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I would like to protect the batteries from over discharge, so I want to make a protection ciruit.

I made this:

I want the following: 2x 9V battery in series, so total voltage is 18V. The regulator is 8V, the mosfet is N-channel. If the first battery voltage is greater than 8V, the op-amp output will high, so the N-channel mosfet gate voltage will be high and the mosfet will open.

If the first battery voltage drop below 8V, the op-amp lower the output, so the mosfet is off, and no more power consuption from the battery.

The problem is the following:

The mosfet doesn't turn off when the battery voltage drops below 5V. (in this example only, I didn't find 8V regulator in LTSpice)

The opamp output not 0V else its ~2v output so the mosfet doens't turn off.

Lobi
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1 Answers1

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Are you really using an RH1028 opamp? It is completely unsuitable - it doesnlt work off 5v.

In particular you need one that can bring the output down to the negative rail.

Something cheap and simple such as an LM324/LM358 would be suitable.

You also cannot connect the positive power rail to the same point as the negative input. You can power the opamp directly from the 18v rail.

The negative opamp supply needs to come directly from the ground - currently you are switching off the opamp and asking that its output pin go below its ground pin. Even when the opamp output is as low as it can go it will still be at the same potential as the drain of the FET.

Also because you have the negative side of your reference LDO connected to the FET as soon as the FET turns off the LDO and the negative input to the opamp will float towards the +18v rail - it will be difficult to turn on again even if the battery goes above 5v.

That will mean you cannot switch off the power to the opamp.

Overall this circuit will not work well and needs rethinking.

Drive the FET gate from a PNP transistor that in turn is driven from the opamp (need to correct polarity). Put a resistor from the gate to ground. When the PNP turns off so will the FET,even if there is 18v across the transistor the current can still be extremely low (sub microamp).

Kevin White
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    You need to connect the opamp negative power to ground, not the drain of the FET. You are asking the opamp output to be below its negative rail. – Kevin White Jun 15 '17 at 00:29
  • Do you need the centre tap of the battery? Or do you just need 18v? – Kevin White Jun 15 '17 at 04:32
  • Try this circuit https://electronics.stackexchange.com/questions/213304/tl431-low-battery-cut-off and https://electronics.stackexchange.com/questions/19714/adding-voltage-cutoff-to-a-circuit and don't forget to add a small hysteresis (100k resistor from mosfet drain to TL431 ref input) – G36 Jun 15 '17 at 09:35