# COMMUNITY : FORUMS : TECH

## Problem with low battery indicator

Hi guys,

I have built a low battery indicator circuit for a 6V battery source. It will light up a led when the voltage reaches below 5V.
The 6V battery source powers a servo motor, which is controlled by a microcontroller.

The problem I have is that when the servo motor rotates, the led lights up. And when the servo stops rotating, the led goes off. This happens even when the battery is above 5V. Any idea what is wrong? And how I could rectify the problem?

I have attached the circuit of the low battery indicator, if it helps.

mpilchfamily4 years ago
Nothing wrong with your indicator. It's doing the job it was designed to do. Your just taxing the battery quite a bit causing a temporary voltage drop.
Orngrimm4 years ago

Every battery also has an internal resistance. The more you empty a battery the higher this internal resistance becomes (normally).
Now: If you draw current from the battery, this internal resistance also causes a voltage drop WITHIN the battery in the order of
Uloss = Idrawn * Rinternal
Lets say you draw 1 amp (Not uncommon for a motor) and your interal resiatance is around 1 Ohm, you loose 1 Volt within the battery. If your voltage-divider is set up in a way that it detects your (now by 1 volt lower) batteryvoltage during the motor-action, it will trigger since at the moment you are drawing current, that IS the voltage of the battery.
If you stop drawing current, the Current is 0 again and whatever your internal resistance is, if you multiply it with 0 amp you get 0V drop again.
So summarised:

Motor OFF:
Internal resistance in the battery: 1 Ohm
Current drawn: 0A
Voltage: 5V
Detector-level of the circuit: 4.2V (wild guess)
State of circuit: Not lighted.

Motor ON:
Internal resistance: 1 Ohm
Current drawn: 0A
Voltage with motor ON: 4V
Detector-level of the circuit: 4.2V (wild guess)
State of circuit: LIGHTED

Often people think a battery has a voltage which may drop a bit over the course of its lifetime, but stays constant no matter what current you draw from it. Not true. Not at all. :)