Re: why is my full wave bridge overheating?

In article <1155516665.533011.71980@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, linnix
wrote:

Back on earth, I found a cement block power resistor in my box, a 10W
2.7 ohm. I put it in series between the bridge's positive DC output

If you circuit is drawing 200 mA, that will give you approx 5V drop.
That's meaningless when compared to the 25V drop on the regulator.

2.7 ohms at 200 mA will drop .54V and read that much to a voltmeter.
(And if this is true will decrease 7805 regulator heating in the
previously described application by 2.5%.)

If the resistor is upstream of the positive terminal of the main filter
capacitor, it can drop instantaneously and even charge-weighted-average
several times that. However, I would agree that more is needed by a
factor of at least 10, maybe 20 or somewhat more. (The
instantaneous/average current ratio will decrease as this resistor value
is increased to decrease current.)

and the regulator. Surprisingly the cement resistor does not even get
warm (but it obviously is not blown because the circuit powers up). I
think I may have this wrong as I have seen a load resistor on some
schematics that connects to ground. Should it go one end to the bridge
and one end to ground or in series as I have it now?

Try a 10 ohm, it will be hot.

10 ohm may produce noticeable heat, maybe get noticeably quite warm, and
probably do so while taking maybe a mere fraction of the heat from the
regulator and some heat from the transformer windings.
That is, if the resistor is between the rectifier and the filter
capacitor. If the resistor is downstream from the filter capacitor, it
will experience lower instantaneous current and lower RMS current (by
experiencing only the DC load cuirrent and regulator low leg loss current
in the form of steady DC) and drop so much less voltage as to heat up less
despite passing current more continuously.
If you have some need for this added resistor to be downstream of the
main filter capacitor, it will need to be of a higher value than if it is
upstream of the main filter capacitor.
Downstream location will have the resistor take heat only away from the
regulator barring malfunction. Upstream from the main filter capacitor
will reduce heating of both the regulator and the transformer.
Keep in mind that transformer output voltage before any resistive and
diode losses times average current, with voltage appropriately weighted
for appropriate weighting in such an average (likely to be close to
1.2 times the open-circuit AC voltage), is power consumption in the sum of
the load, resistors, rectifier, and transformer secondary winding.
If you draw 100 mA DC from a 24 VAC transformer supplying a
rectifed-regulated DC power supply with 5 VDC output, plan on
close to 2.5 or even maybe closer to the worst-case roughly 3 watts being
generated overall - not including heat generated by the load nor the
transformer core nor in the transformer's primary winding. Multiply by
about 3.42 to get BNTU per hour.

There is a reason I'm using a 24v AC transformer. Mainly it's because
I don't want to run two power supplies and the item ultimately being
powered via relay requires 24v AC. I have to be cautious as to how
much I reveal of the entire circuit because -sooner or later it seems
like someone always winds up saying "just go out and buy one off the
shelf, Pokey, you idiot" and I'd like to make this work.

In that case, go for a switching regulator. A typcial 90% switching
regulator will be on 10 % to 20%. You will save lots of heat and
energy. For example, try the lm3485 circuit at http://linnix.com/smps

- Don Klipstein (don@xxxxxxxxx)
.

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