1.

If channel length modulation is present, what is the total impedance at node X?(a) Rd || ro1 || 1/gm2(b) Rd || ro1 || 1/gm2 || ro2(c) Rd || 2ro1 || 1/gm2 || ro2(d) Rd || ro1 || 2/gm2 || ro2The question was asked by my school teacher while I was bunking the class.I'm obligated to ask this question of MOSFET Amplifier with CG Configuration topic in section FET Amplifiers of Analog Circuits

Answer»

Correct choice is (B) Rd || RO1 || 1/gm2 || ro2

To explain: We develop a Thevenin’s equivalent w.r.t. node X and ground. But this is LENGTHY process. But then again, we can use the process as follows. For Thevenising, firstly we set Vdd and VIN to ground. Thereafter, we see that Rd and ro1 are in parallel to each other since they are connected from node X to ground. Next, we see that the impedance provided by M2 is 1/gm2 || ro2 but it again occurs from Node x to ground. Hence, all the impedances are parallel to each other and the total impedance at node X is Rd || ro1 || 1/gm2 || ro2.


Discussion

No Comment Found

Related InterviewSolutions