OP AMP AC Simulation

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vleam13

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Hi,

Usually I use large resistor (>1M ohm) to break the loop between output/ inverting input (feedback path) of op-amp. Large Capacitor also added (1F) at inverting input.
AC + DC level source added at non-inverting input to run simulation. Output over Non inverting input used for gain/phase calculation.

from my understanding. Big resistor used to break the loop because we cannot leave it open, we need DC operating point. Big Capacitor to allowed AC signal pass over.
Please correct me if I am wrong.

Question:
1) what is the different if I put AC + DC source at inverting (between big capacitor and ground) or non-inverting input.
2) What is the consideration to set capacitor and resistor value? How to decide the optimized value?
3) why sometime we replace big resistor with big inductor?

Thanks.

regards.
 

vleam13 said:
Hi,

Usually I use large resistor (>1M ohm) to break the loop between output/ inverting input (feedback path) of op-amp.
Click to expand...

...to break te loop? I rather would say: ...to establish a feedback loop.
 
Hi,

if you run a simulaton, then maybe you have a schematic to show us?

Klaus
 



attached is the simplified diagram. Thanks.

- - - Updated - - -

to establish "open" feedback loop. ;p
 

Hi,

1) what is the different if I put AC + DC source at inverting (between big capacitor and ground) or non-inverting input.
Click to expand...
* we need to know the +IN connection
* we need to know AC and DC source impedance.

--> with the signal on th +In pin you know +Vin AC voltage and -Vin AC voltage (estimated to be zero)
With the output AC voltage you may calculate open loop gain at this frequency, as slong as output is not clipped.

2) What is the consideration to set capacitor and resistor value? How to decide the optimized value?
Click to expand...
R is set to generate about zero load current to the output. C is set to generate about zero AC voltage at -VIN.

3) why sometime we replace big resistor with big inductor?
Click to expand...
see 2)
the inductor combines with the C to a second order low pass filter. It is lower impedance for DC (improved DC bias adjust), but it is higher impedance for high frequencies, so it generates lower load current.

Klaus
 


Klaus - wouldn`t you think that the C makes the circuit unstable? And a combination with an inductor rather would cause a highpass-effect (however, unstable).
 

The capacitor is 1F, achieving a stable loop with regular OPs (loop gain roll-off befire the first OP pole).

Not explicite said, but you apparently want to "measure" OP open loop gain in a simulation. The setup can basically work, a DC bias would be only needed for single supply OP configuration.

RC and LC filters to break the AC feedback can both work, dimensioning according to device impedances and intended frequency range. A more general method using voltage (or current) injection has been suggested by Middlebrook, there are various similar threads refering to it at Edaboard, e.g.
https://www.edaboard.com/threads/72440/
 

Hi,

Klaus - wouldn`t you think that the C makes the circuit unstable? And a combination with an inductor rather would cause a highpass-effect (however, unstable).
Click to expand...

I don´t think the C at the -Vin makes the circuit unstable (in the meaning that the circuit oscillates without input signal) . But the lower the input frequency, the higher the gain of the OPAMP (according open loop gain chart). For low frequencies one need very low input voltages to avoid opamp clipping.

Klaus
 

I don´t think the C at the -Vin makes the circuit unstable
Click to expand...
A 1 µF capacitor possibly would, at least it results in about zero feedback phase margin and doesn't work work for the intended purpose. But 1F is O.K.
 

Some very old opamps like the 741 have a fairly low input resistance (minimum 300k ohms) so their feedback resistor value should not be higher than 15k or 30k ohms.
If you short AC feedback frequencies to ground with a capacitor at the inverting input then the AC gain will be almost unlimited and full of hiss.

One megohm feeding one Farad will take such a long time to charge that I don't want to calculate it.
 

KlausST said:
Hi,
... But the lower the input frequency, the higher the gain of the OPAMP (according open loop gain chart). For low frequencies one need very low input voltages to avoid opamp clipping.
Klaus
Click to expand...

Klaus - wouldn`t you see 100% feedback (unity gain) for very low frequencies? And - larger gain for higher frequenicies (decreasing feedback)?

- - - Updated - - -

FvM said:
A 1 µF capacitor possibly would, at least it results in about zero feedback phase margin and doesn't work work for the intended purpose. But 1F is O.K.
Click to expand...

OK - I didn`t realize the value of 1E6 µF = 1F. A rather uncommon value.
 

I assume this is to plot the open loop gain function of the op amp.
You have it rather backwards.
The resistor is used to close the loop and provide DC feedback so the op amp stays in the linear mode.
The capacitor is used to attenuate the AC feedback so the AC open loop gain can be measured.

1) Putting the input signal at the inverting input will attenuate the signal and give an incorrect open loop gain measurement.
2) The RC value determines the low frequency point at which you are no longer measuring the open loop gain.
The RC rolloff must be more than the open loop gain at the lowest frequency of interest for the measurement to be correct.
For typical op amps, this can be well below 1 Hz.
There is no "optimum" value, you just make the roll-off frequency low enough so is doesn't affect the lowest frequency of interest.
3) An inductor performs the same function as the resistor but introduces no DC offset due to any op amp bias current.

Why are you adding a 1Vdc offset to the input signal?
 
Hi,

Klaus - wouldn`t you see 100% feedback (unity gain) for very low frequencies? And - larger gain for higher frequenicies (decreasing feedback)?
Click to expand...
Yes you are right for the complete circuit with the external R and C.

I assumed that the OP wanted to measure the Open Loop Gain of the OPAMP IC itself. Therfore he chose this circuit.
So i referred to the OPAMP (internal) gain, the open loop gain.

As long as I understand the circuit right it is a DC feddback just to stabilize operation point. Then maybe above 50 times cutoff frequency (of the externel RC, which is 50 x 1.59Hz = 8Hz @ 1uF) the OPAMP acts like "open loop"

Klaus
 

Why are you adding a 1Vdc offset to the input signal?[/QUOTE]

Closer to actual operating condition. If no 1Vdc, AC voltage will swing between +VE and -VE voltage.
 

If the opamp output swings between +VE and -VE then it must be a "rail-to-rail" type with a very high load resistance AND the input signal level is too high OR the opamp is amplifying its own noise.
 

FvM said:
A 1 µF capacitor possibly would, at least it results in about zero feedback phase margin and doesn't work work for the intended purpose. But 1F is O.K.
Click to expand...



1uF is not working for me. I use 1F capacitor.
 

vleam13 said:
..............
1uF is not working for me. I use 1F capacitor.
Click to expand...
1µF with 1megΩ gives a rolloff of about -16dB at 1Hz. This will not give the correct op amp open loop gain at that frequency, which is typically much higher than 16db.
1F with 1megΩ has a rolloff of about -136dB at 1Hz. Since this is likely much more than the op amp open loop gain, then the gain measurement should be accurate to well below that frequency.
 
crutschow said:
1µF with 1megΩ gives a rolloff of about -16dB at 1Hz. This will not give the correct op amp open loop gain at that frequency, which is typically much higher than 16db.
.
Click to expand...

1µF with 1megΩ gives a high pass corner frequency of about w=1 [1/s] equivalent to 0.16 Hz.
I think, that is quite sufficient to cover the maximum gain which will start to decrease certainly not before 10 Hz.
(By the way: What is a "rolloff of about -16dB"? Do you mean attenuation?)
 

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