I have an op-amp which is supposed to work as a current to voltage converter. The feedback resistor that i have is 25 mega ohm while my input current is about 1 nanoamp. when i was looking at the output voltage, it was oscillating too much, but after i added a capacitor (10 pF) in parallel to the feedback resistor, i saw that the output voltage got very stable. can anyone explain why this thing is happening? i mean how a capacitor can stabilize a non-stable output voltage?
thanks

 

Short version: The inputs of an opamp have capacitance. The resistor you connected to that capacitance forms a frequency point. The 10 pf capacitor completely swamps the effect of the input capacitance.

 

Short version: The inputs of an opamp have capacitance. The resistor you connected to that capacitance forms a frequency point. The 10 pf capacitor completely swamps the effect of the input capacitance.

Thanks a lot,
can you please make it more clear to me?
does this mean that whenever we have a capacitor and a resistor in series we will end up with having a frequency point?
and am i right that by having that capacitor we basically make a filter to get rid of that oscillation?

 

no, yes, no.
No, I can't stay long enough (today) to expect to make this very clear. There are others here that can take this as far as you want to because it is an old problem with an old solution.
Yes, any RC connection has a frequency associated with it.
No, by having that FEEDBACK capacitor (not the input capacitance) we change the frequency of the filter enough to get it out of the uncontrolled range (oscillation).

This document is 13 years old, but I could swear I read the explanation about 30 years ago.

 

thanks #12,
your comment and the attached file were very helpful.

 

Holden_C: The effect of the feeedback capacitor becomes clear if you look at the corresonding BODE plot.
To see the loop gain response you can plot two different curves:
1.) Curve 1: Open-loop gain Aol of the opamp
2.) Curve 2: 1/beta with beta=Feedback factor

The DIFFERENCE (in dB) between both curves gives the loop gain response (magnitudes)
A,loop=Aol(dB)-1/beta(dB) corresponding to A,loop=Aol*beta.

For stability both curves (Aol and 1/beta) must NOT cross each other with a net slope of -40 dB/dec or more (stability criterion). Introducing a capacitor across the feedback resistor INCREASES beta for rising frequencies and, hence, REDUCES the slope of 1/beta, thereby reducing the net slope of Aol(dB)-1/beta(dB). For ideal resistive feedback the slope of 1/beta would be zero.

Note 1: Aol and 1/beta both have negative slopes.
Note 2: Similarly, the effect of a parasitic input capacitance (as mentioned by #12) can be verified. Without the feedback capacitor the slope of 1/beta even would INCREASE for rising frequencies. Thus, with a feedback capacitor this increase can be compensated.

 

Holden_C, to visualize my explanation, please have a look on the attached pdf-file (Bode plot).
Error correction: The mentioned slope values should be in dB/decade (not dB)

 

The feedback resistor and input capacitance cause a delay in the signal being fed back and turn the negative feedback into positive feedback and then you have an oscillator.

 

The feedback resistor and input capacitance cause a delay in the signal being fed back and turn the negative feedback into positive feedback and then you have an oscillator.

Then the feedback capacitor changes the delay time and gets you back into the stable range for that opamp and resistor. The phase shift approach is entirely valid, but it requires a higher level of understanding on your part. As you get better at this, you will change to thinking about phase shift because it is more versatile.

 

Below is a simulation of a current-voltage op amp circuit with an assumed (arbitrary value) input stray capacitance of 50pF .

The response with no feedback capacitor (green) shows a high peak in the response and a large change in phase shift at about 35kHz. This will cause a large amplification of noise at that frequency or possible oscillations.

The response with 10pF feedback capacitance (blue) shows no peaking or large phase-shift change and smoothly rolls off the response above about 200Hz (a smaller feedback capacitor will raise the corner frequency while still preventing the peak if you need a higher frequency response).

In general you should always limit the frequency response of a circuit to no more than somewhat above the highest frequency of interest to minimize circuit noise and instability.

Edit: Note that to get the maximum frequency response with stability, Cfb may need to be less than one pF which such a large feedback resistor. This small value can be made with a "gimmick" capacitor consisting of two short pieces of insulated wire twisted together. The optimum capacitance can be found by adjusting the length of the wire and changing the number of twists.

 

Then the feedback capacitor changes the delay time and gets you back into the stable range for that opamp and resistor. The phase shift approach is entirely valid, but it requires a higher level of understanding on your part. As you get better at this, you will change to thinking about phase shift because it is more versatile.

This reminds me on the working principle of an oscilloscope probe.
If a suitable feedback capacitor is selected the whole feedback path (including the parasitic input cap) acts simply as a frequency-compensated voltage divider.

 

In general you should always limit the frequency response of a circuit to no more than somewhat above the highest frequency of interest to minimize circuit noise and instability.



View attachment 65409

Thank you very much crutschow for your analysis.
In fact my frequency of interest is about 2-3 KHz. This circuit is made for an Scanning tunniling microscope. If i want an image with a scan rate of 4Hz and 512 sample per line i will end up with 2028Hz and my breaking frequency with 25 Mohm and 10 pf will be 4 kHz which is well above the frequency that i need at 4 Hz.

may i ask how did you simulate those curves? i mean is there a special software to do that?

thanks

 

I think crutschow uses LTSpice.
Free download. Lots of support forums. Not exactly easy when you first start out learning it.

 

In fact my frequency of interest is about 2-3 KHz. This circuit is made for an Scanning tunniling microscope. If i want an image with a scan rate of 4Hz and 512 sample per line i will end up with 2028Hz and my breaking frequency with 25 Mohm and 10 pf will be 4 kHz which is well above the frequency that i need at 4 Hz.

may i ask how did you simulate those curves? i mean is there a special software to do that?

10pF and 25Mohm gives a -3dB rolloff of 636Hz (equals 4k rad/sec which is apparently what you calculated), as you can see from the simulated blue line plot. For a -3dB rolloff of 4kHz you need C equal to 1.6pF.

Yes, it was simulated with LTspice, a free downloard from Linear Technology. If you are interested in doing the simulation yourself you can download the software and I will post the .asc simulation file for you to run.

 

1.6 pf is so close to zero that it makes me wonder if you'll get back into oscillation. Best to assume crutschow is ahead of me on this one.

 

1.6 pf is so close to zero that it makes me wonder if you'll get back into oscillation.......................

That is a very small capacitance but it is still has a reactance that significantly affects the rolloff, being paralleled with such a large resistance. Since the capacitance is so small and even the stray capacitance across the resistor can have an effect, the desired capacitance can probably be best provided by a gimmick capacitor such as I mentioned in post #10.

 

10pF and 25Mohm gives a -3dB rolloff of 636Hz (equals 4k rad/sec which is apparently what you calculated), as you can see from the simulated blue line plot. For a -3dB rolloff of 4kHz you need C equal to 1.6pF.

Right, i was wrongly calculating w instead of f.
in fact i started with 100 Mohm and i changed the the value of the capacitor from 1 nf to 1 pf and even at 1 pf the circuit was unstable, therefore i had to decrease the value of the feedback resistor to stop oscillation. So may be i have to decrease this value less than 25 Mohm to get into the desired breaking frequency.

I will appreciate if i can have the simulation file.

Thanks,

 

Are you sure that the inherent noise generated in such large resistors (25-100 MΩ) is not swamping the signals of interest. Sounds like your circuit is operating below the noise floor at room temperature.

http://en.wikipedia.org/wiki/Johnson–Nyquist_noise

100 MΩ and 10 kHz gives about 128 uV of noise voltage just from the resistor regardless of the voltage across it.

 

Good part? You're learning the limitations of real op-amps. Learning the theory of "Ideal" op-amps seems to require about an equal amount of time (or more) to unlearning it and adjusting to real IC's and physical circuit layouts.

 

Are you sure that the inherent noise generated in such large resistors (25-100 MΩ) is not swamping the signals of interest. Sounds like your circuit is operating below the noise floor at room temperature.

Actually a higher feedback resistor improves the signal to noise ratio since the signal gain goes up directly with the value of the resistance but the RMS thermal noise only goes up as the square-root of the resistance. And the feedback resistor noise is not amplified by the gain. That's why you normally want the input stage to have as high a gain as possible.