B
Bob Engelhardt
Guest
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
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