## Saturday, October 21, 2023

This instructable is about a universal short circuit protection that I've designed to use in bench power supplies. I've designed it to fit in most power supplies circuits. In order to fit this circuit in your bench power supply, you will need to do some calculations, but don't worry, I'll explain everything on the next steps.

The circuit is really easy to understand.

A resistor of low value (the resistor value will be explained later) 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 (stage 2).

The way it works is really simple, you just need to follow this rule:

If the voltage on the non-inverting output is higher than the inverting output, then the output is set to “high” level.

If the voltage on the non-inverting output is lower than the inverting output, then the output is set to “low” level.

I put quote marks on "high" and "low" for an easier understanding of the op amp operation. It has nothing to do with logical micro controllers 5 volts levels. When the op amp is at "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 close to 0v.

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

So, we have a resistor with a variable voltage that is determined according to the current that flows through it and a reference voltage. Does this ring any bell on your mind? We're almost finished with the theory be brave and follow me.

As the voltage drop on the resistor in series with the power supply is too small, we need to amplify it a little bit because some op amps are not too accurate when comparing low voltages like 0.5 volts or lower. And that's why the first stage (stage 1) of this circuit is an amplifier using another op amp. A 3 to 4 times amplifications is more than enough in this case.

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 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. For example, if your power supply is based on a LM317, you can simply use the relay to physically disconnect the LM317 output pin from the power supply, as we are using the relay normally closed pin (I've uploaded a picture to better describe this example).

The PNP transistor on stage 3 just act like a seal to keep the relay turned on after the short circuit so you can press a button to disarm it. Why I didn't use the relay itself to do this? It's because the relay is too slow to do it.

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 happen, the relay contacts did not have enough time to complete its course and connect to the other contacts to close the seal. The behavior of the circuit if I used the relay itself to close the seal would be the relay madly trying to turn off the output, but without success. I know I could have used a capacitor to supply enough current to the relay, but I would need a big capacitor and no one can grant that it would work 100% of the times the output of the power supply is shorted. Electrolytic capacitors fail over time, and failure is not a good option in this circuit.

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 doesn'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 toBe clear, do not connect this circuit supply pins directly with your 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.

I've created a separated step for this because this series resistor is the most crucial part of the circuit. As I've said before, this resistor is connected in series with the power supply output. As current starts to flow through it a small voltage drop will appear on it.

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 acts 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. In this part I can't help you, you've got to know the maximum current your power supply can supply and therefore dimension your series resistor value.

Let's say your power supply can supply 3 amps (The voltage of your power supply does not matter in this case). So, we've got R= 0.6V/3A. R = 0.2 Ohm. If you calculated the resistor and the result is not a commercial value, don't worry. Just get a commercial value resistor that is close to your calculations results.

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.

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 peaks. As I've said, it takes some time to calibrate it.

I've uploaded a small video of it working on my power supply that I've built some weeks ago.

The picture attached on this step is the protection circuit installed inside my bench power supply (you can check it here).

And that's it. Anything that I can help just ask on the comments and if you based on my project to make a new one please comment too, I'm very curious to know if you did a really nice upgrade to this project.

This instructable is about a universal short circuit protection that I've designed to use in bench power supplies. I've designed it to fit in most power supplies circuits. In order to fit this circuit in your bench power supply, you will need to do some calculations, but don't worry, I'll explain everything on the next steps.

The circuit is really easy to understand.

A resistor of low value (the resistor value will be explained later) 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 (stage 2).

The way it works is really simple, you just need to follow this rule:

If the voltage on the non-inverting output is higher than the inverting output, then the output is set to “high” level.

If the voltage on the non-inverting output is lower than the inverting output, then the output is set to “low” level.

I put quote marks on "high" and "low" for an easier understanding of the op amp operation. It has nothing to do with logical micro controllers 5 volts levels. When the op amp is at "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 close to 0v.

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

So, we have a resistor with a variable voltage that is determined according to the current that flows through it and a reference voltage. Does this ring any bell on your mind? We're almost finished with the theory be brave and follow me.

As the voltage drop on the resistor in series with the power supply is too small, we need to amplify it a little bit because some op amps are not too accurate when comparing low voltages like 0.5 volts or lower. And that's why the first stage (stage 1) of this circuit is an amplifier using another op amp. A 3 to 4 times amplifications is more than enough in this case.

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 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. For example, if your power supply is based on a LM317, you can simply use the relay to physically disconnect the LM317 output pin from the power supply, as we are using the relay normally closed pin (I've uploaded a picture to better describe this example).

The PNP transistor on stage 3 just act like a seal to keep the relay turned on after the short circuit so you can press a button to disarm it. Why I didn't use the relay itself to do this? It's because the relay is too slow to do it.

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 happen, the relay contacts did not have enough time to complete its course and connect to the other contacts to close the seal. The behavior of the circuit if I used the relay itself to close the seal would be the relay madly trying to turn off the output, but without success. I know I could have used a capacitor to supply enough current to the relay, but I would need a big capacitor and no one can grant that it would work 100% of the times the output of the power supply is shorted. Electrolytic capacitors fail over time, and failure is not a good option in this circuit.

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 doesn'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 toBe clear, do not connect this circuit supply pins directly with your 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.

I've created a separated step for this because this series resistor is the most crucial part of the circuit. As I've said before, this resistor is connected in series with the power supply output. As current starts to flow through it a small voltage drop will appear on it.

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 acts 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. In this part I can't help you, you've got to know the maximum current your power supply can supply and therefore dimension your series resistor value.

Let's say your power supply can supply 3 amps (The voltage of your power supply does not matter in this case). So, we've got R= 0.6V/3A. R = 0.2 Ohm. If you calculated the resistor and the result is not a commercial value, don't worry. Just get a commercial value resistor that is close to your calculations results.

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.

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 peaks. As I've said, it takes some time to calibrate it.

I've uploaded a small video of it working on my power supply that I've built some weeks ago.

The picture attached on this step is the protection circuit installed inside my bench power supply (you can check it here).

And that's it. Anything that I can help just ask on the comments and if you based on my project to make a new one please comment too, I'm very curious to know if you did a really nice upgrade to this project.