Low voltage indicator based on LM339

Could some one please explain to me how they came up with the value of 430 for R3.

Based on V = IR I can't figure out how it allows for a current of 10mA.

I get 12 mA, assuming 12 V for the battery, but the battery voltage and Zener current would drop as the battery discharges.

jremington:
I get 12 mA, assuming 12 V for the battery, but the battery voltage and Zener current would drop as the battery discharges.

If the battery was to be 5V or 7.4V how would you calculate the value for R1?

Or doesn't it matter that much over such a narrow voltage range?

I think you have to look at the zener in question and see what current it needs. That said I also think that as long as you have a few mA it doesn't matter (I'm prepared to be corrected here).

Also this will not be a very accurate circuit as it depends on the exact characteristics of the particular zener and R1/2.


Rob

Graynomad is right, the actual Zener current doesn't matter very much. You just need to have it be in the general vicinity of the current at which that particular Zener diode's reverse voltage is specified, which is often 5 mA.

The actual Zener voltage has a tolerance of typically 5 or 10%, and the current can vary by a factor of two or more without changing the Zener voltage much. What is more critical is the choice of R1/R2 voltage divider values, which sets the switch point for the circuit. If you want to use this circuit with a lower voltage battery, you may need to use a different Zener diode. The switching point is Vsw and when this drops below the Zener voltage, the circuit gives a low battery indication.

Vsw = Vbatt*R2/(R1+R2)

You want Vsw to be greater than the Zener voltage for healthy battery indication.