Short circuit protection for (almost) any power supply

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This instructable is about a universal short circuit protection that They ‘ve designed to use in bench power supplies.

1.Understanding the circuit

A resistor of low value  is connected in series with the output of the power supply. As current starts to flow through it, a small voltage drop will appear on it and we will use this voltage drop to determine whether the power supply out put is overloaded or short circuited.The “heart” of this circuit is an operational amplifier (op amp) configured as a comparator.

They put quote marks on “high” and “low” for a easier understanding of the op amp operation. It has nothing to do with logical micro controllers 5 volts levels. When the op amp is in “high level”, its output will be very approximate of its positive supply voltage therefore, if you supply it +12V, the “high output level” voltage will approximate to +12V. When the op amp is in “low level”, its output will be very approximate of its negative supply voltage therefore, if you connect its negative supply pin to ground, the “low output level” will be very near to 0v.

When they use op amps as comparators,  usually have an input signal and a reference voltage to compare this input signal.

The op amp gain(av) is determined by the formula: av = (RF/R1)+1

In this case we’ve got 3.7 times of gain: av = (2700/1000)+1 = 3.7

The third stage of the circuit is the protection itself. Its a relay that you can connect directly with the output of your power supply if you are dealing with low current (2A) or you can connect it to a bigger relay if you are dealing with bigger current or even shut down a previous stage of you power supply forcing the output to shut down. This will vary with the power supply you’ve got.

Just think about it: At the moment the relay turns off the output of your power supply, the short circuit does not exist anymore and the comparator goes from high level to low level. As there is no more current flowing at the NPN transistor base, there is no more current flowing through the relay coil as well. When all these steps happens, the relay contacts did not had enough time to complete its course and connect to the other contacts to close the seal. The behavior of the circuit if  they used the relay itself to close the seal would be the relay madly trying to turn off the output, but without success.

To disarm the circuit a normally closed switch is connected in series with the base of the NPN transistor. By pressing this normally closed switch, it would open its contact and disconnect the base of the NPN transistor from the rest of the circuit breaking the seal and resetting the power supply output.

The 1uF capaciton on the NPN transistor base is just a threshold so a little peak consumption don’t trigger the protection.

You can feed this circuit 9V to 15V. Just be careful to correctly choose your relay voltage and the capacitors voltage. And just to be clear, do not connect this circuit supply pins directly with you power supply output or it will be useless. Just imagine, if your output is shorted, there won’t be enough voltage to supply the protection circuit. You will need to connect it on a stage before the output, maybe a dedicated voltage regulator just for it. A LM7812 will be more than enough.

2.Choosing series resistor value

You need to choose a resistor that the voltage drop on it is around 0.5~0.7 volts when the overload current is passing through it. The overload current is the point that the protection circuit actuates and shuts down you power supply output to prevent damage on it.

You can choose a resistor by using ohms law: V=R*I. In this case we’re going to use: R= V/I.

The first thing you need to determine is the overload current of your power supply.

Let’s say your power supply can supply 3 amps(The voltage of your power supply does not matter in this case). So, you have got R= 0,6V/3A. R = 0.2

The next thing you must do is calculate the power dissipation on this resistor, so it does not burn when current is flowing through it. You can calculate the power dissipation by using the formula: P=V*I.

If we use our last example we would get: P=0.6V*3A. P=1.8W a 3W or even a 5W resistor would be more than enough.

3.Bill of materials

  • To build a board like mine, you will need:
  • 1 – TL082 (dual op amp)
  • 2 – 1N4148 (diode)
  • 1 – TIP122 (NPN transistor)
  • 1- BC558 PNP transistor. You can use a BC557, BC556 or equivalent.
  • 1 – 2700 ohm resistor
  • 1 – 1000 ohm resistor
  • 1 – 10Kohm resistor
  • 1 – 22Kohm resistor
  • 1 – Series resistor
  • 1 – 10Kohm potentiometer
  • 1 – 470uf capacitor
  • 1 – 1uf capacitor
  • 1 – Normally Closed Momentary Switch
  • 1 – Relay model T74

4.Designing the board

This circuit is not really that big so I’ve fit it on a 5cm x 5cm board.

Note: The board has an optional led so you can know when the protection circuit have disarmed the output of your power supply. If you don’t want to use a led, you must short the pins that the led would be connected or else the circuit won’t work. If you want to connect the led, the square pin is the anode and round pin is the cathode. You can connect any LED but high brights ones.

5.Etching and soldering the board

  1. Print the circuit using a good laser jet printer.
  2. Use coated paper to print your circuit.
  3. Make sure your Iron can get as hot as 170ºC~200ºC (338ºF~392ºF).
  4. Before starting your tone transfer, clean the board using a thin steel wool. Your board will get really shinny and clean.
  5. You can always use YouTube to see new methods and how people make their transfer.
  6. By following these tips, you will definitely get a good toner transfer.
  7. After the tone transfer you can etch you board using your favorite etching method. I used iron perchlorate.
  8. They did not took any picture while drilling the holes to mount the components, sorry about that. A drill with 1mm will do just fine.
  9. If the output current of your power supply is higher than 2A, you should reinforce the lines of the series resistor and relay.

6.Testing and calibrating the circuit

To turn this circuit on, you will need to supply it a voltage that can be from 9V to 15V. See the attached picture for more information about the input.

To calibrate the circuit, measure the voltage on the op amp inverting input and turn the potentiometer. As you turn it the voltage will increase or decrease according to the side you are turning it. The value you need to adjust this potentiometer is the gain of the input stage times 0.6 Volts (something around 2.25 to 3 volts if your amplification stage is like mine).

This procedure takes some time and the best method to calibrate it is trial and fail. You may need to adjust a higher voltage on the potentiometer so the protection does not trigger on peaks.

Source : Instructables

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