Capacitor Smoothing

If you use a capacitor to smooth a 'Full-Wave' rectified AC input, why does it allways (on the falling edge of the wave) release just the right amount of current to return the output voltage (allmost) back to a constant amount?

Posted by Tom Green (hatespam) on 08/10/2003, at 14:29 GMT

How a smoothing capacitor works

It is probably easiest to understand this if you think about the voltage across the smoothing capacitor. The voltage is proportional to the charge held by the capacitor. When the supply voltage starts to fall, charge starts to flow from the capacitor so that its voltage matches the supply. The faster the supply voltage falls, the faster charge flows from the capacitor giving a greater current.

The supply voltage is controlling the charge (and current) taken from the capacitor (that is why it is 'just the right amount') but note that there must be a fall in voltage when the capacitor loses charge so the supply voltage cannot be kept perfectly constant. The aim is to choose a sufficiently large capacitor so that it can supply the required charge (and therefore current) with a relatively small drop in voltage.

The same principle applies in reverse to re-charge the capacitor when the supply voltage rises.

You may find it helpful to think of the electrical circuit as a water supply system with a reservoir. Voltage is like the water pressure, charge is like the water and the smoothing capacitor is like the reservoir. When demand exceeds supply of water (falling pressure) water flows from the reservoir and this helps to maintain the pressure. When supply exceeds demand (rising pressure) water flows into the reservoir.

Posted by John Hewes on 09/10/2003, at 13:45 GMT