Calculate transfer function of transconductance amplifier circuit

Question

How can I model/calculate the behavior from Vin to Iout of the following circuit? (Credits of the picture go to Analog Devices):

Essentially, I am interested in the large signal response, such as a step at Vin.

Using the standard scheme for single loop feedback control systems, I could calculate \$\rm V1/V_{IN}\$, getting \$\rm I_{OUT}\$ from \$\rm V1\$ means then just using Ohm´s law.

Two problems here:

1. The transfer function should take the form $$T(s) = \frac{\rm{A_0}}{1 + \rm{A_0}B(s)},$$

   with \$\rm{A_0}\$ being the open loop gain of the opamp. I wonder, though, how to obtain \$\rm B(s)\$, which is essentially the transfer function form the MOSFET Gate to its source.

2. Since the MOSFET is clearly non-linear, I also assume that the transfer function is then only a useful information for a given DC bias point at the gate with an overlay of very small AC signal variations.

Yes, I can simulate the behavior very easily using some spice program, but at least I would like to compare simulation results with some analytical solution. So: Any ideas on how to obtain such as "large signal transfer function" that is valid during the input voltage step?

Answer 1

see correction in the last line:

I don`t know if you are asking for the following analyses. Nevertheless, here it comes:

gm=Transistor transconductance; Vo=opamp output voltage; Acl=opamp closed-loop gain; k=feedback factor

Iout=VGS*gm

VGS=Vo-V1

V1=Iout*R1

Vo=Vin*Acl

Acl=Ao/(1+kAo)

k=gmR1/(1+gmR1) >> source follower

This gives (for infinite Ao): Acl=(1+gmR1)/gm*R1.

Now you can combine all the equations - starting at the top (simply insert the succeeding expressions):

The result is: Iout=Vin/R1

EDIT: In case, the real and frequency-dependent open-loop gain Ao(s) is to be considered, we arrive at the following expression (same set of equations, however, without setting Ao to infinite):

CORRECTION: There was a computational error (1/gm was missing in the denominator)

Iout=Vin/[R1 + (R1+1/gm)/Ao(s)]=Ao(s)Vin/[1/gm + R1 + R1*Ao(s)]

T(s)=Iout/Vin=Ao(s)/[1/gm + R1 + Ao(s)*R1]

Answer 2

The open Loop gain and transfer function of the MOSFET can be neglected if you follow basic rules for DC bias.

• for (RdsON+R1)*Iout < Vo < Vcc (Vo is current sink voltage)
• for Vgs>Vth for gate threshold Vth
• OpAmp-out swings from Vgs (+Iout*R1) to above Vcc (by Vgs)
• for R1 << RdsOn to give more gain and voltage swing on MOSFET
• for Vin=0 to Vmax ( AC with DC offset=Vpk)

Then you get full supply range swing for Iout

• limited only by your power source and Op Amp swing to drive gate and RdsOn and heat sink to dissipate Vds*Iout

Iout = Vin / R1 and has nothing to do with Av, RdsOn, Vth etc.

• however bandwidth will depend on GBW of OA.

Answer 3

The mosfet transfer function Vs/Vg is going to be very close to 1. This will be true so long as you don't run the the circuit to the point you where the fet or opamp are saturated or cutoff. There are capacitors in the fet model Cgs, Cgd, and some others that would give you S terms. If you can ignore the frequency terms from the amplifier you can probably ignore the fet capacitors too.

One of the great things about feedback is that it reduces the effect of nonlinearity. So lets assume during the step response the gain of the fet changed from 1 to 0.9. The large gain of the opamp is going to force the output to almost the same value in either case. The 10% change in gain will hardly be noticed.