From: mkfeil@bcrki9.bnr.ca (Max Feil)
Subject: Re: Battery Discharge Circuit For Tx/Rx
Organization: Bell-Northern Research
Date: Fri, 22 Jan 93 17:13:20 GMT

In article <1993Jan22.025009.20035@a.cs.okstate.edu> worley@a.cs.okstate.edu (WORLEY LAWRENCE JA) writes:
>                                                             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.

Right on.

>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 are forgetting the zener diode. In all circuits where there is a
predictable voltage drop (diodes, zener diodes, base-emitter junctions, etc)
you must subtract the diode voltage from the source voltage, so the correct
formula is:  R = (pack voltage - zener voltage) / current. 

In your example, this gives a resistor value closer to 2 ohms. Remember that
the zener diode is dissipating most of the power (dropping around 3.6 volts)
and the resistor drops the voltage by only another volt or so.

>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 accomp-
>lished by 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.

It's easier to think of this in terms of current. Power dissipated is
proportional to current squared, so you want to divide the current
with a current divider so that each half of the divider is only
dissipating one quarter of the original power. For a 10 ohm final value,
all you need is two 20 ohm values in parallel, where the 20 ohms is
made up of two 10 ohm resistors. This is essentially what you said. 
However since we know we only need a 2 ohm resistor, the total power it
needs to dissipate is only .5 watts - which is easy to find, i.e. only
one resistor is needed.

>                                                             Be sure that
>the diode can handle about one amp of current with no problem, or else it
>will blow!

Yes. You want at least a .5x3.6 = 1.8, let's say 2 watt zener. It is dissipating
most of the total 2.5 watts.

>Good luck-

Max
--
Max Feil  mkfeil@bnr.ca | What do I know? I'm just a Nerd on the Big Ranch.
Bell-Northern Research  |------------------------------------------------------
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