# Single-cell lithium battery powered comparator circuit with hysteresis

I recently was reviewing a circuit I made and began to question its feasibility. The circuit cuts off a battery's connection to a load after the voltage has fallen below a certain threshold.

To eliminate ON/OFF cycling around the cut-off point I added hysteresis. I followed this design guide, scroll to figure 5.

There is one key difference between my circuit and figure 5: Vcc is a battery supply. It is not constant, meaning Voh (Voltage output high) is also varying.

So I decided to graph the Vth and Vtl (threshold high and low respectively) and varying Vbat. I found that if the battery voltage decreases, so do Vtl and Vth.

Does this mean that the comparator will never make a transition?

Also, why does Vin in the diagram never appear in the equations? Is my circuit independent of the non-inverting input's voltage? This seems unlikely.

My overall question is: at what battery voltages will the circuit transition from high to low happen? I have provided some preliminary resistor values below.

My target is to cut the battery supply at 3 V and reconnect it once above 3.2 V.

Here is a diagram showing how the battery voltage affects the hysteresis threshold voltages.

## 1 Answer

There is one key difference between my circuit and figure 5: Vcc is a battery supply. It is not constant, meaning Voh (Voltage output high) is also varying.

That is ok in most cases, but both + and - input terminals of the opamp need to be much lower than Vcc (which is Vbat). Also the output of the opamp will never get higher than Vcc (Vbat) so keep that in mind. I'd probably run the Vref and the Vin of the hysteresis circuit lower than the mid range of the lowest battery voltage (so if your Vbat minimum is 3.0V run the Vref around 1.5V)

It will, because the resistor divider is dividing Vcc/Vbat. If you want it to be fixed, you'll need a reference. You could use something as simple as this circuit (with U2 and R3 and R4 to provide a better stable voltage)

Also, why does Vin in the diagram never appear in the equations? Is my circuit independent of the non-inverting input's voltage? This seems unlikely.

It's because you are comparing Vin to Vtl and Vth. so Vtl and Vth change the state of the comparatator. In essence Vin is being split into Vtl and Vth

• The low voltage shutdown circuit posted is very helpful. But I am unsure if you are saying the circuit I posted will or will not transition? Oct 8, 2022 at 1:40
• It will transition but the transition will change with voltage of the battery, so that's why you need a fixed voltage reference instead of a voltage divider on the positive port of the opamp. I've built one before, but I remember it was a bit finicky to get it to transitions at the right points, if I remember right I had to adjust the resistor values a bit from the predicted values of the equations. Oct 8, 2022 at 5:46