I can do DC analysis and draw the small signal equivalent circuit just fine. I generally have trouble calculating gain. I'm having trouble understanding the answer to this question.

This is the question. Question 4.36: The Question

This is my working. I hope it is clear:

enter image description here According to the solution manual this is the wrong answer. The correct answer is $$A_v = \frac{V_o}{V_o + 1}$$: enter image description here

I don't understand. \$V_i\$ is in parallel with \$V_{sg}\$ therefore \$V_i\$ must equal \$V_{sg}\$ right? Where did I make a mistake?


1 Answer 1


No, Vi is not Vgs, look carefully at the circuit, it's a common drain ! Vi = Vo - Vgs

Also: your small signal equivalent circuit is for a common source while the assignment 4.36 is for a common drain! Note that it is a PMOS with the arrow (indicating the source) at the resistor Rs side.

  • \$\begingroup\$ Actually Vgs is DC and it is AC coupled \$\endgroup\$ Nov 30, 2017 at 15:21
  • \$\begingroup\$ @TonyStewart.EEsince'75 Your comment is confusing. Vgs is not DC only. Vgs consists of a DC component (the biasing, this makes a quiescent Id flow) plus an AC component determined by both Vi (via an AC coupling: Cc1) and Vo (also via an AC coupling: Cc2). \$\endgroup\$ Nov 30, 2017 at 15:28
  • \$\begingroup\$ @TonyStewart.EEsince'75 I think it is correct. Care to enlighten us why you think it is incorrect? "It doesn't look right" is not much to go by. \$\endgroup\$ Nov 30, 2017 at 15:57
  • \$\begingroup\$ \$G_i=R_G/R_L (approx)~~~ g_mR_L>>1\$ \$\endgroup\$ Nov 30, 2017 at 16:43
  • \$\begingroup\$ consider \$G_v=\dfrac{1}{1+\dfrac{R_L+r_o}{g_m~R_L~r_o}}\$ what does \$V_o\$ have to do with small sig gain. and it is a common source not common drain. \$\endgroup\$ Nov 30, 2017 at 17:04

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