Simple Low Voltage / Low Battery Indication Circuit

Started by Rob Strand, October 10, 2022, 01:18:53 AM

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Rob Strand

Can't get any simpler, just add a second resistor across the LED.

It's a simple "better than nothing" solution but the series zener solution is
generally better.

I strongly recommend testing the circuit to make sure it works with your LEDs.

Tips:
-Tune R1 for the LED current with the normal supply voltages.
-Tune R2 so the LED or goes out (or is low brightness) at the lowest supply voltage.
-Don't try to set the off voltage too high.  I think 6V is a good upper limit, 5.4V wastes less power.

Here's some details with some indicative part values.
Notice how the R2 value varies for the LED color.  The R2 value is touchy.  That's why it's best not to use this circuit to detect battery voltages over 6V, in fact 5.4V is much more reliable.
I haven't added a set of values for Blue or White LEDs.


Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Lino22

Rob how dos that resistor in parallel with the LED make it different from just a LED and a limiting resistor? Will the LED light die faster than without it?
When the core started to glow and people started yelling, he promptly ran out the door and up a nearby hill.

Rob Strand

#2
Quote from: Lino22 on February 28, 2024, 11:34:11 AMRob how dos that resistor in parallel with the LED make it different from just a LED and a limiting resistor? Will the LED light die faster than without it?

In essence it does make the LED light die faster.

In simple terms:

For the normal series resistor case the LED current is IL = (Vcc-VLED)/R1.   That means current will keep flowing until the supply drops to VLED.   So the LED will on down to a low voltage albeit with low intensity.

With the added resistor:
- the LED has/requires a certain terminal voltage to turn on. 
- If the voltage across the voltage divider Vd=Vcc*R2/(R1+R2) is less than that the LED cannot turn on. 
- The point where divider applies just enough voltage to turn the LED on is  Vd = VLED.  If think of the supply as varying with R1 and R2 already chosen that point will be,  Vd = Vcc*R2/(R1+R2)= VLED
or Vcc = VLED * (R1 + R2)/R1 = VLED*(1 + R2/R1).   Since 1+R1/R1 is greater than unity the supply threshold where the just LED turns on is greater than the LED voltage and can be set by R2.

The main difference between the two cases is what happens near the end.   The change in threshold makes the LED turn off at a higher supply voltage.

As far as working out the resistors there's some finer points.  For example the threshold isn't a precise on/off. It's defined as when the LED current is low enough to consider the battery is low.

Keep in mind this is only a better than nothing solution.   Using a series zener like on Boss and Ibanez pedals will provide a more rapid transition around the threshold.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Lino22

That's a good idea, LEDs normally over-live the dying sound quality. This battery indication LED can even be mounted separately.
When the core started to glow and people started yelling, he promptly ran out the door and up a nearby hill.