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Power Supply Design Question

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?

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

D from BC

Reply to
D from BC 

--
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.
Reply to
John Fields

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

Reply to
Ken S. Tucker

"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

Reply to
Phil Allison

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