Re: Using a higher rated DC transformer


"Tom Bruhns" <k7itm@xxxxxxx> wrote in message
news:1163021499.404178.230740@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Mike Monett wrote:
"Tom Bruhns" <k7itm@xxxxxxx> wrote:

> Mike Monett wrote:

>> 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:)

> 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

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:)

...

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

I had responded to your other posting, but it never showed up. What I
mentioned is that this is a real Bell System transformer. I believe it had
to withstand a short circuit without self destructing. It was called
reluctance protection, or something like that. My guess is that there is a
lot of inductance involved, but it is all in the way the transformer is
wound. This was probably not a good example for the OP.

I also measured a Nokia cellphone charger. Claims it is 3.6V, but measures
7V at no load. Could well be the same technology. These people have to meet
all kinds of safety and fire code requirements.

Tam


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