I have a circuit that's supposed to cut off at <4V. It instead caps the voltage at 3.3-ish and reduces the voltage linearly below that. Zener - 3.3v Source - 5V +-5V


  • \$\begingroup\$ If your wondering, I used iCircuit \$\endgroup\$ Commented Aug 29, 2015 at 22:57
  • \$\begingroup\$ What is the purpose of the cutoff? I am asking because it will help guide what type of answer to give. The cutoff circuit will draw some power from the supply, even after cutoff. If the supply is a battery, we need to know that so we can minimize the current draw. Also, what is the normal operating current (when the cutoff is not active)? \$\endgroup\$
    – user57037
    Commented Sep 1, 2015 at 18:07
  • \$\begingroup\$ @mkeith Top operating current is .3A \$\endgroup\$ Commented Sep 3, 2015 at 13:53

1 Answer 1


A zener diode is not a perfect stop for current, there is still some leakage. Further because you are using a darlington pair type setup there is a very high beta allowing just even a little bit of current is still enough to tun them on.

Also this is just my best guess, your question is not very clear.

  • \$\begingroup\$ So how should I do UVLO (Under Voltage Lock Out)? \$\endgroup\$ Commented Aug 30, 2015 at 0:07
  • \$\begingroup\$ What is the purpose of the UVLO? To prevent drawing current from a low battery, or to protect the downstream circuitry from operating when the voltage is too low? \$\endgroup\$
    – user57037
    Commented Aug 30, 2015 at 3:06
  • \$\begingroup\$ @mkeith Both, really. But the main one is to prevent the under-volting of ICs \$\endgroup\$ Commented Sep 3, 2015 at 13:52
  • \$\begingroup\$ The easiest way is to use a brown out reset chip. For example take a look at the MAX6326. The chip senses the voltage and will drive the pin high or low in response. You can get a Pchannel fet and use one of these chips with an active high output to drive it. When the battery voltage drops below the preset voltage the chip will shut off the fet and hold the fet off until the voltage comes back up. The chip also has very low current consumption. \$\endgroup\$
    – vini_i
    Commented Sep 3, 2015 at 16:37

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