I am teaching electronics. I have come across this formula for drain current in the syllabus that I am teaching. However, I have never seen it before. My question is where is this number -3 from and is it even correct?

Id = Gm · (Vgs - 3)

You can take a look at this syllabus - it's on page 15.

  • \$\begingroup\$ In the linked document, the formula under discussion is on page 15, (for MOSFET). \$\endgroup\$
    – LvW
    Feb 1, 2022 at 11:04
  • \$\begingroup\$ The formula cannot be correct. At first the expression (Vgs-3) is garbage . Secondly, there must be a quadratic relationship between Id and Vgs. \$\endgroup\$
    – LvW
    Feb 1, 2022 at 11:12
  • \$\begingroup\$ Thank you for confirming this. \$\endgroup\$
    – Jarvel
    Feb 1, 2022 at 11:38
  • \$\begingroup\$ Have a look at Q9 on this exam paper: pastpapers.download.wjec.co.uk/O20/o20-5490-01.pdf \$\endgroup\$
    – Jarvel
    Feb 1, 2022 at 12:02
  • \$\begingroup\$ And now look at the solution on P15 of the mark scheme: pastpapers.download.wjec.co.uk/A20/a20-C490UA0-1-ms.pdf \$\endgroup\$
    – Jarvel
    Feb 1, 2022 at 12:03

1 Answer 1


For saturation region, and ignoring \$\lambda\$ effects, let

$$Id = \frac{\beta}{2}(Vgs-Vt)^2$$ where $$\beta = (k')(W/L)$$ now taking the partial derivative of \$Id\$ with respect to \$Vgs\$ we get, $$g_{m} = \beta(Vgs-Vt)$$

Since we can isolate \$\beta\$ as $$\beta = \frac{2I_{d}}{(Vgs-Vt)^2}$$

Substitute this \$\beta\$ into \$g_{m}\$ eq. above gives, $$g_{m} = \frac {2I_{d}}{(Vgs-Vt)}$$ then $$I_{d} = \frac{g_{m}}{2}(Vgs-Vt)$$ If \$V_t=3\$ then $$I_{d} = \frac{g_{m}}{2}(Vgs-3)$$

*from slide p. 15

"(h) define gM as the gradient of an ID-VGS graph."

I am confident it is off by a factor of 1/2, or they will argue that they absorbed the factor somewhere else (inside \$g_m\$ for example).

  • \$\begingroup\$ But "3" is still missing a unit. It should be "3 V". \$\endgroup\$ Feb 4, 2022 at 10:16

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