I'm confused about the power specifications on wall warts. The current it delivers varies depending on the load, right? I know that's the case with batteries, but the confusion arises because the wall wart specifies a DC current. I have a 12V/200mA wall wart and only need 90mA to power a circuit. So... does 200mA mark the MAXIMUM current that the wall wart can deliver without damaging itself, or is that a constant value that is delivered no matter what? Thanks for the help!

 

It's rated to deliver 12v when a 200mA load is attached (a 60 Ohm 5W resistor would be a good test.)

With a 90mA load, the output voltage of the wall wart will likely be somewhat higher. You can test it by using a resistance of 133 Ohms (2W rating) to see what the voltage will be. Eight 1k Ohm 1/4W resistors in parallel will be around 125 Ohms and 2W, which should be close enough.

 

Oh, so as long as the load does not draw more than 200mA, I can be reasonably assured that the voltage will remain greater than or equal to 12V? Then it seems that I won't fry the circuit as long as I use the proper resistors to ensure 90mA. I suppose that if I used a 12V/1000mA wall wart instead, the voltage would greatly exceed 12V at all currents less than 1000mA, which would simply waste power if not used, so a lower power rating would be more efficient.

 

Oh, so as long as the load does not draw more than 200mA, I can be reasonably assured that the voltage will remain greater than or equal to 12V?

Yes. But, the output from wallwarts isn't typically regulated and filtered - just rectified. If you want accurate voltage regulation, you'll need filter capacitors and a voltage regulator.

Then it seems that I won't fry the circuit as long as I use the proper resistors to ensure 90mA.

It depends upon what your circuit is.

I suppose that if I used a 12V/1000mA wall wart instead, the voltage would greatly exceed 12V at all currents less than 1000mA, which would simply waste power if not used, so a lower power rating would be more efficient.

It might.

I'd measure the output under load before connecting it up to a circuit.

 

The name plate data on the transformer states the maximum current it is designed to deliver. Draw much more then that and often the primary winding is designed to burn open thus making the transformer useless.

As others here have stated, the biggest problem with these transformers is that when drawing less then their rated maximum current, their output voltage will be higher then the name plate data, often destroying ICs when the voltage exceeds their maximum allowed value. Most electronic circuits that use these transformer have an internal regulator or voltage clamping circuit to protect it's component from over-voltage.

Here is a useful bulletin from Sparkfun about using these wall wart transformers:

http://www.sparkfun.com/commerce/tutorial_info.php?tutorials_id=103

Lefty

 

Thanks to both of you, I appreciate the help.