# DC analysis of common source MOSFET circuits to find gate voltage using different power supplies simulate this circuit – Schematic created using CircuitLab

strong text

I am solving problems on DC analysis of MOSFET circuits. We are supposed to find various parameters like Id(Drain Current) and voltages at various nodes like Vg(Gate Voltage).

The MOSFETS used are both NMOS(n-channel enhancement type)

In my book, where circuits of first type are concerned, Vg is given by

Vg=[R3/(R2+R3)]*Vdd ____________ eqn 1 (I figure this is because resistors R3 and R2 form a voltage divider network. Correct me if i am wrong)

However for circuits of second type, Vg is calculated by

Vg=[R3/(R2+R3)]*(Vdd-(-Vdd)) - Vdd _______________ eqn 2

I don't understand why Vdd is being subtracted. I scanned all the pages of my book looking for a one-line explanation but didn't get any. I also tried joining the 2 nodes to form a third loop and then use KVL(Kirchoff's Voltage Law) but I just don't get it.

Can someone explain me the equations 1 and 2?

PS: This is my first question on this site so general tips about asking specific,easy to read questions and courting quick answers will be appreciated. Have a good day.

The Vg eqn for the 1st circuit is correct.

The 2nd circuit has a negative rail (also called Vdd) and this is where the confusion lies. You could regard the voltage at the gate as being twice the 1st circuit but relative to -5V.

Vdd-(-Vdd) = 2*Vdd = 10V because Vdd on its own is 5V. This and the standard potential divider formula gives you a gate voltage relative to the -5V rail.

But, because the gate voltage is more sensibly referenced to 0V, then Vdd needs to be subtracted to make Vg relative to 0V rather than -5V.

You'll actually notice those two equations ARE the same after a little cleaning up.

Instead of 0, -5, or -Vdd we'll call the lower voltage Vss. Now this is what the equation should look like:

$$Vg=\frac{R3}{R2+R3} \times (Vdd-(Vss)) + Vss$$

The first part of the equation applies a voltage divider to the total potential of Vdd-Vss. In the second case you can see 5v to -5v is a 10v difference across the rails and that 10v is what gets divided across R2 and R3. That extra Vss on the end is because whatever voltage is across R3 will get added on top of the reference voltage, Vss.

If you substitute 0 or -Vdd for Vss respectively you should get equations 1 and 2.

• Thank you. It is surprising how the smallest of things can escape one's attention. :) – Pratik Mar 8 '15 at 6:43