: IMO, that's largely because your requirement was vaguely specified in
: your original post. It wasn't until 4 days later that you became more
--Yeah, well, I'm still workin' on the "basics"...
: Doesn't that imply that, smoothed or pulsed, a 12V ('120V'!) output
: from your Variac will be inadequate? I'm guessing that you really want
: a heavy duty DC supply capable of delivering about 26V, preferably
: switchable to say 15V for charging your motorcycle batteries if they
: are 12V (you still haven't specified that).
--Yes. I've been turning the dial on the Variac until the
output, as measured on a voltmeter, has been around 14 to 15v, then
I've hooked that up to the battery I'm trying to charge. I've *never*
cranked the dial past this, for fear of damaging the battery, the
house, myself, etc.
There's nothing wrong with that method, but you neeed to monitor a
couple of things; the voltage across the battery when it's charging
and the current going into it while it's charging, like this:
VARIAC FWB
+--------+ +-----+
ACHOT>---O--|C|<--O----|~ +|---[AMMETER]--+------< <---+
| |O| | | | | |
| |I| | | | | |
| |L| | | | [VOLTMETER] [BATTERY]
| | | | | | |
ACNEUT>--O---+----O----|~ -|--------------+------< <---+
+--------+ +-----+
If you don't have a couple of meters which you can set up to take the
readings simultaneously, and since you have a VARIAC and can use it to
adjust the current to whatever you want, you can do the readings
sequentially (without having to tear down and rebuild anything) by
substituting a resistor for the ammeter (actually, I prefer this
method) and using the voltmeter to measure the voltage drop across it
to determine the current going through it, (and into the battery) like
this, :
VARIAC FWB
+--------+ +-----+
ACHOT>---O--|C|<--O----|~ +|--+--------------------< <---+
| |O| | | | | |
| |I| | | |V O<--O----------+ |+
| |L| | | | |+ [BATTERY]
| | | | |I O [VOLTMETER] |
| | | | | | | |
ACNEUT>--O---+----O----|~ -|--+--[RESISTOR]--+-----< <---+
+--------+ +-----+
To size the resistor, decide what value of current you want the
voltage indicated by the meter to represent, and the use Ohm's law.
Let's say that you want the meter to read 1 volt when there's one amp
flowing through the resistor. Then you would write:
E 1V
R = --- = ---- = 1 ohm
I 1A
You need to be concerned about the wattage, because the higher the
current the hotter the resistor will get.
For typical motorcycle batteries I think the initial xcharging current
is in the range of 2 to 3 amps, so if you were to use a 1 ohm resistor
you'd have:
P = I²R + (3A)² * 1R = 9 watts
being dissipated by the resistor when the current was first set at 3
amps. The current would decrease as charging progressed, but it would
be a good idea to have a hefty resistor in there. A really cheap way
to get 20 watts worth of resistance would be to get four 1 ohm 5 watt
"cement" resistors and wire then like this:
+-----+---->A
| |
[1R] [1R]
| |
+ +
| |
[1R] [1R]
| |
+-----+---->B
or like this:
+-----+---->A
| |
[1R] [1R]
| |
+-----+
| |
[1R] [1R]
| |
+-----+---->B
In either case, the resistance from A to B will be one ohm, and either
array will dissipate 20 W if the resistors are mounted properly. For
the cheap "cement" resistors, I'm pretty sure just letting them
radiate into ambient air will be fine.
Since your NiMH battery packs may want to be charged from a current
source, the sizing of the resistor may go differently from that for
lead acid. What can you tell us about the battery packs?
Also, what can you tell us about your VARIAC, in terms of how much
current it's rated to put out?
Also notice that one of the mains inputs to the VARIAC is labelled
"ACNEUT". That means "AC NEUTRAL" and you need to be ABSOLUTELY,
POSITIVELY, SURE that the rig is connected to the mains that way. If
you don't, you could be killed if you came in contact with what you
though was "ground" while you were grounded.