Power Supply Design Question

Oct 27, 2007 4 Replies
N

I'm designing a power supply for a circuit.



My circuit uses 3.3 volts.



The voltage regulator I've chosen has a maximum dropout of 1.5V.



The bridge rectifier I've chosen has a maximum forward voltage of 1V.



Question: What minimum voltage do I need on the secondary of my transformer?



On the face of things, I'd need 3.3 + 1.5 + 1 = 5.8V.



However, the transformer's secondary is AC and is measured in RMS. Do I multiply the transformer's secondary voltage by sqrt(2) to get the DC voltage when doing this calculation?


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D

Voutrms off the xformer will drop depending on the current drawn. And..you need to select your tolerated ripple.

D from BC

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J

Yes, or the converse; divide your 5.8V by sqrt(2) and your secondary voltage should be 4.1VRMS. You also need to consider that mains are specified as nominal, plus or minus 10%, so with that being case you\'ll need 4.1V RMS with the mains at 108V, which translates to about 4.6V with 120V mains. A good choice, then, would be a transformer with a 5V secondary, which is readily available.
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K

Before math, give us the watts needed, that's why it's called a "power supply". Batteries might be better and cheaper at 3.3 V. Ken

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P

"Nevo"

** That would be 1 V "per diode".
** Well, + 1V more for the second diode that is always involved, then allow 0.5V for ripple voltage on the filter cap.

So you need to see about 7.5 volts peak from the transformer secondary -

  • under load* .

So about 5.3 volts rms under load or say 7 volts off load.

...... Phil

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