I just received a battery charger in the mail, but I forgot to purchase the power supply. I was wondering if it was possible to use a transformer from an old lamp to supply the charger instead:

The charger takes 11-17 volts as input, 50 W as max charge and charge current of 0.1A to 6A [Link]

The transformer specification are:

enter image description here

My textbook says that the current in the primary circuit is related to the current in the secondary by:

$$ I_{1rms} = \frac{N_2}{N_1} I_{2rms} $$

So N2/N1 must be: $$ V_{2rms} = \frac{N2}{N1} V_{1rms} => \frac{N_2}{N_1} = \frac{V_{2rms}}{V_{1rms}} = \frac{12V}{230V} = \frac{6}{115} $$

Is it possible to increase the load resistance so that I can draw 6A (the maximum of the charger) through the transformer?

$$ I_{1rms} = \frac{6}{115} 6A = 0.31 A $$


In short, no. If you want more current into a load, you must decrease load resistance. If you want more current through that transformer and start pulling more than what it's rated at, it will overheat, the insulation will melt, the wires will short, and you'll end up with a nice big, smokey, smelly brick. (I would know, I've done this a couple times before.)

The only way you could possibly draw more current from the transformer than what it's rated at (20VA) is to take it apart, take out the secondary winding wire, replace it with thicker wire while still maintaining the same number of loops so that you maintain voltage. Even then, I doubt you'd be able to fit that many loops in there with the thicker wire because the transformer core itself has been designed from the ground up to support only 20VA. You may be able to get marginal increase in current by that technique and keeping it actively cooled. Even that would only give you 10-20% increase in current capacity though.

There's no way you'd get the 3.6X gain you're looking for.

On the other hand, there's no reason why this transformer won't work with your charger. The transformer will supply what it can and the battery charger will consume what it can. It will make your batteries charge slower, but shouldn't cause any problems. The only thing you should ensure is that your charger doesn't try and consume much more than 20VA at any given time. If so, you'll either want to add a low value resistor in the loop to limit the charger's ability to consume current, or you'll want the transformer more actively cooled.

  • \$\begingroup\$ Thank you for your explanation horta. I won't try to modify the transformer. It looks really packed already, so I don't think it would be possible at all. If i understand this correctly: It seems the charger can charge a maximum of 1A per cell, so if I used the transformer with the charger and only charged a one-cell battery, then there would be no difference in charging time? If i were to try and charge a two-cell battery, would the charger be able to pull more current than the transformer is supplying? \$\endgroup\$ – Attaque Jun 19 '14 at 18:32
  • \$\begingroup\$ @Attaque What you said seems roughly accurate. Until you check the voltage and current yourself while it's charging, you won't know for sure. 20VA (20 Watts) is 2/5 of 50Watts, so if the charger is designed to do two batteries simultaneously, and you only charge one, then you'll still be trying to consume more current than the transformer wants to give by 5 watts. Not horrible, just something you should monitor more closely. \$\endgroup\$ – horta Jun 19 '14 at 18:39
  • \$\begingroup\$ This may not even be possible. To drive 6A output it does also need a very large idle current too. Which will again you may need a bigger core size. \$\endgroup\$ – Standard Sandun Jun 19 '14 at 18:46

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