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Standard disclaimer: I'm not an EE, just a hobbyist. I could be missing some fundamental concept here.

Though not limited to regulators this is how this question started for me. When looking at linear regulator datasheets you always see something like this (values and regulator model are just for show): Example regulator

I get the point of the capacitors is to bypass: filter high frequency noise and provide fast response to spikes (with larger caps). I will restrict the thread to the first question I have.

Regarding CIN, it's supposed to shunt high frequency noise. I know it works but I don't understand how this setup actually doesn't cause issues. If Zcin = 1/(2*pi*f*Cin) and f = 10khz and Cin = 220uF then Zcin = 0.07 ohms (ignoring ESR). I'm not sure if there's an integration trick to calculate the noise current but simplistically I'm thinking if the amplitude of the noise is 100mv then I = 1.38A. That's a ton to draw, no?

I must be thinking about this wrong because I've also tried another experiment. I connected a signal generator with a sine wave with an amplitude of 0.1V on top of a DC 1V offset to the power rails of a breadboard. I placed a 47uF cap on the rails as well and the scope showed an almost fully attenuated signal. However if the logic I applied above was sound I would have burnt out the scope/generater but I didn't. Someone please explain this to me and put me out of my misery :)

Thanks!

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  • \$\begingroup\$ Can your signal generator source Amps of output signal? Most likely not and it will have significant output impedance. With the power levels it can generate, it is unlikely you’d damage it. As well, your 47uF cap has ESR. So, whilst your calc is valid, you’re not creating those conditions in your test. Get yourself a 100W amplifier, feed it 10kHz and put a 220uF cap on the output and see what happens. \$\endgroup\$
    – Kartman
    Commented Feb 25, 2021 at 8:48
  • \$\begingroup\$ A ripple of 100mV at 10 kHz is a lot but it can happen at the output of a switching DCDC converter. What you're forgetting though is that only few DC power sources can deliver 1.38 A at 10 kHz. I mean, the output impedance of the supply will be such that it simply cannot deliver 1.38 A. So this current will be much smaller and the 100 mV ripple will be attenuated a lot by the supply's output impedance and Cin. \$\endgroup\$
    – Bimpelrekkie
    Commented Feb 25, 2021 at 8:57
  • \$\begingroup\$ In old days when the capacitors were worst in parameters you would see placed a series inductor to limit this current (in high capacity cap). Now, at the 100mV ripple the current rush can be neglected. \$\endgroup\$
    – user208862
    Commented Feb 25, 2021 at 10:29
  • \$\begingroup\$ I think you are confusing bypass capacitors with tank capacitors… the idea is similar but the frequency and impedences are way different. Oh, the output one is also a loop stability cap! \$\endgroup\$
    – Lorenzo Marcantonio
    Commented Feb 25, 2021 at 15:47

2 Answers 2

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CIN, it's supposed to shunt high frequency noise.

That it does, too but it's not its primary purpose (if it was, it wouldn't be a 1µF, often done with an electrolytic cap, but something like a 100 nF ceramic capacitor).

It's stabilizing the input of the regulator: The input voltage source doesn't have zero impedance, and neither do the conductors connecting the regulator to that. Especially if any of that is inductive, a fast change in current draw by the load can result in a fast change in current draw by the regulator, which will lead to a voltage drop in Vin, which the regulator has to compensate by reducing its inner effective resistance (so that there's less voltage difference between in and out) if linear, or by drawing more current on average (higher duty cycle) if a switching regulator, but then the source catches up too late, the input voltage rises, the regulator overcompensates, and things start to wobble.

This input capacitor makes sure that transient spikes in current draw are quickly dealt with, thereby stabilizing the whole system.

if the amplitude of the noise is 100mv then I = 1.38A. That's a ton to draw, no?

Well, that's exactly the point. The noise source wouldn't be able to supply that, so noise voltage would be lower!

Also, for transient things to come out of inductors and capacitors, it's not that much.

I must be thinking about this wrong because I've also tried another experiment. I connected a signal generator with a sine wave with an amplitude of 0.1V on top of a DC 1V offset to the power rails of a breadboard. I placed a 47uF cap on the rails as well and the scope showed an almost fully attenuated signal

Check the output impedance of your signal generator! (and of your scope, too. If you can, set it to "high-Z" input mode, so that it has megaohms of input resistance. High frequency instruments often are 50Ω-terminated.)

Draw the schematic of your system including that output resistance in series to your AC+DC voltage source.

schematic

simulate this circuit – Schematic created using CircuitLab

Now you've got your Rsource--C1 voltage divider. Calculate again!

Things get even uglier in reality. You'd have to add an inductor in series to Rsource, too, signifying the inductive properties of the wires and rails on your breadboard. Then, you get the voltage over C1 as voltage divider composed of (Rsource+I_parasitic) and C1.

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  • \$\begingroup\$ Thanks for the detailed clarification and circuit. I will simulate it later as I need to learn how to use CircuitLab or copy it to Proteus. My first follow up question: Am I right in understanding that the input impedance along with the cap form a low pass filter which is why this works? If theoretically there was no input impedance then this would not work properly? \$\endgroup\$
    – MisterM
    Commented Feb 25, 2021 at 9:28
  • \$\begingroup\$ You've got it exactly right! And this beautiful RC lowpass, it only works as unloaded voltage divider as long as the oscilloscope has infinite input resistance: if any current flows through VM1, then that will also change the frequency response, which is why I mentioned the oscilloscope input mode setting. \$\endgroup\$
    – Marcus Müller
    Commented Feb 25, 2021 at 9:32
  • \$\begingroup\$ Thanks. If I also understood correctly everyone is saying this won't happen because the PSU just can't supply that ripple at such a high frequency. I checked the scope again and sure enough the frequency peaked at only 100Hz. My follow up probably dumb question is how did everyone know that it's very highly unlikely that a PSU can supply 100mv ripple at 10khz? \$\endgroup\$
    – MisterM
    Commented Feb 26, 2021 at 2:48
  • \$\begingroup\$ that is far from a dumb question! In fact, the answer is kind of complex, and boils down to "experience, that those smaller switch-mode power supplies are usually designed that way" \$\endgroup\$
    – Marcus Müller
    Commented Feb 26, 2021 at 9:06
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If the amplitude of the noise is 100mv then I = 1.38A. That's a ton to draw, no?

Yes, that's the point. You probably don't have 1.38A of noise, so the capacitor forces the noise amplitude to be a lot less than 100mV.

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