I just started learning about transistors, so I was wondering if I could build a circuit where short circuiting across the load caused a transistor to cut off power (like a fuse).

My idea was that when the resistance across the load diminished (short circuit), the current flowing through the reference resistor would diminish thus turning off the transistor controlling the circuit.


Forgive me if this question seems to simple, but I searched the forum and can't find anything. I tested this circuit and it does seem to limit the current so I'm wondering why not use this instead of a fuse that gets destroyed by the current?

  • 3
    \$\begingroup\$ You can build a short circuit detection circuit that will use transistors to automatically shut off the power, but it is not done as in your schematic. It will be more complicated than a fuse, though, which means it will have more modes of failure. Fuses are usually a last-resort way to stop things from catching on fire if something is going very wrong with your circuit. \$\endgroup\$
    – Justin
    Commented Oct 9, 2019 at 18:27
  • 2
    \$\begingroup\$ If your device is powered from the mains (plugged into the wall outlet) then it can draw hundreds of amps for a very short time. Mains-powered devices therefore always have a fuse. \$\endgroup\$
    – rdtsc
    Commented Oct 9, 2019 at 18:32
  • 2
    \$\begingroup\$ electronicdesign.com/power/… \$\endgroup\$ Commented Oct 9, 2019 at 18:33
  • 1
    \$\begingroup\$ Please use the built-in circuit editor, I can't see what's on that website. \$\endgroup\$
    – pipe
    Commented Oct 9, 2019 at 18:33
  • 1
    \$\begingroup\$ @Benjamin You can use a transistor (or more commonly, an SCR/thyristor) in what's known as a "crowbar" configuration to protect circuitry. High-current SCRs are used to de-excite generator field windings or shunt away fault current. \$\endgroup\$
    – Bort
    Commented Oct 9, 2019 at 18:40

1 Answer 1


Of course you can use a transistor as a fuse. I once built such a circuit for a low power AC/DC supply (I can not show you the full circuit, but you should get the idea):

enter image description here

Here's how it works: The sense resistor (R16) turns on a transistor (Q6) when the supply return current through R16 is > 0.5A. Then the shutdown signal (SHDN*) is set low and can be used to switch off the load with another transistor (not shown here).

Ok, I admit that this circuit needs at least 2 transistors to replace the fuse. So this circuit shown here is only an overcurrent detector.

The time constants (defined by R15/C19 & R9/C18 in the circuit) will define how long the circuit is switched off before it turns on again.

  • 4
    \$\begingroup\$ @CalebReister No, a crowbar circuit prevents from overvoltage, while this one cuts off the load when there is too much current (=fuse). \$\endgroup\$ Commented Oct 9, 2019 at 18:56
  • 1
    \$\begingroup\$ @StefanWyss In order to behave like a fuse it would need to be configured as a switch in series with the load. As it is now, it looks more like some kind of shunt regulator. \$\endgroup\$ Commented Oct 9, 2019 at 19:01
  • 3
    \$\begingroup\$ As I mentioned in my answer, this is not the complete circuit. The part of the circuit shown here only generates the SHDN* signal which can be used to cut off the current with an additional transistor. \$\endgroup\$ Commented Oct 9, 2019 at 19:01
  • 1
    \$\begingroup\$ @StefanWyss I see. Maybe you should show more of the circuit. \$\endgroup\$ Commented Oct 9, 2019 at 19:02
  • 1
    \$\begingroup\$ @CalebReister I'm sorry. I can not show more of the circuit. But I can tell you: There is a switching DC/DC regulator connected to the right that is shutdown with the SHDN* signal in case of an overcurrent situation. \$\endgroup\$ Commented Oct 9, 2019 at 19:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.