Re: How to charge a 9V Ni-MH

Fred Bloggs wrote:


ehsjr wrote:

Fred Bloggs wrote:



ehsjr wrote:

sdeyoreo@xxxxxxxxxxx wrote:

How to make a 9V Ni-MH charger?

Battery is 9V, 150mAh. I read the charging current should be C/10 or
15 mA. Charging voltage should be ~ 1.41 v per cell. Also read that
NiMH is 1.2v/cell. The package for the battery says fully charged will
read 8.4 v. So that's 7 cells, 8.4/1.2. So charge at 7*1.41v = 9.9v @
15mA?





No, it's 6 cells at ~ 1.4 each.


So, what do I use for a current limit resistor? (9.9-8.4)/15mA or
100 ohms?





If you decide to use a resistor to limit the current
to 15 mA, you *must* use a regulated voltage source.
I prefer a regulated current charger (shown below) to
charge the cells.

This is a charger circuit that will hold the current
to very close to 15 mA. The input voltage must be
at least 3 volts higher than the full charge voltage
of 8.4, so 12 volts is shown, because the LM317 chip
needs close to 3 volts of "headroom", minimum, to
operate properly. You can use any 12 volt "wall wart"
with this circuit:

-----
+12 ---|LM317|---+
----- |
| [83R]
| |
+------+------- To NiMh +

Gnd --------------------- To NiMh -

Change the 83 ohms to 167 ohms if you want to use
the C/20 that Henry mentioned.







He can add a 9.1V zener reference diode and automatically taper it off to float charge at full charge of 1.4V/cell, like so, the series diode prevents battery discharge on loss of line power.



That's a nice addition. I do that a little differently
stealing current with a TL431. I set the TL431 for the
desired end-of-charge voltage and have it light an LED
with the stolen current. When the LED comes on, you know
the charge is complete. The zener can be used the same
way, of course, but the TL431 is crisper and allows you
set to different end-of-charge voltages. The R for setting
the charge rate is computed based on C/10 + Ipot. A
resistance can be placed in parallel with the LED & 1K
if you want to steal more current.


+ ---[LM317]--[R]--+
| |
+----------+--->|---+---- to batt +
| |
[1K] |
| |
[LED] /
| \
[TL431]---->/
| \
| |
Gnd ---------------+--------+---- To batt -


You don't want that.

I think you missed a line in my post:
"A resistance can be placed in parallel with the LED & 1K
if you want to steal more current. "

The TL431 set for Vbatt=8.4 makes the difference junction 8.4+Vdiode=9V say, then the TL431+VF,LED =2.5+2=4.5V, making the 431 saturation current (9-4.5)/1K=4.5mA, leaving 10.5mA through the battery; you can't reduce it more than that. You would want to allow a saturation current of say 20mA to keep the '431 linear,

The TL431 is linear under 1 mA Ik (about 400 uA), per the data***.

changing that 1K to 4.5V/0.02=220R. But even then, you still don't know the battery taper current at 8.4V.

It's the parallel R that allows you to set the taper rate.
You set the taper current rate to what you want with the
parallel R mentioned in the post, stealing the "excess".
Using your numbers, there is 10.5 mA left for the battery
when there is 9 volts before the diode. If you wanted
a taper rate of C/50 you'd need to steal 7.5 mA. So
a parallel R across the LED and its R would be 867 ohms.
Of course in practice, an 866 would be close enough, or
even an 820, yielding about C = 1/58




For simple, you can't beat the 1 chip + 1 R solution.


Sure you can, just use a regulated plug-in supply and single R. He doesn't even need a circuit board if he puts the R inline with the (+) wire.

I *knew* that was coming. :-)

I was drawing a comparison between the 1 chip + 1 R
versus the 1 chip + 1 R + TL431 circuit.



For nice, the taper idea is great. Thanks for bringing
it up.


The taper idea is essential if you want both a 12-hour charge and plug-in and forget it capability.

Yes!

Ed


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