Discussion in 'General Electronics Discussion' started by t44florida, Jun 11, 2012.

1. ### t44florida

7
0
Jun 11, 2012
Building a project and using an ac/dc adapter as power source. The adapter spec plate states:

Input: 120Ac, 60Hz, 100mA
Output: 12vdc, 500mA

yet when I measure the amp output with reliable amp meter the output actually reads 2amps... not the 0.5amps as spec plate states. Can anyone tell me why the difference? Is this normal for adapters?
Thanks....

2. ### Harald KappModeratorModerator

12,358
2,938
Nov 17, 2011
How do you measure? Not by placing the ammeter across the output pins, you don't, do you? An ammeter is to be used differently. You connect a load to the adapter (e.g. a lamp or whatever you want to operate) and put the ammeter between the load and the adapter. The ammeter will measure the acual current drawn by the load.
An ammeter is not meant to measure the current capability of a voltage source.

The specs state that at 500mA output current the adapter will supply 12 V. If you draw any current from 0 mA up to 500 mA the output should be 12 V (+- some tolerance). If you increase the load current over 500 mA, the output voltage will drop to less than 12 V.

Harald

3. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,505
2,849
Jan 21, 2010
You have just discovered that the supply doesn't blow up when you short the output. (i.e. you were lucky)

You should NEVER try to measure the current from a voltage regulated supply.

The voltage is what IS supplied by the adapter, the current is the MAXIMUM is can SAFELY supply. The load determines the current at a given voltage. Your load (the meter) essentially demanded infinite current and the power supply maxed out at 4 times it's nominal maximum. If you persisted, it may have blown a fuse, caught fire, gone *pop*, or just got very very hot.

4. ### CocaCola

3,635
5
Apr 7, 2012
How are you measuring the amps? And is the load you are using supposed to pull 2 amps, if so why are you testing it like that?

The adapter will likely supply over it's specified current for a short time, but it's on a path of short term self destruction...

5. ### t44florida

7
0
Jun 11, 2012
As you can figure I am a novice in electronics... I re-tested amps correctly (ampmeter in-line) and the load (LEDs) read 0.97 amps. Thanks.

I take it that the adapter output of 500mA is too small. The load are several small LED strips, each 3 LEDs with built-in resistors. The strip light specs states 12vdc and .4 amps/meter. I suspect I do not have enough amp output to power the load because the converter gets warm (not hot) after 15-30 minutes. I need a 12v converter with enough amp output but not too much that I fry resistors.
There is more... I will be adding additional lights: (10) single 3v, 20mA LEDs with resistors for 12 volt supply and (10) 6v 70mA incandescent bulbs with resistors for 12V and a small animated fountain that I don't know the exact specs for yet, only that it operates on 12-16 volts. I would prefer all electrical from one output. How do I approach sorting out the correct supply output?

6. ### Harald KappModeratorModerator

12,358
2,938
Nov 17, 2011
You get the amperage you need by multiplying the ratings of the LED strips by the length:
I = L*0.4A/m where L = length in meter, I = current in amps.
From that you can determine the rating of the power supply. It needs to be able to supply at least as much current as you calculated.
Don't get fooled by your measurement. Unless you verified that the voltage was 12 V, chances are that you measured 0.97 A at less than 12 V because the adapter may have had a reduced output voltage due to overload.

And don't be afraid of "frying" the resistors. As long as the voltage is 12 V, the resistors will be fine (assumingh the developer of the LED strips did his job right).
It is a common misunderstanding that the rated amperage of a power supply is the current that the supply will drive into a load. Common power supplies (that is almost all) are voltage sources. This means they supply an output voltage. The current rating is given only as the maximum current (Imax) the supply can deliver. The actual current is defined by the load according to
I= U/R
where I = current, U=voltage, R=load resistance.
You can draw any current from 0 to Imax. (as rated) from the supply while is still should deliver 12 V.

Harald

Last edited: Jun 11, 2012