ac -> dc convertor

[scaesic]

when choosing a transformer do you need a secondary with a larger ratio of winds than you actually need or more?

what i mean is: if i have a 15 secondary, i get 15v ac on the output of the secondary, but in the coversion to dc you have to use the rms (and minus the diode drops fro the br)

i need a +/-20v supply, and initially i thought id need a 15-0-15 secondary, but im thinking i may need a 30-0-30 because iv been messing up with the 1.414 factor...

help?

[R.G.]

Like many of the questions here, the answer is at GEO - http://www.geofex.com.

From 11/15/01 - Power Supplies Basics http://geofex.com/Article_Folders/Power-supplies/powersup.htm

[scaesic]

iv read that but i was kind of confused about what secondary windings give what dc voltage   

sorry if its already been discussed.

So how do you figure out how much secondary voltage you need to have to get a specified output voltage? Easy. Let's say we want 12Vdc out at the filter cap. If that's 12V before ripple voltage (because ripple you have with you always) you need 12Vdc plus any ripple voltage you choose to allow, plus two diode drops (for full wave rectification). That's the peak of the AC secondary wave:

Vsecondary(peak) = (Voutdc +Vripple + 2*Vdiode)

To convert that to AC RMS voltage, we divide by 1.414. That factor is the observed ratio of the peak to the RMS value of a sine wave; so Vsecondary(rms) = Vsecondary(peak)/1.414

i know its in this line somewhere but i just cant get my head round it.

[R.G.]

You can get to the right answer two ways.

Here's one way: You want +/-20V, which is a total of 40Vdc. Ignore the center tap for a minute. We know that the peak value of a sine wave like the AC power line is bigger than the RMS value. The peak is 1.414 times the RMS value. So if the RMS value is 30V, the peak value is 30*1.414.

Because we have the DC and want to find the RMS, we take the same relationship, ( Vpk = 1.414 * Vrms) and work it backwards.

We know Vpk, and we want to know Vrms, so Vrms = Vpk/1.414. So in the case you're dealing wth, you want 40Vdc, which we get by using diodes to pick off the peak value into a filter cap. So the Vpk is 40V, and Vrms = 40/1.414 = 28.29Vrms. We know there's a centertap in there, so you want 28.29Vct.

The other way is to look at each half separately. Both the + and - sides of the DC supply rectify one half of the transformer voltage because the centertap is grounded. Again, we figure Vrms = Vpk/1.414, or Vrms = 20 / 1.414 = 14.14Vrms, and we need two of them, so we get 28.28.

You were right at first - a 15+15V secondary will rectify to +/-20V.

[scaesic]

thanks very much, the inforation is invaluable.