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Modifiying a rechargable battery-powered device - some questions |
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| message from D.M. Procida on 9 May 2004 |
I'm cannibalising a fairly simple battery-powered machine. It has a
motor, a switch, and three 1.2V rechargable cells, wired in series. As
it stands, the motor spins much too fast - I want it to spin at about
1/3 of the speed. So presumably I could remove two of the batteries -
would that work?
I don't even know whether they're NiCad or NiMH - they don't say.
Could I wire them in parallel to make them run longer, or would that be
the kind of thing which makes them catch fire?
The current charger puts out 5.3VDC at 140mA. I'm surprised it's being
used for a 3.6V battery, but then I don't know much about rechargable
batteries. I presume that connecting it up to a single cell, or even
two, instead of the three it currently services, would not be a good
idea. Is that right?
Finally, how do these chargers work? I believe this thing was originally
intended to be left plugged in all the time to recharge. How does the
charger know when to stop charging the batteries? Could I substitute a
lower-voltage AC adaptor for this one, or is the charger a bit more
complicated than an AC adaptor I might find off-the-shelf?
Thanks for any help.
Daniele
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| Ian Stirling replied to D.M. Procida on 09 May 2004 |
Maybe.
Is this a cordless screwdriver?
If so, you may be dissapointed.
My 2.4V screwdriver will not reliably start on one cell.
A quick test would be to try and see if the motor ran with only one
cell.
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| D.M. Procida replied to Ian Stirling on 9 May 2004 |
In fact it won't run with only one cell, but runs with two at just about
exactly the right speed.
If I replace the three AA cells with two C cells and use the existing
charger, am I likely to damage them?
Is it a good guess that the charger that I already have is a slow
trickle-charger, since it is intended to be left plugged in all the
time?
And what's a charger rated at 5.3V/140mA likely to do with to two C
cells instead of three AA cells?
Daniele
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| Ian Stirling replied to D.M. Procida on 09 May 2004 |
Maybe.
It could do anything from explode them, to not charge them.
Something in between is likely.
Do you own a multimeter?
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| D.M. Procida replied to Ian Stirling on 10 May 2004 |
Absolutely. In fact I was going to see what the charger was putting out
with nothing attached, and with fully-charged batteries attached, to
provide more information on what I've got here.
Is checking for current with the meter in series going to short-circuit
the charger, or are they usuallly designed to cope with that kind of
state?
Daniele
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| Dave Plowman replied to D.M. Procida on 10 May 2004 |
In article
Measuring the off load state is pointless.
This would be a very good idea. Many cheap chargers consist of nothing
more than an unregulated DC source with a series resistor between it and
the battery to vaguely limit the current. You can make a constant current
charger with one transistor, and a couple of diodes and resistors for
under a quid. The input voltage is then not critical.
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| Ian Stirling replied to D.M. Procida on 10 May 2004 |
Don't measure the current across the charger.
First, measure the charge current when charging three cells.
(use the "10A" range) Now, measure the current when using two cells.
The meter should be in series with the battery and charger.
If the current exceeds 140ma, you probably need to add a resistor.
If you don't care about charge length, then a resistor alone will be
fine.
NiCd cells are fine being charged forever at a rate that will charge
them in 14 hours (C/10, for a capacity of 700mah, 70ma).
NiMH cells seem to require shut-off for reasonable service life (years)
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| D.M. Procida replied to Ian Stirling on 10 May 2004 |
With three cells, it's 170mA (though on the 200mA range it read 150mA).
With two cells, it's 220mA (though on the 200mA range it read 180mA).
With three cells, it's 280mA (no reading on the 200mA range).
Any suggestions for what sort resistor would be suitable for charging
two cells then?
Thanks,
Daniele
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| Ian Stirling replied to D.M. Procida on 10 May 2004 |
On the last one, I'm assuming that you mean with one cell.
What does the charger say about how long should you charge the cells?
Are these AA cells?
As a guess, 170ma at 1.2V will be an equivalent resistance of 7 ohms or
so (1.2V/.17A).
This will drop 1.2V at 170ma, and effectively replace one cell.
So, either 7 ohms 1/4 watt (or higher) or 14 ohm 1/2 watt.
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| D.M. Procida replied to Ian Stirling on 10 May 2004 |
Yes.
They are, but I was thinking of replacing them with C cells.
I'm not sure how you worked that out - the output of the cells is a
nominal 1.2v, but that's not what the charger is doing. Or are you able
to assume that the potential across each cell will be 1.2v - in which
case, how are you able to know that?
Daniele
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| Pete C replied to D.M. Procida on 10 May 2004 |
Hi,
If the charger is not rated to supply 220mA and you want to charge 2
cells with 170mA, one way to do this would be to put 2 diodes (eg
1N4001 from Maplins) in series with the charger.
This would create a voltage 'drop' of 1.2V across the diodes, so the
load on the charger would be as if three batteries were attached to it
instead of two.
If using the diodes, the line on the body should point towards the
positive terminal of the batteries (or away from the negative terminal
of the batteries), as they only conduct in one direction.
Or, to make the motor run slower with 3 batteries, put one or two
diodes in series with the motor. The line on the body of the diode
should go away from the positive terminal of batteries in this case
(or towards from the negative terminal of the batteries).
cheers,
Pete.
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| Mike Faithfull replied to D.M. Procida on 10 May 2004 |
"D.M. Procida" <real-not-anti-spam-address@apple-juice.co.uk> wrote in
If you replace your AAs with 'C' size cells (which are rated at 2.5Amp-hours
if my failing memory is right for once), 200 - 250mA is probably about
right. So unless I've misunderstood your reply, it would appear you do not
need a series resistor for two cells. I'm not sure how the charger will
cope with delivering 220mA for 14 hours though, given that it was originally
intended to supply only about two thirds of that current ...
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| Dave Plowman replied to Ian Stirling on 10 May 2004 |
This isn't so. They're tolerant, but mustn't be left on a C/10 charge
permanently.
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| Ian Stirling replied to Dave Plowman on 11 May 2004 |
In practice, they last several years.
NiMH don't.
Taking them off charge when full is better practice.
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| Mike Faithfull replied to D.M. Procida on 9 May 2004 |
"D.M. Procida" <real-not-anti-spam-address@apple-juice.co.uk> wrote in
Without going into a lot of unnecessary technical detail, I have the
following observations for you:
1) wiring cells in parallel is generally not a great thing to do - if you
want more capacity for a given voltage, it's better to use larger cells.
2) Nickel Cadmium cells should be charged from a constant current source, so
your charger should show a high off-load voltage - say 15 volts - which will
drop to just a little more than the cell's voltage as it 'limits' to the
pre-set current appropriate for the cell being charged. Many (most?)
chargers that come with small tools/appliances etc. are rarely that
sophisticated, however, and a built down to a cost price.
3) Nickel Metal Hydride cells are very similar to NiCads but are a little
more "efficient" for a given size and weight.
4) If you simply try to re-use what you've got, it's likely your cells will
have a shorter life than you would wish for.
(Purists please don't nitpick - I've tried to keep it simple and not
confuse!)
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| D.M. Procida replied to Mike Faithfull on 9 May 2004 |
OK, then, perhaps a C-cell is what I need then, instead of three AAs.
What do they tend to do then?
Thanks,
Daniele
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| John Stumbles replied to D.M. Procida on 9 May 2004 |
"D.M. Procida" <real-not-anti-spam-address@apple-juice.co.uk> wrote in
Overcharge and cook the cells, shortening their life. You can charge a NiCd
at 1/10th its capacity (e.g. 50mA for a 500mAH cell) forever[1] with no ill
effect, but it'll take about 14 hours to recharge fully from flat. Higher
(/faster) charge currents (/times) without 'intelligent' circuitry to
terminate the charge when the battery's fully charged are bad news (and
common to cheap cordless tools).
[1] to a first approximation :-)
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| Ian Stirling replied to D.M. Procida on 10 May 2004 |
Because it is.
Due to the chemistry of the cell, the majority of the time it's on
charge it's at around 1.2-1.3V.
If the current with 3 cells is 170mA, and all cells are at 1.2V, if
you replace one with a resistor that drops 1.2V at 170ma, the overall
current won't change.
In that case, I'd pick a slightly larger resistor, in order to keep the
current below the charger nameplate current.
Some 140ma, perhaps 10 ohms/ 1/2W if you are using it on a 2 cell device.
This should take some 16 hours to charge 1600mAh cells fully.
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