# TL431 low battery cut-off

Could somebody please explain the operation of the attached diagram?

The circuit was proposed here and here as a low voltage cut-off circuit, which switches off the MOSFET when the voltage drops below the reference voltage set by the voltage divider.

What I don't understand is how the TL431 keeps the MOSFET on when the input voltage is above the reference voltage. How is the MOSFET switched off when the voltage drops below the preset voltage? Will the value of the set cut-off voltage influence the working of the circuit? For example, can the reference voltage be set too low to be able to switch off the MOSFET?

I read through the datasheet, but it didn't help me understand this phenomenon. It did help me to choose my resistor values, though. • Just as a future hint: Its better to wait some time before accepting an answer. This encourages more users to provide additional answers and to review existing ones.
– Rev
Jan 25, 2016 at 11:08
• The TL431 is featured in www.badbeetles.com it does have some pitfalls . Devices from different manufacturers are not the same. Jan 25, 2016 at 19:21

The TL431 is being used as a comparator and shunt. If you look at the functional diagram, you'll notice that all it is is a voltage reference, comparator and n-channel transistor/fet: When VRef, set by R1 & R2, is Higher than the reference voltage of 2.5 Volts, the comparator's output goes high, which enables the transistor. This pulls the cathode towards ground. The cathode is connected to R3 and the gate of the P-Channel Q1. So when the cathode is pulled towards ground, Q1 is turned ON.

When VRef is Lower than 2.5 Volts, the comparator's output goes low, which turns the transistor off. The cathode node is then pulled high by R3, which pulls Q1's gate high, disabling Q1.

As for VRef being too low, yes. If the voltage divider of R1 and R2 is chosen so it's mid point is always below 2.5V, even if the battery is fully charge, Q1 will never turn on. The opposite is true too. If the midpoint never goes below 2.5V (before the battery gets drained too low), then Q1 will never turn off.

• Thank you @Passerby. Your explanation makes perfect sense and is much appreciated.
– Egon
Jan 25, 2016 at 8:02

If the battery voltage exceeds Vref*R2/(R1+R2) then the TL431 cathode goes low (to < ~2V anyway), which turns Q1, a p-channel MOSFET, on.

But this is not a very good disconnect. The value of R3 has to be quite low since the TL431 needs as much as 1mA to work properly, and the MOSFET should be off (for example, if 0.6V is the maximum voltage for the MOSFET to be acceptably off then R3 should be 600 ohms).

Also there is no hysteresis in this circuit so it will tend to oscillate (because the battery voltage will rise with the load disconnected) or the MOSFET will get hot if there is a lot of current draw as you hit the cutoff value.

The MOSFET needs to be fully on for the lowest 'on' input voltage and the TL431 at a couple of volts, so probably a logic-level MOSFET for a 12V battery.

If there is any surge on the battery or reversal the circuit will die. The MOSFET gate will typically be rated for 20V or less, so even a small inductive load will kill it.

Russell M. at least suggested a TLV431 which takes about 1/10 the current.

• Good job in pointing out the weaknesses of the circuit.
– Rev
Jan 25, 2016 at 8:21
• Thank you for pointing this out. It now seems that I would need to look for a different, foolproof, design.
– Egon
Jan 25, 2016 at 8:35