-1
\$\begingroup\$

Does anyone know how a Bypass Capacitor work in it's physical operation? How does a bypass capacitor filter out the ac signal in the device structure and physical operation?

\$\endgroup\$
  • \$\begingroup\$ What has this question got to do with MOSFETs? Do you know how capacitors work? \$\endgroup\$ – Andy aka Apr 22 '15 at 7:29
  • \$\begingroup\$ Are you talking about a particular circuit? A diagram would be helpful in that case. \$\endgroup\$ – Dmitry Grigoryev Apr 22 '15 at 7:45
  • \$\begingroup\$ A simple research on google would have given you the answer. \$\endgroup\$ – Dallas Carter Apr 22 '15 at 8:23
3
\$\begingroup\$

It is the same as any other capacitor. If you put DC through it, current will flow, but not for long; charge accumulates on its plates until no more can flow. At the very beginning, it acts as a short circuit. But as the charge on it increases, the voltage across it increases, to the point where no more current flows.

So when using a capacitor as a bypass capacitor, it is connected as in the diagram on the left. And when used as a coupling capacitor, it's done so as in the diagram on the right:

schematic

simulate this circuit – Schematic created using CircuitLab

To understand the circuits, I would suggest you read about RC filter circuits (Resistance-Capacitance filter circuits). The maths is simpler than other types, such as LRC or LC (Inductor-Resistance-Capacitance or Inductance-Capacitance), although you will still need some maths to appreciate how the capacitor works.

You can simulate this circuit below - it will show you how it works in practice. Click the "simulate this circuit" link below it.

schematic

simulate this circuit

To run the simulation, go through these steps:

  1. Click the "Simulate this circuit" link just below the diagram.
  2. Click on the "Simulate" button on the bottom left
  3. Click on the "Time domain" option
  4. Click "Run Time-Domain Simulation" button.

You should see a graph like this:

enter image description here

What you are seeing here, is the input and the output. The input is a 1V, 10Hz AC signal, with a 1V DC offsest. (So the AC signal goes from 0V to 2V peak-to-peak). After the bypass capacitor, the AC signal is largely removed, leaving just the 1V DC signal. The larger the capacitor value, the less the ripple.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.