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I was reading on Capacitive Coupling a bit and become confused when reading the Parasistic section, specifically regarding two claims.

If a high-gain amplifier's output capacitively couples to its input it often becomes an electronic oscillator.

I imagine this to be similar to the situation when you have input from a microphone and output through an amplifier. The output is then coupled back into the microphone until the output of the amplifer rails, clips, or whatever happens that may cause the loud ringing (what is happening exactly?).

My only problem with the electrical example is regarding how the system may couple its output back into its input. Is this a simple case of simple feedback? How does simple feedback relate to capacitive coupling?

It seems that the output would need some ripple voltage to begin with that would be amplified. Without a clear example, I imagine an unstable amplifier with a small margin of ringing to begin with, but through feedback eventually oscillates much more significantly. Is this correct? As in:

enter image description here

One rule of thumb says that drivers should be able to drive 25 pF of capacitance which allows for PCB traces up to 0.30 meters.

What does this driving requirement have at all to do with capacitive coupling? To give some context, the article was talking about the parasitic coupling that exists between PCB traces, breadboard metal strips, and PCB traces to ground. I imagine that type of coupling to exist like this between traces:

enter image description here

And this between a trace and ground:

enter image description here

What exactly would "..drivers should be able to drive 25 pF of capacitance.." be defined as in terms of specifications for a driver, and how can one verify compliance?

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My only problem with the electrical example is regarding how the system may couple its output back into its input. Is this a simple case of simple feedback? How does simple feedback relate to capacitive coupling?

An 1 Mohm 0805 resistor will have self capacitance of about 0.2pF and this, when applied in the feedback loop of an op-amp will reduce the gain from the 3dB point of 795 kHz. This means that the transimpedance amplifier you are designing will not work well at 10 MHz because there is too much capacitance - and that's just from a single surface mount resistor - imagine what capacitance there is between tracks and if this is not properly catered for by good PCB design, you'll be lucky to get 100 kHz flat operation from the above circuit.

The above is the case of negative feedback causing a signal to become smaller on the output due to parasitic capacitance.

It seems that the output would need some ripple voltage to begin with that would be amplified.

The minutest noise will be enough to trigger oscillation if the feedback is positive. Noise is present in all components at temperatures above absolute zero.

Regards driving tracks having capacitance this is not directly related to the possible cause of oscillations and additionally the rule of thumb in your question doesn't take into account many, many factors such as the power capabilities of the driver, and the track dimensions and whether the track is terminated. At low frequencies there is no real rule at all.

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My only problem with the electrical example is regarding how the system may couple its output back into its input. Is this a simple case of simple feedback? How does simple feedback relate to capacitive coupling?

Take - as an example - a simple transistor amplifier in common emitter configuration. There is an unwanted capacitive effect (internal to the transistor) between collector (output) and base (input) that is even enlarged due to the Miller effect. Other coupling may arise due to a collector-emitter parasitic capacitance. Other unwanted coupling effects are caused by a bad layout.

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  • \$\begingroup\$ What is the source of this parasitic capacitance? As in, why does it exist between collector/base and collector/emitter? If the parsitic capacitance is very small, one must then conclude there to be an approximate short circuit between base-collector and collector-emitter for high frequencies (considering the reactance of the capacitor). At which point it must be further concluded that with such a short for high frequency would cripple the application of the transistor in amplifier configurations, right? (1uF @ 10k Hz = 15 Ohms). How does on reconcile this?What is a typical parasitic C value? \$\endgroup\$
    – sherrellbc
    Jul 30, 2014 at 12:11
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There are two subjects here. First, if a system is unstable it's figuratively speaking sitting on a knife edge. A tiny bit of noise (which any amplifier has to one degree or another) will have some frequency content at the frequency the system would like to oscillate at.. that noise is preferentially amplified and the oscillation builds up (more or less linearly) until something nonlinear happens to limit the growth (amplifier hitting the rails, fues blowing, massive explosion,etc.). The feedback via capacitive coupling is what can induce the oscillation.

The second subject refers to the well-known tendency of negative feedback amplifiers with a capacitive load to oscillate. In that case, the feedback is already there, via a resistor network or something like that. The capacitive loading (coupling to ground) in conjunction with the open-loop output Z of the amplifier, is an RC low-pass filter wot delays the feedback, and oscillation can occur. One way to test compliance is to load the amplifier with (say) 50pF and hit the input or output of the amplifier with a step or impulse and look at the ringing on the output. If there is sufficient phase margin the ringing should be nonexistent or minimal and decrease by a large factor from one cycle to the next.

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  • \$\begingroup\$ How is it that only preferred frequencies of the noise get amplified? Also, how could it be that the amplification is linear? Say, for example, you have an amp with a gain of 2. It would seem to me that after a gain of 2, feedback forces another gain of 2, and then again. So then it seems to me that the output will have gone through a 2^n exponential gain sequence until the output has railed or something else happened as you say to hinder its continuation. Finally, the one question I had was how capacitive coupling might occur from output to input. In your first example, how does it happen? \$\endgroup\$
    – sherrellbc
    Jul 30, 2014 at 12:17
  • \$\begingroup\$ As for your second paragraph, what is the open-loop Z you speak of when the op-amp has negative feedback? I thought the feedback network removes the notion of "open-loop". The output Z of the amplifier would then be a part of the output capacitive loading and feedback path to the input, right? What exactly is meant by "delaying the feedback"? Must you first charge the capacitive load? What significance does this have to start oscillation? As for your finaly comment, why is the impulse/step input significant and what would a ringing output indicate? So many questions .. \$\endgroup\$
    – sherrellbc
    Jul 30, 2014 at 12:22
  • \$\begingroup\$ @sherrellbc There is still something like a few hundred ohms effectively inside the op-amp. That is effectively reduced by the closed-loop feedback so you don't normally see it, but when you load the op-amp output capacitively it can be a problem. Ringing output indicates close to oscillation. Search "second order system" for a 1st year Uni diff-eq's view of it. You'll also see the transfer function curves that illustrate the peaking at the resonant frequency that answers your first question. \$\endgroup\$ Jul 30, 2014 at 12:41

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