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Hello, everyone!

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 this circuit fit in your bench power supply, you will need to do some calculations, but don't worry, I'll explain everything on the next steps.

Step 1: Understanding the Circuit

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 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 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 with 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 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 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 fails 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 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.

Step 2: Choosing Series Resistor Value

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 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. 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 near 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.

Step 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. They are all fine for this application)

1 - 2700ohm resistor

1 - 1000ohm resistor

1 - 10Kohm resistor

1 - 22Kohm resistor

1 - Series resistor (see previous step)

1 - 10Kohm potentiometer

1 - 470uf capacitor

1 - 1uf capacitor

1 - Normally Closed Momentary Switch (see attached picture. Any normally closed momentary switch will work fine)

1 - Relay model T74 (This is a very common relay model. Easy to find in
eBay. Just try to search for "G5LA-14" on eBay. There are many coil voltages and contacts amperage. If you dont want to use this model, make sure to change the PCB layout)

Step 4: Designing the Board

I've used Express PCB to design the board and the file I uploaded here is free for you to edit as you like. Edit it as you wish to fit the components you have. A PDF version of the board has been uploaded too if you don't even want to edit it or generate the PDFs yourself.

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 (see PCB picture for more details, I've left a note there) 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.

I've tested it on a bread board and you should test it too so you can know for sure if the protection circuit will work with our power supply.

Step 5: Etching and Soldering the Board

I used the toner transfer method to transfer the printed circuit to my PCB.

Tips for a good tone transfer:

- Print the circuit using a good laser jet printer.

- Use coated paper ( see: http://en.wikipedia.org/wiki/Coated_paper ) to print your circuit.

- Make sure your Iron can get as hot as 170ºC~200ºC (338ºF~392ºF).

- Before starting your tone transfer, clean the board using a thin steel wool. Your board will get really shinny and clean.

- You can always use YouTube to see new methods and how people make their transfer.

By following these tips, you will definitely get a good toner transfer.

After the tone transfer you can etch you board using your favorite etching method. I used iron perchlorate.

I did not took any picture while drilling the holes to mount the components, sorry about that. A drill with 1mm will do just fine.

If the output current of your power supply is higher than 2A, you should reinforce the lines of the series resistor and relay. See the attached PCB picture for more information.

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

Step 7: The End

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. Any thing 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.

<p>It does not work and the LED shines with little intensity.</p>
<p>do you have proteus simulation</p>
<p>i don't understand the pinout of the screw header, can you explane this?</p>
<p>hi Azul, i'm trying to make this circuit but I can't get it to work. every time i connect my 12v supply to power the circuit the relay turns on. Even when I turn the potentiometer all the way to the left or right. Can you also explain to me what you mean with those two negative psu connection. I don't know where to connect them. As op-amp i'm using an ne5532 as transistors an bc557 and bd679. As the reference resistor i'm using aan 0,22 ohm 5 watt resistor. In the pictures you will find my breadboard layout. Thanks for your help.</p>
<p>Thats happening because because you did not connect your psu ground pin to the sensor resistor therefore the opamp input is getting some noise and amplifying it.</p><p>The ground output of your psu should be connected to your 0.22 ohm resistor. On your third picture is the wire on the right of the resistor. </p><p>I would also change this bd679. Its a darlington pair and the way the circuit works could change a bit. Try another general use NPN transistor like a bc548 or a TIP122 or even a 2n2222. just be careful with the pinout if you use my design.</p><p>Sorry for the delayed answer.</p>
Hi azul, thanks for your reply. I will go and change those things. If I don't get it to work i will ask for more help. Thanks again for your reply!
Nice, Must have 'Ible!
<p>you can connect a transistor base emmiter junction across series resistor. When the voltage across the series resistor reaches 0.7 volts the relay connected to collector will operate. If this 0.7 volt drop isnot permissable then this circuit canbe used.</p>
Hi thanks for the circuit. I build it but when I turn on the power on, the relay just working and the LED light on and I have no power at the out put. I build it using your PCB design and all the boom list from above. Trying build another PCB about three time and replace the parts as I worried if the PCB damage or the parts broken. No succeed. Any suggestion please? Thanks.
<p>Aadjust the potentiometer. If it is not adjusted it will cause the relay to turn ON. Turn it all the way to one side and turn ON the power supply. If the relay clicks than turn Off the power supply and turn the potentiometer to other end. Now when you turn the power supply on again the relay it shouldn't turn ON. That's the way i made it work.</p>
<p>Hi dimitart, thanks for you guide, finally I got it working. That do the tricks, before I try to adjust the inverting input to 2.5 - 2.8 volts but still not working. Thanks for your help..</p>
<p>Nice circuit! Well done. Works nicely BUT without that 1uF capacitor. I couldn't get it to work on a breadboard and removed that cap. Now wrks a treat. :) I've implemented this in my diy power supply. Thank you! </p>
<p>I have spent many hours on this, but I cannot seem to get it to work. The relay section seems to work, in fact if I touch a wire just before the 10k resistor it switches the relay and stays like that till reset. But my OP amp part doesn't seem to be working. I have tried two OP seperate dual OP amps thinking it was damaged but the same thing. It seems like my output doesn't change no matter what load, any of the output on the OP amp. What points would I test on the OP amp to tell if it's working as supposed. I have looked at the schematic and checked it so many times, I even eyeballed your breadboard design too, as both my OP amps (LM393p and LM358) and I traced the PCB circuit. Everything seems right. I have tried multiple resistors aswell. I have tried adjusting the pot but it seems to do nothing to the output. How do I troubleshoot this?</p>
<p>I am using a 8550 (http://alltransistors.com/pdfview.php?doc=ss8550.pdf&amp;dire=_fairchild_semi) as my PNP transistor and a pn2222a (http://alltransistors.com/pdfview.php?doc=pn2222a_3.pdf&amp;dire=_philips2) for my relay switch. The relay switched when I touch the wire before the 10k resistor like I said. Maybe my transistors are bad? What kind of output should I be getting out of my OP amp</p>
<p>Hi! Can I see a picture of your setup? If you want to test the opamp part of this circuit you can remove all the transistors, conect a 2K ohm resistor right after the optional led and the other side of the resistor you conect to ground. </p><p>Try with a NE5532 opamp. </p><p>What voltage do you have on the potentiometer middle pin? What resistors did you use on stage one opamp? </p>
<p>I don't know if I could get a picture that would help of my setup as I don't know if it's neat enough, I am still new to breadboard circuit design. I haven't set the voltage on my pot. If I am correct it is used to calibrate the protection circuit, when it kicks in? If so I tried shorting my load supply and adjusting it while shorted, as it is only 6v for now (only testing, I will be running an adjustable linear regulator as my load supply ranging from 22 to 1.2 volts) and nothing has happened in the circuit. It has never once tripped due to short or overload. It is quite trial and error for me as I cannot just order parts, I would need a batch before I order as it costs too much. I am using a 5k pot and I have tried a 15k, as those are the closest options I have to 10k, I do not have any matching transistors as well, but they seem to be fine for this circuit, correct me if I am mistaken. I have tried multiple different resistors, resistors with much higher rating such as 10 ohms to increase the voltage drop, but to no avail. If I understand the equation R= V/A correctly, and I aimed to have a 0.6 volt drop at just 200 milliamps (purely for testing) I would divide 0.6 by 0.200, giving me 3 ohms resistor, is that correct? In the photo the transistors are changed from the ones mentioned earlier, giving the same effect as before. I will test the OP amp as you suggested. Your help is very appreciated. Please excuse the 2 darlington array ICs on the right </p>
<p>You must adjust the pot, or it wont work.<br>I cant see the other resitor that is connected to the op amp due to an wire that is over it, but you should set the voltage diveider on that potentiometer to 0.6*(the opamp gain).<br>Measure the potentiometer midle pin (its the pin you connected to the opamp) and turn it until you get 1 volt (1 volt just for testing purposes ) on that pin and test it.</p><p>What may be happening is that the compare voltage is too high and your psu doesn't have enough current to trigger the protection.</p><p>Also how are you testing the circuit? Remember that the current must flow through the sensor resistor and I dont see any wire connected to it. The left side of that resistor should go to ground and it is connected to nothing.</p><p> Are you using another psu to supply voltage to the protection circuit?</p>
<p>I completely rewired the circuit on my breadboard again, and it works! I guess many different types of transistors can be used just as long as it can handle the power and current of switching a relay</p>
<p>Glad to hear that. Yeah any general use transistors will work. you dont have to use the same as I did. The ones I used I got from my drawer. That TIP122 is certainly a overkill on this project. A BC548 would do the job.</p>
<p>Will check all the things you said and report back, thanks so much. I removed the the wires for the picture, but yes I do know that the ground of the load supply must be before the resistor and then the load after the resistor, as on your schematic. My pot is connected properly, I tested the leads multiple times. I am using a seperate power for powering the protection, should they share a ground? My load supply and my protection supply?</p>
Yes, they should share ground. This is a big problem since the op amp does not have any reference with the voltage droped on the resistor. <br>
<p>I do not understand how to test the OP amp, what must I look for? And what leads must I be testing from? If I test the output of the output of both OP amps and +Vcc I get 11.94v which is the same I am supplying my protection circuit with</p>
Check the things I mentioned and see the results. you also shoud see that potentiometer wiring. From the picture I think it is not conected correctly. If both the opamp's output are with the same voltage you are supplying it, something is wrong with its input and the things I mentioned can be a big problem.
<p>I think it's helpful to learning something about op-amp</p><u><strong>THE 741 IS USED IN TWO WAYS</strong></u><strong>1. An inverting amplifier</strong>. Leg two is the input and the output is always reversed.<br><br>In an inverting amplifier the voltage enters the 741 chip through leg two and comes out of the 741 chip at leg six. If the polarity is positive going into the chip, it negative by the time it comes out through leg six. The polarity has been &lsquo;inverted&rsquo;.<br><br><strong>2. A non-inverting amplifier.</strong> Leg three is the input and the output is not reversed.<br><br>In a non-inverting amplifier the voltage enters the 741 chip through leg three and leaves the 741 chip through leg six. This time if it is positive going into the 741 then it is still positive coming out. Polarity remains the same.
<p>How this circuit can be used for a 230V Ac power supply?</p>
<p>This circuit is not designed to work with AC, only DC.</p>
<p>thanks bro. One more thing where do i have to connect those nc pins of the relay? How the relay helps from protection?</p><p>Maybe these are silly questions. Thanks in advance.</p>
<p>It will depend on your PSU. I designed this protection to be universal, the way you connect it to your PSU will depend on your PSU design. Check the example on step 1.</p>
<p>I have a simulation of this but when I turned on the simulation, the relay gives output when the switch is open and didn't give output when the latter one is close.</p>
<p>The LED is also not working. I am talking about the switch on the right side. Please tell me if I am doing it the wrong way.</p>
<p>I rarely use simulation software and I don't know why it is not working. It looks like you assembled the circuit correctly though. </p><p>I prefer to assemble the circuits on protoboard to test them. It's much more fun and you will know fore sure if it works or not. You don't need to short circuit your PSU to test this circuit, you can use low values resistors to get high currents on its output.</p>
<p>ok bro thanks for your help. :-)</p>
<p>Good design, thanks. You could also put the switch on the collector of Q2. The output will then be protected even if the switch is pressed!</p>
<p>The switch place was intentional. Sometimes I deal with high capacitive loads and they have a high current peak that can trigger the protection. Pressing the button will bypass the protection when these cases happens.</p>
<p>You could also use a smaller shunt resistor value in conjeunction with a very low input offset opamp, such as the <a href="http://ww1.microchip.com/downloads/en/DeviceDoc/20005127B.pdf?utm_source=MicroSolutions&utm_medium=&utm_term=&utm_content=&utm_campaign=MCP6V3x+series+datasheet" rel="nofollow">MCP6V31</a> (7uV offset). This would reduce the dropped voltage to a few mV and the disappated power to a few mW. You know, just in case you need a more precise output voltage ;)</p><p>However you can't use such a low input offset opamp as a comparator, it's described in the datasheet if you want to know more.</p>
<p>I forgot to mention: this is allready damn cool as it is right now. Congrats to the circuit!</p>
<p>Thats cool. You could switch off a really high current supply with an external relay. You might try loading the output of your amplifier stage with a 10K resistor in series with a 1 micro farad cap to ground. That should swamp some of the noise transients. If you added a voltage reference on board it could be self powered. Look at the LM10C op amp, it has an internal reference and is designed for single voltage supplies. Here is an electronic fuse circuit I have used that is kind of neat. <a href="http://circuitsbook.com/electronicfuse.html " rel="nofollow">http://circuitsbook.com/electronicfuse.html </a> </p>
<p>Very clever design, and useful too, thanks.</p>
<p>Thank you! </p>

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