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I am using a 15 A circuit breaker. To detect the trip, I have designed a P-MOSFET based circuit.

enter image description here

The idea is to connect the load signal of circuit breaker to the gate of MOSFET, which is pulled down to ground. When the ciruit function is normal the Vgs will be close to zero. In the event the breaker switch trips, the Vgs will be -12 V and will alert the user. This is the first time I am using a P-MOSFET. On paper I believe this circuit should work. Are there any downsides to this application?

Thanks Deepak

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    \$\begingroup\$ I would place a zener diode between GS to protect gate. Breaking can cause a lot of spikes. \$\endgroup\$ Commented Feb 19 at 12:56
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    \$\begingroup\$ Depending on what your load is, it could perhaps even be simplified to just LED+resistor straight across your breaker. It will however bleed current into your load after trip which could be no issue or detrimental depending on the application. \$\endgroup\$
    – winny
    Commented Feb 19 at 17:27
  • \$\begingroup\$ @winny, The circuit breaker is used in a motor driver circuit to protect the source from overcurrent. I did not realize it was that simple. I dont think few mA of current will be that detrimental. Will this work if I were to add a buzzer as well? \$\endgroup\$ Commented Feb 20 at 5:44
  • \$\begingroup\$ @Michal, Thanks for zener diode suggestion. I will definitely add it. \$\endgroup\$ Commented Feb 20 at 5:45

1 Answer 1

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You don’t need a PMOS for that: “In the event that the breaker switch trips” you can just use the resistor and LED as shown in the simulation below.

enter image description here

However, if you still want to use the PMOS, it will be important how the Load and the Circuit Breaker are connected.

If the circuit breaker sits between the load and the ground, it will work in reverse as the simulation below shows. The LED will stay ON and it will turn OFF when the circuit trips. In this case you will need to add the capacitor C1.

enter image description here

I will assume that you want the Circuit Breaker to sit between the Power supply and the load as it is normally done. The simulation below shows the result and you still need the capacitor C1.

enter image description here

Thanks.

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  • \$\begingroup\$ Thanks for elaborate analysis. I did not realize adding an led across the breaker would do the job. However, I am also considering to add a buzzer. Will it still work? \$\endgroup\$ Commented Feb 20 at 5:48
  • \$\begingroup\$ @DeepakReddy Same as the LED, just more current which bypasses the breaker which your load must accept. If it's a hard short, no issues. Could anyone flip the breaker in order to turn off the load despite no short? \$\endgroup\$
    – winny
    Commented Feb 20 at 10:11
  • \$\begingroup\$ The breaker is a push button reset type that is normally closed. So no one can flip it. So I tested it yesterday and worked. Thank You all for taking time to help me out \$\endgroup\$ Commented Feb 21 at 6:21

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