🚀 go-pugleaf

alitonto wrote:
> 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
> 


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.

In article <1142391196.601886.28100@i39g2000cwa.googlegroups.com>, 
alitonto@hotmail.com says...
> 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

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.

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

Ol' Duffer wrote:

>> 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

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.

> 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...

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...

Peter

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:
> Ol' Duffer wrote:
>
> >> 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
>
> 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.
>
> > 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...
>
> 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...
> 
> Peter

["Followup-To:" header set to aus.electronics.]
On 2006-03-15, alitonto <alitonto@hotmail.com> wrote:
> 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
>

yes parallel combinations see equal voltage and share the current

with series combinations it's the other way around,


Bye.
   Jasen

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.