Analysis of Common-Source (CS) Amplifier

Analysis of Common-Source (CS) Amplifier

• In this section, we will see the common-source amplifier. We will analyze several basic common-source circuits and will determine small-signal voltage gain and input and output impedances.

 

1. Basic Common-Source Configuration

• The Fig. 7.4.1 shows the common source circuit with voltage divider biasing and coupling capacitor. The MOSFET is biased near the middle of the saturation region by R₁ and R₂ resistors to work as an amplifier.

• Assume 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 vs, is in series with an equivalent source resistance R_si.

• Here R_si should be much less than the amplifier input resistance, R_i = R₁ || R₂ in order to minimize loading effects.

• Fig. 7.4.2 shows the resulting small-signal equivalent circuit.

• The input gate-to-source voltage is

• So the small-signal overall voltage gain is,

• Since Rsi is not zero, the amplifier input signal v4 is less than the signal voltage. This is known as loading effect. It reduces the voltage gain of the amplifier.

• The input resistance is R_is = R₁ || R₂

• The output resistance is R_o = R_D || r_o

• We can also relate the a.c. drain current to the a.c. drain-to-source voltage, as

V_ds = - I_d(R_D).

• Fig. 7.4.3 shows the d.c. load line, the transition point, and the Q-point, which is in the saturation region.

 

Ex. 7.4.1 For the circuit shown in Fig. 7.4.4, the MOSFET parameters are : VT = 1.5 V, = OB mA/V² and λ = 0.01 V^-1, determine the small signal voltage gain, R_i , and R_o.

 

Sol. :

Step 1 : D.C. analysis : The d.c. or quiescent gate-to-source voltage is,

 

Ex. 7.4.2 Describe the small-signal voltage gain, input and output resistance of a common source amplifier. For the circuit shown below in Fig. 7.4.5 the parameters are V_DD=10 V, = 70.9 kΩ, R2 = 29.1 kΩ, RD = 10 kQand RL =10kΩ. The transistor parameters are Vm = 1.5 V, Kn = 0.5 mA/v² and X =0.01 V-1,

Sol .:

Step 1 : D.C. Analysis

 

Since VDSQ > VGSQ – VT, The MOSFET is biased in the saturation region.

Step 2 : A.C. Analysis

 

Ex. 7.4.3 A common source amplifier using EMOSFET is shown in Fig. 7.4.6. Assume for this device I DON) = 200 mA at VGS(ON) = 4 V. VT = 2 V and gm = 23 mS, Vin = 25 mV

i) V_GS ii) I_D iii) V_DS iv) a.c. output voltage.

 Sol. :

Step 1 : Calculate V_GS, I_DQ, V_DS

 

2. Common-Source Amplifier with Source Resistor

• Fig. 7.4.7 shows the common source amplifier with source resistor.

 

• The source resistor is introduced to stabilize the Q-point against variations in the MOSFET parameters. However, like in BJT circuits, a source resistor reduces the signal gain.

 

Ex. 7.4.4 Consider the circuit in Fig. 7.4.7. The MOSFET parameters are V_T = 0.8 V. K= 1 mA/V2, and 7=0. Calculate the voltage gain of the amplifier.

Sol. :

Step 1 : DC analysis

Step 2 : Draw the small signal equivalent circuit Fig. 7.4.8 shows the small signal equivalent circuit, for the common source amplifier circuit with source resistor.

 

3. Common-Source Circuit with Source Bypass Capacitor

• A source bypass capacitor connected across the source resistor in the common-source circuit will minimize the loss in the small signal voltage gain, while maintaining Q-point stability.

• The Q-point stability can be further increased by replacing the source resistor with a constant-current source.

• The Fig. 7.4.9 shows common source circuit with source bypass capacitor. It acts as short circuit for a.c. signal and open circuit for d.c. signal.

 

Ex. 7.4.5 For the circuit shown in Fig. 7.4.9, the MOSFET parameters are : V_T = 0.8 V, K = tmA/V², and λ = 0. Calculate the voltage gain.

Sol. :

Step 1: DC analysis

Since the d.c. gate current is zero, the d.c. voltage at the source terminal is V_s =- V_GSQ, and the gate-to-source voltage is determined from

The quiescent drain-to-source voltage is

Since V_DS > V_gs - V_T = 1.51 - 0.8 = 0.71 V, transistor is biased in the saturation region.

Step 2: Draw the small signal equivalent circuit.

g_m = 2K (V_GS -V_T) (V_GS-V_T)

= 2(1) (1.51 0.8)

= 1.42 mA/V

 

Important Concept

Comparing the small signal voltage gain of - 9.656 in this example to the - 5.36 calculated in example 7.4.4, we see that the magnitude of the gain increases when a source bypass capacitor is included.

Ex. 7.4.6 For the common source amplifier circuit shown in Fig. 7.4.11.

Determine :

i) g_m ii) Av iii)  R_i R_i iv) A_vs v) R_o and R_'o

Given :

K = 04 mA/V², V_T = 3 V, I_D =111 mA V_GS = 466 V, r_o = 40 k Ω

Review Questions

1. Draw and explain the common source amplifier with source resistor.

2. Draw and explain the common source amplifier with bypassed source resistor.

3. With neat circuit diagram, perform ac analysis for common source using equivalent circuit NMOSFET amplifier.