I've looked at many places and everyone seems to give a qualitative explanation of the filtering process, which does explain some part but I think fully understanding the stuff needs some equations too.

Basic rectifier circuit

In the circuit above, it seems to me that the input and output voltages should he same at all times but it is evidently not. Why?

  • \$\begingroup\$ You can find a the answer in wikipedia using 'rectifier'. \$\endgroup\$ – Oldfart Feb 19 '18 at 3:09

Above circuit-diagram represents the use of a smoothing capacitor in a rectified output. For sake of convenience, let's assume that the output is generated from a full-wave rectifier, hence supplying a varying DC output in the entire cycle with double the frequency than that of its AC source.

The output voltage in case of using a load resistance alone follows the input voltage from peak to zero. But in this case, we have a capacitor attached in parallel to the load resistance, which charges during the rising-edge of the waveform, and since it cannot discharge quickly (remind yourself of the RC time constant here), it will slowly discharge during the falling-edge, but even before it reaches zero, it meets the next rising-edge of the waveform. [Worth noting here is that the slope of charging and discharging curves aren't equal].

This is the initial phase when you first switch-on the circuit and Capacitor takes time to stabilize to a particular value:

enter image description here

(Image credits and suggested further reading: http://www.skillbank.co.uk/psu/smoothing.htm)

And this is how it follows:

enter image description here

(Image credits: @JohnFu's answer here: How does a capacitor smooth energy?)

So, this continues on and on, creating a near-uniform DC voltage output, which is now, of course, not following the input voltage value but has its own value to the load resistance.

As far as the equation is concerned:

Ripple Voltage Vr = The degree of smooth DC output (by the capacitor in this case) = the peak-to-peak voltage in the DC output by the capacitor

Vr = I(across load) / (frequency x Capacitance)

| improve this answer | |
  • 3
    \$\begingroup\$ Note that for audio power amplifiers the rule of thumb is 2,000 uF per amp of current consumed, hence the reason why we see 10,000 uF capacitors (or larger) in power amplifiers. There is an attempt to keep ripple under .1% of DC supply voltage. \$\endgroup\$ – Sparky256 Feb 19 '18 at 5:00

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.