I'm studying the following circuit with a MOSFET

enter image description here

Now for analyzing this circuit, my book came out with various equations (which I totally understand and have no doubts about it).

Because the gate current is zero we have:

(1) $$V_G=V_{DD} \frac{R_2}{R_1+R_2}$$

Assuming the MOSFET is working on saturation we also have:

(2) $$I_D=k(V_{GS}-V_t)^2$$

And applying KVL:

(3) $$V_{GS}=V_G-R_SI_D$$

Substituting equation (3) on equation (2) we have:

(4) $$I_D=k(V_G-R_SI_D -V_t)^2$$

Now comes to the point I'm not understanding. To show us the dependence of the drain current over the threshold voltage, the book takes the derivative of the expression:

(5) $$\frac{dI_D}{dV_t}=\frac{d}{dV_t}k(V_G-R_SI_D -V_t)^2$$

and then it writes:

(6) $$\frac{dI_D}{dV_t}=\frac{-2\sqrt{kI_D}}{1+2R_S\sqrt{kI_D}}$$

How on earth did they go from equation (5) to equation (6) by taking the derivative? What kind of substitution are they making?

  • 2
    \$\begingroup\$ Granger - Hi, (a) Since you referred to the equations by number, I've added numbers to each of them, to make it easier for readers. :-) I hope I've numbered them as you intended; if not, please correct the numbering. (b) Please can you add a reference to the book which you mention (title, author, page number etc.). Thanks. \$\endgroup\$ – SamGibson Feb 14 '18 at 2:11
  • \$\begingroup\$ differentiate 5 then sub 2 for Vt into result \$\endgroup\$ – Tony Stewart EE75 Feb 14 '18 at 4:35
  • \$\begingroup\$ 6 is the derivative of 4 (I didn't verify this). 5 is saying that "dId/dVt is the derivative of this thing" but without actually having worked out the derivative. \$\endgroup\$ – user253751 Feb 14 '18 at 4:40
  • \$\begingroup\$ looks ok to me. \$\endgroup\$ – Tony Stewart EE75 Feb 14 '18 at 4:43
  • \$\begingroup\$ I obtained $$-2k(V_G -R_SI_D-V_{GS}+ \sqrt{\frac{I_D}{k}})$$. What should I do next? \$\endgroup\$ – Granger Obliviate Feb 14 '18 at 11:15

$$\frac{dI_D}{dV_t}=\frac{d}{dV_t}k(V_G-R_SI_D -V_t)^2$$ $$\implies \frac{dI_D}{dV_t} = 2k(V_G-R_SI_D -V_t)(0-R_S\frac{dI_D}{dV_t}-1)$$ $$\implies \frac{dI_D}{dV_t} = 2k(V_G-R_SI_D -V_t)(0-R_S\frac{dI_D}{dV_t}-1) ---(1)$$ Given that: $$I_D=k(V_G-R_SI_D -V_t)^2$$ $$\implies (V_G-R_SI_D -V_t) = \sqrt{I_D/k} $$ Therefore (1) \$ \implies\$ $$\frac{dI_D}{dV_t} = 2\sqrt{kI_D}(-R_S\frac{dI_D}{dV_t} -1)$$ $$\implies \frac{dI_D}{dV_t} +2\sqrt{kI_D}R_S\frac{dI_D}{dV_t} = -2\sqrt{kI_D} $$ $$\implies\frac{dI_D}{dV_t}(1+2\sqrt{kI_D}R_S) = -2\sqrt{kI_D} $$ $$\implies \frac{dI_D}{dV_t} =\frac{-2\sqrt{kI_D}}{1+2R_S\sqrt{kI_D}} $$

  • 1
    \$\begingroup\$ Thank you very much! I got what I was doing wrong. I was assuming a constant current I_S, which is obviously wrong. \$\endgroup\$ – Granger Obliviate Feb 14 '18 at 14:06
  • \$\begingroup\$ Yes Id is a dependent variable on Vt. \$\endgroup\$ – Mitu Raj Feb 14 '18 at 14:08

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