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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:

enter image description here

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?

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  • \$\begingroup\$ 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. \$\endgroup\$ – abdullah kahraman Jun 4 '13 at 7:46
  • \$\begingroup\$ 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. \$\endgroup\$ – shampoo Jun 4 '13 at 10:22
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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. enter image description here

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.

enter image description here

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.

enter image description here

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.

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  • \$\begingroup\$ 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? \$\endgroup\$ – shampoo Jun 4 '13 at 11:17
  • \$\begingroup\$ 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. \$\endgroup\$ – Adam Lawrence Jun 4 '13 at 12:32
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    \$\begingroup\$ 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 \$\endgroup\$ – JIm Dearden Jun 4 '13 at 16:08
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Hopefully this will help you understand the issue: -

enter image description here

This is a basic type of regulator called a shunt-zener regulator. The 12V zener diode prevents voltages greater than 12V appearing across its terminals. It does this by taking as much current as is necessary through the feed resistor (514 ohms in this example) to keep the output at 12V.

Should the load require more than 46mA however, due to the 514 ohm resistor and the load current, the voltage across the output cannot be maintained at 12V and the circuit begins to current-limit. Short circuit current is of course 36V/514 ohms = 70mA.

This is a workable and practical circuit but has limitations. For instance, on no-load there will be a current into the zener of (36-12)/514 = 46mA and this will produce heat. That heat is power dissipated in the zener which is 46mA * 12V = 0.552W - it will get warm and you'll need a heat-sink.

Also, the power in the 514 ohms is large too - on no-load it will dissipate 1.12W therefore total power consumed on no-load is 1.67W. Imagine if this was a 1A capable supply!!

Modern circuits use power-efficient switching regulators and more sophisticated current limits.

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