[bhjazz]

So the parts are beginning to show up at the door for my Pimeta. Pretty exciting to arrive home and see those packages waiting.

One item that arrived was the wallwart form Jameco. I found an unregulated 24VDC 500mA unit for somewhere around 9 bucks. I am a little short on time to start soldering tonight, but thought I would measure this wallwart with my Fluke 112.

The voltage seems right, and shows about 29.25VDC. I realize that this is likely due to it being unloaded.

But in measuring the current, I got some odd readings. The first number that appeared on the DMM was 5.002 mA (it was autoranging). But then as I left the leads on the jack, the numbers started to descend. I disconnected the meter and scratched my head. hmm. I reconnected (yes, I'm a monkey. can't keep my hands off) and the meter read +/- .0001. The unit wasn't hot, but slightly warm on the front. That seemed reasonable.

So what is going on? I know an unregulated wallwart doesnt have a whole lot of parts inside, but all I can figure is that there is some kind of capacitor in there...or I burned something up already.

 

[SubRosa]

I could be completely wrong but...

The PSU only outputs the current that's required by the device it's powering, I doubt your DMM would be pulling much current. Put a dumby load across it and then you'll be able to measure current.

 

[DaKi][er]

How are you measuring the current? Putting the terminals across the power supply? That will just make a direct short and you'll blow the fuse in the meter

 

[robzy]

Hehe, someone needs to go review their V=IR laws

The "voltage rating" of a PSU can be measured, and it always outputs that voltage. (Ideally)

The "current rating" is merely the maximum current it can supply, differing loads will draw different currents. Connecting your multimeter accross it will just short the output and draw much more than its maximum current. You've probably blown a fuse in your multimeter.

Rob.

 

[Garbz]

When measuring current it's also not sufficient to plug the metre in in series with the load. This is the correct way to do it but what happens is when you turn things on an initial inrush current which is very large but short will charge the capacitors and in the process again blow your fuse.

If you want to measure the current your equipment is using you need to plug it in as per normal. Turn it on. Place your DMM in series with the voltage line you are measuring, i.e. one lead to the wallwart's output, the other lead just down the line a little further, and sever the connection between them.

 

[bhjazz]

I've blown nothing. Not sure where that came up. Meter is working fine, and the Fluke 112 does have current reading capability. I had no initial inrush of current, it was merely right up to what I expected it to be, then slowly descended. I'll opt toward Garbz's idea and just wait to test again: when the wallwart is actually powering the Tread.

Is there a capacitor in an unregulated power supply? If so, it could have been charged up during manufacturing/testing of the unit.

 

[threepointone]

just a guess, but maybe there's one of those resettable fuses in there and when the current went too high, you tripped the fuse? if so, (and you're still willing to risk frying your meter) when it cools down a bit, it should work and do the same thing again when you try to measure current

 

[bhjazz]

OK, let me be clear here: the Fluke 112 can handle 10A of current when measuring current. As the wallwart in question only puts out .5A (500mA), this should read as a no-brainer. My DMM is fine. Let's move on.

I'll still go with Garbz on the explanation of this and wait to test it when it has a real load attached. I was just posing the real question of where the initial 500mA came from when I first connected it. I did a bit of research and found that most 'warts have a filter cap of some sort inside, so I can assume that this was charged during intial testing by the manufacturer. As the DMM presented no load when connected, it simply read the slowly draining value of the capacitor.

 

[nikongod]

most wallwarts have "protection" circuitry. if noting else, a self-resetting fuse.

too much current load on a transformer is not fun.

the inital 500mA was the "turn on rush" and the wallwart suplying power until the internal fuses were tripped. after that, the SMALL current was what leaked through the protection circuitry.

 

[mono]

We need a clear description of exactly what you were measuring, the entire circuit.

If you only hook up a meter to a wart and there is no load, no current could be flowing... but surely you hooked the second meter probe up to "something". If you hooked it up to a Linear supply like a Tread, it would be entirely understandable to get a very low, few mA reading. If/when you use a very high range like the 10A range, your reading wouldn't be very accurate.

Your meter should have at least two current ranges. Your amp (Pimeta) certainly uses less current than the lower range on your meter so when the time comes to measure current, the lower range, not the 10A range, is the one to use.

The majority of unregulated DC warts do have a capacitor in them. Typically there is:

Transformer-> 4 diodes (like 1N4002 as a full wave bridge) -> 50-63V (in this case, being a 24V wart) electrolytic smoothing cap.

It's not manditory, but is far more common for it to have at least one (or rarely another one or two small HF caps around diode bridge ) bulk filter cap in it even on the very cheapest warts. I mean DC, not AC warts.

 

[robzy]

bhjazz: You are missing some fundamentals here. A wallwart current rating is the maximum it can output, it will output whatever it has to - up to that value.

If your circuit draws 10 mA, your wallwart will output 10mA.
If your circuit draws 20 mA, your wallwart will output 20mA.
If your circuit draws 30 mA, your wallwart will output 30mA.
If your circuit draws 100 mA, your wallwart will output 100mA.

(Always at, more or less, the same voltage)

Lets go into some maths now:

V = I * R

V is the voltage accross your circuit (be it an amp or a light, I is the current flowing through your circuit, and R is the effective resistance.

V is, ideally, always a constant. Your plugpack will always output 24v (Or, close to 24v).

-----
Lets say you go and take a 100 ohm resistor directly accross it, this is what we do:
V = I * R
24 = I * 100
I = 0.24 Amps = 240mA
This means that your plugpack will be outputing 24 volts at 240mA
-----
Lets grab a 740 ohm resistor now and connect it accross it:
V = I * R
24 = I * 740
I = .032 Amps = 32mA
This means that your plugpack will be outputting 24 volts at 32mA
-----

A current meter, essentially, as very low resistance. This means that hooking it up directly to the plugback to measure current will result with something like this: (Assuming the current meter has 1 ohm of resistance)

24 = I * 1
I = 24Amps

Because this is WELL above the rating of the plugpack, all types of crazy things can happen.

If you dont understand what i have said, can i recommend that you go and do some reading on it elsewhere?

Rob.

(Unless i misunderstand what you are saying)

 

[Garbz]

Quote:

Originally Posted by bhjazz OK, let me be clear here: the Fluke 112 can handle 10A of current when measuring current. As the wallwart in question only puts out .5A (500mA), this should read as a no-brainer.

If there is a seprate 10A cable then this isn't simply a no-brainer

My metre is 10A capable but blows fuses at 200mA depending on the setting, this is to protect the device as well as the metre.

A DMM would present a short circuit when measuring current (<0.5ohm typ). This is the highest type of load you could possibly think of. Assuming the wallwart isn't damaged, and the metre doesn't blow you'd see a maximum output current from the transformer. If you had done this with a large torroid you'd likely have burnt it out if the metre didn't expire first

nikongod's explaination is possible. The other one is that the metre was working fine and the transformer started slowly dying when it was trying to supply waay more current then it could.

 

[threepointone]

nikongod's probably right--it's most likely a fuse inside the supply. the transformer was not slowly dying; the fuse was probably either just burning out or (if it was one of the resettable ones) tripping when the current went too high. If you want, you can try again with a resistor in series, and see if it starts working again. If it does, it's probably one of those resetting switches and everything's probably fine.

 

[mono]

Quote:

Originally Posted by threepointone nikongod's probably right--it's most likely a fuse inside the supply. the transformer was not slowly dying; the fuse was probably either just burning out or (if it was one of the resettable ones) tripping when the current went too high. If you want, you can try again with a resistor in series, and see if it starts working again. If it does, it's probably one of those resetting switches and everything's probably fine.

Maybe, but most of the smaller unregulated DC warts I've seen had only a thermal fuse inside the transformer windings, no other protection. A simple current test (even one that was a direct short circuit) wouldn't heat it up very quickly from cold power-on.

 

[bhjazz]

My apologies. I think my posts seem a little mean-spirited. I'm not here to do that or act indignant. I respect all of you, and appreciate your help, so please understand that.

Okay, you all bring up good points, and they all focus in a different place: the supply, the meter. Hmmm. Robzy, your calculations do make sense. I suppose I just didn't plan on doing the math before I measured a power supply. That could have been a mistake!

The best I can do at this point is head home tonight and remeasure, however quickly. I'll throw a fat resistor inline just in case. I may check some other resistors and small parts just to asssess the health of my meter as well.

One more question, then: if I avoid measuring current, can I assess the health of the unit by simply measuring the voltage again?

Thanks guys!