DC Motor: Speed, Torque, & Power

Demonstrative Video

Dc Motor Basic Operational Video

Back EMF in DC Motor

$I$ carrying conductor $\Rightarrow$ $B$ $\Rightarrow$ $T$ induce $\Rightarrow$ rotates conductor $\Rightarrow$ cuts $\Phi$ $\Rightarrow$ EMF induce
Direction of induced EMF is opposite to $V$ $\Rightarrow$ Counter or Back EMF
$E_b$ series with $V$ but opposite in direction, i.e. oppose $I$ which causes it

Advantage of Back EMF

Electrical to Mechanical energy ( $E_bI_a$ ) conversion is possible only because of $E_b$ .
$E_b$ makes dc motor self-regulating $\Rightarrow$ $I=\dfrac{V-E_b}{R_a}$
- No load $\Rightarrow$ requires small $T$ $\Rightarrow$ for controlling friction and windage loss $\Rightarrow$ withdraws less $I$ $\Rightarrow$ $E_b \downarrow$
- load $\uparrow$ $\Rightarrow$ $\left(T_m<T_L\right)$ $\Rightarrow$ $N \downarrow$ $\Rightarrow$ $E_b \downarrow$ $\Rightarrow$ $I \uparrow$ $\Rightarrow$ $T \uparrow$
- load $\downarrow$ $\Rightarrow$ $\left(T_m>T_L\right)$ $\Rightarrow$ $N \uparrow$ $\Rightarrow$ $E_b \uparrow$ $\Rightarrow$ $I \downarrow$ $\Rightarrow$ $T \downarrow$

Condition for Maximum Power, $P_m$

\begin{aligned} P_{m} & = V I_{a} - I_{a}^{2} R_{a} \\ \Rightarrow \frac{d P_{m}}{d I_{a}} & = V - 2 I_{a} R_{a} = 0 \Rightarrow I_{a} R_{a} = \frac{V}{2} \\ ∴ V & = E_{b} + I_{a} R_{a} = E_{b} + \frac{V}{2} \Rightarrow E_{b} = \frac{V}{2} \end{aligned}

$\begin{aligned} P_{m} & =VI_{a}-I_{a}^{2}R_{a}\nonumber\\ \Rightarrow\dfrac{dP_{m}}{dI_{a}} & =V-2I_{a}R_{a}=0 \Rightarrow I_{a}R_{a} =\dfrac{V}{2} \\ \therefore V & = E_{b}+ I_{a}R_{a} = E_{b}+\dfrac{V}{2} \Rightarrow \boxed{E_{b} =\dfrac{V}{2}}\nonumber \end{aligned}$

Torque Equation of a DC Motor

\begin{aligned} P_{m} & \Rightarrow E_{b} I_{a} = T_{a} ω \\ \Rightarrow (\frac{Φ P N}{60} \times \frac{Z}{A}) I_{a} = T_{a} (\frac{2 π N}{60}) \\ \Rightarrow T_{a} = \frac{1}{2 π} \times \frac{Φ P Z}{A} \times I_{a} \\ T_{a} = K_{a} ϕ I_{a} \\ T_{s h} & = T_{a} - mechanical losses \end{aligned}

$\begin{aligned} P_{m} & \Rightarrow E_{b}I_{a}=T_{a}\omega\\ & \Rightarrow\left(\dfrac{\Phi PN}{60}\times\dfrac{Z}{A}\right)I_{a}=T_{a}\left(\dfrac{2\pi N}{60}\right)\\ & \Rightarrow T_{a}=\dfrac{1}{2\pi}\times\dfrac{\Phi PZ}{A}\times I_{a}\\ & \boxed{T_{a} =K_{a}\phi I_{a}}\\ T_{sh} & =T_{a}-\mbox{mechanical losses} \end{aligned}$

For series motor: $\Phi \propto I_a \Rightarrow T_a \propto I_a^2$

For shunt motor: $\Phi =$ constant $\Rightarrow T_a \propto I_a$

Speed of a DC Motor

\begin{aligned} E_{b} & = V - I_{a} R_{a} \\ \Rightarrow \frac{Φ P N}{60} \times (\frac{Z}{A}) & = V - I_{a} R_{a} \\ \Rightarrow N & = \frac{V - I_{a} R_{a}}{Φ} \times (\frac{60 A}{Z P}) \\ \Rightarrow N & = \frac{E_{b}}{Φ} \times (\frac{60 A}{Z P}) \\ \Rightarrow & N = K \frac{E_{b}}{Φ} \end{aligned}

$\begin{aligned} E_{b} & =V-I_{a}R_{a}\\ \Rightarrow\dfrac{\Phi PN}{60}\times\left(\dfrac{Z}{A}\right) & =V-I_{a}R_{a}\\ \Rightarrow N & =\dfrac{V-I_{a}R_{a}}{\Phi}\times\left(\dfrac{60A}{ZP}\right)\\ \Rightarrow N & =\dfrac{E_{b}}{\Phi}\times\left(\dfrac{60A}{ZP}\right)\\ \Rightarrow & \boxed{N =K\dfrac{E_{b}}{\Phi}} \end{aligned}$

and

\begin{aligned} N = K \frac{E_{b}}{Φ} \\ T_{a} \propto Φ I_{a} \end{aligned}

$\begin{aligned} & \boxed{N =K\dfrac{E_{b}}{\Phi}}\\ & \boxed{T_{a} \propto\Phi I_{a}} \end{aligned}$

Relation between Torque and Speed

$T_a$ is function of $\Phi$ and $I_a$ but independent of $N$
$N$ depends on $T$ and not vice-versa
$\Phi \uparrow \Rightarrow N \downarrow$ but $T_a \uparrow$ not possiblebecause $T$ always tends to produce rotation. If $T \uparrow$ , $N$ must $\uparrow$ rather than $\downarrow$

Following Sequence happens:

Suppose $\Phi$ is $\downarrow$ by $I_f \downarrow$
$\left[E_b \propto N\Phi\right] \downarrow$ , $N$ = constant (inertia of heavy armature)
$E_b \downarrow \Rightarrow \left[I_a = (V-E_b)/R_a\right] \uparrow$ . Moreover, $\Phi \downarrow \Rightarrow I_a \uparrow \uparrow$
$T_a \propto \Phi I_a$ , $\Phi \downarrow$ is counterbalanced by $I_a \uparrow \uparrow$ , result $T_a \uparrow$
$T_a \uparrow \Rightarrow N \uparrow$

Speed Regulation

% speed regulation = \frac{N . L s p e e d - F . L s p e e d}{F . L s p e e d} \times 100

$\% \mbox{speed regulation} = \dfrac{N.L~speed - F.L~speed}{F.L~speed} \times 100$

Change in speed when the load on the motor is reduced from rated value to zero, expressed as percent of the rated load speed