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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.

Diagram/Schematic

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?

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    \$\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
    Oct 9, 2019 at 18:27
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    \$\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
    Oct 9, 2019 at 18:32
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    \$\begingroup\$ electronicdesign.com/power/… \$\endgroup\$
    – Tony Stewart EE75
    Oct 9, 2019 at 18:33
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    \$\begingroup\$ Please use the built-in circuit editor, I can't see what's on that website. \$\endgroup\$
    – pipe
    Oct 9, 2019 at 18:33
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    \$\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
    Oct 9, 2019 at 18:40

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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.

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    \$\begingroup\$ @CalebReister No, a crowbar circuit prevents from overvoltage, while this one cuts off the load when there is too much current (=fuse). \$\endgroup\$
    – Stefan Wyss
    Oct 9, 2019 at 18:56
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    \$\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\$
    – Caleb Reister
    Oct 9, 2019 at 19:01
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    \$\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\$
    – Stefan Wyss
    Oct 9, 2019 at 19:01
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    \$\begingroup\$ @StefanWyss I see. Maybe you should show more of the circuit. \$\endgroup\$
    – Caleb Reister
    Oct 9, 2019 at 19:02
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    \$\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\$
    – Stefan Wyss
    Oct 9, 2019 at 19:07

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