Draining maximum power from a transformer

I have a transformer. The only data I know about this transformer is as below:
- Conversion ratio is 220V:12V
- Power rating is 4VA
- Rated frequency is 50Hz

I want to connect a heater (resistor) across the secondary windings of this transformer, so that it will drain maximum power from the transformer.

I know that, according to the Maximum Power Transfer Theorem, impedance of the load must be complex conjugate of impedance of the source. But how do I calculated internal impedance of a transformer?

Can you please suggest me a method that will either calculate internal impedance of the transformer, or the value of the load impedance which would drain maximum power from the transformer?

I want to connect a heater (resistor) across the secondary windings of this transformer, so that it will drain maximum power from the transformer.

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OK so you have a 4 VA transformer or volts * amps = 4 and you know the secondary voltage. So we get 4 / 12 = 333.333 mA or roughly .3 amp. 12 volts / .3 amp = 40 Ohms. Therefore to load the transformer close to its maximum current rating you want about a 40 ohm load which at 12 volts will draw about .3 amp. Next you have .3 amps * 12 volts = 3.6 watts so you want about a 40 ohm 5 watt resistor for the load.

Why are you concerned with the impedance? The link you listed pretty much covers power transfer and you may also want to give this link a read. While it deals with old radio the principal of impedance matching for power transfer holds true.

Ron

^ Rated secondary voltage is 12V, but that doesn't mean that it will always give 12V. I measured it to see that when there is no load, secondary voltage is about 13.6V. And when I connect a 10Ω resistor, it drops down to 7V. Because of that, I needed a more precise way to calculate it.

Would this be by any chance a simple wall wart transformer? Yes, when you place a 10 ohm load across the secondary you well exceed (by about four times) the current the transformer is rated for. What exactly would you expect to happen? Something has to give. A transformer's secondary output (voltage) is based on the turns ratio which is explained in the link I posted. The secondary current is based on the gauge of the wire in the windings.

Conversion ratio is 220V:12V So what do we get if we say 220 / 12? You have a turns ratio of about 18 to 1. When you measured the 13.6 volts what was the primary voltage?

If this is in fact an inexpensive (cheap) wall wart transformer then all bets are off. Wall wart transformers typically output a voltage well in excess of their nameplate data.

Rated secondary voltage is 12V, but that doesn't mean that it will always give 12V.

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Based on ratio it does. If I connect a good transformer to a variac and slowly increase the voltage to the primary while measuring the secondary when I have exactly 1.0 volt RMS on the secondary I can measure the actual primary and see the ratio.

Maybe someone else has better thoughts than I have?

Ron

hkBattousai said:

I want to connect a heater (resistor) across the secondary windings of this transformer, so that it will drain maximum power from the transformer.

I know that, according to the Maximum Power Transfer Theorem, impedance of the load must be complex conjugate of impedance of the source. But how do I calculated internal impedance of a transformer?

Click to expand...

The Maximum Power Transfer Theorem is normally only used for RF circuits. A power transformer does not operate in that mode since you will be dissipating half the power in the transformer which will overheat it. Power transformers are designed with a low impedance so that most of the power is dissipated in the load for good efficiency, not in the transformer.

A power transformer should never be loaded with more than its rated load.