With normal loads, the power transistor is in saturation and so the regulated +12V line is directly available (minus only the Vcesat of the PNP transistor). If the output is accidentally shorted, the npn transistor turns off and removes the base drive of this transistor.
This circuit is intended for protection from short circuits. It might be possible to apply a load that brings the series transistor out of saturation without removing base drive to it: such a situation will cause heating and might cause it to fail short-circuit. Those who wish to draw significant current from this circuit will be well advised to check that the transistor is indeed in saturation with their intended load.
That said, this is a simple way to draw a few tens of milliamps at 12 V from your PC without danger to either load or the PC.
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Signing UpStep 1: The circuit
A small signal NPN transistor is fed with the output voltage through two resistors connected as a potential divider. Should the output voltage become less than a certain value, this transistor will turn off.
The collector of this transistor supplies the base current of the power PNP transistor. So the two transistors will always switch on or off together.
The power transistor (BD140) is supplied with a certain current via the green LED and 470 ohm resistor. It has a certain gain, around 50 or so, and its collector can supply only the gain times base current before it drops voltage (comes out of saturation).
So the maximum current draw can be changed by appropriately dimensioning the base resistor of the PNP power transistor.
The voltage divider at the output determines the voltage at which the current supplied by the transistor drops to zero.
The 470 ohm resistor in parallel with the PNP transistor supplies a certain amount of current so that the circuit will switch on when first powered on.













































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I run this circuit with the 5V off my USB port. I didn't have to make any modifications except to make sure the transistors remain biased (changing value of the R5 pot).
Any comment from experts ? :-)
I did an article about a current limiter to use with the PC supply at the following link. This limiter has a very sharp knee and would work with any input voltage.
http://www.uchobby.com/index.php/2006/11/12/current-limited-pc-power/
The circuit you show here would limit the current, based on the beta of the pass transistor, I dont think it would have a sharp cut-off.
This circuit works as a self-resetting fuse.
Can I find it empirically from the voltage at R1 ends ? (But then a gain of 50 in Q1 and the value you give will mean: Vr1= 470*.22/50 = 2V but that look a bit too few)
By the way what is the role of R5/(R4+R5) ? It look important to make the whole thing work but you did not say why :"That resistor at the output, originally 10K, changed to 36K, had to be changed to 22K." (in 8)
The regulator does have overtemp and short circuit protection but the input current would still spike.
I describe a different current limiting circuit on my site in case you are interested.
http://www.uchobby.com/index.php/2006/11/12/current-limited-pc-power/
Little fuse