On Sun, 24 Jun 2018 11:43:29 -0700, you wrote:
Hi Harvey,
I think you (and Dave before) forget here that the supplies are protected by foldback current limiting. In case of a nearly dead short the current will be much smaller than in normal circumstances, and I guess that also the power dissipation in the series pass transistor will be smaller (the reduced current "wins" from the increased Vce so to speak). The bulk cap voltage will be higher than normal.
True, and I've made some of them. However, as I remember it, the
voltage across the sense resistors should be the VBE drop of the
trigger transistor. I'm also going on general theory without looking
at the specific schematic as well.
I prefer (if possible and reasonable) to talk about the theory, and
let the owner apply it. Maybe a little more difficult, but once
that's done, then the theory and practice are both in hand.
For the theory, the "standard" current limiting says that if the
supply is a 1 amp supply, it will try to deliver 1 amp into the load,
even if shorted. The pass transistor has to get rid of all the power
(say for an 18 volt bulk supply running at 1 amp, 18 watts).
Foldback current limiting triggers at the current limit (say for both
supplies, 1.5 amps), but effectively reduces the short circuit current
to, (say), 300 ma. Thus the pass transistor has to dump .3 * 1.8, or
5.4 watts into a dead short.
Harvey
Albert
On Sun, Jun 24, 2018 at 08:11 am, Harvey White wrote:
So, look for the bulk voltage, you'll find it to be a bit low. Now
look at the voltage drop on the pass transistor (that's the variable
resistor function). If there's a short on the output (I'd suspect
one), then the transistor drops the whole bulk supply voltage and
should be rather warm. There's a current sense resistor in there, say
0.1 to 2 ohms or so, generally with a transistor base/emitter junction
across it. Look for .5 to .7 volts across this resistor. If so, then
the supply is trying to current limit to protect itself.
