Question

In the circuit diagram shown below, NMOS is in saturation region, \( \mu_n C_{\text{ox}} = 200 \, \mu \text{A/V}^2 \), width \( W = 40 \, \mu \text{m} \), length \( L = 1 \, \mu \text{m} \), the threshold voltage is 0.4 V, and the ratio of body-effect transconductance (\( g_m b \)) to transconductance (\( g_m \)) is 0.1. A small input voltage \( v_{\text{in}} \) is applied at the bulk-terminal to produce a small change in the output voltage \( v_{\text{out}} \). The dc gain for \( v_{\text{out}} / v_{\text{in}} \) is \(\underline{\hspace{2cm}}\). (Neglect channel-length modulation for NMOS and all intrinsic capacitors.) 

• A. -0.4
• B. -4.0
• C. -4.4
• D. -3.6

Hint:

In small-signal analysis of NMOS amplifiers, the voltage gain depends on the transconductance and the load resistance. If the body-effect transconductance is provided, use the relationship \( g_m = 0.1 \times g_m b \).

Answer 1

The Correct Option is A

In this question, we are tasked with finding the dc gain \( \frac{v_{\text{out}}}{v_{\text{in}}} \) for the given NMOS amplifier circuit. The given parameters include NMOS in the saturation region, the mobility factor \( \mu_n C_{\text{ox}} \), dimensions of the transistor, and other necessary values.

Step 1: Understanding the parameters.
- Threshold Voltage (Vth): 0.4V
- Transconductance (\( g_m \)): It is the change in drain current (\( I_D \)) with respect to the gate-source voltage (\( V_{GS} \)) in saturation, calculated as:
\[ g_m = \frac{2 I_D}{V_{GS} - V_{th}} \] - Body-effect transconductance (\( g_m b \)): It is related to the change in \( g_m \) with respect to the bulk-source voltage (\( V_{BS} \)), and the ratio of \( g_m b \) to \( g_m \) is given as 0.1.

Step 2: Determine the small-signal model.
For a small signal analysis, we use the equivalent small-signal model of the NMOS transistor. The DC gain \( \frac{v_{\text{out}}}{v_{\text{in}}} \) is given by the expression for the voltage gain in the presence of resistances and transconductance:
\[ A_v = - g_m \cdot R_{\text{load}} \] Where:
- \( g_m \) is the transconductance.
- \( R_{\text{load}} \) is the load resistance seen at the output. In this case, \( R_{\text{load}} \) is given as \( 1 \, \text{k}\Omega \).

Step 3: Calculate the DC gain.
We know the transconductance is related to the current and voltage conditions, but in this case, we are directly given the relationship between body-effect transconductance and transconductance. Hence, \( g_m = 0.1 \cdot g_m b \), where \( g_m b = 200 \, \mu A/V^2 \), which gives us: \[ g_m = 0.1 \times 200 \, \mu A/V^2 = 20 \, \mu A/V \] Now, calculating the voltage gain: \[ A_v = - g_m \cdot R_{\text{load}} = - 20 \times 10^{-6} \times 1 \times 10^3 = -0.4 \]

Final Answer: (A) -0.4