Need help with circuit....

[prabh94]

the circuit i made is attached below : see attachment.


Let me explain the circuit. So the output from the 3.7 V battery , is fed to the (-) of the op-amp. i've used a 5V zener and a potential divider to get a potential of 0.37 V (10% of 3.7 V) to (+) of the op-amp.

Now if the voltage given by the battery is greater than 0.37 V ,

> The input to the load = output of the battery
> The IC reads input=1 and does not activate the LED and Alarm

and if the voltage given by the battery is less than or equal to 0.37 V ,

> The output to the load = 0V . battery stops discharging.
> The IC reads input=0 and activates the LED and Alarm.


will this work?? any problem with the op amp and zener usage??? plz help

[pebe]

I am afraid your circuit will not work. The supply voltage is only 3.7V and that will not operate a 5V zener.

Here is a circuit that should work OK. Use an op-amp that will operate with a supply voltage down to 3V. Make sure it has enough output to drive your load.

The battery voltage rises slightly when the amp disconnects the load and that would make the amp try to turn the load on again. The process would repeat so the load would be continually switched on and off. To prevent that, the resistor, R2, is used to provide positive feedback to give some hysteresis. If it is not enough, reduce the value of R2.

When the battery has run down to the level for the alarm to trigger, set P1 so that the op-amp output goes low. Because of the hysteresis, there will be different settings of P1 for when the amp output goes from high to low, or from low to high.. The one you want is when the output is going from high to low.