# Heat vs derating power resistor Lithium battery discharge load tester

I am making a Lithium battery discharger/tester to test some 18650 batteries that I found in some old laptop cells because I am too poor to buy some brand new battery which cost more than 5 USD each. Using Two 20 ohm 1 w 2512 chip resistors, it gives me a sweet little tiny 10 ohm 2W resistor. A lithium battery at full charge is 4.2v. With my calculation, the maximum current should be less than 4.2v/10 ohm = 420mA; or With Power <=RI^2= 10 0.42^2= 1.76 watt. So a 2w resistor should be able to deal with a 1.76w load. Unfortunate, it's not the case. My resistors are extremely hot to touch (hot glue melting kind of hot). After running for a hour or so, even the other side of the PCB is burning hot to touch. But to my release, other than being super hot (oh boy, I wish my gf is that hot too), I don't find any problem. First of all, the current for my testing battery is constant at 360ma, so this really tells me that the resistance is not changing much. Second of all, nothing is burning and after removing the load and read the value of those 1% accuracy resistors, they still read 10 ohm.

Here we go, after my boring story, my first question, can heat really kill resistors (especially this kind of resistor) when operated below its rated wattage? To my own understanding, heat does kill the resistor eventually, but it should last pretty long time? Third question, what is an ideal load for discharging a 4.2v lithium battery @ about 500ma?

After soldering a huge nail to one size of the pad as a heat-sink as others have suggested, this helps a little if not at all. The resistors themselves are still super hot, while the whole nail is hot to touch. So My conclusion is that huge PCB trace is simply not enough to dissipate the heat generated by these two tiny resistors.

• My guess is that these resistors can dissipate 1 W if you keep them cool enough. Usually they are soldered with direct contact to a large copper area on the PCB which then acts as a heatsink. Your wires are very small and will have very limited heatsinking capacity. If you have some pieces of copper, for example 2x2 cm, solder these to the resistor's contacts to act as a heatsink. In the long run the lifetime of the resistors will be less when you make them this hot. Also the solder joints might deteriorate. I would keep 90 degrees Celcius as a maximum. Mar 13, 2017 at 15:08
• If you are not already, try directing a fan at the board to help cool the board and resistors. Also, make sure both sides of the PCB are exposed to the air flow. Lift it up off of the table a bit, or rotate it so the PCB is vertical. Even if you don't have a fan, exposing both sides of the PCB will help. Mar 13, 2017 at 15:22
• After soldering a big nail to the resistor's contacts, the resistors are still very hot (able to melt hot glue. Probably cooler than before), while the nail is also hot to touch after 5 minutes. Maybe I need to use more resistors? 5x 50 ohm 1w to form a 5w, 10 Ohm resistor?
– user83582
Mar 13, 2017 at 18:48

You are running the 1 W resistors at 925 mW. That's within spec, but barely. They will get quite hot, but should be OK as along as you keep the board open to ambient air as you show in your picture. Normal office air temperature should be OK.

Resistors aren't silicon semiconductors that stop working at 150°C. Almost certainly those resistors are getting hot enough to boil water. However, that's OK. They should be able to handle it.

Again, make sure they remain open to the air, and everything should be fine. These resistors are meant to operate at high temperature at their rated power. You can point a small fan at your board if you don't want the board and things around it to get so hot.

An ideal load for discharging a battery at a constant level is a constant current sink (or constant current load). Usually built with an opamp and a transistor (which gets hot) and a current sense resistor and a voltage reference.

Resistors can be destroyed even when operating below the rated wattage if the cooling is not good enough. The copper connection to the resistor plays a big role in cooling. So a fat trace or even a plane can help to cool the resistor down.

In a datasheet you'll find typical values to calculate the temperature of your resistor based on certain copper areas connected to the pads. (actually quite hard to find)

Most resistors are able to handle their rated wattage at an ambient temperature of 70 °C, so they are designed to get burning hot. Vishay power resistors for example are able to handle 170 °C.

FR4 is able to handle 115-140 °C, so I'd still try to spread the heat with some copper areas around the resistor pads (connected to the pad) to reduce the temperature of the FR4. If you can, areas on both sides connected with vias will help even more to spread the heat and increase the cooling effect.

• Seems like it will always be cheaper to dissipate power in a resistor rather than in silicon, even in a constant current load. In a transistor you will always be up against the hard limit of the maximum junction temperature. Just use the active circuitry to control the current, but put most of the voltage drop across a resistor. Mar 13, 2017 at 15:26
• @mkeith yeah that's probably true. If you have a specific input and load scenario you probably can adjust the design in a way to reduce the power of the transistor. An all purpose constant current sink based on resistors will be difficult? (wasn't asked for here, but came to my mind immediately) Mar 13, 2017 at 15:32
• Yes. I agree. I have great admiration for the people who design electronic loads and power supplies which can sink current. Mar 13, 2017 at 15:43