Help please

This circuit was given to me an the person who gave it claimed it worked...well it does sorta. The LED that comes on when the battery voltage is low is extremely dim. I need help figuring out how to make it brighter. Please help. The goal was to swith on the LED when the battery gets low and swith off the load (the 100ohm resistor R6) Thanks for the help.

The way it is now, the LED current is only about 6mA when the battery voltage is 11.3V and less when the battery voltage is less.
If you replace R3 with a piece of wire then the LED current will be about 21mA when the battery voltage is 11.3V.

Well if the battery's voltage is dropping then it is probably unable to produce the current needed to make the led really bright. That being said there is a resistor in front of the LED that I assume is for controlling the current...but if the voltage is dropping then the value probably can be lowered to accommodate the new voltage.


(Just my thoughts)

I have done the old wire trick... and that causes the LED to be on even when the battery is not low (it starts at around 27 volts). Once the other transistor switches off, the circuit only draws current via the LED so the voltage remains fairly steady on the batteries.

hi,
You could try to modify the LED drive transistor to a constant current source.
If the comparator is capable of switching from close to 0V and +Vs.

The voltages marked in Green are when the output of the Comp is near 0V.

Eric

The output voltage of the opanmp does not go high enough to turn off the transistor. So add another resistor at the base of the transistor to turn it off when the opamp's output is high.

Downloaded the LM1492 datasheet, I agree with 'agu' the maximum high output voltage from the LM1492 is about +3.5/+4v, which will leave enough overhead to switch on the LED drive transistor.

Use the potential divide he recommends, you could also try the constant current solution. [it also requires the 4K7 divider, as shown in 'agu's drawing.

Eric

EDIT: The above LM1492 max high output was quoted for a Vs of +5V, it should be actually quoted for 1 to 2 volts lower than the Vbty/Comp supply voltage.