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Prince Umer Jamil asked How to calculate the value of C for the rectification purpose?
I am actually working on a project to build a power supply. I have a transformer with following specs: Input Voltage: 220V Output Voltage: 15V Power: 24W Current Rating = 24/15 = 1.6 amps Here are my Questions: 1) How can I calculate the value of C for Rectification Purpose? 2) I actually want to get 1 amps current at the output of transformer. Is it possible with this transformer or should I buy a new one?
And got the following answer:
If using a bridge rectifier with a capacitor as the filter, the capacitor size depends on the load current (1A) and the percentage of ripple that is acceptable, as well as the frequency, and some other relatively unknown circuit values. The capacitor charges to the peak voltage less 2 diode drops and some drop for series resistance of 2 diodes and transformer and wiring. The first link gives one approach to this: C = 0.7(I) / ΔE(f) Lets say the transformer delivers 15V RMS at 1A. As it is 220V primary assume 50Hz so 100Hz ripple voltage. Allow about 7% ripple, so 1V peak to peak. C = (0.7 * 1A) / (1 * 100) = 7000uF This seems reasonable. It does depend on resistance of wires, transformer, equivalent series resistance (ESR) of capacitor, even the mains voltage and waveform, so is always a bit uncertain. Also electrolytic capacitors can vary considerable with temperature and aging. Any value from 4000uF to 10000uF is likely to be suitable, depending just how much ripple is acceptable. If low ripple is required a series regulator (3 terminal regulator) will reduce ripple very significantly, so long as the dropout voltage of the regulator is not exceeded at the minimum input voltage to the regulator during the cycle, and at maximum current load. The dropout voltage is the minimum voltage across a series regulator to achieve proper regulation, typically a few volts, and obtained from the data sheet. WIth 15V RMS sine wave, expect about 21V peak, and 19V after the bridge rectifier. The ripple might drop the voltage further to 18V at part of the cycle, so this is fine for a 12V regulator, and may be marginal for a 15V regulator unless it is a suitable low dropout type. The regulator needs a heat-sink as it dissipates around 7W. Note the rectifier diodes need to be 3A or even 10A size to reduce heating.