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How to determine back EMF voltage?

Posting this question publicly so that others may benefit but is there a way to, on paper, calculate the expected peak reverse voltage from a set of motors that start to idle? I've Googled around and there's a whole bunch of stuff around Lenz law and induction but I can't seem to nail a theoretical way to do this.

The reason for asking is that using the back EMF for LED effects is pretty cool but it'd be useful to know what sort of voltage to expect from motors for spec'ing auxiliary components like resistors. When we did Skuzzer's Desolator, we just shoved a 1kR in there and called it day. It seems to work and we've not had a problem since but it was a bit of a kludge.

I know I could go and get the Oscilloscope out and determine it experimentally but if there is a way, it'd be hand to have it around. If not, I'll go get the scope and publish a table of peak voltages and battery combinations. Smile

Back EMF increases with motor speed - it's effectively just the electrical potential that the motor generates as it spins (motors and generators are effectively one and the same - they may be designed/optimised for one role or the other but they are still capable of performing as either). This is basically Faraday's Law which states that the EMF induced is equal to the rate of change of magnetic flux through the circuit - you spin the motor faster and the coils "cut" through more lines of flux (between the magnets) per unit time so the EMF induced is greater. On top of that Lenz's Law states that the EMF induced opposes the change that produced it - in other words if you're powering a motor with a forward voltage the motor will generate a voltage in the opposite direction to oppose that forward voltage - this is basically what "back EMF" is and at no-load it'll be almost the same as the supply voltage (and, to be honest, for resistor calcs that's probably all you need to know).

You can calculate it more precisely by using the stall current and supply voltage to find the coil resistance (since at stall there is no back EMF generated) and then using no-load current (or the current at a given load) and coil resistance to find the "effective voltage" across the coil at that load. The back EMF generated will be the supply voltage minus the "effective voltage". The only time the back EMF generated will be greater than the supply voltage is if you suddenly open a switch and the current flowing through the motor coils has no way of dissipating - the coil is an inductor and will try to resist the sudden change in current resulting in a voltage spike and/or arcing across contacts/inside the FET. If you use a flyback diode the current has a path to freewheel and safely dissipate which prevents this spike from occuring.

Using Hellcats as an example...

At a nominal 11.1V:
Stall current = 22.02A
No-load current = 0.46A

From R=V/I:
Coil resistance = supply voltage / stall current = 11.1V / 22.02A = 0.504Ω

From V=I*R:
"Effective voltage" at no-load = no-load current * coil resistance = 0.46A * 0.504Ω = 0.23V

From Eb=V-I*R
Back EMF at no-load = supply voltage - "effective voltage" at no-load = 11.1V - 0.23V = 10.87V

That's using one motor but if you use two identical motors in parallel the resistance will be halved and the current will be doubled meaning the back EMF generated will be the same.

Like I say though, that's only 0.23V less than supply voltage and in the grand scheme of things when calculating LED series resistance I'd argue it simply doesn't matter. The difference between nominal voltage and fresh-off-the-charger voltage for most packs far exceeds that difference, as does the variation in voltage drop as a result of the (commonly 5% or 10%) tolerance in resistor values.

If you use a 3S pack to power the motors aswell as some 2V 20mA LEDs the difference between fully charged and nominal is 1.5V. On top of that a 530Ω resistor (assuming you aim for 12.6V as a worst case for resistor calcs - ((12.6V-2V)/0.02A=530Ω) with a 5% tolerance band could actually have a resistance of anywhere between 504Ω and 557Ω - assuming a 20mA constant draw from the LED that's a difference in voltage drop of up to 1.06V from two supposedly identical resistors.

The back EMF at no-load will vary a bit depending on the motors you use but, for the most part, aiming for supply voltage should be sufficient (if you want the LEDs to turn on with the motors, and slowly fade out when you release the switch, they will have to be specced to run at supply voltage anyway).

For comparison:

Rhinos no-load back EMF at 11.1V = 10.73V
Honey Badger no-load back EMF at 7.4V = 6.79V
Honey Badger no-load back EMF at 11.1V = 10.19V
Wolverine no-load back EMF at 7.4V = 7.25V
Wolverine no-load back EMF at 11.1V = 10.87V
FK180SH-3240 no-load back EMF at 7.4V = 7.19V

Huh, so the 800V flyback diodes we use are a touch over kill then? Very Happy Thanks for that, saved me a lot of time. Remind me to buy you a drink if we're ever at a war.

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