Analogue circuit analysis

[Craig405]

Hi,

I would really appreciate it if someone could help me solve this problem. The circuit is for a single stage amplifier with an active load.
I am having immense headaches trying to calculate the quiescent drain current in the NFET, I make the P-channel device current equal to 0.274uA.

How can i assume what region the lower mosfet is in to apply the correct drain current equation?

I must be overlooking somthing?

[Bonsai]

Graig405, the quickest way to get a grip on this type of problem is to use spice

LTSpice tutorial

Hope this helps.

[Craig405]

Thanks Bonsai, actually I have been asked to compare the theoretical performance to that of HSPICE, which I believe is similar to LTSPICE.

Any ideas on applying theory to this?

what confuses me is the feedback network of resistors, this makes it more difficult to analyse. Would I be correct in saying that the feedback factor can be considered from the ratio of these when putting togeather a transfer function?

[rtarbell]

Craig405,

I believe you could do it by iteration:

1) assume that there is NO current flowing through the R1-R2 branch (all of the 274nA are flowing through the NMOS)

2) calculate the Vgs of the NMOS at this current

3) The Vgs of the NMOS is the same as the voltage across R2 - so calculate the current through R2

4) Subtract this R2 current from the original 274nA

5) RECALCULATE the Vgs of the NMOS (it will now be different because not all of the 274nA is going through the drain)

6) repeat steps 3 through 5 until your answer doesn't change very much (it should only take about 4 to 5 iterations to get a very close approximation)

Having said all of these steps, I would agree with Bonsai - Spice will be much faster.

[Craig405]

Thanks a lot rtarbell, i will be simulating in spice once ive made some calculations, the iteration looks like a good option for the hand calcs.

I just need to compare numbers to spice, hopefully a one off excersise

[rtarbell]

Some equations of the circuit:

(1) I(r1) = I(r2) = 274nA - Id(nmos)

(2) Id(nmos) = (K/2)*(W/L) * (Vgs - Vt)^2

(3) Vgs = R2 * I(r2)

Plugging in equation (1) into (3):

(4) Vgs = R2 * (274nA - Id(nmos))

Now, plug in (4) into (2), and solve for Id(nmos), as it appears on both sides of the equation. (You will probably get two answers due to the quadtratic equation, but only one of those two answers will make sense, as Vgs must be greater than Vt for the NMOS to conduct any current).

[Craig405]

Thanks, youve been a great help!
Im about to tackle this today after I have finished coffee and laundry hehe
Ill post the analysis here once ive got it (hopefully) right in case others are interested.