Using a higher rated DC transformer

Nov 06, 2006 30 Replies

Good morning, all.



I have an electronic device that requires a DC Wall transformer rated for 12VDC at 600 mA. Unfortunately, I have lost the power supply and I cannot find another one with that exact rating. My friendly Radio Shack representative tells me that the device only "pulls" what it needs from the transformer, the transformer does not "push" that much to the device. He tells me I can use a 12vdc at 1A for the device and it will work.



Is this true? I do not want to fry my device by "pushing" 400mA more to it than it needs. If he is right, may I use a transformer rated for



12vdc at 2.5 amps protected by the same principle? You can tell I'm not an electrical engineer. :) I would appreciate any thoughts. Thank you!

Matthew



Unregulated supplies will have higher voltages if used with lower current draw. Your 600 ma divice will see more voltage. Can't tell what you have unless you look at with a scope. Some DC supplies just have diodes and no filtering. Its likely that a filtered DC supply would work in any case. You can buy some 12 volt regulated supplies, from Jameco that will put out 12 volts regardless of current. Since you didn't say what the device is, we would hope any noise from a switching supply would cause interference.

greg

I don't really know what that means, but the power supply I am thinking of using says this:

DC OUTPUT:

12.0VDC 2.0A MAX

Since it says 2.0 amps MAX, does the MAX imply that a device rated for less would be ok? I take it to mean, "the most you can use this transformer with is a device that pulls 2.0 amps" which implys (to me) that it CAN be used with lower rated devices. Any thoughts about that? Thank you!

If it is a regulated 12V supply, there is no reason you couldn't use a 100 Amp source. If it is unregulated, the 1 A transformer may put out a slightly higher voltage when running at 600 ma. The instrruction book on your "device"may specify what the maximum input voltage is. I would buy a transformer with a regulated output.

The RS 273-1667 will work, but you will still have to figure out which polarity you need. Should be shown on your device input jack.

Tam

In addition to the cautions in the other replies, make sure that the polarity of the new wall wart matches the requirement of your device.

Thanks all, for your input. Can you tell me if the word "MAX" on my

12vdc 2.5a transformer (as mentioned above) would imply that it IS regulated?

No. It just means that you can draw up to 2.5 A from it without having the thing catch fire.

Here is what a typical non regulated supplyu might do. I have a 12V, 300ma AT&T telephone transformer. At no load it puts out 16V. At 10 ma it puts out

14V. At 300 ma it puts out 12V.

Tam

Interesting. The internal resistance decreases as the load increases.

At 10 mA, it's (16 - 14) / 10e-3 = 200 ohms.

At 300 mA, it's (16 - 12) / 300e-3 = 13.33 ohms. You made that up, didn't you:)

Regards,

Mike Monett

Antiviral, Antibacterial Silver Solution:

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No. It means it will die if you draw more current than 2.5 A

No, I didn't make it up. I am actually using the thing as a charger for a riding lawn mower. The first year I had the mower I destroyed the battery because it got left uncharged over the winter. The ATT transformer works great, because it will float the battery at about 14V. Regular lead acid battery; about the size you would have in a very small car. I don't keep it plugged in all the time; mostly when I think about it, which ends up being for a day or so every week or two.

I think the reason the no load voltage is so high is that the diode drops go to near 0V

Tam

Mike Monett wrote: ...

Gee, that's just what I'd expect. The minimum dynamic resistance should be just a bit more than the transformer's secondary DC resistance, plus the primary's DC resistance transformed by the square of the turns ratio, plus any other resistances in there. But that's only at high load, where the diodes are on hard almost all the time. Of course, the diode's dynamic resistance is high when the current is low.

Interestingly, though, even if the diode switched cleanly between zero resistance when forward biased and infinite resistance when reverse biased, the output dynamic resistance would vary with load. That's because the diodes, when off, disconnect the transformer from the load, and the average resistance is inversely proportional to the diode's conduction duty cycle. At light loads (assuming a capacitor that holds the voltage up), the diodes are off most of the time.

Cheers, Tom

"Tom Bruhns" wrote:

I can see the internal resistance changing with the diode conductance angle, but that amount seems excessive. I tried modeling it in SPICE but could not find any combination of leakage inductance, source voltage and resistance, and filter cap that produced those voltages. Maybe one of the data points is in error.

Here - you try:)

Version 4 SHEET 1 880 680 WIRE -592 144 -672 144 WIRE -464 144 -512 144 WIRE -336 144 -384 144 WIRE -272 144 -336 144 WIRE -208 144 -272 144 WIRE -32 144 -144 144 WIRE 16 144 -32 144 WIRE 192 144 16 144 WIRE -672 176 -672 144 WIRE 16 176 16 144 WIRE 192 176 192 144 WIRE -336 208 -336 144 WIRE -272 240 -272 144 WIRE -208 240 -272 240 WIRE -96 240 -144 240 WIRE -96 256 -96 240 WIRE 16 288 16 240 WIRE 192 288 192 256 WIRE -672 336 -672 256 WIRE -336 336 -336 272 WIRE -336 336 -672 336 WIRE -272 336 -336 336 WIRE -208 336 -272 336 WIRE -32 336 -32 144 WIRE -32 336 -144 336 WIRE -272 432 -272 336 WIRE -208 432 -272 432 WIRE -96 432 -144 432 WIRE -96 448 -96 432 FLAG 16 288 0 FLAG 192 144 Vout FLAG 192 288 0 FLAG -96 256 0 FLAG -96 448 0 SYMBOL cap 0 176 R0 SYMATTR InstName C1 SYMATTR Value 10000µ SYMBOL voltage -672 160 R0 WINDOW 123 24 134 Left 0 WINDOW 3 9 108 Left 0 SYMATTR InstName V1 SYMATTR Value SINE(0 17 60) SYMBOL diode -208 128 M90 WINDOW 0 0 32 VBottom 0 WINDOW 3 32 32 VTop 0 SYMATTR InstName D1 SYMATTR Value MURS120 SYMBOL CURRENT 192 176 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName I1 SYMATTR Value 300m SYMBOL diode -208 320 M90 WINDOW 0 0 32 VBottom 0 WINDOW 3 32 32 VTop 0 SYMATTR InstName D2 SYMATTR Value MURS120 SYMBOL diode -144 224 R90 WINDOW 0 -9 34 VBottom 0 WINDOW 3 32 32 VTop 0 SYMATTR InstName D3 SYMATTR Value MURS120 SYMBOL diode -144 416 R90 WINDOW 0 -9 34 VBottom 0 WINDOW 3 32 32 VTop 0 SYMATTR InstName D4 SYMATTR Value MURS120 SYMBOL ind -480 160 R270 WINDOW 0 32 56 VTop 0 WINDOW 3 5 56 VBottom 0 SYMATTR InstName L1 SYMATTR Value 5mh SYMBOL cap -352 208 R0 SYMATTR InstName C2 SYMATTR Value 2n SYMBOL res -496 128 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R1 SYMATTR Value 2.2 TEXT -320 24 Left 0 ;'Full Wave Bridge Rectifier TEXT -288 64 Left 0 !.tran 0 20 0 200u

Regards,

Mike Monett

Antiviral, Antibacterial Silver Solution:

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OK, there is more to it than that. It has a REAL US made ATT transformer. I believe these things had to withstand operating into a short circuit without self destructing. Reluctance protected?

Also, had measured a Nokia 3.6V cellphone charger. That put out around 7V at no load.

Tam

I think that would be leakage inductance. It doesn't seem to help.

The problem appears to be the 14 volt, 10mA data point. It's easy to get

16V with 0.16uA drain (a 10 meg dvm), and 12V at 300mA. But this gives an output voltage of about 15.37V at 10mA. This is an internal resistance of (16 - 15.37) / 10e-3 = 63 ohms, which seems much more realistic.

I tried adding an internal bleed resistor, but that didn't help.

It would be nice if you were drawing a lot more current at 14V.

Regards,

Mike Monett

Antiviral, Antibacterial Silver Solution:

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...

I agree that for a common wall-wart type supply, the 10mA voltage sounds too low, given the open-circuit and 300mA load output voltages.

16V OC, ~15.5V @ 10mA, and 12V @ 300mA seems more like what I'd expect. I can think of a couple possible ways that you could see such voltages, but they aren't all that plausible, given that it's a simple wall-wart. One way is that the supply uses a choke input filter. Then you get an open-circuit output voltage nearly as high as the sine input peak voltage, and the output at full load is considerably less: diode drops and I*R drops lower than the average of the absolute value of a sine, which is 2/pi times the peak voltage, or sqrt(8)/pi = 0.9*Vrms. The output voltage at 1/30 full load could well be mid-way between the full load and open circuit values in that case. A second way is to put an NTC thermistor in the output path: it's a moderately high resistance at 10mA, but at 300mA it heats up and drops to a low resistance. -- Hey, I did say it's not very plausible, didn't I?? ;-) Another thing to consider: the line voltage may not be very sinusoidal. I've seen some pretty ugly ones, but probably not ugly enough to give results like that.

So--inquiring minds would like to know: just what's inside that particular wall-wart, to give results like that?

Cheers, Tom

Hee Hee. Now we get to examine different ways of opening wall-warts:)

I just whack it with a hammer then glue the pieces back together with PVC/CPVC Transition Cement. That glues just about every plastic except polyethylene. Just dab a bit on first. If it softens the plastic, away you go!

Regards,

Mike Monett

Antiviral, Antibacterial Silver Solution:

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SPICE Analysis of Crystal Oscillators:
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Noise-Rejecting Wideband Sampler:
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Well, we need to get Tam to do the autopsy (or perhaps just exploratory surgery), or to ship one of us the wall wart in question. I was going to measure one that I have that I've noticed in the past has a large drop from OC to full load, but I haven't found that Round Tuit yet.

Cheers, Tom

Got any neat trick to open a TV remote? I've got a JVC TV that's outlived three remotes... the ON/OFF always fails :-(

...Jim Thompson

| James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | | http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food.

Gently squeeze it in a vice/vise, preferably with plastic jaws. The plane described by the joint should be parallel to the jaw gripping surfaces. You may have to turn it a bit and squeeze again. It should crack apart almost entirely at the line, and it should be possible to glue it back together again so that you can barely tell it was opened.

Don't blame me if you electrocute yourself or someone else or start a fire due to improper modifications.

Best regards, Spehro Pefhany

"it\'s the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com

Would putting a 12V voltage regulator on the wall wart's outputs help any?

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