inverting amplifier/mixer: small cap in feedbackpath?

[fonik]

hi all,

we often see a small cap in parallel to the resistor in the feedback path of inverting OPA stages. the values vary approx. between 2pF and 22pF.

what is the use of these? how do i calculate the value?

[Uncle Krunkus]

From what I've worked out, they allow complete negative feedback of high frequency oscillations which sometimes are a problem when you've got an op-amp configured for a gain of 10 or more. This thereby actively eliminates them from that gain section. Someone jump in and tell me if that's off track.
I'm not sure about how to decide what kind of value to use, but because of the way they sap the oscillations it makes sense (and I've heard before) that any value from 10pF to 100pF will do the trick. I figure if you take the value to high you'll start to audibly lose top end response.

[fonik]

[Uncle Krunkus]

Not as far as I understand it.
Thanks Fonik, you've actually given me the opportunity to display some very specific knowledge which I have gained solely by hanging out on this forum. I feel all kind of warm and fuzzy inside. (could be the beer? )

[frijitz]

[funkyfarm]

Discreet, but this is surely the best topic of the year !!

fonik rules !

(btw, i put 10 or 15pf everywhere)

[zipzap]

a good probe could be the thing, cause when i turn my scope up to very high amplification, i see oscillation everywhere. even if i measure nothing or ground. The universe seems to be oszillating...
on the scope i sometimes have difficulties to juge if what i see is good or bad.
So normally, in our analog opamp world, what is it we are looking for? what range of frequency and amplitude?

[yusynth]

Hi

Here are some infos.

Yes you are right the feedback capacitor is only required when the gain is greater than, say 3. Now how do one calculate the value of this cap ?
If the maximum frequency you allow to pass is F and the feedback resistor is R, then the value of C is :

C = 1 /(2xPIxFxR), for example if F= 30kHz and R = 150k then C is 35pF.

[jhaible]

As I see it, the main reason to add these capacitors is to compensate the input capacitance of the opamp.

The opamp creates a phase shift towards higher frequencies.
Also with increasing frequency, the opamp's (open loop) gain is decreasing. This leads to the infamous phase margin: how much of additional phase shift would be required for the whole thing (opamp plus feedback path) to oscillate. This depends on the opamp type and the closed loop gain, i.e. how much feedback you have. Normally tighter feedback is worse - lower gain of the (closed loop) amp will be more prone to oscillating. That's why some fast opamps will only work down to a certain closed loop gain (like 3, or 5), and not to unity gain.
As this is specific for each opamp, you normally find the relevant information in the data sheet. It shows a way how to build an amp with good "safety margin" against oscillation.

Now everything that creates an extra phase shift, in addition to the normal data sheet applications, makes this phase margin smaller. You'll get more overshot/ringing, and for severe extra phase shifts, even self oscillation.

One way to get increased phase shift in the feedback circuit of an opamp is to use large resistor values. A simple resistor from output to input will create a low pass filter together with the opamps own (parasitic) input capacitance. (And with the stray capacitance of the copper tracks on the pcb, of course.) This low pass filter will add some phase shift, thus decreasing the phase margin of the circuit.

You can cure this by connecting a small capacitor across the feedback resistor (parallel connection). Now, at "very high" frequencies, where your ringing and oscillations are bound to occur, that capacitor dominates the impedance of the feedback_resistor, capacitor parallel connection: the cap has a much lower impedance there. This also means that the combined impedance is (almost) capacitive in nature. Now, still looking at highe frequencies, in combination with the opamp's input capacitance you don't have a low pass filter anymore (series R and shunt C); now you have a simple voltage divider (series C and shunt C). The whole divider is capacitive in nature, but as both the series and the shunt arm of the divider are capacitive, there is no additional phase shift!

How does this relate to noninverting unity gain amplifiers?
If your feedback path is a straight wire, you obviously don't need a capacitor. But in specific cases you may have a single resistor from the opamp output to the inverting input even in a unity gain application (for instance, to match a source impedance at the positive input, in order to do a bias current compensation); then, if the resistor value is large, you may consider connecting a cap across this as well.

Does this method help to compensate the effect of vcapacitive loading of the opamp output? - I can't see any way it would compensate this directly, but capacitive loading creates phase shift from the opamp's output resistance and the capacitive load (another low pass filter!), so we might be well advised to minimize the phase shift elsewhere (at the input).
Maybe that's why that cap across the feedback resistor also helps with capacitive loading, to a certain degree.

JH.