Electronic Circuit Breaker

241

03:38:43

When the current exceeds 2.2 amps the breaker will trip. When shorted out it will sustain a current of 100 amps for 200uS. This can start with a capacitive load of up to 400uF (not 470uF) and not trip thanks to the 200uS delay provided by the 2.2uF capacitor. The delay time, and to some extent the trigger current are dependent on the transistor’s gain. At a gain of 150 (vs. 100) the trip current is reduced to 2 amps. Also, at 150 gain, the trigger time when shorted is reduced by approximately a third, and therefore is the maximum capacitance of the load that can be tolerated without tripping the breaker.

published 8 years ago

thebugger

8 years ago

The problem with this kind of protection is that it uses a low level circuit breaker, which means the ground rail of the device you're trying to protect is not referenced at 0V, but is lifted by the MOSFET, not able to sink all the current without developing a voltage drop. In regular non demanding circuits, this shouldn't pose a problem, but sensitive circuits like audio amplifiers might develop a ground loop problem and start buzzing. Voltage regulators would become unstable, and thw output voltage would vary with the load. It's better to use a high side protection.

hurz

8 years ago

Amused

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