the saturation region in order to produce a linear amplifier.

2. Replace each element in the circuit with its small-signal model, which means replacing the
transistor by its small-signal equivalent circuit.

3. Analyze the small-signal equivalent circuit, setting the dc source components equal to zero, to
produce the response of the circuit to the time-varying input signals only.

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EE212 Electronics I Lecture # 4 4.1 THE MOSFET AMPLIFIER

The previous discussion was for an n-channel MOSFET amplifier. The

same basic analysis and equivalent circuit also applies to the p-channel
transistor. Figure 4.8(a) shows a circuit containing a p-channel MOSFET.

transistor and (b) corresponding ac equivalent circuit

Figure 4.8 (a) Common-source circuit with PMOS
Note that the power supply voltage VDD is connected to the source.
(The subscript DD can be used to indicate that the supply is connected
to the drain terminal. Here, however, VDD is simply the usual notation
for the power supply voltage in MOSFET circuits.) Also note the change
in current directions and voltage polarities compared to the circuit
containing the NMOS transistor. Figure 4.8(b) shows the ac equivalent
circuit, with the dc voltage sources replaced by ac short circuits, and all
currents and voltages shown are the time-varying components.
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EE212 Electronics I Lecture # 4 4.1 THE MOSFET AMPLIFIER

In the circuit of Figure 4.8(b), the transistor can be replaced by the equivalent circuit in Figure
4.9. The equivalent circuit of the p-channel MOSFET is the same as that of the n-channel
device, except that all current directions and voltage polarities are reversed.

The final small-signal equivalent circuit of the p-channel MOSFET amplifier is shown in Figure
4.10. The output voltage is

Vo = gmVsg(roǁRD) (4.19)

The control voltage Vsg, given in terms of the input signal voltage, is

Vsg = - Vi (4.20)

and the small-signal voltage gain is Figure 4.10 Small-signal equivalent circuit
of common-source amplifier with PMOS
transistor model
Av = Vo / Vi = −gm(roǁRd ) (4.21) Lecture # 1 31
EE212 Electronics I Lecture # 4 4.1 THE MOSFET AMPLIFIER

This expression for the small-signal voltage gain of the p-channel MOSFET amplifier is exactly
the same as that for the n-channel MOSFET amplifier. The negative sign indicates that a 180-
degree phase reversal exists between the ouput and input signals, for both the PMOS and the
NMOS circuit.

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EE212 Electronics I Lecture # 4 4.1 THE MOSFET AMPLIFIER

We may note that if the polarity of the small-signal gate-to-source voltage is reversed, then the
small-signal drain current direction is reversed. This change of polarity is shown in Figure 4.11.
Figure 4.11(a) shows the conventional voltage polarity and current directions in a PMOS
transistor. If the control voltage polarity is reversed as shown in Figure 4.11(b), then the
dependent current direction is also reversed. The equivalent circuit shown in Figure 4.11(b) is
then the same as that of the NMOS transistor. However, the author prefers to use the small-
signal equivalent circuit in Figure 4.9 to be consistent with the voltage polarities and current
directions of the PMOS transistor.

4.1.3 Modeling the Body Effect

Note: We’ve skipped this section (4.1.3). Please read it! You are not responsible for this
section! Lecture # 1 33
EE212 Electronics I Lecture # 4 4.2 BASIC TRANSISTOR AMPLIFIER CONFIGURATIONS

4.2 BASIC TRANSISTOR AMPLIFIER CONFIGURATIONS

Objective: Discuss the three basic transistor amplifier configurations.

As we have seen, the MOSFET is a three-terminal device. Three basic single-transistor amplifier
configurations can be formed, depending on which of the three transistor terminals is used as
signal ground. These three basic configurations are appropriately called common source,
common drain (source follower), and common gate.

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EE212 Electronics I Lecture # 4 4.2 BASIC TRANSISTOR AMPLIFIER CONFIGURATIONS

The input and output resistance characteristics of amplifiers are important in determining
loading effects. These parameters, as well as voltage gain, for the three basic MOSFET circuit
configurations will be determined in the following sections. The characteristics of the three
types of amplifiers will then allow us to understand under what condition each amplifier is
most useful.

Initially, we will consider MOSFET amplifier circuits that emphasize discrete designs, in that
resistor biasing will be used. The purpose is to become familiar with basic MOSFET amplifier
designs and their characteristics. In Section 4.7, we will begin to consider integrated circuit
MOSFET designs that involve all-transistor circuits and current source biasing. These initial
designs provide an introduction to more advanced MOS amplifier designs that will be
considered in Part 2 of the text.
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EE212 Electronics I Lecture # 4 4.3 THE COMMON-SOURCE AMPLIFIER

4.3 THE COMMON-SOURCE AMPLIFIER

Objective: Analyze the common-source amplifier and become familiar with the general
characteristics of this circuit.

In this section, we consider the first of the three basic circuits—the common-source amplifier.
We will analyze several basic common-source circuits, and will determine small-signal voltage
gain and input and output impedances.

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4.3.1 A Basic Common-Source Configuration

Figure 4.14 shows the basic common-source circuit with

voltage-divider biasing. We see that the source is at ground
potential—hence the name common source. The signal from
the signal source is coupled into the gate of the transistor
through the coupling capacitor CC, which provides dc
isolation between the amplifier and the signal source. The dc
Figure 4.14 Common-source circuit
transistor biasing is established by R1 and R2, and is not with voltage divider biasing and
coupling capacitor
disturbed when the signal source is capacitively coupled to
the amplifier.

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EE212 Electronics I Lecture # 4 4.3 THE COMMON-SOURCE AMPLIFIER

If the signal source is a sinusoidal voltage at frequency f, then the magnitude of the capacitor
impedance is |ZC| = [1/(2πf CC)].

For example, assume that CC = 10μF and f = 2 kHz. The magnitude of the capacitor impedance
is then .

The magnitude of this impedance is generally much less than the Thevenin resistance at the
capacitor terminals. We can therefore assume that the capacitor is essentially a short circuit to
signals with frequencies greater than 2 kHz. We will also neglect, in this chapter, any
capacitance effects within the transistor.

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For the circuit shown in Figure 4.14, assume that the transistor is biased in the saturation region
by resistors R1 and R2, and that the signal frequency is sufficiently large for the coupling capacitor
to act essentially as a short circuit. The signal source is represented by a Thevenin equivalent
circuit, in which the signal voltage source vi is in series with an equivalent source resistance RSi . As
we will see, RSi should be much less than the amplifier input resistance, Ri = R1 ǁ R2, in order to
minimize loading effects. Figure 4.15 shows the resulting small-signal equivalent circuit. The small-
signal variables, such as the input signal voltage Vi , are given in phasor form.

Figure 4.15 Small-signal

equivalent circuit,
assuming coupling
capacitor acts as a short
circuit.

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EE212 Electronics I Lecture # 4 4.3 THE COMMON-SOURCE AMPLIFIER

Since the source is at ground potential, there is no body effect. The output voltage is

Vo = −gmVgs(roǁRD) (4.27)

The input gate-to-source voltage is (4.28)

so the small-signal voltage gain is

(4.29)

We can also relate the ac drain current to the ac drain-to-source voltage, as Vds = −Id (RD).

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Figure 4.16 shows the dc load line, the transition point

(that separates the saturation bias region and
nonsaturation bias region), and the Q-point, which is in the
saturation region. In order to provide the maximum
symmetrical output voltage swing and keep the transistor
biased in the saturation region, the Q-point must be near
the middle of the saturation region. At the same time, the
input signal must be small enough for the amplifier to
Figure 4.16 DC load line and
remain linear. transition point separating saturation
and nonsaturation regions

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EE212 Electronics I Lecture # 4 4.3 THE COMMON-SOURCE AMPLIFIER

The input and output resistances of the amplifier can be determined from Figure 4.15. The
input resistance to the amplifier is Ri = R1 ǁ R2. Since the low-frequency input resistance
looking into the gate of the MOSFET is essentially infinite, the input resistance is only a function
of the bias resistors. The output resistance looking back into the output terminals is found by
setting the independent input source Vi equal to zero, which means that Vgs = 0. The output
resistance is therefore Ro = RD ǁ ro.

EXAMPLE 4.3

Objective: Determine the small-signal voltage gain and input and output resistances of a
common-source amplifier.

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For the circuit shown in Figure 4.14, the parameters are: VDD = 3.3 V, RD = 10 kΩ, R1 = 140 kΩ,
R2 = 60 kΩ, and RSi = 4 kΩ. The transistor parameters are: VT N = 0.4 V, Kn = 0.5 mA/V2, and λ =
0.02 V−1.

Solution (dc calculations): The dc or quiescent gate-to-source voltage is

The quiescent drain current is IDQ = Kn (VGSQ − VT N) 2 = (0.5)(0.99 − 0.4)2 = 0.174 mA

and the quiescent drain-to-source voltage is VDSQ = VDD − IDQ RD = 3.3 − (0.174)(10) = 1.56V

Since VDSQ > VGSQ − VT N , the transistor is biased in the saturation region.

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Small-signal Voltage Gain: The small-signal transconductance gm is then

and the small-signal output resistance is

The input resistance to the amplifier is

Ri = R1 ǁ R2 = 140 ǁ 60 = 42 kΩ

From Figure 4.15 and Equation (4.29), the small-signal voltage gain is

or Av = −5.21

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Input and Output Resistances: As already calculated, the amplifier input resistance is

Ri = R1 ǁ R2 = 140 ǁ 60 = 42 kΩ

and the amplifier output resistance is

Ro = RD ǁ ro = 10 ǁ 287 = 9.66 kΩ

Comment: The resulting Q-point is not in the center of the saturation region. Therefore, this
circuit does not achieve the maximum symmetrical output voltage swing in this case.

Discussion: The small-signal input gate-to-source voltage is

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EE212 Electronics I Lecture # 4 4.3 THE COMMON-SOURCE AMPLIFIER

Since RSi is not zero, the amplifier input signal Vgs is approximately 91 percent of the signal
voltage. This is called a loading effect. Even though the input resistance to the gate of the
transistor is essentially infinite, the bias resistors greatly influence the amplifier input
resistance and loading effect. This loading effect can be eliminated or minimized when current
source biasing is considered.

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EE212 Electronics I Lecture # 4 4.3 THE COMMON-SOURCE AMPLIFIER

DESIGN EXAMPLE 4.4

Objective: Design the bias of a MOSFET circuit such that the Q-

point is in the middle of the saturation region. Determine the
resulting small-signal voltage gain.

Specifications: The circuit to be designed has the configuration

shown in Figure 4.17. Let R1 ǁ R2 = 100 kΩ. Design the circuit such
that the Q-point is IDQ = 2 mA and the Q-point is in the middle of
the saturation region. Figure 4.17 Common-source
NMOS transistor circuit

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Choices: A transistor with nominal parameters VT N = 1 V,

k'n = 80 μA/V2, W/L = 25, and λ = 0.015 V−1 is available.

Solution (dc design): The load line and the desired Q-

point are given in Figure 4.18. If the Q-point is to be in
the middle of the saturation region, the current at the
transition point must be 4 mA. The conductivity
Figure 4.18 DC load line and transition point
parameter is for NMOS circuit shown in Figure 4.17

We can now calculate VDS (sat) at the transition point. The subscript t indicates transition point
values. To determine VGSt , we use IDt = 4 = Kn (VGSt − VT N)2 = 1(VGSt − 1)2 which yields VGSt = 3V.
Therefore VDSt = VGSt − VT N = 3 − 1 = 2V
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If the Q-point is in the middle of the saturation region, then VDSQ = 7 V, which would yield a 10
V peak-to-peak symmetrical output voltage.

From Figure 4.17, we can write VDSQ = VDD − IDQ RD or

We can determine the required quiescent gate-to-source voltage from the current equation, as
follows: IDQ = 2 = Kn (VGSQ − VT N)2 = (1)(VGSQ − 1)2 or VGSQ = 2.41V

Then

which yields R1 = 498 kΩ and R2 = 125 kΩ

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EE212 Electronics I Lecture # 4 4.3 THE COMMON-SOURCE AMPLIFIER

Solution (ac analysis): The small-signal transistor parameters are

and

The small-signal equivalent circuit is the same as shown in Figure 4.7. The small-signal voltage
gain is or Av = −6.58.

Comment: Establishing the Q-point in the middle of the saturation region allows the
maximum symmetrical swing in the output voltage, while keeping the transistor biased in the
saturation region.

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4.3.2 Common-Source Amplifier with Source Resistor

A source resistor RS tends to stabilize the Q-point against

variations in transistor parameters (Figure 4.19). If, for
example, the value of the conduction parameter varies from
one transistor to another, the Q-point will not vary as much if
a source resistor is included in the circuit. However, as shown
in the following example, a source resistor also reduces the
Figure 4.19 Common-source circuit
signal gain. with source resistor and positive and
negative supply voltages

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EE212 Electronics I Lecture # 4 4.3 THE COMMON-SOURCE AMPLIFIER

The circuit in Figure 4.19 is an example of a situation in which the body effect should be taken
into account. The substrate (not shown) would normally be connected to the −5 V supply, so
that the body and substrate terminals are not at the same potential. However, in the following
example, we will neglect this effect.

The circuit shown in Figure 4.20(a) is a PMOS version of the common-source amplifier with a
source resistor RS included.

EXAMPLE 4.5

Objective: Determine the small-signal voltage gain of a PMOS transistor circuit.

Consider the circuit shown in Figure 4.20(a). The transistor parameters are Kp = 0.80 mA/V2,
VTP = −0.5 V, and λ = 0. The quiescent drain current is found to be IDQ = 0.297 mA. (Show it!)
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The small-signal equivalent circuit is shown in Figure 4.20(b). To sketch

the small-signal equivalent circuit, start with the three terminals of
the transistor, draw in the transistor equivalent circuit between these
three terminals, and then sketch in the other circuit elements around
the transistor.

Solution: The small-signal output voltage is Vo = +gm Vsg RD . Writing a

KVL equation from the input around the gate–source loop,
we find Vi = −Vsg − gm Vsg RS or

Figure 4.20 (a) PMOS circuit for Example 4.5, and (b) small-signal
equivalent circuit
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Substituting this expression for Vsg into the output voltage equation, we find the small-signal
voltage gain as

The small-signal transconductance is

We then find the small-signal voltage gain as or Av = −2.48.

Comment: The analysis of a PMOS transistor circuit is essentially the same as that of an NMOS
transistor circuit. The voltage gain of a MOS transistor circuit that contains a source resistor is
degraded compared to a circuit without a source resistor. However, the Q-point tends to be
stabilized.

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EE212 Electronics I Lecture # 4 4.3 THE COMMON-SOURCE AMPLIFIER

Discussion: We mentioned that including a source resistor tends to stabilize the circuit
characteristics against any changes in transistor parameters. If, for example, the conduction
parameter Kp varies by ±10 percent, we find the following results.

With a ±10 percent variation in Kp, there is less than

a ±1.8 percent variation in the voltage gain.

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4.3.3 Common-Source Circuit with Source Bypass Capacitor

A source bypass capacitor added to the common-source circuit with a

source resistor will minimize the loss in the small-signal voltage gain,
while maintaining the Q-point stability. The Q-point stability can be
further increased by replacing the source resistor with a constant-
current source. The resulting circuit is shown in Figure 4.21, assuming an
ideal signal source. If the signal frequency is sufficiently large so that the Figure 4.21 NMOS common-
source circuit with source
bypass capacitor acts essentially as an ac short-circuit, the source will be
bypass capacitor
held at signal ground.

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EE212 Electronics I Lecture # 4 4.3 THE COMMON-SOURCE AMPLIFIER

Example 4.6

Objective: Determine the small-signal voltage gain of a circuit biased with a constant-current
source and incorporating a source bypass capacitor.

For the circuit shown in Figure 4.21, the transistor parameters are: VT N = 0.8 V, Kn = 1 mA/V2,
and λ = 0.

Solution: Since the dc gate current is zero, the dc voltage at the source terminal is VS = −VGSQ,
and the gate-to-source voltage is determined from IDQ = IQ = Kn(VGSQ − VT N)2 or 0.5 =
(1)(VGSQ − 0.8)2 which yields VGSQ = −VS = 1.51V

The quiescent drain-to-source voltage is VDSQ = VDD − IDQ RD − VS = 5 − (0.5)(7) − (−1.51) =

3.01V
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The transistor is therefore biased in the saturation region.

The small-signal equivalent circuit is shown in Figure 4.22.

The output voltage is Vo = −gm Vgs RD

Figure 4.22 Small-signal equivalent
Since Vgs = Vi , the small-signal voltage gain is circuit, assuming the source bypass
capacitor acts as a short circuit

Comment: Comparing the small-signal voltage gain of 9.9 in this example to the 2.48 calculated
in Example 4.5, we see that the magnitude of the gain increases when a source bypass
capacitor is included.

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EE212 Electronics I Lecture # 4 4.4 THE COMMON-DRAIN (SOURCE FOLLOWER) AMPLIFIER

4.4 THE COMMON-DRAIN (SOURCE-FOLLOWER) AMPLIFIER

Objective: Analyze the common-drain (source-follower) amplifier and

become familiar with the general characteristics of this circuit.

The second type of MOSFET amplifier to be considered is the

common-drain circuit. An example of this circuit configuration is
shown in Figure 4.26. As seen in the figure, the output signal is taken
Figure 4.26 NMOS source-
off the source with respect to ground and the drain is connected follower or common-drain
amplifier
directly to VDD. Since VDD becomes signal ground in the ac equivalent
circuit, we have the name common drain. The more common name is
source follower. The reason for this name will become apparent as we
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4.4.1 Small-Signal Voltage Gain

The dc analysis of the circuit is exactly the same as we

have already seen, so we will concentrate on the
small-signal analysis. The small-signal equivalent
circuit, assuming the coupling capacitor acts as a short
circuit, is shown in Figure 4.27(a). The drain is at signal
ground, and the small-signal resistance ro of the
transistor is in parallel with the dependent current
source. Figure 4.27(b) is the same equivalent circuit,
but with all signal grounds at a common point. We are
again neglecting the body effect. Figure 4.27 (a) Small-signal equivalent circuit of NMOS source
Lecture # 1
follower and (b) small-signal equivalent circuit of NMOS
61
source
follower with all signal grounds at a common point
EE212 Electronics I Lecture # 4 4.4 THE COMMON-DRAIN (SOURCE FOLLOWER) AMPLIFIER

The output voltage is Vo = (gm Vgs)(RS ǁ ro) (4.30)

Writing a KVL equation from input to output results in the following:

Vin = Vgs + Vo = Vgs + (gm Vgs)(RS ǁ ro) (4.31(a))

Therefore, the gate-to-source voltage is (4.31(b))

Equation (4.31(b)) is written in the form of a voltage-divider equation, in which the gate-to-
source of the NMOS device looks like a resistance with a value of 1/gm. More accurately, the
effective resistance looking into the source terminal (ignoring ro) is 1/gm. The voltage Vin is
related to the source input voltage Vi by (4.32) where Ri = R1 ǁ R2 is the
input resistance to the amplifier.

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Substituting Equations (4.31(b)) and (4.32) into (4.30), we have the small-signal voltage gain:

(4.33(a)) or (4.33(b))

which again is written in the form of a voltage-divider equation. An inspection of Equation

4.33(b) shows that the magnitude of the voltage gain is always less than unity.

EXAMPLE 4.7

Objective: Calculate the small-signal voltage gain of the source-follower circuit in Figure 4.26.
Assume the circuit parameters are VDD = 12 V, R1 = 162 kΩ, R2 = 463 kΩ, and RS = 0.75 kΩ, and
the transistor parameters are VT N = 1.5 V, Kn = 4 mA/V2, and = 0.01V−1. Also assume RSi = 4 kΩ.

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Solution: The dc analysis results are IDQ = 7.97 mA and VGSQ = 2.91 V. The small-signal
transconductance is therefore gm = 2 Kn(VGSQ − VT N) = 2(4)(2.91 − 1.5) = 11.3 mA/V and the
small-signal transistor resistance is ro= [IDQ]−1 = [(0.01)(7.97)]−1 = 12.5kΩ.

The amplifier input resistance is Ri = R1 ǁ R2 = 162 ǁ 463 = 120 kΩ. The small-signal voltage gain
then becomes

Comment: The magnitude of the small-signal voltage gain is less than 1. An examination of
Equation (4.33(b)) shows that this is always true. Also, the voltage gain is positive, which
means that the output signal voltage is in phase with the input signal voltage. Since the output
signal is essentially equal to the input signal, the circuit is called a source follower.

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Although the voltage gain is slightly less than 1, the source follower is an extremely useful
circuit because the output resistance is less than that of a common-source circuit, as we will
show in the next section. A small output resistance is desirable when the circuit is to act as an
ideal voltage source and drive a load circuit without suffering any loading effects.

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EE212 Electronics I Lecture # 4 4.4 THE COMMON-DRAIN (SOURCE FOLLOWER) AMPLIFIER

DESIGN EXAMPLE 4.8

Objective: Design a source-follower amplifier with a p-channel

enhancement-mode MOSFET to meet a set of specifications.

Specifications: The circuit to be designed has the configuration

shown in Figure 4.28 with circuit parameters VDD = 20 V and
RSi = 4 kΩ. The Q-point values are to be in the center of the
load line with IDQ = 2.5 mA. The input resistance is to be Ri =
200 kΩ. The transistor W/L ratio is to be designed such that the
small signal voltage gain is Av = 0.90.

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EE212 Electronics I Lecture # 4 4.4 THE COMMON-DRAIN (SOURCE FOLLOWER) AMPLIFIER

Choices: A transistor with nominal parameters VT P = −2 V, k'p = 40 μA/V2, and λ = 0 is available.

Solution (dc analysis): From a KVL equation around the source-to-drain loop, we have VDD =
VSDQ + IDQ RS or 20 = 10 + (2.5) RS which yields the required source resistor to be RS = 4 kΩ.

Solution (ac design): The small-signal voltage gain of this circuit is the same as that of a source
follower with an NMOS device. From Equation (4.33(a)), we have

which yields

We find that the required transconductance must be gm = 2.80 mA/V. The transconductance
can be written as . We have which yields

Kp = 0.784 × 10−3A/V2.

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EE212 Electronics I Lecture # 4 4.4 THE COMMON-DRAIN (SOURCE FOLLOWER) AMPLIFIER

The conduction parameter, as a function of width-to-length ratio, is

which means that the required width-to-

length ratio must be .

Solution (dc design): Completing the dc analysis and design, we have IDQ = Kp(VGSQ + VT P)2 or
2.5 = 0.784(VSGQ − 2)2 which yields a quiescent source-to-gate voltage of VSGQ = 3.79 V. The
quiescent source-to-gate voltage can also be written as

Since we have

The bias resistor R1 is then found to be R1 = 644 kΩ.

Since Ri = R1 ǁ R2 = 200 kΩ, we find R2 = 290 kΩ.

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Comment: In order to achieve the desired specifications, a relatively large transconductance is

required, which means that a relatively large transistor is needed. A large value of input
resistance Ri has minimized the effect of loading due to the output resistance, RSi, of the signal
source.

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4.4.2 Input and Output Impedance

The small-signal input resistance Ri as defined in Figure 4.27(b), for example, is the Thevenin
equivalent resistance of the bias resistors. Even though the input resistance to the gate of the
MOSFET is essentially infinite, the input bias resistances do provide a loading effect. This same
effect was seen in the common-source circuits.

To calculate the small-signal output resistance, we set all independent small signal sources
equal to zero, apply a test voltage to the output terminals, and measure a test current. Figure
4.30 shows the circuit we will use to determine the output resistance of the source follower
shown in Figure 4.26. We set Vi = 0 and apply a test voltage Vx.

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Since there are no capacitances in the circuit,

the output impedance is simply an output
resistance, which is defined as (4.34)

Writing a KCL equation at the output source

Figure 4.30 Equivalent circuit of NMOS source follower, for
terminal produces (4.35) determining output resistance

Since there is no current in the input portion of the circuit, we see that Vgs = −Vx . Therefore,
Equation (4.35) becomes (4.36(a)) or

(4.36(b)). The output resistance is then (4.37).

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From Figure 4.30, we see that the voltage Vgs is directly across the current source gm Vgs . This
means that the effective resistance of the device is 1/gm. The output resistance given by
Equation (4.37) can therefore be written directly. This result also means that the resistance
looking into the source terminal (ignoring ro) is 1/gm, as previously noted.

EXAMPLE 4.9

Objective: Calculate the output resistance of a source-follower circuit.

Consider the circuit shown in Figure 4.26 with circuit and transistor parameters given in
Example 4.7.

Solution: The results of Example 4.7 are: RS = 0.75 kΩ, ro = 12.5 kΩ, and gm = 11.3 mA/V. Using
Figure 4.30 and Equation (4.37), we find or Ro = 0.0787kΩ =78.7Ω

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Comment: The output resistance of a source-follower circuit is dominated by the

transconductance parameter. Also, because the output resistance is very low, the source
follower tends to act like an ideal voltage source, which means that the output can drive
another circuit without significant loading effects.

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EE212 Electronics I Lecture # 4 4.5 THE COMMON-GATE CONFIGURATION

4.5 THE COMMON-GATE CONFIGURATION

Objective: Analyze the common-gate amplifier and become familiar with the general
characteristics of this circuit.

The third amplifier configuration is the common-gate circuit. To determine the small-signal
voltage and current gains, and the input and output impedances, we will use the same small-
signal equivalent circuit for the transistor that was used previously. The dc analysis of the
common-gate circuit is the same as that of previous MOSFET circuits.

4.5.1 Small-Signal Voltage and Current Gains

In the common-gate configuration, the input signal is applied to the source terminal and the
gate is at signal ground.

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The common-gate configuration shown in Figure 4.32 is biased with a constant-current source
IQ. The gate resistor RG prevents the buildup of static charge on the gate terminal, and the
capacitor CG ensures that the gate is at signal ground. The coupling capacitor CC1 couples the
signal to the source, and coupling capacitor CC2 couples the output voltage to load resistance
RL.

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EE212 Electronics I Lecture # 4 4.5 THE COMMON-GATE CONFIGURATION

The small-signal equivalent circuit is shown in Figure 4.33. The small-signal transistor resistance
ro is assumed to be infinite. Since the source is the input terminal, the small-signal equivalent
circuit shown in Figure 4.33 may appear to be different from those considered previously.
However, to sketch the equivalent circuit, we can use the same technique as used previously.
Sketch in the three terminals of the transistor with the source at the input for this case. Then
draw in the transistor equivalent circuit between the three terminals and then sketch in the
remaining circuit elements around the
transistor.

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EE212 Electronics I Lecture # 4 4.5 THE COMMON-GATE CONFIGURATION

The output voltage is Vo = −(gm Vgs) (RD ǁ RL ) (4.38)

Writing the KVL equation around the input, we find Vi = Ii RSi − Vgs (4.39) where Ii = − gm Vgs .

The gate-to-source voltage can then be written as (4.40)

The small-signal voltage gain is found to be (4.41)

Also, since the voltage gain is positive, the output and input signals are in phase. In many cases,
the signal input to a common-gate circuit is a current. Figure 4.34 shows the small-signal
equivalent common-gate circuit with a Norton equivalent circuit as the signal source. We can
calculate a current gain. The output current Io can be written (4.42)

At the input we have (4.43) or (4.44)

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The small-signal current gain is then (4.45).

We may note that if RD >> RL and gm RSi >> 1, then the current gain is essentially unity.

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EE212 Electronics I Lecture # 4 4.5 THE COMMON-GATE CONFIGURATION

4.5.2 Input and Output Impedance

In contrast to the common-source and source-follower amplifiers, the common-gate circuit has
a low input resistance because of the transistor. However, if the input signal is a current, a low
input resistance is an advantage. The input resistance is defined, using Figure 4.33, as

Since Ii = −gmVgs , the input resistance is

This result has been obtained previously.

We can find the output resistance by setting the input signal voltage equal to zero. From Figure
4.33, we see that Vgs = −gm Vgs RSi , which means that Vgs = 0. Consequently, gm Vgs = 0. The
output resistance, looking back from the load resistance, is therefore Ro = RD (4.48)

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EE212 Electronics I Lecture # 4 4.5 THE COMMON-GATE CONFIGURATION

Example 4.10

Objective: For the common-gate circuit, determine the output voltage for a given input
current.

For the circuits shown in Figures 4.32 and 4.34, the circuit parameters are: IQ = 1 mA, V+ = 5 V,
V - = −5 V, RG = 100 kΩ, RD = 4 kΩ, and RL = 10 kΩ. The transistor parameters are: VTN = 1 V, Kn =
1 mA/V2, and λ = 0. Assume the input current in Figure 4.34 is 100 sinωt μA and assume Rsi =
50 kΩ.

Solution: The quiescent gate-to-source voltage is determined from IQ = IDQ = Kn(VGSQ − VTN)2 or
1 = 1(VGSQ − 1)2 which yields VGSQ = 2V. The small-signal transconductance is

gm = 2 Kn(VGSQ − VTN) = 2(1)(2 − 1) = 2mA/V

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EE212 Electronics I Lecture # 4 4.5 THE COMMON-GATE CONFIGURATION

From Equation (4.45), we can write the output current as

The output voltage is Vo = IoRL , so we find

or Vo = 0.283 sinωt V.

Comment: The MOSFET common-gate amplifier is useful if the input signal is a current.

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 4.6 THE THREE BASIC AMPLIFIER CONFIGURATIONS:
EE212 Electronics I Lecture # 4 SUMMARY AND COMPARISON

4.6 THE THREE BASIC AMPLIFIER CONFIGURATIONS: SUMMARY AND COMPARISON

Objective: Compare the general characteristics of the three basic amplifier configurations.

Table 4.2 is a summary of the small-signal characteristics of the three amplifier configurations.

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EE212 Electronics I Lecture # 4 SUMMARY AND COMPARISON

Voltage Gains:

The common-source amplifier voltage gain magnitude is generally greater than 1. The voltage
gain of the source follower is slightly less than 1, and that of the common-gate circuit is
generally greater than 1.

Input Resistances:

The input resistance looking directly into the gate of the common-source and source-follower
circuits is essentially infinite at low to moderate signal frequencies. However, the input
resistance of these discrete amplifiers is the Thevenin equivalent resistance RTH of the bias
resistors. In contrast, the input resistance to the common-gate circuit is generally in the range
of only a few hundred ohms.
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 4.6 THE THREE BASIC AMPLIFIER CONFIGURATIONS:
EE212 Electronics I Lecture # 4 SUMMARY AND COMPARISON

Output Resistances:

The output resistance of the source follower is generally in the range of a few hundred ohms or
less. The output resistance of the common-source and common-gate configurations is
dominated by the resistance RD. In Chapters 10 and 11, we will see that the output resistance
of these configurations is dominated by the resistance ro when transistors are used as load
devices in ICs.

The specific characteristics of these single-stage amplifiers are used in the design of multistage
amplifiers.

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EE212 Electronics I Lecture # 4 4.7 SINGLE-STAGE INTEGRATED CIRCUIT MOSFET AMPLIFIERS

Note: We have skipped

4.7 SINGLE-STAGE INTEGRATED CIRCUIT MOSFET AMPLIFIERS

4.7.1 Load Line Revisited

4.7.2 NMOS Amplifiers with Enhancement Load

4.7.3 NMOS Amplifier with Depletion Load

4.7.4 NMOS Amplifier with Active Loads

CMOS Common-Source Amplifier

CMOS Source-Follower Amplifier

CMOS Common-Gate Amplifier

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 4.8 MULTISTAGE AMPLIFIERS – 4.9 BASIC JFET AMPLIFIERS –
EE212 Electronics I Lecture # 4 4.10 DESIGN APPLICATION

4.8 MULTISTAGE AMPLIFIERS

4.8.1 Multistage Amplifier: Cascade Circuit

4.8.2 Multistage Amplifier: Cascode Circuit

4.9 BASIC JFET AMPLIFIERS

4.9.1 Small-Signal Equivalent Circuit

4.9.2 Small-Signal Analysis

4.10 DESIGN APPLICATION: A TWO-STAGE AMPLIFIER

Please read them but you are not responsible for the Exams!

Lecture # 1 86