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You are here: Home / Power Supply Circuits / LM317 with Outboard Current Boost Circuit

LM317 with Outboard Current Boost Circuit

Last Updated on November 3, 2020 by Swagatam 103 Comments

The popular LM317 voltage regulator IC is designed to deliver not more than 1.5 amps, however by adding an outboard current boost transistor to the circuit it becomes possible to upgrade the regulator circuit to handle much higher currents, and upto any desired levels.

You might have already come across the 78XX fixed voltage regulator circuit which are upgraded to handle higher currents by adding an outboard power transistor to it, the IC LM317 is no exception and the same can be applied for this versatile variable voltage regulator circuit in order to upgrade its specs for handling massive amounts of current.

The Standard LM317 Circuit

The following image shows standard IC LM317 variable voltage regulator circuit, using a bare minimum of components in the form of a single fixed resistor, and a 10K pot.

This set up is supposed to offer a variable range of zero to 24V with an input supply of 30V. However if we consider the current range, it's not more than 1.5 amps regardless of the input supply current, since the chip is internally equipped to allow only up to 1.5 amps and inhibit anything that may be demanding above this limit.

LM317 regulator circuit

The above shown design which is limited with a 1.5 amp max current can be upgraded with an outboard PNP transistor in order to boost the current on par with the input supply current, meaning once this upgrade is implemented the above circuit will retain its variable voltage regulation feature yet will be able to offer the full supply input current to the load, bypassing the IC's internal current limiting feature.

Calculating the Output Voltage

For calculating the output voltage of a LM317 power supply circuit the following formula could be used

VO = VREF (1 + R2 / R1) + (IADJ × R2)

where is = VREF   = 1.25

Current ADJ can be actually ignored since it is usually around 50 µA and therefore too negligible.

Adding an Outboard Mosfet Booster

This current boost upgrade can be implemented by adding an outboard PNP transistor which may be in the form of a power BJT or a P-channel mosfet, as shown below, here we use a mosfet keeping things compact and allow a huge current upgrade in the specs.

adding an outboard PNP transistor

In the above design, Rx becomes responsible for providing the gate trigger for the mosfet so that it's able to conduct in tandem with the LM317 IC and reinforce the device with the extra amount of current as specified by the input supply.

Initially when power input is fed to the circuit, the connected load which could be rated at much higher than 1.5 amps tries to acquire this current through the LM317 IC, and in the process a proportionate amount of negative voltage is developed across RX, causing the mosfet to respond and switch ON.

As soon as the mosfet is triggered the entire input supply tends to flow across the load with the surplus current, but since the voltage also begins to increase beyond the LM317 pot setting, causes the LM317 to get reverse biased.

This action for the moment switches OFF the LM317 which in turn shuts off the voltage across Rx and the gate supply for the mosfet.

Therefore the mosfet too tends to switch OFF for the instant until the cycle perpetuates yet again allowing the process to sustain infinitely with the intended voltage regulation and high current specs.

Calculating Mosfet Gate Resistor

Rx may be calculated as given under:

Rx = 10/1A,

where 10 is the optimal mosfet triggering voltage, and 1 amp is the optimal current through the IC before Rx develops this voltage.

Therefore Rx could be a 10 ohm resistor, with a wattage rating of 10 x 1 = 10 watt

If a power BJT is used, the figure 10 can be replaced with 0.7V

Although the above current boost application using the mosfet looks interesting, it has a serious drawback, as the feature completely strips off the IC from its current limiting feature, which can cause the mosfet to blow-of or get burnt in case the output is short circuited.

To counter this over-current or short-circuit vulnerability, another resistor in the form of Ry may be introduced with the source terminal of the mosfet as indicated in the following diagram.

The resistor Ry is supposed to develop a counter voltage across itself whenever the output current is exceeded above a given maximum limit such that the counter voltage at the source of the mosfet inhibits the gate triggering voltage of the mosfet forcing a complete shut off for the mosfet, and thus preventing the mosfet from getting burnt.

LM317 outboard boost mosfet application circuit

This modification looks pretty simple, however calculating Ry could be little confusing and I do not wish to investigate it deeper since I have a more decent and a reliable idea which can be also expected to execute a complete current control for the discussed LM317 outboard boost transistor application circuit.

Using a BJT for Current Control

The design for making the above design equipped with a boost current and also a short circuit and overload protection can be seen below:

LM317 outboard boost transistor with short circuit protection

An couple of resistors, and a BC547 BJT is all that may be required for inserting the desired short circuit protection to the modified current boost circuit for the LM317 IC.

Now calculating Ry becomes  extremely easy, and may be evaluated with the following formula:

Ry = 0.7/current limit.

Here, 0.7 is the triggering voltage of the BC547 and the "current limit" is the maximum valid current that may specified for a safe operation of the mosfet, let's say this limit is specified to be 10amps, then Ry can be calculated as:

Ry = 0.7/10 = 0.07 ohms.

watts = 0.7 x 10 = 7 watts.

So now whenever the current tends to cross the above limit, the BC547 conducts, grounding the ADJ pin of the IC and shutting off the Vout for the LM317

Using BJTs for the Current Boost

If you are not too keen on using  mosfet, in that case you could probably apply BJTs for the required current boosting as shown in the following diagram:

LM317, LM338 current boost using outboard transistors

Courtesy: Texas Instruments

Adjustable Voltage/Current LM317 High Current Regulator

The following circuit shows a highly regulated LM317 based high current power supply, which will provide an output current of over 5 amps, and a variable voltage from 1.2 V to 30 V.

In the figure above we can see that the voltage regulation is implemented in the standard LM317 configuration through R6 pot which is connected with the ADJ pin of the LM317.

However, the op amp configuration is specifically included to feature the useful a full scale high current adjustment ranging from the minimum to the maximum 5 Amp control.

The 5 amp high current boost available from this design can be further increased to 10 amps by suitably upgrading the MJ4502 PNP outboard transistor.

The inverting input pin#2 of the op amp is used as reference input which is set by the pot R2. The other non-inverting input is used as the current sensor. The voltage developed across R6 through the current limiter resistor R3 is compared with the R2 reference which allows the output of the op amp to become low as soon the maximum set current is exceeded.

The low output from the op amp grounds the ADJ pin of the LM317 shutting it off and also the output supply, which in turn quickly reduces the output current and restores the LM317 working. The continuous ON/OFF operation ensures that the current is never allowed to reach above the set threshold adjusted by R2.

The maximum current level can be also modified by tweaking the value of the current limit resistor R3.




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About Swagatam

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!

You'll also like:

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  • 3.  Add this Short Protection Circuit to Your Power Supply
  • 4.  Adjustable 3V, 5V, 6V, 9V,12V,15V Dual Power Supply Circuit
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  1. Search Related Posts for Commenting

  2. salturi says

    Hi wanting to put two mosfets in parallel, how much does the mosfet activation resistance rx affect with the second mosfet in parallel

    Reply
    • Swagatam says

      Hi, there will no effect since FETs are high input impedance devices, so the same Rx can work with the 2 devices.

      Reply
  3. Ervin says

    Dear Sir,

    please could you help me – can I use in the last circuit MOSFET IRF5210PBF for make a regulated power supply 1,2-30 V and 0-30 A DC? Or it is better to use a pair of BJT transistors in parallel (for example MJ15004). And I have o lot of LM338, it is better to use LM338 instead of LM317? (I don´t know – for stability?).

    Reply
    • Swagatam says

      Hello Ervin, you can try it. R1 may need to be modified a bit, may be increased to 100 ohms. First try with a BJT, once the working is confirmed, you an replace the BJT with a MOSFET and modify R1 accordingly

      Reply
  4. Marat says

    Hi,
    Do I need to install LM317 on a heatsink in this current boosting circuit? I am assuming that all the current will go through the MOSFET therefore LM317 won’t heat up at all or not that much. Am I right or wrong? If heatsink is still needed for LM317, how big should it be?

    Reply
    • Swagatam says

      Hi, it will depend on the Rx value, if it is large enough to allow only a few 100mA through the LM317, then heatsink will not be required

      Reply
  5. PaulK says

    Hi, in reference to schema: “Adjustable Voltage/Current LM317 High Current Regulator”:
    Do you consider it as a laboratory power supply? I mean, could you tell me how accurate it can be setting for example voltage/current off load and testing it on load? I consider powering it with transformator 230V->35V with bridge rectifier and capacitors but I am looking for information if it could be better than my old one https://www.homemade-circuits.com/wp-content/uploads/2020/11/bench-power-supply-compressed.jpg . The old one has low current capabilities (max 1A – but i have never got so much from this) and it is vulnerable to noises.
    By the way, good job. I really appreciate your work. I couldn’t find anything what would be so useful and versatile as your article.

    Reply
    • Swagatam says

      Hi, yes definitely you can use it as a lab power supply

      Reply
  6. Aloke Ghosh says

    Hi, In your post a circuit for “High current adjustable regulator circuit ” is published using LM317, TIP73, 2N2905, in the diagram other passive component values are clearly marked except two resistors one variable other fixed. Please do specify .
    Also for my use I need to make an Bench Power supply of 0-30v ,0-10A regulated ; Please suggest suitably .

    Reply
    • Swagatam says

      Hi, you can use 2.2k for the right side resistor, and 10k preset or pot for the variable resistor.

      for the work bench power supply you can try the following concept:

      https://www.homemade-circuits.com/lm317-variable-switch-mode-power-supply/

      For 10 amps you will have to upgrade the outboard transistor with a TIP36 transistor, and also use an appropriately calculated thicker wire for the shown inductor.

      Reply
  7. suat kaleli says

    Hi Mr Swagatam;

    my input voltage is about either 12 volt dc or 14 volt. I need to boost those to 20-25 or 30 volts dc and max. 5 ampheres. Please advise an circuit I should use. Thanks

    Reply
    • Swagatam says

      Hi Suat, if voltage increases then the current will go down proportionately.

      to get 5 amps at 25V, the input 12V should be rated at 10 amps.

      Reply
      • suat kaleli says

        Hi Swagatam;
        My input voltage is 14 v and amphere is about 20 amphere(from the source amt-s240w-12 model) no problem. I think that is enough to support 5 ampheres at output-Thanks again

        Reply
        • Swagatam says

          Suat, you can try the first circuit from this article:

          https://www.homemade-circuits.com/12v-car-laptop-charger-circuit-using/

          Reply
  8. Dare says

    Swagatam, thanks for this great article
    I am struggling to understand Texas Instrument’s original high-current implementation. It works, but there is no current limit in there, am I right? I tried to simulate their circuit but I can’t get wrap my mind around what they are trying to achieve with this. Your implementations seem very logical and clear – unlike that.

    Reply
    • Swagatam says

      Thank you Dare, yes, the second last circuit doesn’t seem to have a current control protection. But the last circuit has it.

      Reply
  9. Prithwiraj Bose says

    Swag, sorry for my lack of understanding. In your 547 based overcurrent and short circuit protection circuit, I see that the transistor emitter is connected to 0v. But the circuit also uses ground. What is the difference. I mean, if I use a Center Tapped 12-0-12 trafo and a bridge rectifier as my input DC Stage, which would give an input voltage of roughly 15 volts or so, where would I get the 0v transformer and where would the ground terminal be… Can you please help…

    Reply
    • Swagatam says

      Prithwiraj, the ground symbol only denotes the common DC negative line and has nothing to do with any external grounding or earthing. You can use your transformer as 0-12V supply, or as 0-24V supply by appropriately configuring the wires with the bridge rectifier, and use the output from the bridge to power the indicated circuit designs.

      Reply
    • NerdyNerd says

      Prithwiraj Bose, I think he’s adding a low-side shunt between transformer’s 0V and the GND of the rest of the circuit (Power GND). You do not connect 0V from the transformer directly to the ground, but you do it via the shunt, which develops voltage drop as the current through the connected circuit increases. In short – bad idea, if you ask me.

      Reply
      • Swagatam says

        For a center tap full wave rectifier power supply, the center 0V wire can be connected to the point shown as 0V, and the common cathode of the diodes to the point marked +.
        However, using a bridge rectifier is the recommended design, since transformer heat dissipation is minimum in a 4 diode bridge topology.

        Reply
  10. Sreenivasulu Dammu says

    swagatam sir do u have any igbt or mosfet 12v dc to 40vdc 20amp ckt diagrams with help of atx transformer. note my requirment voltage 12v dc to 40v dc 20amp maximum.

    Reply
    • Swagatam says

      hello Shree, you can try the first diagram from this article, but please note that you will have to optimize the coil correctly through trial and error, until you get the specified 48v from…remember, to get 40V 20 amp, your input must be 12 V 66 amp

      https://www.homemade-circuits.com/12v-car-laptop-charger-circuit-using/

      you will have to replace the 24V zener with a 40 V zener diode

      Reply
  11. Sreenivasulu Dammu says

    swagatam sir i am requesting u in want dc to dc converter with atx transformer mosfets or igbts ckt diagram for 48vdc e bike. input 12v dc 26amp . and output 48v 26amp required .

    Reply
    • Swagatam says

      Hello Sree, sorry I do not have this boost converter design at this moment with me. However if the voltage is increased the current will go down proportionately, in your case the 48V current will drop to 6.5 amps

      Reply
  12. Garry Kraemer says

    Swagatam, I’d like to use the ” BJT for current control” to power a glow plug. Typical glow plugs are 1.5VDC and it takes about 3-4 amps to make them glow. I have a PIC16F628 that generates a PWM signal on an output pin. The max current the PIC can source is 20mA. Will this circuit work with the output of the PIC being at 3.3 or 5V? I want to use the PIC to sense 1000 to 2000 mS pulses from the receiver and turn on or off the PWM signal to the BJT current control input signal.
    Thanks
    Garry

    Reply
    • Swagatam says

      Gary, for a BJT current controller driver, you can try the circuit which is explained in the following article. The LM317 design cannot be used for your application
      https://www.homemade-circuits.com/universal-high-watt-led-current-limiter/

      Reply
  13. david says

    hi sir,
    regarding the last circuit using bjt, cant it drive at least 12 amperes?

    Reply
    • Swagatam says

      Hi David, for 12 amps you can replace TIP73 with two TIP142 in parallel and check the response….

      Reply
  14. reem says

    hi, I’m a student
    I need help to solve this problem
    Q1: design 12v current boosted 3 terminal regulator consumes 8mA with current limitation at 3 A.
    Q2: design positive adjustable regulator using LM317 500mA to supply load that varies in range (48-96V)

    Reply
    • Swagatam says

      Hi, THe current limit can be achieved by implementing the second last concept, but I am not sure how to get 96V using a LM317

      Reply
      • reem says

        ok.thank you so much

        Reply
  15. Aravindhan Nagaraj says

    hi brother,
    i have doubt how negative volt will produce across to bias pnp transitor…..please explain…….

    Reply
    • Swagatam says

      Hi, the (-) voltage is developed across Rx, due to the load connected at the output.

      Reply
  16. Adeyemi says

    Good day sir, Swag, please i need a fixed voltage regulator to handle the ffg specs:
    Input voltage-36v, output 5v
    Output current 10amps.

    I was thinking of using the 317 outboard circuit but it may vary the output voltage depending on input voltage. It will not produce fixed output voltage

    Thanks

    Reply
    • Swagatam says

      Adeyemi, LM317 will not vary the output even if the input is changed, but it must not exceed 36V. However at 5V the IC may get very hot…

      Reply
    • Adeyemi says

      Thanks Swag for your help, please the charging for phone is slow, can I also add the npn transistor in parallel.

      Reply
      • Swagatam says

        Hi Adeyemi, please try the last circuit and see whether it speeds up the charging or not.

        Reply
      • Adeyemi says

        That’s what I used, is getting hot on heatsink can I add more npn transistors

        Reply
        • Swagatam says

          Yes you can more NPN in parallel or you can use a Darlington instead

          Reply
          • Adeyemi says

            Please how do I do darlington with this circuit

            Reply
            • Swagatam says

              You can use a ready made Darlington TIP122 on heatsink, it will allow 3 to 4 amp current max.

            • Adeyemi says

              Thanks for your help, please in the last circuit where will I place tip122 and which one should remove.

            • Swagatam says

              TIP122 is NPN so it will be replace TIP73 in the last circuit

  17. Esaie Sichela Ngounpemeu says

    Hi Swagatam i’m right now doing a schema for a 900 mA current limiter whith a LM 317 but it’s not working. so can you please help with a schema for a 18V voltage 900 mA curent limiter? thanks

    Reply
    • Swagatam says

      Esaie, please try the following circuit:

      LM317 voltage regulator power supply with current control

      Use Rc = 0.7 / 0.9 = 0.77 Ohms 1 watt

      R1 = 120 ohms,
      Pot = 10K pot, linear

      Reply
  18. Everett says

    Hi Swagatam:
    I was able to follow everything you did up until the circuit shown in 8.3.12. I understand what you di with resistor Rx in the earlier circuits but now there’s an additional resistor on the base of the 2N2905 and a 500 ohm resistor between the base and emitter of the TIP73. COuld you take a minute to explain their function and how their values are determined?
    Thanks for a super site!
    EBS

    Reply
    • Swagatam says

      Hi Everett, the diagram is per the datasheet of the IC. The concept is similar to the one which is explained at the top, except the the extra NPN which reinforces the PNP to generate more current and an evenly distributed heat dissipation. They both complement each other to produce more current and less heat, that’s all, no other special feature is available from the last configuration

      Reply
  19. issam says

    Hi swagatam first congratulation for your hard work .
    i’m a hobbyist and i want to ask stupid question 🙂 with your design can i upgrade currant i mean from transformer 4 ampere can i have approximately 20 ampere ?
    If not you have project or solution for simple project to have big currant low voltage ? 20amp transformer is so big not practical for the project i want to make thank good jobs br!

    Reply
    • Swagatam says

      Thank you Issam, voltage and current can be changed using a transformer but the wattage will always remain the same. For example if you have a 4 amp 25 V supply, it can be changed into 5 V 20 amp. If you multiply the two V x I you get 100 watts on both sides, so it works in this way.

      Reply
  20. Shaker says

    Hi Swag
    Can I replace N-channel instead P-Ch,, and how ? Please

    Reply
    • Swag says

      Hi Shaker, N channel won’t work, i am sorry.

      Reply
    • Sergey says

      Can I put several Mosfet irf9540 transistors in parallel?

      Reply
      • Swagatam says

        Yes that’s possible…

        Reply
        • Sergey says

          Changes in the scheme are needed?

          Reply
          • Swagatam says

            No changes are required for parallel mosfets, but please confirm with a single mosfet first, and only then proceed with more numbers.

            Reply
  21. Solomon says

    Hello sir, i two questions to ask pls:
    1. can i use TIP127? If yes how can i connect the pins to the circiut?
    2. While the transistor increases the current, is the output current of the circuit equal the transistor’s current or it will be plus the ICs current? Thanks

    Reply
    • Swag says

      1) You can use TIP127, pin connections will be similar to what is shown in the diagram.take the help of the datasheet of the transistor
      2) The output current will be equal to the sum of both IC and the transistor

      Reply
  22. jan says

    Hi Swagatam,
    I would like to confirm with you the calculation of Rx & Ry.
    I require 3A max ( even though the TIP34C are rated at 10A) from the power supply. I use a BJT (TIP34C) instead of a Mosfet in my circuit.
    Therefore to calculate Rx & Ry I used the following calculations:
    Rx = TIC34C Trig. V/LM317V Optimal Current
    Rx = 0.7V/1A=0.7R
    W = 0.7V x 1A=0.7W
    Ry = Trig. V BC547+LED Forw. V/Current Limit TIP34C
    Ry = 0.6V + 3.6V/3A = 1.40R
    W = 4.2Vx3A = 12.6W
    Are my calculations correct to achieve 3A output.
    Your help will be much appreciated
    Kind regards
    Jan

    Reply
    • Swag says

      Hi Jan, referring to the last diagram all your calculations look OK to me, the LED is supposed to be in series with the base resistor of BC547 which is shown as 47 ohm

      Reply
    • jan says

      Hi Swagatam,

      In the following post you said the LED must be in series with the emitter of the BC547. Now you say it must be in series with the base resistor? How do I calculate Ry then?
      Regards
      Jan
      Reply
      Swagatam says
      March 19, 2017 at 7:20 am
      connecting it in series with the emitter will be a better idea and a resistor can be avoided.

      but in that case make sure the Ry formula is modified in the following way:

      Ry = 0.6 + LED fwd Voltage / current limit value

      Reply
      • Swag says

        Hi Jan,
        connect it with base, because connecting it with emitter will keep the ADJ pin always at 3.3V higher than ground, and never allow 0V at the output.

        the calculation will be the same whether the LED is in emiter or base.

        Reply
      • jan says

        Thank you for your assistance. I really enjoy your site.
        Regards
        Jan

        Reply
        • Swag says

          It’s my pleasure Jan!!

          Reply
  23. Amadeu Mendes says

    Hello, Mr. Swagatam.
    Can I change the LM317 30v IC to the 60v LM317HVT to have a higher Vout and maintain the IRF9045?
    Would something like Vout = 0 – 60v and Iout = 10 amp = +/- 600W!?

    Thank you
    Amadeu Mendes

    Reply
    • Swag says

      Hi Amadeu, yes you can implement that, you will be able to get the mentioned amount of power at the output

      Reply
    • Amadeu Mendes says

      Thank you.
      You are the best…
      Amdeu Mendes

      Reply
      • Swag says

        you are most welcome Amadeu.

        Reply
  24. Fajar says

    Hi swagatam, i wanna try this circuit, and i want to ask if is too complicated if i want to make it with adjustable current about 3amp.
    i alway like your design

    Reply
    • Swag says

      Thanks Fajar, It could be achieved by modifying the Rx stage of the last diagram as per the instructions provided in the following article:

      https://homemade-circuits.com/?s=current+resistor

      Reply
  25. Bingo says

    hi Swagatam i really like your blog and the way u answer the questions . Thank you very much i will try this circuit

    Reply
    • Swagatam says

      Thanks Bingo! Wish you all the best

      Reply
  26. Unknown says

    Hi! Thank you for this helpful article!
    I wonder how much power is dissipated by the MOSFET. I think it is U * I where U is the input voltage minus the output voltage and I the current going through the transistor. Do you confirm ?

    Reply
    • Swagatam says

      yes that's correct.

      Reply
  27. GORDON ROBERTSON says

    Thanks for providing such a clear and simple explanation of how the outboard transistor works with the lm317. I couldn't get my head round how the external transistor was able to provide the correct voltage alongside the lm317 now that I know, it seems so simple, thanks.

    Reply
    • Swagatam says

      Thank for visiting my site, I am glad the above post could help you to solve your curiosity. Please keep posting

      Reply
  28. Nikola Risteski says

    Hi Swag,
    I used this circuit, because i tried to reach 5Amps max. Ry=0.14 Ohm.
    But whenever i short circuit the output, the current goes 8-10Amps and the cut-off is not working.
    I did a modification, connecting the Pot (10K) between collector and emitter on BC547 but still the same result. As i observed, whenever i do short circuit i don't have 0.7V on the base of BC547, but 0.55V. And on the collector voltage varies 1.6-2.5 depending of the position of the potentiometer.

    Reply
    • Swagatam says

      Hi Nikola, you can try the second circuit and dimension Ry with some trial and error until the current is restricted at 6amps max.

      Yes the the third design might require some improvement, because the BC547 set-up might not be able stop the mosfet from conducting completely.

      0.55V is sometimes enough to enable a BC547 to conduct…

      Reply
  29. Swagatam says

    No the shown configuration in the last diagram is correct…the actual ground is supposed to be connected to Ry and the emitter of BC547 because the return path of the current must pass through Ry before reaching the transformer ground.

    Reply
  30. Swagatam says

    Hi, sorry I could not quite get your point, can you elaborate differently?

    Reply
  31. The Rocking Time says

    sir i am feeding the input pin of lm317 with 48V dc.. i need an output voltage of fixed 12V and 10A.. what modification i have to do ?
    Moreover i dont have 0.07 ohm resistors with me.. the only resistors i am having are 3 no.s of 0.1 ohm 10 W resistors.. so please suggest a suitable circuit

    Reply
    • Swagatam says

      RT, You can try the last circuit and calculate Ry as per the formula.

      you can use many 0.1 in parallel which may yield 0.07 value approximately.

      No other modification would be required according to me.

      Reply
  32. Em Amador says

    Hi sir,
    i just want to ask ,
    if what is the possible connection.
    to achieve 13.4V with 800mA current limit,
    thanks

    Reply
    • Swagatam says

      Hi Em,

      It is possible, but the following design will be more suitable:

      https://homemade-circuits.com/2012/02/how-to-make-current-controlled-12-volt.html

      Reply
  33. Kanta says

    Hi Swa,
    how are you ?
    If i want to put an LED as short circuit indicator in series with the BC 547 collector, do we need to put a resistor in series to the LED? If so, how big is the value of the resistor?
    Thanks
    Kanta

    Reply
    • Swagatam says

      Hi Kanta,

      No it won't be required according to me.

      Reply
  34. Em Amador says

    Hi sir, can i used a LM2940CT 12 Volt 1 Amp Low Dropout Regulator instead of LM317? thanks sir

    Reply
    • Swagatam says

      Hi Em, yes it will do!

      Reply
  35. Paul Jones says

    could you use a poly fuse for short circuit protection

    Reply
    • Swagatam says

      it's not required

      Reply
  36. Mrunal Ahirrao says

    The last circuit design link seems to be broken, so I cannot see the circuit please upload a new one..And I think heatsink would be required to LM317 as well.

    Reply
    • Swagatam says

      It is perfectly opening and visible to me here, just click on the diagram to enlarge it.

      Reply
  37. shubham kumar says

    sir what is the maximum current which can flow through this circuit.

    Reply
    • Swagatam says

      It'll be as per your desired specs and requirement…

      Reply
  38. victory says

    Is this circuit tested and confirmed

    Reply
    • Swagatam says

      yes…

      Reply
  39. Anil Kumar. K says

    Hi Swagatham
    An LED or a Piezo buzer with internel ocilator can be added to indicate the short circuit condition. Where can I connect it in the last diagram.
    Can I use an N-channel mosfet instead of P-channel after some modifications in the circuit….? If yes, how.

    Reply
    • Swagatam says

      Hi Anil, using an NPN transistor will make the design very complex, so PNP is the only easier option…connecting a buzzer will be difficult because there's no appropriate space for it in the above design, however you can use a red LED in series with the BC547 collector for a short circuit indication.

      Reply
    • Lenar OX says

      resistor for the led on the collector of bc547 can alter the design? what is the voltage on the collector of 547 transistor and how can I calculate the resistor for the led?

      Reply
    • Swagatam says

      connecting it in series with the emitter will be a better idea and a resistor can be avoided.

      but in that case make sure the Ry formula is modified in the following way:

      Ry = 0.6 + LED fwd Voltage / current limit value

      Reply
  40. Anil Kumar. K says

    Hi Swagatam

    Many thanks for this circuit. This circuit will be very useful for every electronic technicians, especially for beginers.

    Reply
    • Swagatam says

      Hi Anil, It's my pleasure and thank you for suggesting me to post this important circuit concept.

      Reply
  41. andy parry says

    Thanks for this. Note thought that near the begging you write "Rx = 10/1mA" but your description says 1A, not 1mA.

    Reply
    • Swagatam says

      OK thanks, that looks like a typo, I'll correct it soon…

      Reply


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