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determination of ratings of voltage regulators and capacitors

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snazzy c

Junior Member level 3
Nov 26, 2010
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thanks for your replies. pls., how do i know the ratings of the capacitors, and voltage regulators to be used in designing any of my dc power supply!

Let me give you an example.
If I have to regulate an output voltage, Vout at 5Vdc, from an input voltage Vin at 8Vdc, the common rule of thumb to use capacitor of suitable rating will be:
5Vdc * 120% = 6Vdc, so I can use 6.3V or 10V that are commonly available.
8Vdc * 120% = 9.6Vdc, so I can use 10V or 16V that are commonly available.
This rating for capacitor is intended for allowance for voltage spike, aging (cap will not always perform as specifiec forever), derating factor (due to operating in higher temperature, cap may not perform constantly as well) and drift (tolerance due to manufacturing process, aging and temperature).

And don't forget about the smoothing capacitor value in a rectifier circuit.
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:

Rload * C >> 1/f

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 if a full wave rectifier is used).

For cases where the ripple is small, compared to the supply voltage, it is possible to calculate the ripple
V ripple = Iload / 2 f * C

For many circuits a ripple which is 10% of the supply voltage is satisfactory and
the required value for the smoothing capacitor

C = (5 * I)/(U * f)

C = smoothing capacitance in farads (F)
I = output current from the supply in amps (A)
U = supply voltage in volts (V), this is the peak value of the unsmoothed DC
f = frequency of the AC supply in hertz (Hz), (50Hz in Europe)

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