I wanted to measure the internal resistance (IR) of my Hobbyking 4S batteries. I went about this two ways.
1) I made up a wooden board with ten 12v 50w lamps, with pairs of lamps in series, five of these pairs in parallel. I paired the lamps because I didn't want the 16.8v of the battery to burn out the 12v lamps. When I give it 17 volts (fully charged lipo) it flows 17.4 amps, so the resistance is very nearly one ohm.
Then I measured voltage sag; on each of eight batteries, that was between 0.9 and 1.2 volts. Ohms law (Vsag/amps) gives me an internal resistance of between 53 and 70 milliohms. That's for a four-cell battery, so it's about 15 milliohms per cell.
2) I got a 7 in 1 Turnigy megameter, costing about $40. I powered it up from one battery, and used the twin-pin probe to measure individual cell IR. I got values of around 100 milliohms per cell, meaning 400 millohms per battery. That's several times as much as measured with the lightbulb method.
So, now some sums.
If the IR is 400 milliohms (as per Turnigy meter), and my bike pulls 20 amps, then that means that there's an 8 volt drop across the battery, and the battery dissipates 160 watts
If the IR is 60 milliohms (as per light bulb test), and my bike pulls 20 amps, then that means that there's an 1.2 volt drop across the battery (I times R), and the battery dissipates 24 watts (I squared times R).
I have a voltmeter on the handlebars, so I *know* I'm not getting an 8 volt drop. Whereas 1.2 volt drop is about right (maybe it's actually a bit higher).
So.
Either I've misunderstood how all this works, or the Turnigy is doing something very clever whereas my light bulbs are measuring pure resistive load, or else the Turnigy gives me readings that are way high compared to reality.
I'm keen to hear other folks opinions on this, especially if I've buggered up my calculations or failed to understand something here.
1) I made up a wooden board with ten 12v 50w lamps, with pairs of lamps in series, five of these pairs in parallel. I paired the lamps because I didn't want the 16.8v of the battery to burn out the 12v lamps. When I give it 17 volts (fully charged lipo) it flows 17.4 amps, so the resistance is very nearly one ohm.
Then I measured voltage sag; on each of eight batteries, that was between 0.9 and 1.2 volts. Ohms law (Vsag/amps) gives me an internal resistance of between 53 and 70 milliohms. That's for a four-cell battery, so it's about 15 milliohms per cell.
2) I got a 7 in 1 Turnigy megameter, costing about $40. I powered it up from one battery, and used the twin-pin probe to measure individual cell IR. I got values of around 100 milliohms per cell, meaning 400 millohms per battery. That's several times as much as measured with the lightbulb method.
So, now some sums.
If the IR is 400 milliohms (as per Turnigy meter), and my bike pulls 20 amps, then that means that there's an 8 volt drop across the battery, and the battery dissipates 160 watts
If the IR is 60 milliohms (as per light bulb test), and my bike pulls 20 amps, then that means that there's an 1.2 volt drop across the battery (I times R), and the battery dissipates 24 watts (I squared times R).
I have a voltmeter on the handlebars, so I *know* I'm not getting an 8 volt drop. Whereas 1.2 volt drop is about right (maybe it's actually a bit higher).
So.
Either I've misunderstood how all this works, or the Turnigy is doing something very clever whereas my light bulbs are measuring pure resistive load, or else the Turnigy gives me readings that are way high compared to reality.
I'm keen to hear other folks opinions on this, especially if I've buggered up my calculations or failed to understand something here.
