Calculating resistor rating for simulated load of X amps.
Using the charging ports, I'd like to use a resistor to simulate a ~35 amp load on each of the 16, ~4 volt li-ion parallel subpacks separately, measuring the voltage drop in order to help isolate any bad cells.
I only know one of the three variables in ohm's law -- the battery voltage. So how do I calculate the resistance needed?
I doubt the charging port is meant to handle the level of current you're talking about.
- David Herron, http://www.7gen.com/
Good point, David. I ran 18 gauge wires to these XLR ports, so I figured it'd be fine for short tests to 30 amps or so. But perhaps like Patrick says, I can get the same comparative results using 10 - 20 amps.
--xyster
moderator
In any case that will be a big resistor. lets see 15 amps at 3.6 volts. R=E/I 3.6/15=.24 ohms and 3.6*15=54 watts. http://www.digikey.com/scripts/dksearch/dksus.dll?Detail?Ref=169700&Row=780895&Site=US two of these in parallel would work out nicely. Kind of expensive at 7 bucks a piece, and becareful theyll get hot.
Joe
Thank you for the input, guys. I realized with a twist of the throttle my plain-jane DVM, connected to wires from an xlr plug plugged to any one of the jacks will perform the same function.
I just received a new order of 30 18650's, so it's time to go to work soldering another 2s15p, he he :-)
--xyster
moderator
power resistors typically have lousy temperature coefficients, so when the resistor gets hot its value will change thus throwing off your measurement. as the battery voltage lowers the current will drop, another source of error. it is very easy to build a load that will be constant and not move around with temp and voltage. all you need is an op amp and a fet that can dissipate the power you need, and a sense resistor. Suppose you have a 10 milliohm sense resistor, that would have 350 mv across it at 35A. you use a fet to control the current through the sense resistor, then you feed the voltage across the resistor to the summing input of the op amp and 350 mv to the other input and you have a programmable load. you can cut down on the power dissipation of the FET(s) by using a series dropping resistor. 4 volts will push 35A through .114 ohms, so if you used a .1 ohm 150W resistor in series with the FET then the FET would only have to dissipate 20W so a small heat sink would be enough. The FET would keep the current at a constant 35A because the sense resistor has a better tempco and it will not heat up anyway.
-bob
Good idea building a constant current source, i didnt even consider the fact that the load wouldnt be constant. Would be a lot more expensive than just using a resistor. I usually just use a zener diode supplying a constant voltage to the base of a transistor to do this.
A bipolar transistor would serve better than a fet using this method. The minimum voltage you can apply to the CC load is the sense resistor voltage plus the gate to drain or base to emitter voltage. In most fets you either have 10 volts or 5 volts gate to drain plus a certain hundred or so millivolts in the resistor. With a bipolar you have 1.2~ volts plus the sense resistor. Most big bipolar transistors also have low gain so a seperate transistor hooked up in darlington will probably be needed. Your method using an op-amp would get rid of this problem but the op-amp would end up needed a power supply other than the batteries being drained themselves, a 9 volt battery would suffice and would power an op-amp for a long time. Just a thought.
Joe
You do know the other variable - you want to draw 35 amps. So, 4/35, or a little more than a tenth of an ohm. Since you'll be pulling 35 amps x 4 volts, that will need to be a 140 watt resistor.
To make things a little easier, I would try for a 10 to 20 amp load. Since you're comparing packs, that should give you just about the same results.