Question

In the circuit, \( I_{\text{DC}} \) is an ideal current source, the transistors \( M_1 \), \( M_2 \) are assumed to be biased in saturation wherein \( V_{\text{in}} \) is the input signal and \( V_{\text{DC}} \) is the fixed DC voltage. Both transistors have a small signal resistance of \( R_{ds} \) and transconductance of \( g_m \). The small signal output impedance of the circuit is:
 

• A. \( R_{\text{out}} = \frac{1}{g_m} \)
• B. \( R_{\text{out}} = R_{ds} + \frac{1}{g_m} \)
• C. \( R_{\text{out}} = R_{ds} \)
• D. \( R_{\text{out}} = g_m R_{ds} \)

Hint:

For transistor amplifiers in saturation: - The small signal output impedance is influenced by both \( R_{ds} \) and the transconductance \( g_m \).
- In the case of a current source circuit, \( R_{\text{out}} \) is the sum of \( R_{ds} \) and the inverse of \( g_m \).

Answer 1

The Correct Option is B

In this circuit, the transistors \( M_1 \) and \( M_2 \) are biased in saturation. For small signal analysis, the output impedance of a transistor is influenced by both the drain-source resistance \( R_{ds} \) and the transconductance \( g_m \). 
- Transistor in saturation: The small signal output impedance \( R_{\text{out}} \) can be approximated by the combination of the intrinsic resistance \( R_{ds} \) and the effect of the transconductance \( g_m \). 
- Looking into the drain of \( M_2 \): The small signal output impedance is the parallel combination of \( R_{ds} \) and the impedance due to \( g_m \), which is given by \( \frac{1}{g_m} \). 
Thus, the total small signal output impedance is: \[ R_{\text{out}} = R_{ds} + \frac{1}{g_m} \]