# Op amp transfer function incorporating output impedance

1. Apr 22, 2012

### jrive

Hello,

I was trying to derive the transfer function of a simple inverting op amp topology, incorporating the output impedance, Ro, of the non-ideal op amp, to see how it, when combined with a load capacitance can induce oscillations (or at least ringing). I haven't been able to get a valid answer --could be bookkeeping errors with the math, or just that my set up is incorrect for deriving the circuit's transfer function. Refer to the attached picture.

Using Laplace, the xfer function I end up with is:
$\frac{vo(s)}{vi(s)}$=$\frac{-A*RF}{(R1+RF)*(s*CL*Ro*RF+(Ro+RF))+A*R1}$

If A is really big...this simplifies to -RF/R1, which is sort-of expected for negligible Ro. However, I expected the frequency dependent component (which includes RO) to be part this result even as A is big, so I must be doing something wrong somewhere.

I suspect it has to do with the derivation of the xfer function using A(Vp-Vn). Basically, I obtain Vn via superposition of the contributions from Vin and Vo as $\frac{(V1*RF +Vo*R1)}{(R1+RF)}$, Vp =0, and then use KVL at the Vo node....That is...

$\frac{-A*Vn-Vo}{Ro}$+ $\frac{(Vn-Vo)}{RF}$=Vo*s*CL....I then sub in the equation for Vn above and solve for Vo/Vin.....

Any help is appreciated...
Jorge

#### Attached Files:

• ###### opamp_ro.png
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2. Apr 23, 2012

### Staff: Mentor

Hi jrive! That looks like it will be a first-order system, so you won't get any oscillations out of that. You should model the op-amp more realistically. At least give it a gain that falls off with frequency, a single pole giving A₀/(1 + jω/ω₀). Typically, ω₀ is something like 100Hz, but if you look at the data sheet you'll see the corner frequency of its open loop gain.

3. Apr 23, 2012

### jrive

yep...you're right NascentOxygen....thanks for pointing that out...I'll give that a shot to see if my math works out.