Using a Zener for reference voltage

voltage-referencezener

I want to use a 9.1V Zener diode for a ref voltage into a op amp comparator so that when the battery voltage (11-12V DC) drops below 9.5V DC it will cut the supply to the circuit via a P-channel MOSFET.

The supply voltage will go through a voltage divider dropping it by 10% so when it hits the magic level it will still be above the Zener voltage and cause the comparator to drive the gate of the MOSFET low and turn it off. Thus saving my batteries from harmful deep discharge.

Will this work or is there something I'm going to get bitten with?

schematic

simulate this circuit – Schematic created using CircuitLab

Best Answer

Dzarda's comment is valid. You have about 29uA's of current flowing through there at 12V supply and even less as your battery voltage drops. The Zener voltage will depend on the current flow through it. The lower of a resistance you have supplying current, the easier it will be to keep the Zener voltage relatively constant/stable.

You'll definitely want to look into Dirceu's comment and make sure the output of the comparator is rail to rail and isn't open-collector or open drain. If it is, adding a resistor as he recommended will work.

I think the main reason this will likely work poorly is because there's no hysteresis in it. Depending on the current draw of the circuit, you'll end up with some voltage drop due to the internal battery resistance. As soon as this limiting circuit cuts off the current, the terminal battery voltage will jump back up because no current is flowing. That wall cause the circuit to turn back on, which in turn will cause this limiting circuit to shut it off, endlessly turning on and off the circuit. This may work for your purpose, but the main circuit may get quite confused and you'll generate a lot of noise. If you have a comparator with hysteresis built into it, or if you add hysteresis to this circuit, that will be avoided.

Your current resistors show that you'll cutoff at about 10.11 volts: 9.1*1000k/900k = 10.11 V.

Lastly, although Ignacio is correct that Zener's aren't great voltage references, you don't need anything terribly accurate for your application.