Question

In the small signal circuit shown, the enhancement mode n-channel MOSFET is biased in saturation with a transconductance \( g_m \). A small signal low-frequency voltage \( v_d \) injected at the supply terminal results in a small signal voltage fluctuation \( v_o \) at the output. If the channel length modulation of the MOSFET is ignored, the small signal gain \( \frac{v_o}{v_d} \) is given by _________

• A. ( \frac{-g_m R_0}{1 + g_m R_0} \)
• B. ( (g_m R_0 + 1)^{-1} \)
• C. ( \frac{-g_m R_0}{1 + 2g_m R_0} \)
• D. ( (g_m R_0 / 2 + 3 / 2)^{-1} \)

Hint:

For small signal analysis of MOSFETs, the voltage gain is typically determined by the transconductance \( g_m \) and the load resistance \( R_0 \), and can be expressed as \( \frac{-g_m R_0}{1 + g_m R_0} \).

Answer 1

The Correct Option is A

Step 1: Understanding the small signal model.
In small signal analysis, the MOSFET operates in its linear region, and the voltage gain can be approximated by the ratio of the output voltage \( v_o \) to the input voltage \( v_d \). For the given circuit, where the channel length modulation is neglected, the voltage gain depends on the transconductance \( g_m \) and the load resistance \( R_0 \). Step 2: Analyzing the options.
- (A) Correct, the small signal gain in this configuration, neglecting channel length modulation, is given by \( \frac{-g_m R_0}{1 + g_m R_0} \).
- (B) Incorrect, this expression does not match the correct gain formula.
- (C) Incorrect, this gain formula includes an incorrect term for the gain.
- (D) Incorrect, this formula does not match the standard expression for small signal gain. Step 3: Conclusion.
Thus, the correct answer is (A).