simulate this circuit – Schematic created using CircuitLab

I need a suggestion to build a minimalist short circuit alarm. With this diagram will light a LED easily. But if I replace with a Buzzer, it will not make a beep. My input voltage will be 12 Volts, alarm should be raised if current draws about 2 Ampere.

One transistor short circuit monitor

The R1 can not go further more Ohms, since I don't want too much current lost on the output. I tried to Darlington Q1, but the buzzer still won't beep.

Edit1: add the correct schematic. Sorry for the messed, this is my first thread on EE Community.

My main goal: I want to have a simple high current alarm by using transistor(s) and to replace LED with buzzer. When high current occurs (>= 1A) it will have constant beep.

  • \$\begingroup\$ The BJT needs current on its collector (the middle terminal on the schematic) how is it getting current? It may be useful to use the circuit editor and redraw this circuit with voltage sources so its clear how you are powering it. Also, this is a Q&A site, could you edit your post and include a question? Thanks \$\endgroup\$ – Voltage Spike Aug 18 '16 at 16:19
  • \$\begingroup\$ @Bianca: If the voltage source is connected at IN and the load is connected at OUT then the orientation of the transistor is wrong. Either reverse IN and OUT side or Base and Emitter of Q1. \$\endgroup\$ – Curd Aug 19 '16 at 10:24
  • \$\begingroup\$ You seem to have swapped the input and output in your new schematic. \$\endgroup\$ – user253751 Nov 1 '18 at 21:54

Actually, I'm surprised that it will light the LED.

You are shorting the power supply to the LED/buzzer when you short V_Out to GND_Out.

This is what you are doing: enter image description here

There's a sort of glaring problem there. V_out shorted to GND_OUT leaves the buzzer with zero volts to work with.

You need a second power supply for the buzzer/LED. @jbord39 made suggestion that might help if all you need is a short "bzzt" when the short occurs.

Try it like this:


simulate this circuit – Schematic created using CircuitLab

The capacitor provides a little power to the buzzer when Q1 connects the buzzer to ground. The diode prevents the capacitor from being discharged by the short from V_Out to GND_Out.

You only get a short buzz, but better than nothing.

Whether you can get a longer buzz or not depends on the power supply.

I've added a simulated battery, and you can simulate the circuit to find out the voltage to the buzzer.

Changing R3 will change how much current the battery can supply.

Changing R2 changes the severity of the short.

If R2 is a dead short (0 Ohm) then all you will ever get is a short buzz.

In short, the wimpier your battery the shorter the buzz. If the battery is really beefy, it can supply the short circuit and the buzzer.

To get a continuous buzz regardless of the severity of the short, you will have to power the buzzer from a separate power supply.

You do realize that it will take a pretty hefty current flow to make this thing trigger, right?

It takes over 1 Ampere to make enough of a voltage difference across the resistor to reach the 0.7V needed for the transistor to turn on.

  • \$\begingroup\$ If the short was momentary enough, you may be able to get away with a crude diode + capacitor to power the buzzer (momentarily) when the short occurs. Also, if the short is quite far downstream the voltage at the detector may still be high enough to power the buzzer. \$\endgroup\$ – jbord39 Aug 18 '16 at 15:45
  • \$\begingroup\$ JRE, OMG. why i didn't noticed this. Seems i need to work out my logic -_- @jbord39 i got your idea. \$\endgroup\$ – Bianca Aug 18 '16 at 15:56
  • \$\begingroup\$ One transistor design is work as expected. A very short beep when short circuit occurs. Actually what i need is kind of simple alarm if current consumption draws above 1 Ampere on 12 Volts. Can I do this with 3 transistors max ? I have done with op-amp voltage comparator, but the end product will be costly. \$\endgroup\$ – Bianca Aug 19 '16 at 8:01

JRE is correct so to get a second supply (albeit temporary) add a diode and capacitor.

enter image description here

  • \$\begingroup\$ I will try to go with this. So it will stop beeping until C1 cleared, right? Do you have another idea if i want B1 continuously beep if high current occurs ? 2-3 transistors no problem. \$\endgroup\$ – Bianca Aug 18 '16 at 16:01
  • \$\begingroup\$ @Bianca As capacitors run out of oomph very quickly you might want to consider some form of rechargeable battery being trickle charged from the supply. Remember to leave the diode in circuit. \$\endgroup\$ – JIm Dearden Aug 18 '16 at 16:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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