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So i've got a question about electrical motors.

I put 5V through a small motor and my lab bench PS reads 150mA. If I calculate this using ohms law, my motor resistance should be approximately 33Ω, but when measured, works out at 2Ω.

Is there an issue with my multimeter or am I in need of some more education?

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  • \$\begingroup\$ Welcome. A free-running motor consumes much less current than a loaded motor, and certainly MUCH less than a stalled motor. \$\endgroup\$
    – user105652
    Jul 13 '20 at 4:16
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If you measure the current with the motor stalled you'll probably get more like 2.5A, at least until the motor begins to smoke or your supply current-limits.

The simplest useful model of a DC motor is a resistance (the one you measured) in series with a voltage source that bucks the applied voltage as the motor spins. That "back EMF" is proportional to how fast the motor is spinning. If you were to spin the motor with another motor, you'd measure that voltage at the open motor terminals (it would act as a generator). The resistance represents the losses in the copper windings mostly.

If the motor was perfect, no friction in the bearings or air resistance on the spinning armature, the current with no load on the shaft would go down to zero as the motor winds up from a start. As it turns out, for your motor, it goes from 2.5A to 0.15A so about 94% of the way there. As you load the motor, the armature RPM drops and the current increases.

If your motor runs at (say) 5,000 RPM with no load, and 5V applied, there is 150mA so the back EMF must 5V-300mV = 4.7V. So back EMF = (RPM/5000)* 4.7V. If you load the shaft so that the RPM drops to 3,000 RPM, the back EMF will be 2.82V and therefore the current will be

I = (5V-2.82V)/2\$\Omega\$ = 1.09A.

Here is a speed/torque/efficiency curve for a random brushed DC motor (Mabuchi RS-555SH):

enter image description here

As you can see, as the speed drops (under load, with fixed voltage applied) the current increases. The \$\eta\$ (efficiency) peaks at relatively low current, but the maximum power output will be close to where \$\eta\$ = 50%.

If you start applying fast-varying voltage (as in microseconds or milliseconds) to the motor you also have to consider the inductance, but for slow-varying 'DC' it doesn't matter.

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  • \$\begingroup\$ This has been very helpful thankyou! I only have one question and that is how did you get the 300mV which you subtracted from 5V for the back EMF part? Is there a formula for this? \$\endgroup\$
    – user257446
    Jul 12 '20 at 6:01
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    \$\begingroup\$ 150mA * 2\$\Omega\$ = 300mV. Beware that handheld multimeters often have a fairly high offset on the lowest range, especially egregious below about 10 ohms. If you short the probes and get, say, 0.3 ohms, you can subtract that from the reading when you are measuring a low resistance, in order to get closer to the real resistance. \$\endgroup\$ Jul 12 '20 at 6:04
  • \$\begingroup\$ Commutation causes each coil to alternate functions as motoring coil or generating coil. Time curve for motor current starts at the stall current, where shaft speed is zero. Current goes down as speed goes up because the back-EMF voltage is produced by speed of motion in generating coils. The back-EMF rises as an exponential curve until it roughly equals the supply voltage to the motor if there is no load on the output shaft. So motor speed increases exponentially and motor current decreases exponentially over time as the motor spins up to no load maximum speed. \$\endgroup\$ Jul 12 '20 at 15:28
  • \$\begingroup\$ How do you tell that the mamimum power happens at ~50%? Is that inferred from the plot somehow or just a general characteristic of brushed motors? \$\endgroup\$
    – Austin
    Jul 13 '20 at 2:44
  • \$\begingroup\$ @Austin It's a general rule like for impedance matching. You can confirm that it works here, from the graph, by looking at the product of torque and RPM, which is, of course, proportional to power output at the shaft. It is maximum around 50% efficiency. \$\endgroup\$ Jul 13 '20 at 2:59
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Your measured value may not be very accurate but the actual value will be much less than 33ohms .The approx 2 ohm value represents the current drawn when the rotor speed and hence back emf are zero .If your power feed including and solid state components is rated at 2.5amps then the electronics will withstand a locked rotor .Remember that if 5VDC is suddenly applied at standstill the peak current could be 2.5 amp tapering down to your steady state value of 150mA after the full rotor speed has been reached .

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V = Eb + Ia.Ra

where

V = Applied voltage

Eb = Back EMF generated by the motor as it rotates (Eb α RPM)

Ia = Armature current

Ra = Armature resistance.

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