# opamp somehow exceeds open loop gain!

Discussion in 'Electronic Basics' started by Asa Cannell, Apr 28, 2004.

1. ### Asa CannellGuest

I have an opamp (OPA348 to be exact) in a non inverting configuration.
Take a standard noninverting configuration (two resistors in series
from output to ground, resitor junction connected to inverting input),
but instead of connecting the resistor network to ground, connect it
to one side of a capacitor, and the other side of the capacitor to
ground. Now we frequency variable gain, (high pass).

In the simulator, if I change the large resistor to 1000k and the
small resistor to 3.16k, and the capacitor to .001125uF, I am getting
a gain of 316 at 12KHz. This exceeds the opamps open loop gain
specified in the datasheet, which is well under 40db (100) at 12KHz.
I have a 100k resistor on the opamps output as a load, per the
datasheet.

How is this possible? Is the opamps phase shifts somehow acting as an
inductor, and along with the capacitor, somehow a series resonant LC
circuit is formed and I am getting voltage gain? (please post
responses in the forum)

Asa

2. ### SoerenGuest

Hi Asa,

Simulators are nice toys, but they cannot be better than the models used
(and the programming).

If you "simulate" with your brain and some real life components...
WYSIWYG.

3. ### Roy McCammonGuest

you are probably getting enough phase shift ( almost 90 for
the op amp and almost 90 for your feedback) to have almost
positive feed back. Its also called regenerative gain.
Bottom line yes you can get extra gain that way
but the actual gain is very sensitive. In effect
you have an almost oscillator.

4. ### Fred StevensGuest

The closed loop gain equation Av = 1 + Zf/Zg (Zg the impedance to
ground) is accurate under the assumption that the open loop op amp
gain is infinity which is clerly not the case with your circuit!
Obviously the simulator you are using is making certain "assumptions"

Fred.

5. ### Asa CannellGuest

By extra gain do you mean gain above open loop gain? (seemingly
violating the datasheet spec). I notice that the inverting terminal is
approximately equal to the noninverting terminal over most of the
freq. band (as expected), but where I have the "extra gain" peak, it
also peaks, and is no longer equal to the noninverting input at that
frequency.

Asa

6. ### Kevin AylwardGuest

Well, the "simulator" doesn't make those kinds of assumptions. It does
what it is told to do, usually. It solves equations very accurately. The
*model* may well have approximations and assumptions that result in
these accurate solutions but have incorrect results.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

7. ### Fred StevensGuest

What I meant is that the simulator uses a model in which certain