How can I modify this circuit to if once the battery turned off, never turn on again or turn only when min. ~0.5V increased?

Question

I made this circuit:

I would like the following:

The LM1 opamp wathching the battery voltage and the output high while the battery voltage higher than 3V. When the voltage drops below 3V, the opamp output go low and turn off the N-channel mosfet, so no more current consuption from the battery (except the watching circuit which is only 1-2mA).

How can I modify this circuit to if once the battery voltage dropped below 3V, never turn off, or turn on olny when the voltage higher than 3.5V?

Now my problem is when turning off because the low voltage, immediately turning on again because the voltage higher than 3.00V.

Answer 1

It sounds like you need an under-voltage-lock-out (UVLO) circuit like this: -

You would select R5 to provide the right amount of positive feedback to ensure the switching points are at 3.0 volts and 3.5 volts. So, once the MOSFET is turned off, U1's output raises the target voltage at point B and this means the Li-ion cell has to "recover" a certain amount before the circuit reactivates the MOSFET and applies power to the load. Once power is re-applied, the op-amp output is low thus setting a lower target voltage at point B hence, the Li-ion has to drop to a lower threshold before the load is disconnected.

You might also be interested in the TLV3011/3012 - it has a voltage reference built in and can be "wired" to have hysteresis.

Answer 2

You are looking for a Schmitt-trigger action. This is accomplished by adding positive feedback to the comparitor.

Figure 1. Op-amp Schmitt trigger. Note that \$ R_{123} = R_1 || R_2 || R_3 \$. Source: HyperPhysics.

There's an online calculator in the link above.

It appears the output logic may need to be inverted for your application.

Answer 3

Note that an LM1 is an opamp from before the beginning of time. If it ever existed knowledge about us is denied to us by the singularitry at the time of the big bang (or whatever really happened). You can substitute most modern opamps for it so I'll assume below that it actually exists.

For interest - here is a 469 page databook from near the beginning of time (1988 in this case) that knows about ICs like the LM1 but not about the LM1.
Fabulous Linear Technology Linear Data Book 'supplement' - 1988

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The circuit below is like unto your one but Rf and Rb have been added to supply "hysteresis".

Add Rf and Rb
Rb maybe 10K
Rf probably in the 100k - 470k range.

Define ratio Rb/(Rb+Rf) = N

V_OA_+ = Vbat + (V_LM1_out - Vbat) x Rb/(Rf+Rb)
ie
When V_LM1_out is above Vbat it adds 1/Nth of difference in voltages to V_LM1_+
When V_LM1_out is below Vbat it subtracts 1/Nth of difference in voltages from V_LM1_+

As Vbat is about 0.5V (by the OP's stated requirement) at the circuit's turn off point, the op amp output will raise the voltage at V_LM1_+ by much more than it will lower it (as the opamp output hi is >> 0.5V but opamp output low is not much lower than 0.5V). Exact amounts depend on ratio N and on output V of LM1 when high and low.

To make calculations somewhat easier you can add a diode in series with Rf so it conducts only when V_LM1_out is HIGH. Then the circuit will raise V_LM1+ above Vbat when V_LM1_out is high BUT will leave it AT Vbat when V_LM_1 is low.

So if N = say 11 (10k, 100k) Then from above:
V_OA+ = Vbat + (V_LM1_out -Vdiode - Vbat) x Rb/(Rf+Rb)
and
When V_LM1_out is above Vbat it adds 1/Nth of difference in voltages to V_LM1_+
So Assume V_LM1_high = 10V.
Assume V_LM1_low = 0V.
Assume diode used and Vdiode = 0.6V.

V_OA+ = Vbat + (V_LM1_out -Vdiode - Vbat) x Rb/(Rf+Rb)
= 0.5+ (10 -0.5 -0.6) x 10k/ (100k + 10k)
= 0.5 + 8.9 /11 = 0.5 + 0.81 1.31V = too high

Change N to say about 100 adds about 0.08 V = 0.58V switching point

Try removing diode and seeing how V_OA_+ is affected by low voltage

V_OA+ = Vbat + (V_LM1_out - Vbat) x Rb/(Rf+Rb)
= 0.5 + (0 - 0.5) /11 = 0.5 - -0.045 = 0.45V
Vbat would need to rise by ABOUT 0.05V to restart the circuit.

The very low voltage of Vbat at turn off makes this arrangement hard but not impossible to use. Addition of an extra resistor and a diode would allow the OA to pull down V_)A_+ via a resistor but to pull it up via a diode and a much larger resistor so the variation are more symmetrical.

Example values:

With Rb = 10K, Rf = 1 megohm and Opamp output swinging 0/10V, when OA is high then OA_+ will rise by about 0.1V. So setting tyhe divider to 0.6V will allow cutoff at about Vbat 0.5V.
The schottky dide (and small schottky, BAT85, 1N414x, ...) causes the opamp to clamp the LM1_+ at about 0.3V. Vbatt will cause restart only for some quite high value of Vbat. To get a lower voltage restart add resistance in series with the diode.

ie the above is a good starting point - play and ask additional questions.

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Question: Do you actually have an LM1 ?