DC Chopper Drive — Advanced Analysis & Design

Step 1 — Rated power and torque

\[P_{\text{rated}} = 220\times100 = 22{,}000\,\text{W}\] \[\omega_{700} = \frac{700\times2\pi}{60} = 73.30\,\text{rad/s}\] \[T_{\text{rated}} = \frac{P_{\text{rated}}}{\omega_{700}} = \frac{22{,}000}{73.30} = 300.1\,\text{N·m}\]

Step 2 — Load current (saturated region)

Since the motor operates in the saturated region, \(T \propto I_a\):

\[T_L = 1.5\times300.1 = 450.2\,\text{N·m}\] \[I_a \approx 1.5\times I_{\text{rated}} = 1.5\times100 = 150\,\text{A}\]

Step 3 — Average voltage and back EMF

\[V_{\text{avg}} = \alpha\,V_s = 0.70\times220 = 154\,\text{V}\] \[E_b = V_{\text{avg}} - I_a(R_a+R_f) = 154 - 150\times0.10 = 139\,\text{V}\]

Step 4 — Motor constant from magnetisation curve

At \(I_f = I_a = 150\,\text{A}\), read \(E_{OC} = 211\,\text{V}\) from table:

\[K\Phi = \frac{E_{OC}}{\omega_{\text{ref}}} = \frac{211}{73.30} = 2.879\,\text{V·s/rad}\]

Step 5 — Operating speed

\[\omega = \frac{E_b}{K\Phi} = \frac{139}{2.879} = 48.28\,\text{rad/s}\] \[N = 48.28\times\frac{60}{2\pi} = 461.2\,\text{rpm}\]

Step 1 — Commutating capacitor

\[C = \frac{I\,t_{\text{off}}}{V_s} = \frac{100\times20\times10^{-6}}{100} = 20\,\mu\text{F}\]

Step 2 — Chopper period

\[T = \frac{1}{f} = \frac{1}{400} = 2.5\,\text{ms}\]

Step 3 — Lower bound on \(L\)

The peak resonant current must exceed the load current \(I\):

\[I_{\text{peak}} = V_s\sqrt{\frac{C}{L}} > I\] \[\Rightarrow\quad L > C\!\left(\frac{V_s}{I}\right)^{\!2} = 20\times10^{-6}\times\left(\frac{100}{100}\right)^{\!2} = 20\,\mu\text{H}\]

Step 4 — Upper bound on \(L\)

The resonant half-period must be within 10% of the chopper period:

\[\pi\sqrt{LC} < 0.1\,T \;\Rightarrow\; L < \frac{0.01\,T^2}{\pi^2\,C}\] \[L < \frac{0.01\times(2.5\times10^{-3})^2}{\pi^2\times20\times10^{-6}} = \frac{0.01\times6.25\times10^{-6}}{197.4\times10^{-6}} = 31.66\,\mu\text{H}\]

Step 1 — Base quantities

\[V_b = 10\,\text{V}\] \[V_n = \frac{V_{\text{source}} - V_{\text{drop}}}{V_b} = \frac{24-1}{10} = 2.3\,\text{p.u.}\] \[I_b = \frac{P_{\text{rated}}}{V_b\,\eta} = \frac{746}{10\times0.785} = 95\,\text{A}\] \[R_{an} = \frac{I_b\,R_a}{V_b} = \frac{95\times0.01}{10} = 0.095\,\text{p.u.}\]

Step 2 — Per-unit duty cycle equation

\[d = \frac{T_{en}\,R_{an} + \omega_{mn}}{V_n}, \qquad T_{en} = 2\,\text{p.u.}\]

Step 3 — At minimum speed (\(\omega_{mn} = 0\))

\[d_{\min} = \frac{2\times0.095 + 0}{2.3} = \frac{0.190}{2.3} = 0.0826\]

Step 4 — At maximum speed (\(\omega_{mn} = 1\,\text{p.u.}\))

\[d_{\max} = \frac{2\times0.095 + 1}{2.3} = \frac{1.190}{2.3} = 0.517\]

Armature time constant and period ratio

\[T_a = \frac{L_a}{R_a} = \frac{1.5\times10^{-3}}{0.04} = 37.5\,\text{ms}\] \[T = \frac{1}{f_c} = \frac{1}{2000} = 0.5\,\text{ms}\] \[\frac{T_a}{T} = \frac{37.5}{0.5} = 75 \gg 1\]

Rectifier DC voltage

\[V_s = 1.35\times V_{\text{line}}\times\cos\alpha = 1.35\times230\times1 = 310.5\,\text{V}\]

Back EMF at 300 rpm

\[E_b = K_b\,\omega = 4.172\times\frac{2\pi\times300}{60} = 4.172\times31.416 = 131.1\,\text{V}\]

Critical duty cycle check

\[d_c = \frac{T_a}{T}\ln\!\left[1 + \frac{E_b}{V_s}\!\left(e^{T/T_a}-1\right)\right]\] \[d_c = 75\ln\!\left[1 + \frac{131.1}{310.5}\!\left(e^{1/75}-1\right)\right] = 0.423\]

Since \(d = 0.55 > d_c = 0.423\): current is continuous. ✓

Exact waveform — minimum current \(I_{a0}\)

Using \(dT/T_a = 0.55/75 = 0.00733\) and \(T/T_a = 1/75 = 0.01\overline{3}\):

\[I_{a0} = \frac{V_s\!\left(e^{dT/T_a}-1\right)}{R_a\!\left(e^{T/T_a}-1\right)} - \frac{E_b}{R_a}\] \[= \frac{310.5\times(e^{0.00733}-1)}{0.04\times(e^{0.01\overline{3}}-1)} - \frac{131.1}{0.04} = 979.0\,\text{A}\]

Exact waveform — maximum current \(I_{a1}\)

\[I_{a1} = \frac{V_s\!\left(1-e^{-dT/T_a}\right)}{R_a\!\left(1-e^{-T/T_a}\right)} - \frac{E_b}{R_a} = 1004.7\,\text{A}\]

Average current and torque (exact)

\[I_{av} = \frac{I_{a0}+I_{a1}}{2} = \frac{979.0+1004.7}{2} = 991.8\,\text{A}\] \[T_{\text{em}} = K_b\,I_{av} = 4.172\times991.8 = 4137.7\,\text{N·m}\]

Averaging method — direct formula

\[I_{av} = \frac{d\,V_s - E_b}{R_a} = \frac{0.55\times310.5 - 131.1}{0.04}\] \[= \frac{170.775 - 131.1}{0.04} = \frac{39.675}{0.04} = 991.9\,\text{A}\] \[T_{\text{em}} = K_b\,I_{av} = 4.172\times991.9 = 4138.0\,\text{N·m}\]

Step 1 — Average armature voltage and back EMF

\[V_a = \alpha\,V_s = 0.65\times200 = 130\,\text{V}\] \[E_b = V_a - I_a R_a = 130 - 40\times0.4 = 114\,\text{V}\]

Step 2 — Operating speed

\[\omega = \frac{E_b}{K\Phi} = \frac{114}{1.8} = 63.33\,\text{rad/s}\] \[N_{FL} = 63.33\times\frac{60}{2\pi} = 604.9\,\text{rpm}\]

Step 3 — Power quantities

\[P_{\text{mech}} = E_b\,I_a = 114\times40 = 4560\,\text{W}\] \[P_{R_a} = I_a^2\,R_a = 40^2\times0.4 = 640\,\text{W}\] \[P_{\text{in}} = \alpha\,V_s\,I_a = 0.65\times200\times40 = 5200\,\text{W}\]

Step 4 — Efficiency

\[\eta = \frac{P_{\text{mech}}}{P_{\text{in}}} = \frac{4560}{5200} = 87.7\%\]

Step 5 — No-load speed and speed regulation

\[E_{b,NL} = V_a = 130\,\text{V} \quad (I_a = 0\,\text{at no-load})\] \[\omega_{NL} = \frac{130}{1.8} = 72.22\,\text{rad/s},\quad N_{NL} = 689.8\,\text{rpm}\] \[\mathrm{SR} = \frac{N_{NL}-N_{FL}}{N_{FL}}\times100\% = \frac{689.8-604.9}{604.9}\times100\% = 14.0\%\]