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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:
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 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.
Hopefully this will help you understand the issue: -
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.
