# Battery Discharger

Discussion in 'Electronic Basics' started by LazarWulf, Jan 7, 2004.

1. ### LazarWulfGuest

I know this is easy, but yet it eludes me.
I am a Software Engineer by trade, currently studying electronics as a
hobby. I also happen to be an avid R/C car racer.

Here is the problem.
We discharge our 3300mah 7.2v NiMh batteries using 15 lightbulbs that
are connected in a series thus drawing about 30amps, and remove them
when the lights dim, to simulate the load of our R/C cars.

I want to make a discharger that discharges at 30amps without
lightbulbs(they break and are bulky), includes led voltage indicators
(i.e. 7v,6v,done), and most importantly cuts off at 5.4 volts.

Next I would like to make an equalization tray. This would drain each
individual cell to .9v and have an led that is lit per cell, while
there is MORE than .9v in the cell.

I can see the whole thing fabricated in my head. But my limited
knowledge is making it impossible for me to do.

Which common resistors do I use to total up to 30amp draw and cuts off
at 5.4v?
BTW they make dischargers that do this, however, they cost \$40 bucks
for the discharger and another \$40 for the equalizer tray!
Thanks.

2. ### happyhobitGuest

Based on your specification it would not be that easy.

It would require a 'smart' discharger.

Three comparators (one for each level indicated) controlling a mosfet that
could handle 30 amps. The total resistance, including the on resistance of
the mosfet, would have to be .24-ohm. The total heat you would have to
dissipate, mosfet and load resister, would be 216 watts. A resister is dumb.
It don't know when to turn off.

The discharger would start at 30 amps and end up at 22.5 amps when the
battery got to 5.4 volts. You could maintain the current at 30 amps but it
would require a constant current load. (More work)

You didn't specify a current draw for the equalizer.

You would need 7 comparators, or a microcontroller with 7 analog inputs and
7 digital outputs driving the indicator LED's. If you want it automatic you'
ll need 7 mosfets controlling the load resisters.

Do you want me to go on?

I don't think \$40 is that much.

Jay

3. ### happyhobitGuest

Oops, make that 6 of everything not 7. Sorry

Jay

4. ### Robert C MonsenGuest

For a fellow software engineer...

Here is a circuit for the discharger. The parts are probably probably \$15 at
mouser.

http://home.comcast.net/~rcmonsen/misc/discharge.gif

current sense resistors (you need 2, put them in parallel):

<http://www.mouser.com/index.cfm?handler=displayproduct&lstdispproductid=234
464&e_categoryid=15&e_pcodeid=58807>

Power mosfet:

<http://www.mouser.com/index.cfm?handler=displayproduct&lstdispproductid=202
098&e_categoryid=94&e_pcodeid=78101>

Big heat sink:

beats me. Get a big hunk of metal, and a fan. You'll be dissipating about
250W.

Regards,
Bob Monsen

5. ### Robert C MonsenGuest

If I understand what you need correctly, this circuit may do it.

It will discharge a cell until it gets down to 0.9V. While the cell is
discharging, it will light an LED. The power mosfet you use should be able
to conduct fully at Vcc-2 volts, since that's how high the opamp's output
can get. Again, you need a separate power supply.

The opamp comes in a DIP package of 4, so you should be able to build a set
of 4 with a single IC. In addition, there are ICs that contain darlington
arrays that can be used instead of the power mosfet. However, you'll need to
redesign the display logic by using a resistor between the opamp output and
the darlington base. Make sure the led is powered from the opamp side of
that resistor.

http://home.comcast.net/~rcmonsen/misc/equalizer.gif

Regards,
Bob Monsen