# How to protect my power supply output?

I want to protect my power supply ouput and I know that there are foldback solutions suggested by LM723 circuits, but I'd like to understand what is happening inside the circuits. Consider the simple concept in the circuit below. I used a simple resistor to protect my output. The problem is shown in the image below:

As you can see in the right circuit I don't have efficient behavior, how should I replace the block on the right to make such a chart?

• You should search Google for current limiting first. Start with this. A transistor acting as a current limiter may work. However, inside the ICs, it is much more complicated than a simple transistor current limiter. Jun 4, 2013 at 7:46
• i searched a lot, it's been 2 days right now. but the answers are far more complicated than the simple solution i was looking for example, i know one of the best solutions is LM723 fold back. i saw the link you mentioned but i didn't understand what "Drive circuit" is in the 1st circuit suggested. Jun 4, 2013 at 10:22

You don't give details of the output voltage of the supply or the required value of the 'regulated' voltage output so the following is a generic answer.

Assume the output (unloaded) is 12V and you require a 9V regulated voltage and a short circuit current of 70mA and you don't want to use a regulator IC.

First step is to fix a reference voltage.

A small current passes through the Zener (say 10mA.) If the Zener voltage is 9V6 then for the assumed values this means the resistor drops (12 - 9.6) = 2.4V and at 10 mA this gives 240R for its value. The Capacitor is there to smooth the voltage and has a value of 10 - 100uF (not critical)

The current available to take from the zener is too small to be used directly so we need to amplify it with a transistor.

The transistor drops about 0.6V between the base and emitter leaving 9V at the output. The problem is that if the output is short circuited it will destroy the transistor so we need to limit the current.

The second transistor is only turned on when the current flowing through the limit resistor produces the turn on voltage (0.6V)

For 70mA this will be 0.6/70 * 1000 = 8R6.

When this voltage is reached the second transistor shuts off the first transistor and the output voltage falls.

IC regulators contain much more sophisticated circuits and for a cost effective solution I would go for them every time.

• thank you for the response. it made the situation so much more clear to me. but in the last circuit, if we have a short circuit. the Tr2Base would be 0+0.6=0.6 and the Tr1Base = Zener = 9.6, so the Tr1BE = 9.6-0.6 = 9, so what causes the Tr1 not to burn? Jun 4, 2013 at 11:17
• SOA. TR1 has to be rated to continuously handle the short-circuit current, or have thermal protection that cuts off the input to the regulator altogether. A fuse could also be used for this. Jun 4, 2013 at 12:32
• Up to the current limit TR2 is effectively out of the circuit. At the current limit (70mA output) TR2 base - emitter voltage is 0.6V (it starts to turn on) allowing current to flow through TR2 collector to emitter, taking current from the base of TR1. (Not shorting it out) This action pulls the voltage at TR1 base lower than the zener voltage and the voltage falls to the point where only a current of 70mA can flow out. As the load resistance gets smaller the voltage at the output drops to maintain the output current at 70mA (the current limit). TR1 must be rated to take more than 70mA Jun 4, 2013 at 16:08