DC motor performance and selection

The idea

There are two kinds of DC motor question. They use exactly the same equations, run in opposite directions:

	Performance analysis	Selection (verification)
The question	”What does this motor actually do?"	"Does this datasheet motor meet the robot’s needs?”
You’re given	Measured V, I, $R_a$, $K_t$, the gearbox and wheel	The requirement (e.g. 20 km/h) and the datasheet (e.g. $P_{rated}$)
You find	Torque, speed, power, efficiency, robot speed	Whether $P_{out}\le P_{rated}$, and by how much
Used for	An existing robot: testing, benchmarking, fault-finding	A new design: choosing parts

The calculation chain

Start from the steady-state model in Electric motors and DC motor types and work along:

$V_R=I_a{R}_a\to E_b=V-V_R\to \omega =\frac{E_b}{K_e}\to n=\frac{60\omega }{2\pi }\to T=K_t{I}_a$

$P_{out}=T\omega P_{in}=V{I}_a{P}_{Cu}=I_a^2{R}_a\eta =\frac{P_{out}}{P_{in}}\times 100\%$

Check: in this model copper is the only loss, so $P_{in}-P_{out}-P_{Cu}=0$. If it isn’t, there’s a slip somewhere.

Then the robot side, through a gearbox of ratio G:1 and a wheel of radius r:

$n_{wheel}=\frac{n}{G}\to {\omega }_{wheel}=\frac{2\pi n_{wheel}}{60}\to v={\omega }_{wheel}r\to v_{km/h}=3.6v{T}_{wheel}=TG$

($T_{wheel}=TG$ assumes an ideal gearbox; a real one gives $TG{\eta }_g$.)

Worked example: performance analysis

A robot’s PMDC drive motor is measured on the bench: 36 V supply, $R_a=0.8$ Ω, $I_a=6$ A, $K_t=0.12$ N·m/A. It drives a 0.09 m radius wheel through a 4:1 gearbox. How fast does the robot go, and how much torque reaches the wheel?

Answer

$V_R=6\times 0.8=4.8$ V → $E_b=36-4.8=31.2$ V $\omega =31.2/0.12=260$ rad/s → $n=260\times 60/2\pi =2483$ rpm $T=0.12\times 6=0.72$ N·m $P_{out}=0.72\times 260=187.2$ W · $P_{in}=36\times 6=216$ W · $P_{Cu}=6^2\times 0.8=28.8$ W $\eta =187.2/216=$ 86.7% · check: $216-187.2-28.8=0$ ✓ Wheel: $2483/4=620.7$ rpm → ${\omega }_w=65.0$ rad/s → $v=65.0\times 0.09=5.85$ m/s = 21.1 km/h Wheel torque $=0.72\times 4=2.88$ N·m

Worked example: selection

A datasheet motor is rated at 250 W, with 48 V, $R_a=0.6$ Ω, $I_a=5$ A and $K_t=0.12$. It drives a 0.1 m wheel through a 6:1 gearbox. The robot must reach 20 km/h. Is the motor suitable?

Answer

$E_b=48-5\times 0.6=45$ V → $\omega =45/0.12=375$ rad/s → $n=3581$ rpm $T=0.12\times 5=0.6$ N·m → $P_{out}=0.6\times 375=225$ W · $P_{in}=240$ W → $\eta =93.8\%$ Power utilisation $=225/250=90\%$ · power reserve $=250-225=25$ W Wheel: $3581/6=597$ rpm → $v=6.25$ m/s = 22.5 km/h ✓ (meets 20 km/h) Verdict: it meets the speed, but at 90% utilisation there’s only 25 W of headroom. A slope, an extra payload or a hot day would push it past its rating. Suitable on paper; a cautious designer would pick the next size up.

The ideas behind the numbers

$K_t=K_e$ in SI units. It’s one number describing the same magnetic coupling from both sides: torque per amp, and volts generated per rad/s. If a datasheet gives $K_e$ in V/krpm, convert it to V/(rad/s) first.

The $K_t$ trade-off:

• Higher $K_t$: more torque per amp, but a lower top speed, because the back-EMF reaches the supply voltage sooner.
• Lower $K_t$: faster, but weaker.

Application	Typical $K_t$ (N·m/A)
Mini line-follower	0.01–0.03
Educational robot	0.03–0.06
Mobile robot (20–100 W)	0.05–0.12
AGV (100–500 W)	0.10–0.30
Robot arm joint	0.20–0.80
Collaborative robot joint	0.30–1.50

The gearbox trades speed for torque. A G:1 reduction divides speed by G and multiplies torque by G, so a fast, weak motor becomes a slow, strong wheel. That’s why nearly every robot drive is geared. But a gearbox can’t create power: $P=T\omega$ is the same on both sides (minus the gearbox’s losses).

Utilisation is not efficiency.

• Efficiency is how much of the input becomes useful output. It decides battery life.
• Utilisation is how close the motor runs to its rated limit. It decides how much margin you have. Somewhere around 70–85% leaves room for slopes, payload and temperature.

Worked example: when gearing can’t save you

The customer now wants the selection motor above to deliver 5 N·m at the wheel and still reach 20 km/h. Can a different gear ratio, or more current, do it?

Answer

Change the gearing: $G=5/0.6=8.33$. The torque is there, but the speed drops to $22.5\times 6/8.33=16.2$ km/h. Fails the speed. Push more current at G = 6: $I=5/(0.12\times 6)=6.94$ A → $E_b=43.8$ V → $\omega =365$ rad/s → $T=0.833$ N·m → $P_{out}=304$ W, which is 122% of rating. Overloads the motor. Why neither works: 5 N·m at 20 km/h on a 0.1 m wheel needs $5\times (5.56/0.1)=278$ W at the wheel, which is more than the motor’s 250 W. The answer is a bigger motor, plus a margin.

Common mistakes

• Forgetting the $I_a{R}_a$ drop and using $\omega =V/K_e$ (that’s only the no-load speed).
• Comparing $P_{in}$ with the rating instead of $P_{out}$.
• rpm and rad/s: $P=T\omega$ needs rad/s. m/s and km/h: × 3.6.
• Dividing torque by G instead of multiplying. A reduction gearbox increases torque.

Check yourself

1. A 12 V motor has $R_a=2$ Ω and $K_t=K_e=0.02$. At 1 A, find its speed in rpm, its output power and its efficiency.

$E_b=12-2=10$ V → $\omega =500$ rad/s = 4775 rpm · $T=0.02$ N·m → $P_{out}=$ 10 W · $P_{in}=12$ W → $\eta =$ 83%.

2. Two motors both meet a speed target. One runs at 95% utilisation, the other at 70%. Which do you choose?

The 70% one: it has margin for extra load, slopes and heat. The 95% one is at its limit and would overheat with any extra demand. (The only argument for it is size, weight or cost.)

3. Why is a robot drive almost always geared?

Small DC motors are most efficient at high speed and low torque, but wheels need low speed and high torque. A reduction gearbox converts one into the other, at the same power.

Next: motors that move in exact steps, in Stepper motors.