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Confirmation needed

I'm going to make a 12v DC power supply with a linear reg. I will half-wave rectify 12v AC and have a smoothing cap. The load is only 10ma.

So: 12v RMS is 17v peak, minus the diode drop of 0.7 is 16.3v peak. Using a 47uF cap, the ripple will be 3.5v p-p. So the min voltage into the regulator will be 16.3-3.5 = 12.8. Ripple calculated from Vpp = i/fC (.010/(60*47e-6).

Am I missing anything?

Thanks, Bob

Reply to
Bob Engelhardt
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snipped-for-privacy@gmail.com wrote: =======================

** Yep two:

  1. The drop out voltage of a standard 12v reg IC is not 12.8 - more like 13.8.

  1. The AC supply varies and may be be 10% low on occasion.

So, you need a 15VAC tranny.

OR you can make a "voltage doubler" supply with two diodes and two electros.

That will give about 33V DC - minus the same deductions.

..... Phil

Reply to
Phil Allison

Thank you - that is helpful in 2 ways: pointing out what I missed & was wrong about, and even more helpful is not having to correct my analysis.

Reply to
Bob Engelhardt

** If your 12v tranny is really small, like say 100mA - it will output about 15V to 16v AC off load.

So would do at a pinch with a puny 10mA DC load .

....... Phil

Reply to
Phil Allison

Thanks again - I will take some measurements on the tranny.

Reply to
Bob Engelhardt

Thanks again - that's good to know. The OC voltage is 15v & it doesn't drop below 12v until loaded to 200ma. That gives me a lot of working room!

Reply to
Bob Engelhardt

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