From: worley@a.cs.okstate.edu (WORLEY LAWRENCE JA)
Subject: Re: Battery Discharge Circuit For Tx/Rx
Organization: Oklahoma State University
Date: Fri, 22 Jan 93 02:50:09 GMT

From article <C16Csu.3LG@news.rich.bnr.ca>, by martyg@bnr.ca (Martin Gallant):
> I have just become the owner of a (new) Futaba 4-channel FM transmitter
> and corresponding receiver.  This is my first R/C radio set.  I would
> like to cycle both battery packs a couple of times before their first
> flight, to get an estimate of their capacity.
> 
> I have also been told this is a good idea for new NiCd's anyway.
> Opinions?
> 
> I need an inexpensive method to discharge the packs, ie. simple resistor
> and miniature light bulb method.  Can anyone supply a schematic and a
> parts list?  Is there an archive for this sort of information?

I have made several different dischargers.  The easiest one, and least 
expensive, consists of a zener-diode, and a power-resistor.  I am of the 
opinion that when discharging a pack of cells connected in series, you 
should not let the average cell voltage drop below .9volts per cell.  For
a 4-cell receiver pack, this works out to 3.6volts. 

I like to discharge 550mA nicads at their rated current, or about 1/2Amp.
This prevents any damage that can occur at higher discharge rates.  The way
to calculate the size of your power resistor is as follows:
Resistor-value(in ohms) = Pack-voltage/Max.current(in Amperes).  So, if you
have a 4-cell pack, it's "resting" voltage is around 4.8-5.5 volts, depending
on its state of charge.  Plug in R=5/.5 (500mA=.5A), you get 10 ohms.  So, 
my discharger for the receiver pack uses a 10 Ohm power resistor.  You must
also make sure your power resistor can dissipate sufficient heat.  This is
calculated as: power-needed = Pack-voltage x discharge-current.  In this
case, 5volts x .5amps = 2.5 Watts.  Now, if you used a 2.5 watt power resistor,
(assuming there was such a thing), it would get REAL hot.  I use the equivelant
of a 60 watt, 10-ohm resistor, to keep the generated down.  This is accomplishedby using four 10-ohm resistors, wiring two sets of two in parallel, and then
those two parallel sets in series.  By wiring two 10-ohm resistors in
parallel, you get a 5-ohm resistor that can dissipate twice as much power.
Wire two of these in series and you get a 10-ohm resistor that can handle
four times the power of just one of these resistors.  I use 15-watt 10-ohm
resistors.

Circuit:


                                                     Resistors
|===========|                                    --|=======|--
|  Battery  |          b                         |           | 
|           |----------/zener/---------------------|=======|--
|           | Pos.                                            | 
|           |                                    --|=======|--
|           |                                |---|           |
|           |                                |   --|=======|--
|           | Neg.                           |    
|===========|--------------------------------|

where --|=====|-- is a resistor, with all four connected in a series-parallel
configuration to give a 10-ohm load.  The zener diode is reverse-biased
(the band goes towards the battery) to only allow current to flow when the 
battery voltage is greater than the value of the zener diode.  In this case,
(4 1.2v cells) use a 3.6 volt or something close (3.2-4.0).  Be sure that
the diode can handle about one amp of current with no problem, or else it
will blow!

You can do without the zener diode if you just watch the battery voltage
while it's discharging.  Just unhook the circuit when voltage reaches
3.6v (4-cells) or 7.2v(8-cells).

Good luck-
Jason Worley