Motor Maths

In a previous post I showed the pitfalls of constant off time choppers like the A3977. Basically you have to set the off time long enough to be able to deliver the lowest current step of the microstepping, otherwise the steps are not equally spaced.

Forrest raised the point of how do you do that without a scope. It is fairly easy to calculate from the target current, supply voltage and motor resistance using nothing more complex than Ohm's law. It did take me a few days to come up with a formula that matched my measurements, but that was because I was accidentally running the chip without synchronous rectification enabled.

The motor current is equal to the reference voltage divided by 8 times the sense resistor. The maximum sense voltage is 0.5V, so a sensible value for the sense resistors is 0.2Ω, giving 2.5A maximum with 4V at the reference pin. My lash up uses two 0.5Ω resistors in parallel giving a 2A maximum.

The minimum current required on the first step of the microstep will be I × sin(π/2n), where n is the number of microsteps. In this case n is 8 so the smallest current step is 19.5% of the full current. To calculate the minimum off time needed we need to be able to work out what the duty cycle will be to get a given current.

Here is what the sense resistor waveform looks like when the current is set to 1A and the motor is stationary. The on period is 3μS and the off period is 20μS. The supply voltage is 12V.



The sloping top of the waveform is actually an exponential curve, but at this scale it is very close to linear and to simplify the calculations I have just used the average value.

So we know that 1A flows from the supply for every 3 out of 23μS. That gives an average current from the supply of 1A × 3 /23 = 130mA. Indeed the supply current measures 260mA as there are two coils energised (I set it to full step mode to make this measurement).

When the chopper is on energy flows into the inductance of the motor, increasing its magnetic field slightly. During the off time the current flows in a loop consisting of the motor and the two low side transistors. Power is dissipated by the motor's resistance, so it loses energy by its magnetic field decreasing slightly. We can calculate the duty cycle by reasoning that the energy going in during the on state must equal the energy coming out in the off state.

The motor is a Lin 4118S-62-07 NEMA17 motor I got from Makerbot. It has a coil resistance of only 0.8Ω. That means the resistance of the sense resistor and the on resistance of the FETs in the chip are significant in the calculation.

During the on state current flows through one top transistor, the coil, one bottom transistor and the sense resistor. All the resistances convert electricity to heat so the power going into the magnetic field is the power drawn from the supply minus the resistive losses in the circuit.

Power = VI or I²R, Energy = PT.

So we have (V_supply × I - I² × (R_motor + R_sense + R_{DS(on) source} +R_{DS(on) sink})) × T_on.

In this example (12 - (0.8 + 0.25 + 0.36 + 0.45)) * 3 = 30.4 μJ.

During the off state the current flows through the motor resistance and two low side transistors, so the energy lost is: I² × (R_motor + 2 × R_{DS(on) sink}) × T_off.

In this example (0.8 + 2 × 0.36) × 20 = 30.4 μJ, so theory matches practice (using typical values from the datasheet for R_DS(on)), always very satisfying.

So if we call the total resistance in the circuit with the switch on R_on and the total when it is off R_off we have: -

R_on = R_motor + R_sense + R_{DS(on) source} +R_{DS(on) sink} = 1.86Ω
R_off = R_motor + 2 × R_{DS(on) sink} = 1.52Ω

Then T_off = T_on (V/I - R_on) / R_off

For our example if we set the minimum T_on (T_blank) to be 1μS, I = 0.195A, so T_off is 39μS.

At 1A T_on will then be ~6μS. So the minimum chopping frequency will be ~22kHz and the maximum will be 25kHz.

C_T = T_blank / 1400 = 714pF, so use 680pF.
R_T= T_off / C_T =58K, so use 62K.

So in conclusion using the simple formulas above it is easy to calculate the correct values for a given motor, supply voltage and minimum current. I wish the datasheet and apps note had included this formula.