[SOLVED] Small Signal of operation amplifier

Hi

Any one know how to do transfer function Vg(s)/Voe(s) of this op amp without drawning small signal. I know some one can do this quick but I do not know the trick. Thanks

The opamp is labeled with a gain of Gm2. Does that mean it is a transconductance amp?

Hi crutshow

Yes

You only have to use transfer formula for the OTA: Iout=Vdiff*gm with vdiff=Voe-Vi (Vi: voltage at inv. terminal) together with Ohm´s law and Kirchhoff´s current law at the output.

Hi LvW

Thanks for your suggestion. I know this system has one zero and 2 poles but the answer is not simple as I think. This is the answer

So - your problem is solved now?

if you need a proof, you can proceed as follow:

As already said, the output curent is: Io=[V(+) - v(-)]*gm = [Voe - v(-)]*gm

let's assume the input current of the transconductance amplifier is 0, then the output voltage will be given by the output current multiplied by the impedance seen by the amplifier, that is the parallel between the impedance Zf due to Cz, Rz and the impedance Z2 due to R2, C2.

Zf=Rz+1/sCz=(SCzRz+1)/SCz
Z2=R2/(SC2R2+1)

then Zo=Zf*Z2/(Zf+Z2)

that is: Vo=Io*Zo, substituting Io:

Vo=[Voe-V(-)]*gm*Zo

But V- is gived by the voltage divider formed by Rz and Cz, then:

V(-)=Vo/(SCzRz+1)

call now k=Vo/(SCzRz+1), then V(-) = k*Vo, substituing back:

Vo=[Voe-k*Vo]*gm*Zo, solving with respect to Vo/Voe:

Vo/Voe= gm*Zo/(1+k*gm*Zo)

To simplify the next calculations let's call Zf=Nf/Df, Z2=N2/D2 and Zo=No/Do, where N. are the numerators and D. are the denominators.

Thus:

Zo=No/Do=[(Nf/Df)*(N2/D2)]/[(Nf/Df)+(N2/D2)]=Nf*N2/(Nf*D2+N2*Df) then:

No=Nf*N2
Do=Nf*D2+N2*Df

and also

k = 1/Nf

Substituting in the equation Vo/Voe we will have:

Vo/Voe= gm*(No/Do)/[1+k*gm*(No/Do)] = gm*No/(Do+1/Nf*gm*No) that is:

Vo/Voe=gm*Nf*N2/(Nf*D2+N2*Df+gm*N2)

substituting now Nf,Df, N2 and D2 with the value aready calculated:

Vo/Voe = gm*(SCzRz+1)*R2/[((SCzRz+1)(SC2R2+1)+SCzR2+gm*R2)