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- Newsgroups: rec.models.rc
- Path: sparky!uunet!cs.utexas.edu!torn!nott!bnrgate!bmerh85!bcrki9!mkfeil
- From: mkfeil@bcrki9.bnr.ca (Max Feil)
- Subject: Re: Battery Discharge Circuit For Tx/Rx
- Message-ID: <1993Jan22.171320.1398@bmerh85.bnr.ca>
- Sender: news@bmerh85.bnr.ca (Usenet News)
- Organization: Bell-Northern Research
- References: <C16Csu.3LG@news.rich.bnr.ca> <1993Jan22.025009.20035@a.cs.okstate.edu>
- Date: Fri, 22 Jan 93 17:13:20 GMT
- Lines: 58
-
- 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 |------------------------------------------------------
- CAD Software Development| "I have a funny feeling I know who he is: Mr. X"
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