Hello,

I had 14.8 V 2200mAh 25C lithium battery. I need to discharge the battery to it cut off voltage lets say 12.5V using resistor. I did various calculations in finding the resistor. Can anyone tell me how to select the resistor value to discharge the battery

HI Sjkrbana,

Welcome to the community!

The way I would go about this is decide what wattage resistor I want to go with and go backwards since then you have two variables which you could then solve for amperage and then after that be able to solve for ohms.

Watts (joules per second) = Voltage (joules per coulomb) x Amperage (coulombs per second)

Say, if you decided to go with a 25W resistor, you have two variables and you can solve for 25W = 14.8V x A

Now solve for A

25W/14.8V = 1.69A

Now use that to calculate resistance in ohms.

Resistance would be 14.8V / 1.69A = 8.75 Ohms

But you don’t want to max out your resistor’s wattage, so derate with say, a 12 Ohm Resistor, which would run at 1.2A, or go with a 15 Ohm 25W resistor, which would be a little under 1A and 15 Watts.

If you don’t want to spend money on a 25W resistor and have more time on your hands decrease the wattage to calculate how much you need to increase your ohms or do some math for resistors in parallel and twist a whole bunch of wire wound resistors together.

So, here we gonna take 25W as just as an assumption and further we can reduce the wattage according to the resistors we had. I see 25C rating given in the data sheet. Do we not to take into account while getting the value of the resistor.

I did in the following way:

Current = 25C x 2.2 A gives 55 A

Power ( watts) = V x I = 55 A x 14.8 V = 814 W

Power = (Current x Current) x Resistance

814 W = (55 x 55) x R

Resistance = 814/ 3025 = 168.12 ohms (Consider 200 ohms)

I am confused in getting the resistance value to discharge the battery because of various math equations to find the resistance.

Hi sjkrbana,

Thank you for your follow up.

A Battery’s discharge rating, like 25C wouldn’t calculate into this equation. 1C is the time it takes the current to draw the battery down to empty in 1 hour. If you’re trying to decide how fast you want to discharge and you’re going with 25C that would be:

60 minutes / 25 = 2.4 minutes

But that’s full discharge. You’re going to 12.5V (not sure what the cutoff voltage is of your battery)

The reason I started with wattage like 25 Watts is we’re talking about using a $3 resistor vs. a $15-40 resistor, and possibly a heat sink/aluminum bar that goes along with the resistor for heat dissipation.

There are different ways to do this equation.

What you’re doing is deciding what amperage to draw and then leaving wattage as your variable which is also acceptable.

Again, for for simpler explaination on calculating amperage.

25W = 14.8V x A

25W / 14.8V = A

25W/14.8V = 1.69A

The process would be the same for calculating wattage and easier because you just multiply V x A.

I should also mention usually batteries when driven at full 25C type ratings do not deliver their full mAh rating. Your battery might go to 1000mAH or 1200mAH if you’re drawing at 25C, it depends upon the battery.

I am very thankful to you for your explanation. So, in the same way I had 4 batteries so I can take the 25W as the minimum watts and can calculate the resistor value to discharge the batteries.

The only thing we choose here 25W resistor because of its low cost when compared to the high wattage resistors.

I do have 4 batteries of 2 LiPo and 2 lead acid.

LiPo:

14.8 V, 2200mAh at 25C with maximum continuous current 55A with 32.6 Wh

11.1 V, 1300mAh at 25C but with 14.43 Wh

Lead acid:

12 V, 1.2 Ah

12V, 1.2 Ah (both having intial currents of 0.36A)

So here I can take 25 W for every battery and can calculate the current and resistor value.

Thanks and Regards

Kranthi Kumar