Good morning
Could somebody please explain the operation of the attached diagram to me?
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? And also how the MOSFET is switched off when the voltage drops below the preset voltage? Also, will the value of the set cut-off voltage influence the working of the circuit (e.g. 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.
Best Answer
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.