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Rechargerable Battery detector...


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Hi,
Welcome to our forum.
A "9V" rechargeable battery is actually 7.2V to 7.5V, or 8.4V to 8.75V, during most of its discharge time, depending on how many NI-CAD or NI-MH cells that it has inside.
A disposeable 9V battery has six carbon-zinc or alkaline 1.5V cells inside. It is called exhausted when its loaded voltage drops to about 6V.
Most rechargeable "9V" batteries also have six cells, but each cell's voltage is only 1.2V to 1.25V, so the battery's voltage is 7.2V to 7.5V. It is called exhausted when its loaded voltage also drops to about 6V.
Some rechargeable "9V" batteries are made with seven cells for them to be a closer match to disposeable 9V batteries, and the total voltage of its seven cells is 8.4V to 8.75V. Each cell is smaller so that they can all fit, so its capacity is less than a six cell battery. It is called exhausted when its loaded voltage drops to about 7V.
A six cell rechargeable battery has a fully charged voltage while still charging of about 8.7V, and a seven cell one 10.15V. The fully charged voltage drops quickly to normal when the battery is removed from the charger.

Your voltage detector circuit could simply be an opamp that is used to drive the 1st LED and as a comparator. One input is from a voltage reference like a biased zener diode, and the other input is from a 2-resistor voltage-divider from the battery being detected. The circuit could be wired so that the 2nd LED is turned on with a resistor, and when the opamp is driving the 1st LED then it is also shorting the 2nd LED through a diode.

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Thanks for your reply but could u pls help me to layout the circuit and what components should i use?


Hi,
Welcome to our forum.
A "9V" rechargeable battery is actually 7.2V to 7.5V, or 8.4V to 8.75V, during most of its discharge time, depending on how many NI-CAD or NI-MH cells that it has inside.
A disposeable 9V battery has six carbon-zinc or alkaline 1.5V cells inside. It is called exhausted when its loaded voltage drops to about 6V.
Most rechargeable "9V" batteries also have six cells, but each cell's voltage is only 1.2V to 1.25V, so the battery's voltage is 7.2V to 7.5V. It is called exhausted when its loaded voltage also drops to about 6V.
Some rechargeable "9V" batteries are made with seven cells for them to be a closer match to disposeable 9V batteries, and the total voltage of its seven cells is 8.4V to 8.75V. Each cell is smaller so that they can all fit, so its capacity is less than a six cell battery. It is called exhausted when its loaded voltage drops to about 7V.
A six cell rechargeable battery has a fully charged voltage while still charging of about 8.7V, and a seven cell one 10.15V. The fully charged voltage drops quickly to normal when the battery is removed from the charger.

Your voltage detector circuit could simply be an opamp that is used to drive the 1st LED and as a comparator. One input is from a voltage reference like a biased zener diode, and the other input is from a 2-resistor voltage-divider from the battery being detected. The circuit could be wired so that the 2nd LED is turned on with a resistor, and when the opamp is driving the 1st LED then it is also shorting the 2nd LED through a diode.

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Hi again,
1) Is this battery voltage indicator circuit used in the battery charger so that the low voltage green LED lights to show that it is charging normally, then the red LED lights to show that the battery is fully charged?
3) Please post the schematic of your battery charger. Since the 14 hours of charging current is so low, this indicator circuit will use all that current leaving none to charge the battery. So this circuit must be powered by the charger's main power supply.
3) What is your normal battery voltage, how many cells does it have?
4) Can you obtain an LM358 dual opamp?

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Thanks for your reply again. But what i want is just a simple circuit that can detect mine rechargeable battery when it fall below 7V~6V and light up mine red led. No charger is connect in here. But when it's above 7V, it will light up mine green led.


Hi again,
1) Is this battery voltage indicator circuit used in the battery charger so that the low voltage green LED lights to show that it is charging normally, then the red LED lights to show that the battery is fully charged?
3) Please post the schematic of your battery charger. Since the 14 hours of charging current is so low, this indicator circuit will use all that current leaving none to charge the battery. So this circuit must be powered by the charger's main power supply.
3) What is your normal battery voltage, how many cells does it have?
4) Can you obtain an LM358 dual opamp?
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Hi again,
Here's a simple circuit for you. The green LED will be on until the battery voltage drops to about 6.1V, then it will turn off and the red LED will turn on from 6.1V down to about 4.2V.
Decrease the value of R1 or increase the value of R2 to raise the voltage's trip point.
Use only this zener diode which operates at the low current of 5mA. Other 3.9V zener diodes require 50mA.
The LEDs are operating at about 5mA, which is fairly dim, to conserve power. The entire circuit will drain a 120mA/hr NI-CAD battery in about 10 hours. If you remove the green LED and R6, the circuit will drain the battery in about 20 hours.

post-1706-14279141784863_thumb.jpg

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Hi Ante,
The old LM358 low-power dual opamp is cheap and available nearly everywhere. It is actually 1/2 of an LM324. It works well at these low supply voltages and has enough output current to drive an LED, unlike many rare CMOS opamps. Its total supply current is typically only 0.5mA, which is one-twentyth of the sum of this dim LED plus the low-current zener. ;D

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Thank you very much for your help and all others. ;D


Hi again,
Here's a simple circuit for you. The green LED will be on until the battery voltage drops to about 6.1V, then it will turn off and the red LED will turn on from 6.1V down to about 4.2V.
Decrease the value of R1 or increase the value of R2 to raise the voltage's trip point.
Use only this zener diode which operates at the low current of 5mA. Other 3.9V zener diodes require 50mA.
The LEDs are operating at about 5mA, which is fairly dim, to conserve power. The entire circuit will drain a 120mA/hr NI-CAD battery in about 10 hours. If you remove the green LED and R6, the circuit will drain the battery in about 20 hours.
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