DC Power Supply Question

Yes that right, with open circuit load, the voltage goes to maximum, once loaded the voltage drop down. It like having series resistor at the output of the transformer.

Low power transformers have very poor regulation. They are rated for the output voltage at full current load. At lower loads, they produce significantly extra voltage. The droop from 16.4 to 12 as the load increases from 0 to 200 mA implys approximately (16.4-12)/.2=22 ohms effective impedance in the output of the supply.

thanks. I will try to calculate the load that is going to be placed on the power supply...

Thanks... I did some quick checks, it does not look like this wall wart would be enough to power up the chips in question...

You were measuring *no load*.

All wall warts behave like that. They are 'unstabilised' power supplies. Try measuring with the rated load.

I'd never trust the voltage on a wall wart label anyway !

Graham

Phil Phil Phil... Calm down!!!!!! It's not the end of the word!!!! Make an appointment with your therapist and let him bring you back to the real world. People often say things that don't meet the criteria of being the absolute truth in every respect, but that's no reason to totally lose your senses.

A more appropriate response from you might have been: "It's true that all unregulated wall-warts exhibit that behavior, but there are some wall-warts that are regulated, and will supply a stable voltage at any current up to their rated capacity."

If you try that approach in the future, you might find that you will be happier and maybe people will like you a little more.

...

When you say, "quick checks", do you mean that you hung a load on the supply and measured the volts, and it drooped? I have a wall wart here that's labeled 7.5V, 1A. It's the one I used in that styrofoam-cutting experiment, and even when I was melting plastic with the nichrome wire, it was still putting out about 10V! (within the accuracy of my little VOM.) If you only have "a few IC chips and stuff", you could calculate what you expect their current load to be, and pick some resistor(s) out of your junque box (R = E/I) and see what happens. :-)

Good Luck! Rich

If you'd read the original post, it would be clear that _this_ wall wart isn't regulated. Perhaps Mr. Bear (Ms. Bear?) could have said, "All _unregulated_ wall warts behave like that".

And the regulated and/or SMPS wall warts I've seen would probably not be lying around in a pile; they're still relatively expensive, and I don't think they're a junque box item quite yet. :-)

It's easy to tell the difference - pick it up and "heft of it" (I learned that expression from some guy from Maine - it means feel how heavy it is). If it's inordinately light for a wall wart, it's a (an?) SMPS. :-) Now, if it _does_ feel like a chunk of iron, that still doesn't say whether it's regulated or not, but if it's rated 12V and you're measuring 16.5, it's not regulated. :-)

Thanks! Rich

I will try to scrounge a "laptop power supply"

How much current do you need to drive the chips, and how much voltage? Maybe they will run on something less than 16 volts. If so, the power supply might run them. Another issue: some wall warts don't have any filter caps in them. That means the voltage drops briefly to zero 120 times per second. In that case, you have to add your own filter cap across the wall wart's output. Here's an idea I had that I would like to put forward. You should measure the ripple. Put a non-polar capacitor and a resistor in SERIES on the output of the wall wart and measure the AC voltage across the resistor. Use a 0.1 uF cap and a ten kilohm resistor. If the supply has an electrolytic filter capacitor of any size at all you should get only the tiniest AC voltage reading across your resistor, less than a volt. If the supply does not have filter caps, you will read something like five or ten volts ac across the resistor. I have never done this test, but just now did a quick back of the envelope thing, could have made a mistake. Let's hear from the pros. Also, I occasionally find SMPS's in garbage bins, but I live in New York -- the dumpster divers' paradise.

Dear

We stiw wuv you, boopy.

WoW - You a sysop on a caffeine burn or something??

"Ignoramus26315"

........ Phil

I think you should do some of this sort of testing.

There is a small problem with it, though you have a good idea buried in there.

If the supply consists only of a transformer and rectifiers, then it can drive current in only one direction. Once the cap in your AC load charges up to the peak voltage, there is no way for current to move in either direction, so the voltage across the resistor will fall to zero.

If you parallel this AC load with a resistor (preferably one of lower resistance than the one in series with the capacitor), it will discharge the cap, at least part of the way, each time a rectifier passes its peak output voltage. Then you will see some AC voltage across the AC load resistor. If the supply contains a filter capacitor that is effective while loaded with these two loads, the ripple voltage will be very much smaller.

No, I simply added up power requirements of two gate drivers and a signal generator chip, they were above the rating of that parrticular wall wart.

I have a decade resistor... But I think that I do not want to take chances wrt driving of a $30 worth of IGBTs...

"Pooh Bear"

Heaps of "wall warts" are REGULATED !!!!!

Been available for at least 30 F****NG years !!!!!!!!!!!!!

Many modern ones are full SMPSs !!!!

......... Phil

In article , Rich Grise wrote: [...]

I've seen ones that were basically a fullwave rectifier driving an LM780X regulator. They were cheaply made. I don't know how much they cost new however.

"DaveM"

Come to your senses.

You are posting a hasty false assessment as fact.

You god ?

The problem is that he is not a rational or competent one.

I short, Stevenson is public menace.

Just like you.

......... Phil

"Rich Grise" Phil Allison

Pooh's post was arrogant, stupid and WRONG.

Like virtually all his others.

........... Phil

Huh? The AC component will pass the capacitor, whether the capacitor is pre-charged to some AC level or not.

In fact, the DC charge will bias the capacitor so that it doesn't have to be non-polar.

Anno