Transformer rated at 1/2 wave

[nashvillebill]

I have a couple of old Chicago Standard (Stancor) power transformers (PA-8421). While one secondary is rated at 6.3VAC, 2A, interestingly they rated the other secondary as : "125 Volts, 1/2 wave 50mA DC" (It's just a two wire 125 VAC secondary, not center tapped)

I've hooked that secondary up to a full wave diode bridge, yielding around 195 VDC with a 100 uF smoothing cap. My question is: how does the current rating change going to a full wave rectification versus the as-rated 1/2 wave? Wouldn't the transformer run cooler in full wave versus 1/2 wave? I'm just curious why they used the 1/2 wave rating method.

[Tom Phillips]

I believe that Stancor listed the two parameters, 1/2 wave & 50mA DC, together because the transformer’s capacity is indeed different depending upon which type of rectifier circuit you use. When you are given the rating for one condition then, if you know the formula, then you can calculate the rating for other conditions.

I have a copy of the 1977/78 Stancor catalog in front of me. A section titled “How to determine secondary AC (RMS) current ratings” includes the following information.

The formula for the relation between secondary RMS current (IAC) which the transformer has to deliver and the D.C. output current taken from the rectifier IDC) is:
IAC = KFF X IDC
Where KFF is the form factor.
KFF for a half wave capacitor input rectifier is 2.3
KFF for a full wave bridge capacitor input rectifier is 1.8

The secondary of your transformer is rated to deliver “50mA DC” with a half wave rectifier. Since they specifically state “DC” then we know that is the IDC figure in the above formula. Doing the math we find that IAC is KFF X IDC = 2.3 X 50 = 115 mA for your transformer. Now that we know IAC we can calculate the IDC rating for your transformer when it is used with a full wave bridge capacitor input rectifier.
IDC = IAC / KFF = 115 / 1.8 = 63.9 mA

As you see, you can pull 28% more current by using the full wave bridge while staying within the published Stancor specification. Or, as you asked, it will run cooler with the FWB if you don’t pull more current. Of course, the ripple will also be lower and the VAC will probably be a little higher depending upon your current draw. There are many tradeoffs to be considered when you design circuits. Musical Instrument amp companies often push the limits. That’s how we end up with amps that get so hot that you can’t keep your hand on the hot transformer.

Cheers,
Tom

[nashvillebill]

Thanks, the formula you quoted provides the answer!