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Add this Short Protection Circuit to Your Power Supply

Add this Short Protection Circuit to Your Power Supply

A rather cheap yet reasonably effective short circuit protection circuit is explained below which can be used for safeguarding power supply circuit


A power supply unit is an indispensable unit for every electronic enthusiast and engineers working in the relevant field. Though today all of us use hi-tech power supply units having built-in protection, there are folks who still rely on ordinary types of power supply units with no protection facility.

The biggest enemy of all power supply units is a possible short circuit that might occur at its output terminals due to an accidental connection or due to the fault with the connected load.

There are various electronic circuits that might be employed with a power supply unit for checking this problem, however these circuits sometimes themselves are at risk of getting damaged due to the limitations with many electrical parameters.

A very innovative way of rectifying this problem has been shown in this article. A single relay is used for sensing as well as tripping the output from the involved malfunction.

Circuit Operation

Referring the circuit diagram, we see that a relay is connected directly to the output of the power supply DC output, however the connection is made through the N/O contacts of the relay. These contacts are also terminated as the output of the unit.

N/O means normally open, that means the contacts are open initially, which in turn keeps the output disconnected from the positive of the power supply.

Now when the shown push button is momentarily pushed, the N/O contacts are bypassed allowing the current to flow across the relay coil.

The relay coil energizes, closing the N/O contacts, which in turn latches and sticks to the position even after the push button is released.

The relay latch maintains this latched position as long as the output is used under normal conditions, but in an event of a short circuit across the output terminals, there may be a sharp drop in voltage, the instant this voltage drops below the coil voltage of the relay, it loses its holding strength and immediately releases the contacts, and trips, cutting OFF the supply to the output and in the course switches OFF the latch preventing the short hazard conditions.

This brings the relay to its initial condition and needs resetting in order to restore power at the output.

The circuit diagram for the power supply short circuit protection is shown below:



About the Author

I am an electronic engineer (dipIETE ), hobbyist, inventor, schematic/PCB designer, manufacturer. I am also the founder of the website: https://www.homemade-circuits.com/, where I love sharing my innovative circuit ideas and tutorials. If you have any circuit related query, you may interact through comments, I'll be most happy to help!

63 thoughts on “Add this Short Protection Circuit to Your Power Supply”

  1. Olá Swagatam,

    Este circuito suporta 50v 10A, ou quais modificacoes devo fazer?

    Poderia postar circuito de fonte 50v 10A, quando regulada para 10A a tensao nao cair ou seja estabilizada.

    • Hi Joao, yes you can use it for 50V 10amp, just add a small value resistor in series with the relay coil, may be a 22 ohm… and make sure the contacts are rated at 20 amps or 15 amps minimum.

  2. Hai sir….*
    I tried this circuit it’s working well…
    I have one adapter type smps…
    O/p 12v 1.5A …

    Its working good…This smps contains short circuit protection…
    If i short it…smps power led goes OFF…if i un short it…smps power led ON…

    I can’t able to find the circuit diagram…fully molded …
    So pls guide me sir…
    Is there any circuit same like…

    Whenever short happen….smps power led off and in normal condition smps power led ON….

    Pls guide me sir….*

  3. Thanks for your suggestion bro, multimeter power supply 9v by adding capacitor in output is not working on my cheap meter . cell phone charger idea help little bit that is while touching the negative by hand gives proper reading , finally I decided to leave on battery power supply with on/off using relay (one COB multimeter gets defective while measuring voltage more than 3v) . thank you …

      • Hai Bro, your help please .
        three dc source with one voltmeter, I need a circuit to measuring voltage of dc source ( common ground) by selecting with one push switch ( 3 way sequential selector switch ) using relay , thank you

        • Hai bro sorry for disturbing you again and again , I thank you once again for your quick response, I tried sequential push selector switch it works perfect but some delay in selection ,then i changed the capacitor between pin 13 &14 with 0.5uf instead of 1uf it works charmingly thank you ..
          I made variable PSU using lm338t plastic package t220 , when output get shorted ic gets heated instantly with in 4 sec ic gets damaged in min/max output ( but using lm317t long term short circuit and work absolutely perfect without ic gets damage ) even though using proper huge heat sink and fan ,please provide any short circuit protection circuit for variable psu ( but while using lm338k its works fine for some limits time approx 1 to 4 mins , I tried your electronic fuse works perfectly for constant voltage 5 to 33v thank you…

          • Bro, LM338 is the 5 amp version of the LM317…and both these ICs are thoroughly protected from overheat and overload situations…..if your LM338 is burning due to overheat that means it is a duplicate IC not original make.

            short circuit protection is absolutely not required for these ICs…you can check the details in the datasheet of the IC.

  4. Sir
    I made your short circuit protection circuit it works fine for transformer power supply , if I connect to battery 24v 10Ah ,circuit not working ( when shorting output voitage across relay not fall to zero ,instead huge spark and melting wires) please give any modification or new circuit.sorry for my bad english

    • Deepak, it’s because a transformer has limited current which is able to drop, whereas a battery may have relatively huge current which may not drop quickly causing melting of wire.

      add a current limiter in series with the battery, and you will find the system working perfectly. You can do this by adding a low value resistor such as a 2 ohm 5 watt etc or you can calculate it using Ohm’s law

      alternatively you can try the following design which is designed for battery use


      • Thanks Bro, I made your new electronic automotive fuse it works fine,one again thank you for your fast response, I made another circuit 24v battery to 12v 200mA circuit using lm317 adjustable but lc get heated ( some power loss approx 3w as heat) please give any efficent simple circuit for 12v 200mA, thank you.

          • Bro another help please, I want to add cheap multimeter ( 2$) as voltmeter for my variable power supply unit , multimeter is 9v battery powered ,I replace it by 7809 ic but while measuring (i.e self powered multimeter) improper reading , please give any simple isolated power source circuit 9v or any suggestion/ideas thank you…

            • sure bro, the 7809 should have worked correctly according to me, first confirm whether or not it is giving an accurate 9V without the meter connected…if it is fine, then you can try connecting a high value capacitor across the output and ground of the 7809 and see if that correts the issue….you can try a 1000uF/16V or even higher. alternatively you an simply use your cellphone charger output and use it for powering the meter, it worked great in my meter.

  5. Hi sanjay bharvad,
    I believe it is possible if you have a step down transformer if it is 50-100HZ ac or high frequency circuit which will step down the 9kv to easy to manage power like 12v. you can then utilize the relay powering circuit from 12v external source and use zener diode to cut out form a cut-off point when the 12v which is directly influenced by the 9kv fluctuates to the level you may want a shut down.
    I have utilized similar circuits in lithium ion short circuit controls for electric cars.

  6. Hello sir ..

    I want to operate 12vdc relay by short circuit 9000 vdc ..please inform me can I use this circuit..for that and which type of changes I have to do

    • Hello Sanjay,

      I don't know how you would configure 9kV with a 12v relay, I think that's not possible with the above circuit.

  7. hello sir as u said above "could be used in variable power supply unite. ." how does relay going to react in case of low voltage is being set eg: 12max out put relay set to 12vminmum if turn down voltage to 2v then I hope relay going to be Delatch .? right sir ?
    Thank you in advance

    • Hello Lokesh, the variable supply stage will need to be after the relay, meaning the relay contacts will be supplying the variable circuit stage, and the output from the variable circuit supply will finally reach the load.

      The relay stage will be associated with the bridge rectifier, and the relay coil will need to be rated as per the bridge output voltage

    • Thank you for replying sir..
      actually I am planning to put buck boost converter
      (LM2577S + LM2596S dual chip modules, DC-DC step up step down module )

      … Instead of 78xx since its too hot to handle. .

      ok as u said if I connect this system before
      buck boost module. .. does it functions same.?

      and more querie does the buck boost module support short circuit cut off.?

    • Lokesh, if you have planned to use a LM2596S circuit then a better and more logical idea would be to use its pin5 for the overload shut down…the above relay method would be too crude for this.

      If possible I'll try to post the whole idea soon in my blog.

    • ho nice.. but I am using ready kit bought from ebay if u type

      "(LM2577S + LM2596S dual chip modules, DC-DC step up step down module )"

      u may get it.. sir in ebay website hope u get an idea about it.. since I am not confirm that's y worried. . hope I can get solution as soon as possible. . 🙂

    • ok I will try sir..Thank you for ur efforts to support me..
      also let me how can I prepare an
      "Add on circuit " for converting
      square wave out put of 230v to sine wave ?

    • Thank you for replying sir.. I have checked the link. .. and I feel that circuit is made for building newly. .but I am looking for circuit which can be added at the end of square wave 230v 45w inverter. . which is presently not supporting any Transformer and capacitive step down type load.. due to square wave out put.. hence I was looking for minimal change in inverter so that I can run above mentioned load too..
      hope my msg is convinced. ..

      Thank you

    • you can try the C1/C2/L1 configuration as shown in the following diagram, and see if that helps:


    • Yes something like this I was looking for thank you sir.. just need to find what the value for 230v 45 w that's ok I will tryout..

      and also let me know. ..

      So eagerly waiting for circuit which will be having following (few or all) features -Low Battery Cut-off
      – Overloading protection
      – Short Circuit protection
      – Reverse current protection
      – Reverse polarity protection
      – Thunder protection
      – Over discharge protection
      – Auto battery shut-off at Low voltage detection
      – Overcharge protection
      – Auto charge stop/ High Volt Detection
      – Battery capacity level display(SOC)

      Making this circuit for underprivileged location as donation for poor via charity So hope I can have one ckt diagram with some or all features mentioned above or lts link.. Looking for ur reply ..
      With full excitement

      Thank you

      If successful I am in plan to put ur & website name on my device As part of tribute to you sir

    • Thank you for replying. ..
      plz also let me know the simple circuit which can be used to charge 24/12/9/3vDC battery ..
      (based one preset settings .. )

      with minimum below options or better then this.:

      1)Disconnect input after (set) full charge.

      2)Disconnect output after (set) minimum level. to avoid Deep discharge

    • Thank you for replying sir..
      but does it supports my 1st conditions
      I.e 1)Disconnect input after (set) full charge.
      if not what simply modifications may required ?

      plz let me know
      thank you..

  8. Hi Swagatham
    Can you publish an LM317 based 1.25V to 24V, 5A variable powersupply circuit. I know 317 can only handle 1.5A, so a power transistor(s) can be used to handle current more than 1.5A.
    But I don't want that bulky power transistor. I prefer an N-Channel power mosfet for enhansing LM317's current capacity. Is it possible….?
    I searched google for such a circuit, not yet found. If you can add output SHORT CIRCUIT PROTECTION, with latching/auto reset after removing short circuit. That will be an added advantage.

    • Hi, Anil,

      although a BJT looks more appropriate as the outboard current booster device since it requires just 0.7V to trigger, let's try this with a mosfet and see how it responds…I'll design and post it soon.

    • hi Anil Kumar. K
      if possible check this by copy pasting below

      "LM2577S + LM2596S dual chip modules, DC-DC step up step down module "

      in ebay.in hope ur requirement gona match

  9. hi swagatam, can you please guide me what component and value should replace in diagram i want to make an variable resistor which is 30v is the max output by using an 24v 6amp transformer

  10. hi swagatam
    can you please guide me i would like to make a variable resistor, which component should i replace and value? i already have transformer which is 24v 6amp. thanks advance

  11. hi swaggatam, how do you make a circuit like above to short circuit protect a lithium bank of 200v?
    Im working on a project and i noted that an accidental increase in current draw would lead to fires,

  12. hi again,
    Thanks for your responses 🙂
    As I said before, I'm placing this circuit after a voltage regulator. The problem now is that when I regulate the voltage to lower than 3.5 volts, the relay opens, so I cant use lower voltages than that. The reason why this appens is quite obvious, I just wanted to ask if you have any suggestion on how to solve it.

    By the way.. have you thought of this circuit as being one bit of digital memory?

    • Hi Daniel, it's a crude circuit for achieving a short circuit control, it can be easily modified for a better response by just adding a BJT with it, i'll try to update it soon…this modification will allow volatge inputs below 1 V.

      yes most latch circuits could be considered as a 1 bit memory circuit…

    • In my prototype the relay was making a lot of noise while switching off while I was testing it with an output short, ……….adding the capacitor instantly solved the chattering issue therefore I had to include the capacitor in parallel with the relay coil.

  13. Hi Swagatam! I've built the circuit and it works great! thanks!
    I just have two questions:
    1 ) is the diode really needed? what is it for? having it means that I will always have a 0.7 voltage drop and I would like to avoid it because i'm puting the circuit after a previously regulated power supply.
    2 ) Is it safe to use this circuit as a short circuit "detector" to repeatedly check solderings, etc. (as I usually do with the buzzer in my multimeter)

    • Thanks Daniel,

      The diode can be removed.

      this circuit is a crude design and will work only for high current short circuit scenarios….it's not suitable for detecting low current soldering faults.

    • Hi Janesh, No modifications would be required, just connect the variable power supply circuit with the output of the above shown design

    • No it won't, you will have to press the push button to reset it back into action. There's no automatic option for this simple design.

  14. The DC comes from the AC after all. So the AC control will be directly proportional to the DC.

    Anyway with a 337 IC you can do the following:

    use a BC557 transistor.
    connect collector to ADJ, emitter to ground-relative to (-).

    connect base to ground via a 100 ohms and put another resistor between the emitter ground connection and base ground connection…meaning the emitter/ground joint and base/ground joint must be bridged with a resistor.

    This resistor decides the current limit.

    find it by using R = 0.6/limit current = Ohms.

  15. you can put it anywhere you like, before or after the regulator….it will do its job as specified.

    by the way if you are using a 78XX regulator then you probably wouldn't need the above arrangement because all modern regulator have built-in short circuit protection and are litrally blow-out proof

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