Clue me in with wall wart power supplies

I am looking to build a few led arrays utilizing the typical wall wart DC power supplies....I have a pretty good assortment of them in all
kind sof output DC volts.....Figureing the resistor is not a problem, but what my problem is is with the actual wall wart itself.
For example I have one that says input 115 VAC output 12 VDC but its out put as neasured with a voltmeter is 17VDC
ANother is rated at 7.5 VDC on its data plate, but it measures 9 VDC output.....
Even brand new ones are somewhat higher than their stated output voltages, so what figure do I Use....My actual measured voltage or the data plate info on the power supply? I finally got thre led stuff figured out, now these power supplies throw me a curve......
Would it be that soome were used in various devices to recharge a battery, and the output is somewhat higher than its rated output, due to it needing to be a bit higher to recharge a battery or?
Any info greatly appreciated..
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~Roy wrote:

Are you measuring that open circuit? The circuits are usually just a transformer, rectifier and sometimes filter caps -- no regulator. This means that the voltage will sag with current. They usually say "V volts at I current". Try loading them with the right resistance to get the rated current at rated voltage, and see what voltage you get.
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So in other words if I put on a load that is equal to say 12 volts at 250 ma, on a wal wart that says its 12 VDC-350ma, thats putting out an open circuit voltage of 17 VDC, it should drop to 12 VDC when its load is applied? Would that so called 350 ma rating be the "Maximum" ma its capable of puttin gout at the rated 12 VDC?
I think I see the light if my above example is right.
Thanks
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snipped-for-privacy@hostmail.com says...

Yes, that is correct.
If you meter one that is putting out (nearly) its rated voltage, it is regulated, and will have much less noise (suitable for audio work).
Mark

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It's because the wall wart is a small and cheap power supply. For these reasons the wire used in winding the transformer secondary will be fairly thin, and so will have significant resistance. So when you draw the rated current you see a significant voltage drop, otherwise known as "poor regulation". The transformer turns ratio is adjusted so that it gives an excessive voltage at no load, but the correct voltage at the rated current. If you need 12 V with widely varying loads, you either need a big transformer or some kind of intelligent power supply. Many transformers are designed with a turns ratio which compensates for losses. I recently stripped a massive 2 kVA unit ready for rewinding. According to the data plate the ideal turns ratio would be 110/230 0.478. In actual fact it was 80/163 = 0.491.
Best wishes,
Chris
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snipped-for-privacy@gmail.com says...

Well, not necessarily. In most low voltage, low current applications, all that's needed is a regulator, and maybe another filter cap or two.
Mark
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I think the kind of regulator chips you're talking about would fall under my classification of "intelligent power supply".
Chris
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On Sun, 12 Feb 2006 20:20:27 GMT, the renowned snipped-for-privacy@hostmail.com (~Roy) wrote:

You can usually assume that the the nameplate voltage is correct when measured at the rated current. For example, if your 12VDC adapter is 12VDC 200mA, it will be close to 12V if you are drawing 200mA.
Of course you should not exceed the 200mA figure. You can find out the approximate voltage by interpolating with a straight line between your measured voltage and the nominal full-load voltage. Suppose your LEDs are going to draw 50mA, then you'd expect around 17 - (17-12)*50/200mA = 15.75V. This is probably close enough for your purposes. Then use that voltage to calculate the resistor values.
Alternatively, you could use a regulator (eg. a LM7808) to give you a fixed 8.0V from the 12-17V input, but it might require a heat sink.
Best regards, Spehro Pefhany
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On Sun, 12 Feb 2006 20:20:27 GMT, snipped-for-privacy@hostmail.com (~Roy) wrote:

Wallwarts are like that; they have some internal resistance and output voltage does droop as load current increases. As a rough estimate, figure the internal resistance as:
{V(noload) - V(rated) } / Irated.
Figure a LED dropping for Vnoload then subtract the resistance calculated above to get the actual resistor value you need.
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On Sun, 12 Feb 2006 16:52:25 -0600, Don Foreman

and SOME wall warts actually have switching power supplies now - that regulate accurately with ANY load up to max. Some regulate very poorly with no load. *** Free account sponsored by SecureIX.com *** *** Encrypt your Internet usage with a free VPN account from http://www.SecureIX.com ***
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On Sun, 12 Feb 2006 20:20:27 GMT, snipped-for-privacy@hostmail.com (~Roy) wrote:

measure the voltage under load, and it will regulate much closer to nameplate voltage. *** Free account sponsored by SecureIX.com *** *** Encrypt your Internet usage with a free VPN account from http://www.SecureIX.com ***
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In some cases, the powered device has a regulator to fix the wall box's poor regulation. Some of the power supplies don't even have a filter cap on board, just a transformer and maybe a half-wave rectifier. For your proposed use, they'll probably do fine if you figure out your current draw, then load the prospective wart with a resistor that draws that much and do some measuring. As another poster has said, they're supplying switching power supplies now for a lot of applications, much better regulation and wider wall voltage and frequency tolerance. Much of a switching power supply can be supplied on one chip now.
If you do a load vs voltage curve, you'll get some interesting results with the usual sort of wall box.
Stan
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In some cases, the powered device has a regulator to fix the wall box's poor regulation. Some of the power supplies don't even have a filter cap on board, just a transformer and maybe a half-wave rectifier. For your proposed use, they'll probably do fine if you figure out your current draw, then load the prospective wart with a resistor that draws that much and do some measuring. As another poster has said, they're supplying switching power supplies now for a lot of applications, much better regulation and wider wall voltage and frequency tolerance. Much of a switching power supply can be supplied on one chip now.
If you do a load vs voltage curve, you'll get some interesting results with the usual sort of wall box.
Stan
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poor regulation won't hurt the LEDs - for long life you don't want to run them at absolute max current anyway. just put in a resistor appropriate for the open circuit voltage and be done with it. If it really bugs you, get a 3 terminal regulator (of your choice) - they cost under a dollar.
what you see is typical of a capacitor input unregulated power supply - this behiavior is not due to transformer resistance per se, it's due to the capacitor charging to the peak voltage when there is no load, and being at about hte RMS voltage under rated load.
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