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Quick questions about transformers


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I just bought a transformer; multi-tapped 0-6-12-24, 12 Amp.

Does it mean it can go up to the max 24 volts rms - 12 Amp ???

Yes, that is right. However that is only true if that is how the secondary terminals are marked.

Some transformers are rated in VA, what does it mean?

Transformers are always rated in VA. In this case you have a teansformer that is 24V @ 12A.

VA = E x I

= 24V x 12A

= 288 VA

I hope this helps,

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I found a site for calculating filter capacitor values...


If we accept the 2.5% ripple as adequate for our purposes then at 13V this becomes 13 * 0.025 = 0.325 Vrms.  The peak to peak value is 2.828 times this value. Vrip = 0.325V X 2.828 = 0.92 V and this value is required to calculate the value of C1. Also required for this calculation is the time interval for charging pulses. If you are on a 60Hz system it it 1/ (2 * 60 ) = 0.008333 which is 8.33 milliseconds. For a 50Hz system it is 0.01 sec or 10 milliseconds.

The formula for C1 is:
C1 (uF) = [ ( IL * t )  / Vrip ] X 106

So for my 24V 12A...

C1 = [ ( 12A X 0.00833 ) / 1.6968V ] X 10^6
C1 = 58910.89 uF  :o ??? :o ???


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That formula isn't "exactly" true. To keep the math simple, consider i=C*dv/dt. Rearrange the terms... (i*t)/C=dv  This tells you that for a given capacitor  of value "C", drawing "i" amps for "t" seconds will result in a voltage dip of "dv". If a square wave was charging the capacitor, the calculation is simple, the capacitor is either being charged, or discharged. With a full wave rectified sinusoid the calculation becomes quite a bit more complicated, as it's not either one state or the other. The "dt" time suggested in the calulation of 1/(2*60) is the "entire period" which isn't accurate. Plus, higher order affects, such as ESR, aren't taken into account. It's easy to see that if you had a ESR of 1 ohm at 120Hz (ESR is frequency dependent), a 24 amp load by itself will cause a 24V dip.

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