I know that electric vehicles have different performances depending on battery and motor, but it's not clear how electrical and mechanical units are related.

Can anybody please help?

Will a 100V motor raise against slopes better than a 50V motor?

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# Electronic – How are current and voltage related to torque and speed of a brushless motor

###### Related Topic

brushless-dc-motormotorspeedtorque

I know that electric vehicles have different performances depending on battery and motor, but it's not clear how electrical and mechanical units are related.

Can anybody please help?

Will a 100V motor raise against slopes better than a 50V motor?

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## Best Answer

The relationship between a motor's electrical characteristics and mechanical performance can be calculated as such (note: this is the analysis for an ideal brushed DC motor, but some of it should still apply to a non-ideal brushless DC motor).

A DC motor can be approximated as a circuit with a resistor, and voltage back-emf source. The resistor models the intrinsic resistance of the motor windings. The back-emf models the voltage generated by the moving electric current in the magnetic field (basically a DC electric motor can function as a generator). It's also possible to model the inherent inductance of the motor by adding an inductor in series, however for the most part I've ignored this and assumed the motor is at quasi steady state electrically, or the motor's time response is dominated by the time response of the mechanical systems instead of the time response of the electrical systems. This is usually true, but not necessarily always true.

The generator produces a back EMF proportional to speed of the motor:

$$ V_{emf} = k_i * \omega $$

Where:

$$k_i = \text{a constant.}$$ $$\omega = \text{the motor speed in}\ \text{rad}/\text{s}$$

Ideally at stall speed there is no back emf, and at no the no-load speed the back emf is equal to the driving source voltage.

The current flowing through the motor can then be calculated:

$$ I = (V_S - V_{emf}) / R = (V_S - k_i * \omega) / R $$ $$V_S = \text{source voltage}$$ $$R = \text{motor electrical resistance}$$

Now let's consider the mechanical side of the motor. The torque generated by the motor is proportional to the amount of current flowing through the motor:

$$ \tau = k_t * I $$

$$k_t = \text{a constant}$$ $$\tau = \text{torque}$$

Using the above electrical model you can verify that at the stall speed the motor has the maximum current flowing through it, and thus the maximum torque. Also, at the no load speed the motor has no torque and no current flowing through it.

When does the motor produce the most power? Well, power can be calculated one of two ways:

Electrical Power: $$ P_e = V_S * I $$

Mechanical Power: $$ P_m = \tau * \omega $$

If you plot these, you'll find that for an ideal DC motor the maximum power comes at half the no-load speed.

So all things considered, how does the motor voltage stack up?

For the same motor, ideally if you apply double the voltage you'll double the no-load speed, double the torque, and quadruple the power. This is assuming of course the DC motor doesn't burn up, reach a state which violates this simplistic ideal motor model, etc.

However, between different motors it's impossible to tell how two motors will perform compared to each other based only on the voltage rating. So what do you need to compare two different motors?

Ideally you'd want to know the voltage rating and stall current so you can design your electronics appropriately and you'd want to know the no-load speed and stall torque so you can calculate the mechanical performance of your motor. You may also want to see the current rating of the motor (some motors can be damaged if you stall them for too long!). This analysis also somewhat neglects the efficiency aspect of the motor. For a perfectly efficient motor, \$k_i = k_t\$, or rather \$P_e = P_m\$. This would cause the power calculations using the two equations to be equal (i.e. electrical power equals mechanical power). However, real motors aren't perfectly efficient. Some are close, some aren't.

p.s. In my calculations I used motor speed as \$\text{rad}/\text{s}\$. This can be converted to

`Hz`

or \$\text{rev}/\text{s}\$ by dividing by \$2\pi\$..