Voltage Amplifier with an OpAmp & two NMOS help! :)

[vidak]

First of all, this isn't really homework, I'm studying for an exam, but I figured this is the most suitable place for such a question. Anyway, please find the circuit in the attachment.

Now, here is the data: I_0 = 100uA, R1=R2=R3=10kOhm, R4=90kOhm, V_T=0.7 V, B=u_n*Cox*W/L=2 mA/V^2, Vdd=-Vss=1.65 V, and no Early effect is present.

Ok, my questions are:

1. Set the + and - (don't know the word for this, are they pins?) for the OpAmp so that there is negative feedback in the circuit.

2. If the OpAmp has a voltage gain of a1=1000, find the voltage gain of the whole circuit, a = v_i / v_u (near the Bias point of course )

The problem I have is that when I draw the small-signal model, I get the voltage gain as: a=-10*gm1/gm2, where gm1, and gm2 are the transistor permitivities (don't know how these are called in english too ), calculated with gm1 = sqrt(2*B*I_1), where I_1 is the DC current near the bias point of transistor 1.

I could say that the main problem I have is calculating the currents I_1 and I_2 near the Bias Point. I said that the + sign is down and the - sign of the OpAmp is above it on the picture, is that correct?

 

[duke37]

I am not into detailed mathematics of circuits but it seems to me that an aproximation can be made.
The op-amp will have its inputs at the same voltage so the two drain currents are the same and equal to half the tail current.
The negative feedback will adjust the voltage on the two gates to be the same, to do this, the output of the amp will need to be 1+R4/R3 as in any other standard op-amp circuit. This is assuming the op-amp gain is infinite, there might be an error of about 1 in 1000.

 

[Harald Kapp]

Duke's explanation about the overall gain looks correct. However, I fear your questions are not answered.

1) The top input of A1 should be "-", the bottom input should be "+".
Look at it this way: If the voltage at the bottom input if A1 falls (more voltage drop across R2 due to more current through M2), you want M2 to conduct less (less current meaning less voltage drop on R2 thus negative feedback). In order to make M2 less conducting you need a lower gate voltage, therefore the output of A1 needs to reduce the output voltage. So the input connected to R2 and the output are in phase. Therefore this input (bottom, connected to R2) should be "+".

As to 2) It is a bit late and I'm not up to doing the calculation. But I think your explanation does seem to miss an important factor: R1 and R2. Apart from the circuit being a difference amplifier a change in gate voltage causes a change in drain current which in turn causes a change in voltage drop across the drain resistor. So the value of the drain resistors R1 and R2 has to factor into the equation.