I need very basic short circuit protection for a discrete regulator. Is this circuit legit? I could only get it to work as a limiter with the amount of short circuit current set by the sense resistor value, but no foldback. Explain to me how the divider works here.
(http://images.elektroda.net/4_1200125223.jpg)
The formulas are in that scan, although too blurry for my eyes. Do you have the original in readable form?
The base idea is to pick the current which is safe for Zero output voltage, then use the 1.5K 15K divider to buck that for higher voltage.
Is a real brain-pain AND is a poor match to the parabolic curve of dissipation AND does not handle the case where cooling has failed (dead fan, clot of dust on the heatsink).
The LM7815 LM317 LM350 include thermal limiting which protect the regulator against current OR heat. Roll-yer-own voltage regulator is pretty old-fashioned today.
Imax=1/Rs[(1+R2/R1)Vbe+(R2/R1)Vreg]
Vout=0: Isc=1/Rs(1+R2/R1)Vbe
ratio Imax/Isc=1+[R2/(R1+R2)](Vreg/Vbe)
It's a recycle project, no fancy ICs for me. Nonetheless, I want to understand the principles and formulas. So first I pick the value of Rs that will determine the short circuit current, then apply the divider for normal voltage operation?
Assume Vbe is 0.6V.
Take 1 Ohm for current sense resistor.
Without foldback frill the limit current is 0.6 Amps.
Assume nominal output is 12V. Assume foldback resistors are 10:1 divider.
At 5V output the foldback divider inserts negative 1.2V in series with the current-sense voltage. Now we won't hit 0.6V at transistor until we have 1.2V+0.6V across the current-sense resistor. The current limit is now 1.8 Amps.
However if the output is shorted to zero, the current limit is 0.6 Amps.
Look what this does to the heat in the pass device. Assume we are making 12V from 22V raw supply. When at-limit with 12V out we have 1.8 Amps dropping 10V so 18 Watts. When shorted, current drops to 0.6 amps, heat is 13.2 Watts. (However if the load hits a critical value, the reg may sit with 6V 1.2A out and dissipate 19.2 Watts.) Without foldback, sense resistor set for 1.8A out, shorted load, we would be dissipating 22V*1.8A= 39.6 Watts.
It's starting to make sense. Thank you PRR. This circuit can only reduce the current by about three, and dropping the divider ratio has really no effect besides lifting both thresholds up, right?
The upper resistor of the divider seems to increase the shorted current limit a little. Can I scale it down to, say, 100 and 1K?
> It's a recycle project, no fancy ICs for me.
I'll point out that LM7805-like parts are readily recycled from old cassette decks and VCRs.
Also you can get ANY foldback line by picking values and connections.
It would be too good to be true. From my measures (not calculations), 1:10, 1:4, 2:1 for the divider had the same effect — Imax ~ 3Isc, just like it was stated in the article.
I only have PC power supplies and random Soviet junk at my disposal. There are a bunch of resettable fuses — what can they do for me?
after the pc power supply does all the chopping and switching and transforming of the mains voltage, it just shoves the result through a 7812 and a 7805 and their negative relatives. they're in there, too, along w/ some nice fast diodes.
Only one of five units had one 7812 hidden inside. There's no way to get 9V out of it. 5 diodes in series? :icon_lol:
I realised later, of course. they use three terminal regs for the supplies rated less than one amp, and more complex regulation for the higher current supply. and, of course, there is no 9V in a comp psu ....