# DC adapter not supplying current

I'm trying to get .5 A of current from a mobile phone's DC adapter, which is within its rated specifications. I am able to detect a voltage of ~5 V DC between the power and ground wires, so it seems like the adapter's working. However, when I apply a load between the two wires, I can't detect any current flowing with my multimeter (it's set to mA, and the terminals are in the correct sockets). I've tried putting a rheostat in between (from 0-1 megaohms) the two wires, as well as a very fine nichrome (resistance) wire - no current flows in either case. Am I doing something wrong, or does the cell phone charger have some sort of safety mechanism built in that prevents it from supplying current to only the phone?

thanks

• Are you using the multimeter in series with your circuit? (just checking :) ) Commented Dec 24, 2011 at 13:13
• Did you put the ammeter directly across the 5 volts and not in series with the circuit? If so, you may have blown the internal fuse of the meter by feeding to much current. Commented Dec 24, 2011 at 13:29
• MOST meters will blow their mA range fuse if shorted across a power supply. Commented Dec 24, 2011 at 13:32

As others suggest, you may have done violence to your metre fuse.
MOST meters will blow their mA range fuse if shorted across a power supply.

Connect, say, 47R across 5V supply.
Measure the voltage.
If the voltage is still about 5V, the supply is just about OK.

• I = V/R
• I = 5/47 ~= 106 mA.
Enough for a test.
• Power = V2/R = 5 x 5 / 47
-~~ 0.5 wATTS.
Use 1W

Or use a smaller still resistor to get larger test currents - but 47R should be enough to check.

• Take care with smaller resistors - the wattage will go up quickly, in order to test your 500mA, you would need a resistor that can dissipate 2.5W (which is quite a lot for a resistor) - you could probably get by with a 2W resistor for a short time to do a test, but it would warm up quickly. Commented Sep 19, 2014 at 3:28

A load resistance of 10Ω should draw 0.5A of current from the supply.

That would dissipate 2.5W of power, so a nice big power resistor would be required.

When measuring the current you want to measure the current through the resistor, not the voltage across the resistor. Make sure your ammeter is in series with the resistor (and connected the right way round):

(+)----(A)---/\/\/\---,
10Ω    |
(-)-------------------'


Forget the meter. You may have blown its fuse as others have said, and it's also not clear how you were connecting it.

At 5V and 500mA max, a 10 Ω resistor is the smallest load resistance within the supply's rating according to what you have said. That will dissipate (V^^2)/R = 25/R = 2.5W. 2.5W will be very obvious. Anything short of a 6 inch long wire wound resistor will heat up easily. 2.5W is enough to blow up a common 1/4W leaded resistor, or at least make it snap, crackle, and pop, and emit noxious fumes. A 2W resistor will get too hot to hold in a couple of seconds.

You don't need exactly 10Ω for this trick. That is the minimum resistance and therefore the maximum power. Just 1/4W thru a 1/4W resistor will make it quite toasty, probably too hot to hold after a few seconds. 100mW thru a small resistor will heat it noticeably. To get 1/4W you need a 25 / (1/4) = 100Ω resistor, and 250Ω will dissipate 100mW.