I am working on an exercise problem (5.18) regarding Discrete MOSFET Amplifiers. I have taken pictures of the question material (Disregard the section under the question) and my latest attempt at a solution:

The question

The given circuit

Alright. Keep in mind that the circuit above is from a different question, they are just re-using it in this one, but they changed the value of VDD from +15V to +5V. Also, they remove RL, but I don't think that matters for my specific issue.

Now, I decided to begin with a DC Analysis to find VOV (Overdrive voltage). I know that I can replace the capacitors with open circuits and then use the MOSFET Saturation current equation and KVL for another current equation, equate them, then solve the resulting quadratic.

However, this method yields an answer different from the book. The book provides VOV = 0.319V, whereas I get VOV = 0.83V.

Is my approach correct? I'm pretty sure the math is sound, so I'm not really sure how to proceed. My solution:

My solution part 1

My solution part 2

  • \$\begingroup\$ I love those "infinite" capacitors. Obviously, they denote "arbitrarily large" and are meant just for DC blocking, but in fact, an infinite capacitor is indistinguishable from a short circuit! \$\endgroup\$
    – Dave Tweed
    Commented Mar 14, 2014 at 0:48
  • \$\begingroup\$ @DaveTweed, to be sure, an "infinite" capacitance capacitor is indistinguishable from an ideal voltage source (and a 0V ideal voltage source is indeed indistinguishable from an ideal short circuit). In fact, once one has the DC solution, replacing a coupling cap with an ideal voltage source of voltage equal to the DC voltage across the cap gives the same result as having infinite capacitance. \$\endgroup\$ Commented Mar 14, 2014 at 1:17
  • 1
    \$\begingroup\$ I want a font made from your handwriting. \$\endgroup\$
    – JYelton
    Commented Mar 14, 2014 at 3:44
  • \$\begingroup\$ Haha! Are you kidding it looks horrible! :P \$\endgroup\$
    – Nevermore
    Commented Mar 14, 2014 at 5:03
  • \$\begingroup\$ Thanks though :) Getting a good pen really helps! \$\endgroup\$
    – Nevermore
    Commented Mar 14, 2014 at 5:04

1 Answer 1


I believe you are misinterpreting the question.

The question is asking you to design the circuit to meet certain specs, which includes choosing the drain resistor Rd. In your analysis, you are assuming that Rd is 10kohms. The answer states that Rd is 78.5kohms.

  • \$\begingroup\$ Wow! I can't believe I misread the question! Thank you so much, I'm going to figure out my next approach now :D \$\endgroup\$
    – Nevermore
    Commented Mar 14, 2014 at 1:31
  • \$\begingroup\$ Alright, I went through the question and answered it completely! Thanks again for your help! :) \$\endgroup\$
    – Nevermore
    Commented Mar 14, 2014 at 2:06

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