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Entwined Darlington Regulator / Schottky Rectifier Idea

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ACharnley

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I had an idea this evening and wanted to run it by you lot.

AC input of 0-100v under no load, 0-6v under significant load. Looking at the pic, the only theoretical disadvantage over a full wave skhottky rectifier is the darlington will drop 0.9v instead of 0.55v. The benefit is when no load there's no heat to shunt (to protect the buck circuit that would be connected to +/-).

The big issue I foresee is the Vceo which is rarely found above 25v and typically 7v. I can't see anyway around it so another schkotty for reverse protection on the emitter would drop a further 0.55v. Ultimately then the voltage drop would be 1v greater than a zener shunt -> full wave schkotty -> load when a load is connected. This would be too great an efficiency loss over the no load benefit.

Any improvements or thoughts?
 

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ACharnley

Member
Perhaps to increase efficiency the 9v from each transistor base can be used to turn on two mosfets which can replace or work alongside the two sckhottys on the left of the diagram. Mosfet breakdown wise I don't know much about the typical reverse drain to source.
 

ronsimpson

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Most Helpful Member
Why the Zener? The B-E voltage will not get above 0.7 volts.
upload_2016-9-14_20-26-14.png
Put a simple diode across the B-E such that forward current will flow B-E and reverse current will flow through the diode.
upload_2016-9-14_20-25-40.png
 

ACharnley

Member
The zener breaks down at 9V (roughly) so the output (+) is limited to that sort of voltage (roughly).

Your idea for the diode from E to B would work (just remove the diode in line with the zener) so long as the B to C reverse breakdown is much higher?
 

ronsimpson

Well-Known Member
Most Helpful Member
The zener breaks down at 9V (roughly) so the output (+) is limited to that sort of voltage (roughly).
The B-E voltage will not get to the 9V of the Zener. So the transistor will save the diode.

There are several other problems but....
 
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