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Motor performance

In this section we’ll dive into more depth on the performance characteristics of DC motors. We’ll derive the torque-speed curve, efficiency, and power output of a DC motor, and see how these depend on the motor parameters.

We begin with the standard DC motor equations from Eq. (1) in the main text:

Ldidt+Ri+Kω=vinJω˙+bωKi=TL\begin{aligned} L \frac{\dd i}{\dd t} + R i + K \omega &= v_\textsf{in} \\ J \dot \omega + b\,\omega - K i &= - T_L \end{aligned}

We consider steady-state operation, so all time derivatives are zero.

Ri+Kω=vinKibω=TL\boxed{\begin{aligned} R i + K \omega &= v_\textsf{in} \\ K i - b\,\omega &= T_L \end{aligned}}

The maximum speed (and minimum current) occur when there is no load torque, i.e., TL=0T_L = 0. From (2), we can solve for the no-load speed and current:

ωnl=KvinK2+bR,inl=bvinK2+bR\boxed{ \omega_\textsf{nl} = \frac{K v_\textsf{in}}{K^2 + b R}, \quad i_\textsf{nl} = \frac{b\,v_\textsf{in}}{K^2 + b R} }

The maximum torque (and maximum current) occur at stall, i.e., ω=0\omega = 0. From (2), we can solve for this stall torque and current:

Tst=KvinR,ist=vinR\boxed{ T_\textsf{st} = \frac{K v_\textsf{in}}{R}, \quad i_\textsf{st} = \frac{v_\textsf{in}}{R} }

If there was no friction (b=0b=0), the the no-load speed would be

ωmax=vinK\boxed{ \omega_\textsf{max} = \frac{v_\textsf{in}}{K} }

which is larger than ωnl\omega_\textsf{nl} in (3). In other words, friction reduces the no-load speed of the motor. The speed ωmax\omega_\textsf{max} is often quoted as the “maximum speed” of the motor, even though it is not achievable in practice when there is friction.

The stall torque and current in (4) are unaffected by friction, since at stall the motor is not moving and thus frictional losses are zero.

Power and efficiency

Now let’s talk about power. The formulas for input power, output power, and motor efficiency are given by

Pin=vini,Pout=TLω,η=PoutPin\boxed{ P_\textsf{in} = v_\textsf{in} i,\quad P_\textsf{out} = T_L \omega,\quad \eta = \frac{P_\textsf{out}}{P_\textsf{in}}}

We care about output power and efficiency, and we want to know when they are maximized and at what operating points (what speed, what load torque). This is essentially an exercise in algebra with a bit of calculus.

Since we are trying to derive the entire range of possible performance for the motor, it’s important to think about what is being held fixed and what is being allowed to vary.

Power-speed curve

Let’s parameterize performance in terms of motor speed ω\omega. Combining (2) and (6), we view this as a system of 5 equations in unknowns (i,TL,Pin,Pout,η)(i,T_L,P_\textsf{in},P_\textsf{out}, \eta), treating vinv_\textsf{in} and ω\omega as fixed parameters. Solving this system, we obtain

i=vinKωR,TL=Kvin(K2+bR)ωR,Pin=vin2KvinωR,Pout=Kvinω(K2+bR)ω2R,η=Kvinω(K2+bR)ω2vin2Kvinω\begin{gathered} i = \frac{v_\textsf{in} - K \omega}{R}, \quad T_L = \frac{K v_\textsf{in} - (K^2 + b R) \omega}{R},\\ P_\textsf{in} = \frac{v_\textsf{in}^2 - K v_\textsf{in} \omega}{R},\quad P_\textsf{out} = \frac{K v_\textsf{in} \omega - (K^2 + b R) \omega^2}{R},\\ \eta = \frac{K v_\textsf{in} \omega - (K^2 + b R) \omega^2}{v_\textsf{in}^2 - K v_\textsf{in} \omega} \end{gathered}

It helps to normalize the equations by defining appropriate dimensionless variables.

ωˉ:=ωωmax,iˉ:=iistall,TˉL:=TLTstall,Pˉin:=Pin14Tstallωmax,Pˉout:=Pout14Tstallωmax,σ:=bRK\begin{gathered} \bar \omega := \frac{\omega}{\omega_\textsf{max}}, \quad \bar i := \frac{i}{i_\textsf{stall}}, \quad \bar T_L := \frac{T_L}{T_\textsf{stall}},\\ \bar P_\textsf{in} := \frac{P_\textsf{in}}{\tfrac{1}{4}T_\textsf{stall} \omega_\textsf{max}}, \quad \bar P_\textsf{out} := \frac{P_\textsf{out}}{\tfrac{1}{4}T_\textsf{stall} \omega_\textsf{max}}, \quad \sigma := \frac{\sqrt{b R}}{K} \end{gathered}

where ωmax\omega_\textsf{max}, istalli_\textsf{stall}, TstallT_\textsf{stall} are given by (5) and (4). The reason for normalizing the power by 14Tstallωmax\frac{1}{4}T_\textsf{stall} \omega_\textsf{max} is that this quantity is the maximum output power, as we will see shortly. The parameter σ>0\sigma>0 captures the effect of viscous friction and resistance on motor performance: the larger σ\sigma is, the more losses the motor has relative to its torque constant KK. An ideal motor has σ=0\sigma = 0.

Substituting (8) into (7), we obtain normalized power-speed relationships:

iˉ=1ωˉ,TˉL=1ωˉ(1+σ2),Pˉin=4(1ωˉ),Pˉout=4ωˉ(1ωˉ(1+σ2)),η=ωˉ(1ωˉ(1+σ2))1ωˉ\begin{gathered} \bar i = 1-\bar\omega, \quad \bar T_L = 1-\bar\omega(1+\sigma^2), \\ \bar P_\textsf{in} = 4(1-\bar\omega), \quad \bar P_\textsf{out} = 4\,\bar\omega(1-\bar\omega(1+\sigma^2)), \quad \eta = \frac{\bar\omega(1-\bar\omega(1+\sigma^2))}{1-\bar\omega} \end{gathered}

We can analyze these equations to find various optimized operating points.

Maximum output power

To maximize output power Pˉout\bar P_\textsf{out}, we take the derivative and set it to zero:

dPˉoutdωˉ=12ωˉ(1+σ2)=0    ωˉ=12(1+σ2)\frac{\dd \bar P_\textsf{out}}{\dd \bar\omega} = 1 - 2\bar\omega(1+\sigma^2) = 0 \implies \bar\omega = \frac{1}{2(1+\sigma^2)}

Converting back to non-normalized variables, we see that maximum power occurs at:

ω=ωmax2(1+σ2)=ωnl2,TL=Tstall2,Pout=11+σ2Tstallωmax4\omega = \frac{\omega_\textsf{max}}{2(1+\sigma^2)} = \frac{\omega_\textsf{nl}}{2}, \quad T_L = \frac{T_\textsf{stall}}{2}, \quad P_\textsf{out} = \frac{1}{1+\sigma^2}\cdot\frac{T_\textsf{stall} \omega_\textsf{max}}{4}

In other words, maximum power is achieved when the motor is loaded to half its stall torque, which causes it to run at half its no-load speed. The maximum output power decreases as friction and resistance increase (as σ\sigma increases), and approaches the ideal value of 14Tstallωmax=vin2/(4R)\frac{1}{4} T_\textsf{stall} \omega_\textsf{max} = v_\textsf{in}^2/(4R) as σ0\sigma \to 0.

Maximum efficiency

Maximizing power output does not necessarily maximize efficiency. To find the speed that maximizes efficiency, we take the derivative of η\eta in (9) and set it to zero. This gives us optimal efficiency at the operating point:

ω=1+η2ωnl,TL=1η2Tstall,η=(1+σ2σ)2\omega^\star = \frac{1+\eta^\star}{2}\,\omega_\textsf{nl}, \quad T_L^\star = \frac{1-\eta^\star}{2}\,T_\textsf{stall}, \quad \eta^\star = \left(\sqrt{1+\sigma^2}-\sigma\right)^2

In other words, the closer to ideal the motor is (the smaller σ\sigma is), the higher the maximum efficiency. Moreover, the motor is most efficient when it runs faster and with less load torque. For example, if the motor has a maximum efficiency of 80%, then it achieves this efficiency at 90% of its no-load speed and loaded at 10% of its stall torque.

Power-speed plot

To see what different performance curves might look like, we can plot the solutions in Eq. (9) as a function of normalized speed ωˉ\bar \omega for different values of the parameter σ\sigma.

Normalized output power and efficiency of a DC motor as a function of normalized speed for different values of the parameter \sigma = \frac{\sqrt{bR}}{K}.

Figure 1:Normalized output power and efficiency of a DC motor as a function of normalized speed for different values of the parameter σ=bRK\sigma = \frac{\sqrt{bR}}{K}.

To get a sense of scale, consider a brushless drone motor. Typical values are given by

R=0.1Ω,K=0.01Nm/A,b=106104NmsR = 0.1\,\Omega, \quad K = 0.01\,\text{Nm/A}, \quad b = 10^{-6} - 10^{-4}\,\text{Nms}

This corresponds to values of σ\sigma ranging from 0.03 to 0.3.

From the plot in Figure 1, we can see that as σ\sigma increases (i.e., as friction and resistance increase), both the maximum output power and maximum efficiency decrease. The shape of the power-speed curve also changes, becoming more skewed towards lower speeds. The efficiency curve also shifts downwards, indicating that the motor becomes less efficient at all operating points.

Power-torque curve

Instead of parameterizing performance in terms of motor speed, we can instead parameterize in terms of load torque TLT_L. Combining (2) and (6), we can view this as a system of 5 equations in unknowns (i,ω,Pin,Pout,η)(i,\,\omega,P_\textsf{in},P_\textsf{out}, \eta), this time treating vinv_\textsf{in} and TLT_L as fixed parameters. We can carry out the same analysis as before and this time plot the results as a function of normalized load torque TˉL\bar T_L.

i=bvin+KTLK2+bR,ω=KvinRTLK2+bR,Pin=vin(KTL+bvin)K2+bR,Pout=TL(KvinRTL)K2+bR,η=TL(KvinRTL)vin(bvin+KTL)\begin{gathered} i = \frac{b v_\textsf{in} + K T_L}{K^2 + b R}, \quad \omega = \frac{K v_\textsf{in} - R T_L}{K^2 + b R},\\ P_\textsf{in} = \frac{v_\textsf{in}( K T_L + b v_\textsf{in})}{K^2 + b R},\quad P_\textsf{out} = \frac{T_L(K v_\textsf{in} - R T_L)}{K^2 + b R},\\ \eta = \frac{T_L(K v_\textsf{in} - R T_L)}{v_\textsf{in}(b v_\textsf{in} + K T_L)} \end{gathered}

As before, we can normalize the equations using (8). Substituting into (14), we obtain normalized power-torque relationships:

iˉ=TˉL+σ21+σ2,ωˉ=1TˉL1+σ2,Pˉin=4(TˉL+σ2)1+σ2,Pˉout=4TˉL(1TˉL)1+σ2,η=TˉL(1TˉL)TˉL+σ2\begin{gathered} \bar i = \frac{\bar T_L + \sigma^2}{ 1+\sigma^2}, \quad \bar \omega = \frac{1 - \bar T_L}{1+\sigma^2},\quad \bar P_\textsf{in} = \frac{4(\bar T_L + \sigma^2)}{1+\sigma^2}, \\ \bar P_\textsf{out} = \frac{4\,\bar T_L(1 - \bar T_L)}{1+\sigma^2}, \quad \eta = \frac{\bar T_L(1 - \bar T_L)}{\bar T_L + \sigma^2} \end{gathered}

This leads us to the following power-torque plot, where this time we vary normalized load torque TˉL\bar T_L instead of speed, for different values of the parameter σ\sigma.

Normalized output power and efficiency of a DC motor as a function of normalized load torque for different values of the parameter \sigma = \frac{\sqrt{bR}}{K}.

Figure 2:Normalized output power and efficiency of a DC motor as a function of normalized load torque for different values of the parameter σ=bRK\sigma = \frac{\sqrt{bR}}{K}.