Shut off circuit when under a certain voltage

batteriesseriesswitchesvoltage

I have three batteries wired in series, in order to produce 9v. If one of those batteries dies, I need the whole circuit to shut off. The best solution I've found is to use a comparator, but the problem with that is that it requires a separate 9v source in order to have something to compare the voltage with. However, if this second 9v source dies off, the circuit will always let power through, which is not what I want.

Is there a way to shut off a circuit when it goes below a certain voltage?

Thanks!

Best Answer

Use something like an 8.2 volt zener diode to conduct current to the base of a BJT when the supply voltage is about 9 volts. Depending on your load current the BJT may be able to drive the load adequately without further transistors. Two resistors are also required; one in series with the zener to limit base current and one to properly turn off the BJT when volts drop below about 8.5 volts: -

schematic

simulate this circuit – Schematic created using CircuitLab

The combined "blocking voltage" of an 8.2 volt zener and the base emitter junction is about 8.9 volts so round about this figure the base is going to start taking a little current and switching on the load. Given that the BJT will have a gain of around 100, 1mA into the base will yield about 100mA into the load.

If this circuit doesn't quite turn on enough at 9 volts then use a lower value zener possibly padded out with a forward connected diode in series.

A 7.5 volt zener will start to feed about 1mA into the base at a battery voltage of about 8.3 volts. At 9 volts I reckon the base will be taking about 8mA and pretty much saturating the transistor for loads that consume less than a couple of hundred milli amps.

Related Topic