I would like the following:

  • 10 X 1.2V battery
  • left mosfet is the ON-OFF mosfet. I can ON and OFF the whole circuit with the 'SW' switch
  • The two LM358 opamps sense the 1st battery voltage
  • IF the voltage drops below 1.11V the right opamp will go high, and the RED LED go light below 1.11V. This is low battery warning led.
  • IF the voltage drops below 1.0V the left opamp turns off the right mosfet, so no more current consuption from the load (in the simulation this is 10R resistor)
  • IF battery voltage drops below 1V only the LT1236 and the two opamp still alive, which only draws 1-2mA from the battery.

THE PROBLEM: - when the SWITCH is OFF, from V+ (12V) to 1st BAT+ ~30-40mA current consuption. Why? What is the problem? In the simulation only ~100uA current consuption when the SWITCH is off.


The mosfets are close correctly, its not a problem. The problem when the switch is off ~30-40mA current flows from V+ (the top of the battery, 12v) to 1st BAT+. In this time to current flow from BAT (-) to left mosfet source.

  • \$\begingroup\$ Here's the datasheet for the LT1236-5. \$\endgroup\$ Commented Aug 7, 2017 at 10:57
  • \$\begingroup\$ Are you sure you got the Drain and Source right for the MOSFET's? \$\endgroup\$ Commented Aug 7, 2017 at 11:00
  • \$\begingroup\$ "from V+ (12V) to 1st BAT+ ~30-40mA current consuption": What does this mean exactly? Where is V+ in your schematic? And what does it mean there is 30-40mA "current consumption" from V+ to the 1st Battery? \$\endgroup\$
    – nickagian
    Commented Aug 7, 2017 at 11:04
  • \$\begingroup\$ Assuming your schematic is correct, try putting a 10k resistor from SW1 to ground. As shown, the gate will float high depending on things like external wiring, static pickup, etc and turn on the left-hand FET. \$\endgroup\$ Commented Aug 7, 2017 at 11:54

1 Answer 1


when the SWITCH is OFF, from V+ (12V) to 1st BAT+ ~30-40mA current consumption. Why? What is the problem? In the simulation only ~100uA current consumption when the SWITCH is off.

Simulations don't always model every characteristic of a component accurately, particularly when operated outside its normal parameters.

When the power switch is turned off in your circuit, the voltage on the 'GND' terminal of the LM358 rises towards +12V. However the inverting input of the second op amp is connected directly to +1.2V, so it will become negative relative to GND.

LM358 inputs are rated for an absolute minimum of -0.3V, but in your circuit it could be getting as much as -10.8V. The LM358 datasheet doesn't say what will happen if you try to pull an input below -0.3V - so I tested it. Current reached 40mA at -1.86V, and increased rapidly as the voltage went more negative. I suspect that in your circuit the LT1236 is dropping about 10V, which is fortunate because that is the only thing stopping the LM358 from blowing up!

To prevent this excessive input current you need to keep the op amp input voltages within their ratings. You could put a 10k resistor in series with the input to limit current, but it will still draw some current when the power is 'off'. A better solution might be to use a voltage divider off battery positive, rather than tapping into the battery at the first cell.

  • \$\begingroup\$ Thank you. Its new for me. When switch turned off why LM358 GND pin rises to +12V? \$\endgroup\$
    – Lobi
    Commented Aug 7, 2017 at 14:55
  • \$\begingroup\$ You are switching GND, and regulator IN is permanently connected to +12V, so for the regulator (and op amp) to be unpowered GND must go up to +12V. In reality the 30-40mA op amp input current creates a voltage drop across the regulator, so GND will be lower than 12V (~1.2V + 1.86V = ~3.1V?). \$\endgroup\$ Commented Aug 7, 2017 at 19:48

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