PC Power 12 V Current Limiter




About: I'm Chandra Sekhar, and I live in India. I am interested in electronics, and building small one-off circuits around tiny chips (the electronic kind).

To safely power something from your PC, the attached loads have to be limited to a safe current in order to avoid the chance that a short circuit might cause inadvertent shutdown and damage to your data. This circuit can be used to safely bring out the 12V line from your computer to supply a limited amount of current to power external gadgets.

With normal loads, the power transistor is in saturation and so the regulated +12V line is directly available (minus only the Vcesat of the PNP transistor). If the output is accidentally shorted, the npn transistor turns off and removes the base drive of this transistor.

This circuit is intended for protection from short circuits. It might be possible to apply a load that brings the series transistor out of saturation without removing base drive to it: such a situation will cause heating and might cause it to fail short-circuit. Those who wish to draw significant current from this circuit will be well advised to check that the transistor is indeed in saturation with their intended load.

That said, this is a simple way to draw a few tens of milliamps at 12 V from your PC without danger to either load or the PC.

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Step 1: The Circuit

Any electronic project starts with a circuit. This one has a PNP power transistor in series with the +12V line from the PC. The collector supplies the external circuitry.

A small signal NPN transistor is fed with the output voltage through two resistors connected as a potential divider. Should the output voltage become less than a certain value, this transistor will turn off.

The collector of this transistor supplies the base current of the power PNP transistor. So the two transistors will always switch on or off together.

The power transistor (BD140) is supplied with a certain current via the green LED and 470 ohm resistor. It has a certain gain, around 50 or so, and its collector can supply only the gain times base current before it drops voltage (comes out of saturation).

So the maximum current draw can be changed by appropriately dimensioning the base resistor of the PNP power transistor.

The voltage divider at the output determines the voltage at which the current supplied by the transistor drops to zero.

The 470 ohm resistor in parallel with the PNP transistor supplies a certain amount of current so that the circuit will switch on when first powered on.

Step 2: Costruction: Board and Connector

A power connector was salvaged from a 3.5" floppy drive. A scrap of copper clad board was dug up from somewhere.

A piece of board slightly wider than the connector was cut to size, edges smoothed and holes drilled to accept the connector.

Step 3: Soldering

The pictures show the connector before and after soldering. The middle two pins are ground. One outer pin is +5 V, which we do not use. The other outer pin is the +12 V we want to use.

The two pins carrying voltage will be isolated by cutting the copper after soldering.

I find that doing things this way allows freehand cutting, because the dimensions are evident after the component is on the board.

Step 4: Mark the Diagram

The diagram has been marked with the components already fitted. This is a great aid in constructing a circuit. You go on soldering and marking, and when you run out of things to mark on the circuit diagram your circuit is probably complete and ready for the first cycle of crashing and burning.

Step 5: Solder Components

The copper on the board has been cut to isolate the live pins and bypass capacitors soldered on. We are not using the +5V line at present, but I fitted a capacitor there anyway.

And the circuit diagram has been marked up to reflect the components on the board.

Step 6: Solder Some More Components

The components around the small transistor has been fitted. The one pictured is not the BC147 in the diagram, it was rescued from some board that had reached the end of its active service and had been thrown away. Like the rest of the components.

In order to know where to cut, the components are first placed on the board, their outline traced, and lines drawn around the perimeter of the desired conductor pattern.

Two closely spaced parallel lines are scored on the copper with a sharp knife, and the thin strip of copper in between peeled away resulting in an insulating channel.

No chemicals are used at all.

Then the components are soldered in, and again the board marked and cut.

I term this "cut and go" because the knife and soldering iron work alternately on the same board.

The circuit diagram has been updated to reflect the components placed on the board. I could get only a 2K0 chip resistor. A calculation resulted in the revised value for the top of the voltage divider - the 10K was changed to 36K, so that the transistor will stop conducting if the voltage drops slightly from the 12V nominal output.

Step 7: Update the Diagram

Mark the diagram in order to match the progress of your work. Make this a habit, as it is easy to overlook a vital component in the heat of the soldering and then spend ages searching for the cause of the resulting malfunction.

I decided to add one more LED to monitor the operation of the circuit. A green one to light when all was hunky-dory and a red one to light when bloopers!!!

A resistor value was also changed, and this has been marked on the circuit diagram.

Now only the power transistor is left to be fitted. Some functional checks were made to ensure that it will all work and not go up in smoke.

Step 8: Test It

All components are now on board and it was tested. Another component change was necessary before it would work. That resistor at the output, originally 10K, changed to 36K, had to be changed to 22K.

With the original 10K, the onset of current limiting was too high - I could draw more than 600 mA from it.

With 36K, it would not turn on when connected to the intended load, a video camera. By trial and error, I determined that 22K would be OK.

The current limit on my prototype was around 220 mA. If the output was short circuited the current fell to about 16 mA. The circuit was judged have fulfilled its design target and will shortly be fitted to my PC. I intend to power a video camera with it. The camera draws around 100 mA. It is at the end of a long cable with many joints (split from flat cable) which occasionally suffers from short circuits.

Step 9: The Annotated Diagram and Pictures

I present the circuit diagram and photos labelled for cross reference.

I have made a circuit board layout in case somebody out there wishes to etch a board in the traditional way. The layout is attached as a pdf file, and the overlay as a gif image.

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    24 Discussions


    4 years ago on Step 9


    What changes must be required if we need to limit the current to about 1.5-1.8 amperes?


    8 years ago on Introduction

    Thanks! I needed this circuit. I put a 10k potentiometer in place of R5 which I think is useful.

    I run this circuit with the 5V off my USB port. I didn't have to make any modifications except to make sure the transistors remain biased (changing value of the R5 pot).


    Reply 11 years ago on Introduction

    It has a sensing resistor in series with the output. It can drop up to Vbe of the sensing transistor - around 600 mV - before current limiting takes place. It passes the full limiting current into a short circuit. This type of current limiting will be appropriate for an audio amplifier, for example. My circuit is intended to be a sort of electronic fuse which cuts off current when a short circuit is applied, and recover automatically when the short is removed.


    Reply 10 years ago on Introduction

    A length of thin wire, or a small resistor (2.2 ohms, 1/4 Watt) will work as a self activating fuse.

    This circuit works as a self-resetting fuse.


    Reply 10 years ago on Introduction

    ah...yes self reseting fuse pretty neat...i might make this for some of my projects


    11 years ago on Introduction

    I like the idea of a trimpot (on R1 ?) to select the max current but I have trouble to find which resistor values are needed. I understood that the maximum current depend of " the gain times base current before it drops voltage ". But it is not clear how to calculate it. Any clue ?
    Can I find it empirically from the voltage at R1 ends ? (But then a gain of 50 in Q1 and the value you give will mean: Vr1= 470*.22/50 = 2V but that look a bit too few)
    By the way what is the role of R5/(R4+R5) ? It look important to make the whole thing work but you did not say why :"That resistor at the output, originally 10K, changed to 36K, had to be changed to 22K." (in 8)

    1 reply

    Reply 11 years ago on Introduction

    10K - onset of current limiting too high 36K - circuit would not turn on with load 22K - current limiting around twice load current. I have modified step 8 accordingly.


    12 years ago on Step 1

    From my experience the PC supply is very noisy and often considerably over voltage. The 5V regulator would give a clean supply if powered from PC 12V. But it would not keep the PC from resetting if you shorted the output.

    The regulator does have overtemp and short circuit protection but the input current would still spike.

    I describe a different current limiting circuit on my site in case you are interested.

    1 reply

    Reply 11 years ago on Introduction

    The computer power supply pre-regulates the power. If you want 5 volts, pull off of a red wire instead of a yellow wire. Red to ground is 5 volts and yellow to ground is 12. The orange wire if available is even regulated at 3.3 for you. The point of his design is not as a voltage regulator but as a very low current overload protector. Fuses are normally rated higher than this and with no way to adjust and have to be replaced. Another advantage to this is that one could change the resistor with a trimpot to make a variable trip overload protector.


    12 years ago on Step 1

    I have to admit to being more than a little puzzled here. Couldn't we drop about four components by using a 7805 regulator, with a series resistor on it's output? That should limit the current and give us a +5v supply. Probably wouldn't even need the bypass caps, I expect the output of a PC power supply is pretty stable.

    2 replies

    Reply 12 years ago on Step 1

    1. I need a 12 V supply. 2. Series resistors impose voltage drops. 3. It has to be protected from short circuits.


    12 years ago

    So this circuit prevents a short from causing a shut down and loss of data in your PC. But if I want more current I can just plug directly into the floppy power connector, right? Some power supplies offer 5v @ 30A for example. So if I want to draw 10 Amps or so, I just have to make sure I never short it otherwise my PC is damaged. Am i getting this right?

    1 reply

    Reply 12 years ago

    Right. But I think if you need 10A on a regular basis you'll be better off investing in a separate power supply. A 12V, 7AH lead acid battery kept charged by a small (500 ma) wall wart is a good option, too, if you need it only intermittently.