undervoltage
Here is schematic of battery charging circuit. The battery voltage at 100% charge is 4.2V and it gradually decrease and is 3.2V, discharging below 3.2V will affect the battery life.
How to make sure, the battery supply is disconnected automatically from the load when it reaches 3.2V or less and when voltage is above 3.2V, it gets connected to Load,
What can be added in the circuit? The circuit should be simple and low on BOM.
Question 1: Option 1. This is a more reliable indicator of the actual battery voltage. The drop out of the regulator will be affected by current.
Question 2: Use the regulator's enable input if it has one. This saves you having to add an extra transistor. This will reduce the amount of current drawn by the regulator.
There are two ways you might implement this:
1) Use a potential divider to create the Enable voltage. When the output voltage of the divider goes below the enable threshold voltage, then the regulator switches off. The problem with this is that the voltage threshold of the enable input is pretty flexible:
The threshold is somewhere between 0.4v and 1.2v! This isn't going to be accurate enough for your purposes.
Instead, I'd suggest using a comparator:
Given the additional information that it is an LT8610, there looks like something that can be done.
The EN/UV input has a 1.0V +/-6% threshold and it has almost no (+/-20nA) bias current. That's with the input rising, it's 40mV less typically with the input falling.
You already have a voltage divider on the input. All you need is a resistor from Vout to EN/UV.
simulate this circuit – Schematic created using CircuitLab
When the UV input is RISING, \$V_{OUT} = 0\$ and thus it will trigger at \$V_{IN}\$ =\$ 1.0V \cdot \$\$(R1 + R2 || R3) \over (R2 ||R3)\$
It is easy\$^{TM}\$ to show that the hysteresis: is \$0.04V\cdot\$ \$R1\over R1||R2||R3\$ + \$V_X \cdot( \$ \$ R1 \over R3\$), where \$V_X \$is the output voltage.
Best Answer
You will need an additional "switch" in series with the load circuit to disconnect the load. Q1 Cannot disconnect the load because of the manner in which it is connected in its present role.
A P Channel FET "Q2" (which could be of the same type as Q1) connected in series with Q1 d-s but with opposite polarity can provide the switching.
A voltage sensitive comparator then drives Q2 gate low when battery voltage is >= 3.2V. A TLV431 would do this with 3 added resistors - 2 form a voltage reference divider for the TL431 reference input and one forms a pullup for Q2 gate to turn it off when the TLV431 is not turned on. The reference input resistors should be as large as bias currents allow to minimise battery drain.
A TLV431 is specified rather than a TL431 as it has lower minimum regulating current requirements. If ultra low batteruy drain bia R1 + R2 is required there are shunt regulators that need only about 1 uA bias currents in R1, R2
simulate this circuit – Schematic created using CircuitLab