Calculate the value of capacitor.......?
Capacitor smoothing basics
The raw DC supplied by a rectifier on its own would consist of a series of half sine waves with the voltage varying between zero and √2 times the RMS voltage (ignoring any diode and other losses). A supply of this nature would not be of any use for powering circuits because any analogue circuits would have the huge level of ripple superimposed on the output, and any digital circuits would not function because the power would be removed every half cycle.
To smooth the output of the rectifier a reservoir capacitor is used - placed across the output of the reciter and in parallel with the load.. This capacitor charges up when the voltage from the rectifier rises above that of the capacitor and then as the rectifier voltage falls, the capacitor provides the required current from its stored charge.
Smoothing action of a reservoir capacitor
It should be remembered that the only way discharge path for the capacitor, apart from internal leakage is through the load to the rectifier / smoothing system. The diodes prevent backflow through the transformer, etc..
Smoothing capacitor value
The choice of the capacitor value needs to fulfil a number of requirements. In the first case the value must be chosen so that its time constant is very much longer than the time interval between the successive peaks of the rectified waveform:
Where:
Rload = the overall resistance of the load for the supply
C = value of capacitor in Farads
f = the ripple frequency - this will be twice the line frequency a full wave rectifier is used.
Rload = the overall resistance of the load for the supply
C = value of capacitor in Farads
f = the ripple frequency - this will be twice the line frequency a full wave rectifier is used.
Smoothing capacitor ripple voltage
As there will always be some ripple on the output of a rectifier using a smoothing capacitor circuit, it is necessary to be able to estimate the approximate value. Over-specifying a capacitor too much will add extra cost, size and weight - under-specifying it will lead to poor performance.
Peak to peak ripple for
smoothed diode rectifier circuit
The diagram above shows the ripple for a full wave rectifier with capacitor smoothing. If a half wave rectifier was used, then half the peaks would be missing and the ripple would be approximately twice the voltage.
For cases where the ripple is small compared to the supply voltage - which is almost always the case - it is possible to calculate the ripple from a knowledge of the circuit conditions:
Full wave rectifier
Ripple current
Two of the major specifications of a capacitor are its capacitance and working voltage. However for applications where large levels of current may flow, as in the case of a rectifier smoothing capacitor, a third parameter is of importance - its maximum ripple current.
The ripple current is not just equal to the supply current. There are two scenarios:
- Capacitor discharge current: On the discharge cycle, the maximum current supplied by the capacitor occurs as the output from the rectifier circuit falls to zero. At this point all the current from the circuit is supplied by the capacitor. This is equal to the full current of the circuit.
- Capacitor charging current: On the charge cycle of the smoothing capacitor, the capacitor needs to replace all the lost charge, but it can only achieve this when the voltage from the rectifier exceeds that from the smoothing capacitor. This only occurs over a short period of the cycle. Consequently the current during this period is much higher. The large the capacitor, the better it reduces the ripple and the shorter the charge period.
Another method
First you have to calculate how much charge the filter capacitor needs to provide between peaks of the rectified waveform. If you've got a full wave rectifier and your line frequency is 50 Hz, the peaks occur at 100 Hz, so the period is 0.01 seconds. By way of example, assume the maximum load is 1 ampere. Then from the formula Q = ∫ i dt, that is, charge equals the integral of current with respect to time, the charge the capacitor will need to provide is 0.01 coulombs Now you've got to decide how much ripple is acceptable. If it is a 5 volt power supply, you might accept 0.1 volts. Then you can determine the minimum capacitor from the formula Q = CV, or charge equals capacitance times voltage. Rewriting the formula as C = Q/V and inserting the above values, you've got C = 0.01/0.1 or 0.1 farads. In more familiar terms that is 100,000 microfarads.
