Suppose we have 150 watt and 300 watts portable photovoltaic system with 13.5V battery (50Ah) and we want the low voltage cut off point to be set at 10V.

Can we achieve this simply by using zener diode?

If yes, then how?

  • 1
    \$\begingroup\$ More detail needed. Cut off what from what? Do you want the battery to stop supplying the load at 10v Battery Voltage? What are the battery specs and types? \$\endgroup\$
    – Russell McMahon
    Aug 11, 2021 at 5:31
  • 1
    \$\begingroup\$ You might want to help those less "up on all thing solar" by explaining that the low voltage disconnect (LVD) helps to protect against over-discharging. And many need to see schematics (I do.) At your level, 50 A, a relay is often used in conjunction with a voltage monitor. You should already know this, though most of us here may not. So you need to spend time educating others on the issue. And no, I don't think a zener diode is the approach. But I'm not interested in even attempting an answer when you haven't provided a well-researched and documented question. I'm ignorant, though. So FYI. \$\endgroup\$
    – jonk
    Aug 11, 2021 at 6:33

3 Answers 3


Case# we have 150Watts & 300Watts portable PV System with 13.5v Battery (50Ah) and we want the low voltage cut off point to be set at 10v. Can we achieve this simply by using zener diode...?


If yes, then how...?

A zener doesn't make a good reference by itself, and it certainty can't control an output.

You could use a zener diode as part of a voltage reference circuit. You could then compare the battery voltage to the voltage reference and turn it off when it dropped below, using a relay or high-side MOSFET. But that's all rather more than 'a zener diode'.

Here's a circuit from a few years ago that does the job for a 12 V supply. original posts here . Alter the resistor divider values R2/3/4 to get a different threshold voltage. With the on/off buttons, it also doubles as the load switch. It uses the voltage referencing function of the TLV431 to set the output trip voltage. It uses M1 and IC1 in a 'thyristor' configuration to form the on/off memory.


simulate this circuit – Schematic created using CircuitLab

If we want to configure this circuit for a 10 V cutoff, then we'd choose components depending on how accurately we wanted it.

The main consideration is the reference voltage of the TLV431. That's nominally 1.240 V, but can be as far out as 1.194 to 1.286 V over temperature and for the lower grade versions.

We also need to make an allowance for the bias current, which is typically 150 nA, but can be as high as 300 nA. If we allow a 10 k impedance at R3, that means a typical offset of 1.5 mV due to bias current, insignificant compared with the possible variations in the reference voltage.

If we want to omit R4 and set R2/3 for the nominal 1.242 V (including the 2 mV for the bias current) at 10 V, then we need a resistor ratio of 8.758/1.242 = 7.0515:1. That can't be made from any two E24 values, but it's possible to use three resistors to get close. R3 = 10k, R2 = 70.515k = 68k + 2.4k gets within the 1% tolerance of typical resistors. You could fine tune the ratio further in testing by adding further resistors (small ones in series or large ones in shunt)

If you want to include R4 to adjust the +/- 50 mV tolerance of the TLV431, then we want at least 1.19 to 1.29 volts across R4. Use for instance R3 = 10k, R4 = 1k, and R2 = 72.5k, though given the adjustment, 75k will probably do. It's good policy to keep the adjustment range as small as possible. If you only have a large value adjustable resistor to hand, it's possible to reduce its range by putting a small value resistor across it to reduce the voltage available to it.

  • \$\begingroup\$ What values of R2/3/4 do I need for 13.5V battery (50AH), to set the cut off point at 10V? \$\endgroup\$ Aug 23, 2021 at 11:57
  • \$\begingroup\$ @ihsan.duet11 added to the answer \$\endgroup\$
    – Neil_UK
    Aug 23, 2021 at 13:19
  • \$\begingroup\$ 0.5 watt or 1watt of R1/2/3/5/6 will be suitable...? what should be the wattage rating for R4 (Potentiometer) ? \$\endgroup\$ Aug 24, 2021 at 7:21
  • \$\begingroup\$ @ihsan.duet11 The power dissipated in a resistor (in watts) is the voltage across it (in volts) times the current through it (in amps), which can also be written as V^2/R or I^2R. I'll leave those calculations as an exercise for you. A grasp of Ohm's Law through practice will be more valuable to you going forward than simply doing the calculations for you. It's good practice to use a resistor power rating of at least twice the expected dissipation. This allows you to pretty much ignore any ambient temperature effects that would have to be considered if less margin was used. \$\endgroup\$
    – Neil_UK
    Aug 24, 2021 at 8:27

You can use a Zener as a reference for a low-voltage cutoff. But Zeners are imprecise, temperature dependent, and may consume too much current from the source even in the cutoff state.

Not the best choice in my opinion. You will probably want a very low quiescent current reference and comparator set up to turn off a high-side p-channel MOSFET.

If you are trying to go to mass production and you need to keep BOM cost under control, it could be worth it to try to engineer something with the Zener reference. But for a one-off just use a fancy reference.


Doing a low voltage cutoff with an z-diode sounds the wrong direction.

  1. The diode will begin to conduct "over" a specific threshold voltage.
  2. The diode won't cut at your exactly needed voltage, it will slightly begin to conduct and produce heat over time.
  3. If it cutoff, there is still a problem with the hysteresis. After the cutoff, your battery will remain without load and will overdrive the diode. That will lead in an on/off cycle for a while and stress you whole system.

There are cheap and easy voltage supervisor IC's which takes care of that kind of application with additional safety functionalities.


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