DC Chopper Drive — Fundamentals

Duty Ratio • Speed Control • Current Ripple • Minimum Inductance — Electric Drives

1. Formula Reference Sheet

Back EMF, Speed and Motor Constant

Back EMF (motoring) \(\displaystyle E_b = V_a - I_a R_a\) \(V_a\) = average armature voltage
Speed–EMF relation \(\displaystyle E_b = K\Phi\,\omega, \quad \omega = \frac{2\pi N}{60}\) \(K\Phi\) in V·s/rad; \(N\) in rpm
Speed scaling (const. flux) \(\displaystyle \frac{E_2}{E_1} = \frac{N_2}{N_1}\) Only valid when \(\Phi\) is unchanged
Motor constant \(K_a\Phi\) from nameplate \(\displaystyle K_a\Phi = \frac{E_{b,\text{rated}}}{N_{\text{rated}}}\) Units: V/rpm (if using rpm directly)

Step-Down (Buck) Chopper — Duty Ratio

Average output voltage \(\displaystyle V_a = \alpha\,V_s, \qquad 0 \le \alpha \le 1\) \(\alpha\) = duty cycle; \(V_s\) = supply voltage
Duty cycle (motoring) \(\displaystyle \alpha = \frac{E_b + I_a R_a}{V_s}\) Numerator = required average voltage \(V_a\)
Switch on-time \(\displaystyle t_{\text{on}} = \frac{\alpha}{f}\) \(f\) = chopper switching frequency (Hz)
Equivalent resistance \(\displaystyle R_{\text{eq}} = \frac{V_s}{\alpha\,I_a}\) Input impedance seen by supply

Armature Current Ripple & Minimum Inductance

Peak-to-peak ripple \(\displaystyle \Delta I = \frac{V_s\,\alpha\,(1-\alpha)}{2\,L\,f}\) Maximum at \(\alpha = 0.5\)
Minimum inductance \(\displaystyle L_{\min} = \frac{V_s\,\alpha\,(1-\alpha)}{2\,\Delta I_{\max}\,f}\) \(\Delta I_{\max}\) = allowable peak-to-peak ripple (A)
Why ripple matters
Armature current ripple causes torque pulsations, increases copper losses (\(I^2R\)), and can lead to discontinuous conduction. The armature inductance \(L_a\) (and any external inductor) must be large enough to keep \(\Delta I\) within the specified limit.

Power Relations

Mechanical output power \(\displaystyle P_{\text{mech}} = E_b\,I_a\) Power converted to shaft
Armature copper loss \(\displaystyle P_{R_a} = I_a^2\,R_a\) Heat dissipated in armature resistance
Input power \(\displaystyle P_{\text{in}} = \alpha\,V_s\,I_a\) = \(P_{\text{mech}} + P_{R_a}\) (energy balance check)
Developed torque \(\displaystyle T_d = K_v\,I_f\,I_a\) For separately excited motor with field current \(I_f\)

2. Speed Range and Duty-Cycle Range

Problem 1.1

Problem Statement

Motor & Drive Data
Supply voltage \(V_s\)120 V
Armature resistance \(R_a\)0.5 Ω
Rated armature current \(I_a\)20 A
Motor constant \(K_a\Phi\)0.05 V/rpm
Objective
A separately excited DC motor is controlled by a step-down chopper. Determine:
  1. The speed range achievable.
  2. The corresponding duty-cycle range \(\alpha\).
Given:

\(V_s = 120\,\text{V}\), \(R_a = 0.5\,\Omega\), \(I_a = 20\,\text{A}\), \(K_a\Phi = 0.05\,\text{V/rpm}\)

Find:

Speed range \(N\) and duty-cycle range \(\alpha\)

Solution

Step 1 — Maximum speed (\(\alpha = 1\), full supply)

At maximum duty cycle the full supply appears across the motor:

\[ E_{b,\max} = V_s - I_a R_a = 120 - 20\times0.5 = 110\,\text{V} \] \[ N_{\max} = \frac{E_{b,\max}}{K_a\Phi} = \frac{110}{0.05} = \mathbf{2200\,\text{rpm}} \]

Step 2 — Minimum speed (motor at standstill, \(E_b = 0\))

At standstill the entire average voltage drives current through \(R_a\) only:

\[ V_{a,\min} = I_a R_a = 20 \times 0.5 = 10\,\text{V} \] \[ \alpha_{\min} = \frac{V_{a,\min}}{V_s} = \frac{10}{120} = \frac{1}{12} \approx 0.083 \]

Below \(\alpha_{\min}\) the average voltage cannot sustain rated current, so the speed range starts effectively from \(N = 0\).

Key Relations Used
\[V_a = \alpha\,V_s\] \[E_b = V_a - I_a R_a\] \[N = \frac{E_b}{K_a\Phi}\]
Why \(\alpha_{\min} \neq 0\)
Even at zero speed the motor still draws rated current \(I_a\). The chopper must supply at least \(I_a R_a = 10\,\text{V}\) on average to maintain that current, requiring \(\alpha = 1/12\) minimum.
▶ Key Results
ParameterMinimumMaximum
Speed \(N\)0 rpm2200 rpm
Duty cycle \(\alpha\)\(1/12 \approx 0.083\)1
\(0 \le N \le 2200\,\text{rpm}\)
\(\dfrac{1}{12} \le \alpha \le 1\)

3. Speed Reduction at Constant Torque & Minimum Inductance

Problem 1.2

Problem Statement

Motor & Drive Data
Supply voltage \(V_s\)250 V
Armature resistance \(R_a\)2.5 Ω
Rated armature current \(I_a\)20 A
Initial speed \(N_1\)600 rpm
Target speed \(N_2\)400 rpm
Switching frequency \(f\)400 Hz
Objective
  1. Find the duty ratio \(\alpha\) to reduce speed from 600 rpm to 400 rpm at constant torque (constant \(I_a\)).
  2. Find the minimum inductance \(L_{\min}\) such that the peak-to-peak armature current ripple \(\Delta I \le 10\%\) of rated current.
Given:

\(V_s = 250\,\text{V}\), \(R_a = 2.5\,\Omega\), \(I_a = 20\,\text{A}\), \(N_1 = 600\,\text{rpm}\), \(N_2 = 400\,\text{rpm}\), \(f = 400\,\text{Hz}\)

Find:

(1) Duty ratio \(\alpha\)   (2) Minimum inductance \(L_{\min}\)

Solution — Part (1): Duty Ratio

Step 1 — Back EMF at rated speed \(N_1 = 600\,\text{rpm}\)

\[ E_1 = V_s - I_a R_a = 250 - 20\times2.5 = 200\,\text{V} \]

Step 2 — Back EMF at target speed \(N_2 = 400\,\text{rpm}\)

Field flux is constant (separately excited), so:

\[ E_2 = E_1 \cdot \frac{N_2}{N_1} = 200 \times \frac{400}{600} = 133.33\,\text{V} \]

Step 3 — Required average armature voltage

Torque is constant, so \(I_a = 20\,\text{A}\) remains unchanged:

\[ V_a = E_2 + I_a R_a = 133.33 + 20\times2.5 = 183.33\,\text{V} \]

Step 4 — Duty ratio

\[ \alpha = \frac{V_a}{V_s} = \frac{183.33}{250} = \mathbf{0.733} \]
Why \(E_2/E_1 = N_2/N_1\)?
For a separately excited motor with constant field flux \(\Phi\): \[E_b = K\Phi\,\omega \propto N\] So back EMF scales linearly with speed. This is the key assumption at constant torque (constant \(I_a\) and \(\Phi\)).
Verification
\[ N_2 = N_1 \times \frac{E_2}{E_1} = 600 \times \frac{133.33}{200} = 400\,\text{rpm}\;\checkmark \]
▶ Result — Part (1)
\(\alpha = 0.733\) \(t_{\text{on}} = \alpha/f = 0.733/400 = 1.83\,\text{ms}\)

Solution — Part (2): Minimum Inductance

Step 1 — Allowable ripple current

\[ \Delta I_{\max} = 10\% \times I_a = 0.10 \times 20 = 2\,\text{A} \]

Step 2 — Apply minimum inductance formula

\[ L_{\min} = \frac{V_s\,\alpha\,(1-\alpha)}{2\,\Delta I_{\max}\,f} \] \[ L_{\min} = \frac{250 \times 0.733 \times (1-0.733)}{2 \times 2 \times 400} \] \[ L_{\min} = \frac{250 \times 0.733 \times 0.267}{1600} = \frac{48.88}{1600} = 30.5\,\text{mH} \]
Ripple Formula Origin
During \(t_{\text{on}}\), the inductor voltage is \(V_s - E_b\). The current rises by \(\Delta I\): \[ \Delta I = \frac{(V_s - E_b)\,t_{\text{on}}}{L} = \frac{V_s\,\alpha\,(1-\alpha)}{2Lf} \] (using \(V_a = E_b + I_a R_a \approx \alpha V_s\))
Design Note
The ripple is worst at \(\alpha = 0.5\). At \(\alpha = 0.733\) the ripple is slightly less than maximum, so \(L_{\min} = 30.5\,\text{mH}\) is specific to this operating point. If the drive must satisfy the 10% limit at all speeds, use \(\alpha = 0.5\) in the formula to get the worst-case \(L_{\min}\).
▶ Result — Part (2)
\(L_{\min} = 30.5\,\text{mH}\) \(\Delta I_{\max} = 2\,\text{A}\)

4. DC–DC Converter Drive: Power, Speed & Torque

Problem 1.3

Problem Statement

Motor & Drive Data
Supply voltage \(V_s\)600 V
Duty cycle \(\alpha = k\)0.60
Armature resistance \(R_a\)0.05 Ω
Back-EMF constant \(K_v\)1.527 V/(A·rad/s)
Armature current \(I_a\)250 A
Field current \(I_f\)2.5 A
Objective
A separately excited DC motor is fed by a 600 V DC–DC converter. Neglecting current ripple, find:
  1. Input power \(P_{\text{in}}\)
  2. Equivalent circuit resistance \(R_{\text{eq}}\)
  3. Motor angular speed \(\omega\)
  4. Developed torque \(T_d\)
Given:

\(V_s = 600\,\text{V}\), \(\alpha = 0.60\), \(R_a = 0.05\,\Omega\), \(K_v = 1.527\,\text{V/(A·rad/s)}\), \(I_a = 250\,\text{A}\), \(I_f = 2.5\,\text{A}\)

Find:

\(P_{\text{in}}\), \(R_{\text{eq}}\), \(\omega\), \(T_d\)

Solution

Step 1 — Input power

\[ P_{\text{in}} = \alpha\,V_s\,I_a = 0.6 \times 600 \times 250 = 90\,\text{kW} \]

Step 2 — Equivalent resistance

The chopper presents an effective resistance to the supply:

\[ R_{\text{eq}} = \frac{V_s}{\alpha\,I_a} = \frac{600}{0.6\times250} = 4\,\Omega \]

Step 3 — Average terminal voltage and back EMF

\[ V_a = \alpha\,V_s = 0.6 \times 600 = 360\,\text{V} \] \[ E_b = V_a - I_a R_a = 360 - 250\times0.05 = 347.5\,\text{V} \]

Step 4 — Motor speed

For a separately excited motor: \(E_b = K_v\,I_f\,\omega\)

\[ \omega = \frac{E_b}{K_v\,I_f} = \frac{347.5}{1.527\times2.5} = \mathbf{91.03\,\text{rad/s}} \]

Step 5 — Developed torque

\[ T_d = K_v\,I_f\,I_a = 1.527\times2.5\times250 = \mathbf{954.4\,\text{N·m}} \]
Motor Model

The motor back-EMF and torque are:

\[E_b = K_v\,I_f\,\omega\] \[T_d = K_v\,I_f\,I_a\]

These are the separately excited machine equations where \(K_v\,I_f\) acts as the effective flux-linkage constant.

Energy Balance Check
\[ \begin{aligned} P_{\text{mech}} &= E_b\,I_a = 347.5\times250 = 86.875\,\text{kW}\\ P_{R_a} &= I_a^2 R_a = 250^2\times0.05 = 3.125\,\text{kW}\\ P_{\text{in}} &= 86.875 + 3.125 = \mathbf{90\,\text{kW}}\;\checkmark \end{aligned} \]
Speed in rpm
\[ N = \omega\times\frac{60}{2\pi} = 91.03\times\frac{60}{2\pi} = 869.4\,\text{rpm} \]
▶ Key Results
QuantityValue
Input power \(P_{\text{in}}\)90 kW
Equivalent resistance \(R_{\text{eq}}\)4 Ω
Motor speed \(\omega\)91.03 rad/s
Developed torque \(T_d\)954.4 N·m
\(P_{\text{in}} = 90\,\text{kW}\)
\(R_{\text{eq}} = 4\,\Omega\)
\(\omega = 91.03\,\text{rad/s}\)
\(T_d = 954.4\,\text{N·m}\)