measuring AC wallwart output?

[dentaku]

I was just looking at a 15V AC wallwart from an old Comtrend DSL box and decided to measure its output on my regular hardware store multimeter and then my old scope.
It's rated for 15V AC, the meter measures it as 18.3V and the scope shows me a decently good 60Hz sine wave at around 50V Peak-to-Peak.
I found this article and it explains RMS etc. pretty well for a "dummies" site.
http://www.dummies.com/how-to/content/electronics-measurements-how-to-measure-alternatin.html
I can see that if the scope was measuring about 52V peak-peak, that's 26V peak * .707 (just assuming it's a good sine wave) which = 18.382V

Question: Would the measurement be different and closer to the manufacturers 15V if there was something connected to it, some kind of load?
If so, what would someone use to put a load on it?

[rs20]

I believe AC wallwarts are typically just plain old transformers, which have no closed loop feedback (or any silicon at all, for that matter), and so they do sag under load a lot more than an SMPS. So if it's a dumb transformer, the output voltage will be a function of output current (i.e., "load"), AND the mains electricity input voltage, which isn't a precisely fixed number either.

So I think your suspicion is correct; the voltage will drop when you put a load on it. How much it drops will depend how much load. You could use a big resistor (say you want to draw 1A and 15V, that's 15 ohms. But that'll burn off 15V * 1A = 15W of heat, most normal resistors can only dissipate 0.25W, so you need a big resistor]), or you could just use the DSL box it's designed to power...

[edy]

I've noticed on some of my walwarts the same thing, where the voltage is higher when measured than what it says on it. I would guess a resistor or a  small light-bulb should do the trick as far as load goes (as long as it is able to handle the current and not burn out by overheating). Then you could measure across the device and see if it is any lower.

[sacherjj]

These are generally used in devices that either aren't too concerned about the voltage (simple motors, etc.) or have internal regulators.

[Alan Liefting]

Wall warts (plug packs on this side of the world) like the one you have are just an iron cored transformer.  Since they are not 100% efficient due to resistive and magnetic losses the output voltage will drop under load.  If you imagine a resistor in series with the output (which is what is in effect how the wall wart looks electrically) it therefore follows that you will get a voltage drop across that resistor when current is drawn through it.

Some wall warts, especially phone chargers, are a switched mode power supply.  They are more efficient and have voltage regulation.

[gotwood]

What is your line voltage....and what was the output basing its input on?

[IanB]

If the power pack said, for example, "Output: 15 V AC, 1 A", then you may expect the measured voltage to be closer to 15 V when supplying a current of 1 A. The unloaded voltage will always be higher to allow for the voltage dip under load.

[dentaku]

I just did a test because I realised that I actually have one of those DSL boxes upstairs to put a load on it.
I plugged it in and measured the AC voltage at the barrel jack and now it's measuring 16.2V instead of 18.3V so it definitely makes a difference when it's plugged in.
There's a bunch of 1.5A,100V bridge rectifiers and electrolytic caps in there plus a 12V regulator so I guess they chose the 15VAC 400ma plug so they could easily get 12V DC.

I'm tired of running stuff off of two 9V batteries and I'd like to build something like this http://www.musicfromouterspace.com/analogsynth_new/WALLWARTSUPPLY/WALLWARTSUPPLY.php
Since I have two identical surplus AC wallwarts I wanted to test them out first.

If the power pack said, for example, "Output: 15 V AC, 1 A", then you may expect the measured voltage to be closer to 15 V when supplying a current of 1 A. The unloaded voltage will always be higher to allow for the voltage dip under load.

[nadona]

If the power pack said, for example, "Output: 15 V AC, 1 A", then you may expect the measured voltage to be closer to 15 V when supplying a current of 1 A. The unloaded voltage will always be higher to allow for the voltage dip under load.

Yes, I got a 8V transformer from Digikey which was measured at 10.3V without load. When inserted into the circuit, it went down to 8.5V, because it is partially loaded, I guess.