# 13.8V 20A linear power supply

Posted on Feb 6, 2014

The design presented here is a little bit unusual in its arrangement, but offers some advantages over the usual designs that I will explain in the following paragraphs. First, let`s start from the basics: A linear power supply has a transformer that steps down the line voltage to some voltage that is higher than what will be required at the regulated output. Then a rectifier and

a filter capacitor transform the low voltage AC into a moderately filtered DC that still is unregulated and has some ripple. Finally, a regulating circuit "burns off" the excess voltage, leaving only the exact amount desired at the output, typically 13. 8V for communication equipment. One typical mistake made by many amateur designers is using a transformer that has a voltage that`s too low for the combination of rectifier, filter and regulator used. The situation is this: You need 13. 8V at the output at all times. Your regulator eats up a certain minimum voltage, which depends on its design. Many regulators need at least 2V across them, so you need 15. 8V minimum at the worst time across the filter capacitor. This is the voltage at the minimum point of the ripple waveform, but the capacitor needs to be charged to the maximum of this ripple voltage. So, the size of the capacitor defines how much additional voltage you need for this. A 60000uF capacitor, used at 20A, and discharging during almost a half cycle at 50Hz (10ms), will drop the voltage by almost 3. 3V. So, you need to charge the capacitor to at least 19. 2V under the worst conditions! If you are using a bridge rectifier made from silicon diodes, which loose about 1. 2V each at peak current, then you end up having two diodes conducting at the time of charging the capacitor, dropping a total of 2. 4V. So, the transformer needs to develop 21. 6V peak voltage. This...

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