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I am in a muddle here, so thanks for making the effort to have a read.

I understand that electrical power going into motor is given by: \$P_e=I^2R\$

I understand that mechanical power out is given by: \$P_m=\tau\cdot\omega\$

Where \$\tau\$ = Torque at motor output shaft

Where \$\omega\$ = Speed at output shaft

I understand that for a DC motor, torque is proportial to current: \$\tau=k_E\cdot I\$

Therefore if current is proportional to voltage via resistance, and electrical power in is proportional to current, power is proportional to current.

Then imagine that a motor is reacting a fixed torque value, and the supply voltage is increased to raise the output speed. A fixed torque and a rising speed means a rising output power. However input power is fixed, since torque is fixed because current is fixed. Graphing this (x-supply voltage, y-Power) gets you a flat line for electrical power in and an angled line for mechanical power out. Therefore the lines must intersect and therefore on one side of the intersection point, output power must be greater then input power, which is impossible.

I am clearly overlooking something so any pointers would be greatly appreciated. Cheers now, all the best.

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  • \$\begingroup\$ The voltage and current are not independent, and the 'resistance' of the motor is not a fixed quantity either. \$\endgroup\$
    – pjc50
    May 28, 2013 at 20:26
  • \$\begingroup\$ Surely if you fix torque, current is then fixed. And assuming resistance stays 'almost' constant, then power is then fixed. So you still get the approx straight power line on the graph. The resistance of the motor would then have to vary massively to prevent the electrical power ever dropping below mechanical power on this graph. The voltage surely needn't come into the power equation - P=I^2*R. And current is directly proportional to torque. \$\endgroup\$ May 28, 2013 at 20:36
  • \$\begingroup\$ V=IR also always holds true (giving you P=VI), so I don't think you can assume the R holds constant. It's determined by the back EMF of the motor. \$\endgroup\$
    – pjc50
    May 28, 2013 at 20:50
  • \$\begingroup\$ Surely net voltage is determined by the back EMF of the motor, not R? I would have thought the only way R can vary is via heating - which I would guess would be slight? \$\endgroup\$ May 28, 2013 at 20:57
  • \$\begingroup\$ Given that net voltage = supply voltage - back emf \$\endgroup\$ May 28, 2013 at 20:57

3 Answers 3

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Some of your basic premise is correct : torque is proportional to current, and power DISSIPATED IN THE MOTOR ITSELF is a constant I^2*R, where R is the (constant) DC resistance of the motor, as measured across its terminals with the motor stationary.

Now let's run the motor at current I. The V required is not IR. (If it is, the motor is stalled so that the back EMF = 0.)

Instead, V = IR + back-EMF.

Now, I * back-EMF is the electrical power delivered to the load as mechanical power, and I * IR is the power wasted in the motor as heat.

Let's increase V and increase speed keeping I constant. Now, input power has increased (IV) but the motor's resistance hasn't changed : therefore IR is the same and I*IR is the same. But what HAS changed is the back-EMF - obviously, since it is proportional to speed (which has increased).

So the power dissipated in the motor as heat is constant; but the electrical power delivered to the load (I * back-EMF) has increased, exactly as the mechanical output power (torque * speed) has.

No magic, and it all adds up correctly.

But what IS interesting is that the efficiency has increased because the wasted power is constant but the useful power has increased. So a general rule is that electrical efficiency is higher in a lightly loaded motor running fast, than a heavily loaded motor running slow and drawing high current.

(There are limits to this : the less you load a motor and the faster you run it, the higher a proportion of power lost to friction in bearings and especially brushes. Bearings (ballraces) are easy : brushes are not, so brushless motors have a big advantage at high speed and high efficiency)

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  • \$\begingroup\$ Ok this adds up now. Thanks very much. It seems a bit counter intuitive though, that P_delivered = V_EMFI. To me it would make more intuitive sense that power delivered be calculated from the net voltage (V_net=V_supply-V_emf). Seems almost like this requires using the returned voltage to calculate the power delivered... Thanks very much for the answer. Nice to see there is no magic. I do hate mathematical magic. To clarify then, P_elec=(IV_emf)+(I^2*R)+Frictional Losses. \$\endgroup\$ May 29, 2013 at 8:24
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The input power to a motor (any motor, not just DC) is always the voltage at the lead wires times the current into the motor (not I2R). The output power is, as you said, always speed times torque.

The difference between the input power and the output power will always be greater than zero and is referred to as the losses. I2R losses will generally cause most of a losses but there are also core losses (eddy current and hysteresis), friction/windage losses, and stray load losses. Many of these losses are dependent on speed (for example, core losses are dependent on the frequency of the magnetic switching in the iron), so even if torque stays constant, it doesn't mean that efficiency will remain constant as speed increases/decreases.

As others mentioned, a DC motor can be modeled electrically as V = IR + Vemf, where Vemf is the back-emf, which is proportional to the speed of the motor. At high speed, V and Vemf are (almost) equal and there for little current flows. At zero speed, you have maximum current flow. If the load torque is constant but the speed is variable, then current flow will be roughly constant with slight changes due to different losses at different speeds and temperatures.

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  • \$\begingroup\$ Hi Eric. Thanks for your answer. If P_elec = I * V, and V = I * R, then P_elec must = I^2 * R too, if we are taking V as the net voltage in the lead wires. It seems the confusion is coming from which V to use - Brian states that P_elec = (I*V_emf)+(I^2*R), which makes sense since the formula for electrical heat power is also I^2*R. What does not make sense is why you would use V_emf to calculate power delivered, since to me V_emf represents the opposing voltage generated by the motor. \$\endgroup\$ May 29, 2013 at 8:34
  • \$\begingroup\$ If V is the voltage across the lead wires and I is the current flowing into the motor, then it is not true that V = I * R. For that to be the case, the model of a DC motor would have to be a resistor. That is not the case. Nobody uses back emf to calculate power delivered. Power delivered is speed times torque. When a motor manufacturer specifies rated power or maximum power on its data sheet, they measured speed and torque at those points. One reason is that, as I said above, there are other loses besides I^2*R losses. \$\endgroup\$
    – Eric
    May 29, 2013 at 13:44
  • \$\begingroup\$ So, I would say for a DC motor V*I=(mechanical power)+(I^2*R losses)+(core losses)+(friction and windage losses)+(stray load losses). \$\endgroup\$
    – Eric
    May 29, 2013 at 13:49
  • \$\begingroup\$ Hi Eric, the motor windings must have a resistance value surely. If V is taken as the net voltage, (supply voltage - back emf), then V must = I*R. Or Ohm's Law would not apply. I know it is not a resistor - because it returns back emf which is proportional to speed. But that is why the net voltage used with ohms law takes this into account, by taking the back emf away from the supply voltage. \$\endgroup\$ May 29, 2013 at 14:40
  • \$\begingroup\$ Yes, you are correct. \$\endgroup\$
    – Eric
    May 29, 2013 at 15:08
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Ok. Sorry to drag this out but I really want to get to the bottom of it. On their website, national instruments claims that electrical power supplied to the armature of a DC motor is given by:

\$P_e=V_b\times I\$

Where:

\$V_b = Supply~ Voltage ~(V)\$

http://zone.ni.com/devzone/cda/ph/p/id/46

I would assume this was a reliable source - and so answers my question. However it still makes no sense to me that you would use the supply voltage in this calculation. Unless the motor is stalled, the supply voltage does not describe the real potential difference across the motor terminals. It therefore does not drive the current which is also used in the calculation. If anyone could shed some light on this I would be really grateful. Thanks.

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  • \$\begingroup\$ This is what I said above. The power supplied to any motor is the voltage across the lead wires (supply voltage) times the current into the motor. The supply voltage is the potential difference across the motor terminals (think of the motor as a 2 terminal black box). It doesn't matter if it is stalled or not. \$\endgroup\$
    – Eric
    May 29, 2013 at 17:47
  • \$\begingroup\$ Ok thanks Eric. If the supply voltage is across the motor terminals - where does the net voltage (supply voltage - back emf) exist? \$\endgroup\$ May 29, 2013 at 18:09

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