Author Options:

How can i adapt this low battery indicator for a 3V battery supply Answered

Picture of

How can i adapt this low battery indicator for a 3V battery supply
so that the red light will turn on when the voltage drops to 1.8V?
without the use of a pot

3 Replies

verence (author)2012-03-18

Without looking at the circuit..

You want the red LED to go ON when the supply voltage drops below 1.8V?

In that case the LED will be on not for long, as the LED needs about 1.6V (rule of thumb for red LEDs, depends on LED and current). So, when the battery drops below 1.6V the LED will stop working.

Select as Best AnswerUndo Best Answer

mpilchfamily (author)2012-03-17
Why did you choose this circuit?

The circuit you have pictured is monitoring the charge lvl of a battery and lights the LED to indicate the battery is fully charged. I found the site you got the schematic from. Read the paragraph above the circuit.

When the battery voltage is above 5.1 volt, zener provides regulated 5.1 volt to pin3 of IC and its output goes high to light LED. It indicates that battery is full.

You need to find a differnt circuit that will act as a low indicator light. Something like this.

If you don't want to use a pot on the final circuit then use a pot temporarily with an AA battery driving the circuit. Adjust the pot till teh LED light then measure the resistance on the pot so you can replace the pot with a single resistor. 


Select as Best AnswerUndo Best Answer

Jack A Lopez (author)2012-03-17

The way this circuit is intended to work is the comparator there (or mabye it is an op-amp being used as a compartor) turns on whenever pin 3 (V+) is at a higher voltage than pin 2 (V-).  

Then there are some other assumptions. 

Vzener, the characteristic voltage of that zener diode there, has to always be less than the voltage of the battery that is supplying it current through R1.   A good choice for the zener would be one that has its Vzener at a voltage about half the lowest battery voltage.

The reason for doing this is to insure that the voltage seen at pin 3 (V+ of the comparator) is always Vzener, and never less than that.

Another assumption is that the battery voltage will always be large enough to turn on the LED. Note that the forward voltage across a red LED is about 1.8V, so that might be too close.  For that reason alone you might want to make this circuit so it turns on at 2.0 V instead.

The voltage at pin 2 (V- of the comparator) is just F*Vbattery, where F is the fractional resistance of that pot , from the wiper to the low side of the battery.  F is a number between 0 and 1. (E.g. if you've got the pot adjusted to 8K of its 10K total, then F =0.8)

Then the battery voltage at which the comparator decides to turn on that LED is given by:

Vzener = F*Vbattery     <==>   Vbattery = Vzener/F

If you can find a zener diode, or something like one, that has a constant characteristic voltage drop of around 1 volt.  Then dial the pot to the middle of its range, so that F =0.5.  Then the voltage where the comparator turns on is:

Vbattery = Vzener/F = (1.0 V)/(0.5) = 2.0 V

It may be desirable to change R2, to something smaller, since the voltage driving current through it and the LED is only 2.0 V, or so, now.

If you don't like pots, or you happen to know the exact value to which it should be set, then you can use the appropriate sized voltage divider (made from two fixed resistors) in place of the pot.

Select as Best AnswerUndo Best Answer