I am about to add an LED to a charger but need further opinions.

I am trying to add an LED to a NiMh charger for 5 AA battery pack (total voltage about 6.93) to be able to see that is actually turned on.

I have seen some chargers that have the dropping resistor and LED in series with each other and then this combination is in parallel with another resistor of lower ohms which goes to the battery pack: examples

(82 ohm and LED) in parallel with (a resistor of 24 ohms) - used for charging 4 AA batteries

(47 ohm and LED) in parallel with (a resistor of 18 ohms) - used for charging 2 AA batteries

I can calculate the resistor for the LED but any thoughts on what the other resistor should be? or what things to take into account in choosing this resistor ?

Thanks Ali

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Is there a resistor in the charger already which is in series with the battery being charged? If so then measure the voltage across it, subtract the LED's forward voltage from that and then use ohms law to determine the resistor you need for the correct current across the LED,

5-10 mA ought to do the trick. In a pinch you can just experiment and see what works.
Reply to
James Sweet

I assume your reason for considering this is to get a visual indication that there is actually current flowing into the battery. A reasonable thing to want, but...

Some of the makers who have done this have had trouble with blowing out the LED's. Consider what happens when you plug in a dead battery, or maybe even worse a bad one that happens to be shorted...

If you have the room and can tolerate more parts, maybe consider something like using the b-e junction of a power transistor that can shrug off the worst case current and drive the LED with a resistor off the collector.

Reply to
Ol' Duffer

Thanks for promp reply. James, does this mean that any resistor in series with the battery being charged should have a voltage drop equal to the total voltage drop across the LED and its dropping resistor? Is it common to have a resistor (in series with the battery to be charged) within all chargers? (I haven't opened the charger yet) Ali

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What I've often seen done is a string of three series diodes in the output lead with a LED and series resistor in parallel with the diode string. Just make sure that the diodes can handle the maximum charger current.

Since a nicad charger is supposed to be constant current, I can't see that being a problem unless whoever "designed" it didn't know what they were doing...


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Just to add further: the battery pack consists of 5 NiMh AA batteries and the actual charger is just a wall plug with 9 volts dc at .500 mA max. The batteries are inside a torch and I am wanting to add this LED circuit inside the torch with a simple diagram. Will the voltage across the resistor/LED combinatiion adjust itself to be equal to the voltage across the resistor in paralell for when the batteries are really flat and therefore higher current will be flowing? Ali

Pete wrote:

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yes parallel combinations see equal voltage and share the current

with series combinations it's the other way around,

Bye. Jasen

Reply to
Jasen Betts

You should calculate everything, the shunt resistor too!

I assume the current to the batteries is constant enough to do this.

Ideally, you would have a wire (0 ohms) feeding the current to the batteries. But you want to tap off just enough current to light the LED, while sending the rest to the batteries as before. There's probably a range of values that will work for you.

Measure the (max) current to the batteries first, to see what you're working with.

Take the voltage drop of the LED, and add just a bit of headroom (for instance, if the LED drops 1.0V during conduction, use 1.5V.) Then use Ohm's law to figure out what series resistor you'll need to get the mA current you need (the LED's rated operating current) through the LED with 1.5V across the whole thing.

Finally, take the current you measured going to the batteries, and use Ohm's law again to determine what size shunt resistor you'll need to get your 1.5V voltage drop. Use Ohm's law again to make sure the shunt resistor is large enough in terms of wattage.

Reply to
Mr. Land

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