# Stall current calculation for DC motor

I'm trying to calculate the stall current for a 12 V DC gear motor (919D1481):

No load speed = 15800 RPM

No load current = 0.52 A

Nominal speed = 13360 RPM

Nominal current = 2.85 A

Nominal torque = 154.4 g·cm

Stall torque = 1000 g·cm 

\begin{align} \Delta I &= 2.85\mathrm{A}-0.52\mathrm{A}\\ &=2.33\mathrm{A} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \end{align}

\begin{align} \Delta\mathcal{T} &= 1000\mathrm{g·cm}-154.4\mathrm{g·cm}\\ &=845.6\mathrm{g·cm} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \end{align}

\begin{align} I_{stall} &= \left(\frac{845.6\mathrm{g·cm}}{154.4\mathrm{g·cm}}\right)*2.33\mathrm{A}\\ &=12.76\mathrm{A} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \end{align}

Can you tell me if this is correct?

Thanks.

For a crosscheck, see the rest of the datasheet for winding resistance R (or measure several times, rotating the shaft very slowly to find the minimum brush resistance) then I(stall) = V/R.

• What would the formula be if the gearbox is taken into account? – somers Jan 2 '19 at 14:48
• No winding resistance value in the datasheet unfortunately mfacomodrills.com/pdfs/gbxbrch1.pdf – somers Jan 2 '19 at 15:09
• How could the gearbox change the DC resistance? Gearing trades higher torque for lower speed. – Brian Drummond Jan 2 '19 at 16:30

The calculated stall current seems reasonable, but the method of computation does not.

The data table is for performance of the motor only. It does not include a gearbox. There is no reason to consider the gearbox in estimating the stall current. If the no-load data was for a motor with a gearbox, the gearbox would be important. The stall current would be the same with or without the gear box.

The no-load loss is the total input power, 12 V x 0.52 A = 6.42 W. If friction is neglected, R = 6.24/0.52^2 = 23.1 ohms.

At maximum power, the input power is 12 V x 2.85 A = 34.2 W. Since the output power is 21.2 W, the total losses are 34.2 - 21.2 = 13 W. If friction is neglected, R = 13/2.85^2 = 1.6 ohms and stall current is 12/1.6 = 7.5 A.

Since neglecting friction gives two very different values for R, friction can not be neglected.

If the stall current is 12.76 amps, the armature resistance is 12 V / 12.76 A = 0.94 ohms, The resistive loss at no-load is 0.52^2 x 0.94 = 0.25 W, making the friction loss 5.99 W. At max efficiency, the resistive loss is 2.85^2 x 0.94 = 7.64 W. That makes the friction loss 5.86 watts.

That makes 12.76 A a reasonable estimate for stall current.

Here is another version of the motor data to be considered:

This is a reasonable approximation if the gearbox is ignored. Some portion of the 0.52A no-load current is due to the torque required to overcome friction, so the actual torque experienced by the stalled motor will be a little higher at stall than the calculated stall torque.

Remember that as you approach stall the motor will be less efficient as the winding losses increase. At stall you are 0% efficient so all of the ~150W of stall power will be converted to heat in the windings. It is likely that the motor cannot be run continuously at loads much above the nominal load.

• How would I go about the calculation taking the gearbox into account? – somers Jan 2 '19 at 14:35
• My understanding is that the gearing won't change the current value - it'll just allow a higher torque for the same current. – somers Jan 2 '19 at 15:18
• Since the table gives only theoretical results (100% efficient gears), a rule of thumb is that a good quality, small gearbox loses about 5% per stage; for example, a three-stage gearbox will be around 85% efficient. This efficiency may be lower for inexpensive gears. – John Birckhead Jan 2 '19 at 17:06