I have a circuit circuit fed by a 48V nominal lithium battery source. I'm using a few simple resistors to divide the voltage down to a suitable range to drive a p-channel mosfet gate. When the battery goes into BMS cutoff, I expected 0V and open circuit would be the conditions, and for that set of conditions, my circuit works perfect in bench tests.

But it turns out the BMS manufacturer delivers 6V during BMS cutoff, and they don't provide any info about the exact nature of that source. They even called the 6V "fake volts", whatever that means. So the end result is that I want to block those low voltages so my circuit behaves like an open circuit below some threshold voltage, say 39V. A zener diode seems like the answer, but they have a minimum current of about 1-3mA needed to go into breakdown, and my circuit requires low power consumption, so it normally only uses 350 uA. I really don't want to waste more power to solve this problem.

Is there any other simple way to block low voltages (< 10V), or only allow high voltages (> 39V) to pass? Key word is simple, and with 250-500 uA power consumption.

If it just won't work this way, I'll redesign the whole thing using a voltage comparator or op-amp, etc, but I'd much rather use my existing circuit. It's very stable, has very low current consumption, has a test function, an LED indicator, and allows for a separate power source to deliver the output signal of + 12V which I use to shut down alternator charging.

Any help is greatly appreciated. The only thing I need to know is how to block low voltages or create an open circuit condition below some set threshold voltage without wasting a lot of power.


Here's a schematic of just the core circuit without the LED indicator on Vout and the pushbutton tester that I use to short the output of R1 to ground. Please note Vstarter and Vhouse share a common ground.

The p-FET is rated for 30V so that it can handle a gate voltage of 16.3V to 24.5V from the divider depending on the output of a 48V nominal battery (40V to 60V). The 12V nominal source can be as high as 15V, so I wanted the gate voltage to be higher when the battery is operational. I was told by another helpful person on this forum that was important. I think he said the FET could be damaged otherwise, but maybe it was only to keep the gate switched off.

I was able to briefly test during a BMS cutoff and the 6V output will power an LED directly, so it does deliver some amount of mA. My circuit also appears to work under that condition. I get the expected 12V at Vout. But is the p-FET okay with 2.4V on the gate input and 12V (can be up to 15V) on the source? I thought @nanofarad explained that having the source voltage higher than the gate was a problem. Is that only for normal operation, but for my case a good thing since I actually want the p-FET to be conducting when the voltage is significantly lower, like the 2.4V it gets during this odd 6V BMS cutoff output?

enter image description here

  • \$\begingroup\$ Why not just get rid of the BMS cutoff when you are adding a circuit that basically does the same thing? \$\endgroup\$
    – DKNguyen
    Jun 14, 2021 at 13:53
  • \$\begingroup\$ This circuit is external to a sealed battery case with an internal BMS. \$\endgroup\$
    – VanGogh66
    Jun 14, 2021 at 14:40
  • \$\begingroup\$ "my circuit requires low power consumption," but "I'm using a few simple resistors to divide the voltage down." Down to what? Using resistors to drop the voltage is a waste of power. \$\endgroup\$ Jun 14, 2021 at 17:29
  • \$\begingroup\$ It's a necessary evil to drive a p-FET from a 48V battery. With 47K and 68K as the divider circuit, there's only 350 uA passing, which is low enough for my requirements. The 47K resistor yields 16.3 to 24.5V across it, which is always higher than the ~15V max starter battery that feeds the source of my 30V rated p-FET. It works perfectly when the battery is at it's usual 51V, and it even appears to work during the unusual 6V output state during BMS cutoff, so I may not need to modify it after all. But I'm not confident in the 6V state because it was designed for a total cutoff. \$\endgroup\$
    – VanGogh66
    Jun 14, 2021 at 18:11

1 Answer 1


it turns out the BMS ...delivers 6V during BMS cutoff... They even called the 6V "fake volts", whatever that means.

The voltage may be caused by leakage through some device, most likely a transistor. They call it "fake" because it has a very high impedance source. It may be picked up by an oscilloscope, or a high impedance voltmeter, or the gate of a mosfet. Or the voltage may be residual voltage from a charged capacitor. However, with any significant load, the voltage will drop to near zero.

One solution to such problems is to use a pull down resistor. However, as you wish to limit the lost current, both when the supply voltage is low, and when it is quite high, some sort of constant current sink might be more appropriate. However, this solution may not be satisfactory if the "fake" voltage turns out to be "not so fake".

An alternative solution involves a series switch and a Zener diode. For example:


simulate this circuit – Schematic created using CircuitLab

You objected to the use of a Zener diode stating:

A zener diode seems like the answer, but they have a minimum current of about 1-3mA needed to go into breakdown, and my circuit requires low power consumption, so it normally only uses 350 uA.

It may be the case that when the current through a Zener is small, it's behavior is not well documented. However, according to the 1N5363B datasheet, the 1N5363B has a maximum reverse leakage current of 0.5uA. I thus suspect that even if it misses it's nominal breakdown voltage, it will breakdown somewhere near where you want it. As I do not know the current requirements from your nominal 48V supply, you will need to choose the series MOSFET in this circuit, and depending upon the max Vgs of your choice, you may need to choose a different Zener diode. But since there is lots of space between your minimum 39V on voltage and 6V that appears when the supply is "off", you should be able to find values that switch off your series mosfet within that range.

Now here is a shunt circuit that may solve your problems. The circuit could definitely be improved.


simulate this circuit

Here is what the current drawn by the circuit is for various voltages.

enter image description here

  • \$\begingroup\$ Interesting. I'll run some tests and see what happens. Since my circuit already has a voltage divider before the mosfet, only ~2.5V of the 6V is reaching the gate. Maybe the current won't be enough to pass the resistors and trigger the gate. But that feels too vague for me to be sure. How would I add a pulldown resistor, and also feed my circuit, which needs 48V during normal operation? If I add a pulldown resistor on the positive output of the battery, I can't then also have a separate parallel path feeding my circuit. The current will be split, or take the lower impedance route, right? \$\endgroup\$
    – VanGogh66
    Jun 14, 2021 at 14:51
  • \$\begingroup\$ Show us a schematic of what you have. \$\endgroup\$ Jun 14, 2021 at 14:53
  • \$\begingroup\$ Okay, but everytime I post a schematic it just opens up a can of worms. I've added a simple schematic above, without the LED or tester button switch, just the core circuit. It works at the battery's usual 51V, and it even appears to work at the 6V output that I was able to briefly demonstrate. That unusual 6V output apparently will supply some amount of meaningful current. It powered an LED with no problems. \$\endgroup\$
    – VanGogh66
    Jun 14, 2021 at 18:12
  • \$\begingroup\$ Also, please read the notes I added in the two paragraphs above the circuit, otherwise the voltages and resistor choices may seem unusual. \$\endgroup\$
    – VanGogh66
    Jun 14, 2021 at 18:28
  • \$\begingroup\$ " everytime I post a schematic it just opens up a can of worms." Could you be more specific? Are you saying that people start criticizing other aspects of the circuit? If so, perhaps you should listen to the same people you want to help solve the problem that you know about. \$\endgroup\$ Jun 14, 2021 at 19:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.