DC Chopper Drive — Braking & Multi-Quadrant Operation

1. Formula Reference Sheet

Two-Quadrant Chopper

Forward motoring \((S_1/D_1)\)	\(\displaystyle V_a = \alpha\,V_s, \quad \alpha = \frac{E_b + I_a R_a}{V_s}\)	\(V_a > 0,\; I_a > 0\) — Quadrant I
Regenerative braking \((S_2/D_2)\)	\(\displaystyle V_a = (1-\alpha)\,V_s, \quad \alpha = 1 - \frac{V_a}{V_s}\)	\(V_a > 0,\; I_a < 0\) — Quadrant II
Braking voltage relation	\(\displaystyle V_a = E_b - \|I_a\|\,R_a\)	Current flows back to supply
Switch on-time	\(\displaystyle t_{\text{on}} = \frac{\alpha}{f_c}\)	\(f_c\) = chopper frequency (Hz)

Four-Quadrant (H-Bridge) Chopper — Bipolar PWM

Output voltage	\(\displaystyle V_a = (2\alpha - 1)\,V_s\)	Ranges from \(-V_s\) to \(+V_s\)
Duty cycle	\(\displaystyle \alpha = \frac{V_a + V_s}{2\,V_s}\)	\(0 \le \alpha \le 1\); \(\alpha = 0.5\) gives \(V_a = 0\)
Required armature voltage	\(\displaystyle V_a = E_b + I_a R_a\)	Sign of \(E_b\) and \(I_a\) determines quadrant

Four-Quadrant Map

Q2 — Forward Regen
\(V_a{>}0,\;I_a{<}0\)

Q1 — Forward Motor
\(V_a{>}0,\;I_a{>}0\)

Q3 — Reverse Motor
\(V_a{<}0,\;I_a{<}0\)

Q4 — Reverse Regen
\(V_a{<}0,\;I_a{>}0\)

Boost (Step-Up) Chopper — Low-Speed Regeneration

Motor terminal voltage (generator mode)	\(\displaystyle V_{\text{mot}} = E_b - I_a R_a\)	\(E_b\) at braking speed; \(I_a\) = braking current
Boost duty cycle	\(\displaystyle \alpha = 1 - \frac{V_{\text{mot}}}{V_{\text{bus}}}\)	Larger \(\alpha\) for lower motor voltage
Bus voltage check	\(\displaystyle V_{\text{bus}} = \frac{V_{\text{mot}}}{1-\alpha}\)	Must equal DC bus voltage
Regenerated power	\(\displaystyle P_{\text{regen}} = V_{\text{mot}}\,I_a\)	Power returned to the bus
Minimum inductance	\(\displaystyle L_{\min} = \frac{V_{\text{mot}}\,\alpha}{\Delta I_{\max}\,f}\)	Boost ripple formula (differs from buck)

Rheostatic (Dynamic) Braking

Chopper output voltage	\(\displaystyle V_{\text{ch}} = R_b\,I_a\,(1-k)\)	\(R_b\) = braking resistor; \(k\) = duty cycle
Power in braking resistor	\(\displaystyle P_b = I_a^2\,R_b\,(1-k)\)	Dissipated as heat — energy not recovered
Equivalent circuit resistance	\(\displaystyle R_{\text{eq}} = R_b(1-k) + R_a\)	Total resistance seen by back EMF
Peak switch voltage	\(\displaystyle V_p = I_a\,R_b\)	Voltage rating required for the switch

Rheostatic vs. Regenerative Braking

Rheostatic braking dissipates kinetic energy as heat in \(R_b\). Regenerative braking (two-quadrant or boost chopper) returns energy to the supply — preferred whenever the DC bus can absorb it.

Power Signs Convention

Motor (back-EMF)	\(\displaystyle P_{\text{motor}} = E_b\,I_a\)	Positive = mechanical power out; Negative = returning energy
Supply	\(\displaystyle P_s = V_s\,(i_s)_{\text{avg}}\)	Negative \(P_s\) = power returned to supply
Copper loss	\(\displaystyle P_{R_a} = I_a^2\,R_a\)	Always positive (dissipated)

2. Two-Quadrant Chopper: Motoring and Regenerative Braking

Problem 2.1

Problem Statement

Drive Data

Supply voltage \(V_s\)	120 V
Motor constant \(K\Phi\)	0.1 V/rpm
Armature resistance \(R_a\)	0.2 Ω
Chopper frequency \(f_c\)	250 Hz
Current	Ripple-free

Two Operating Conditions

(a) Motoring via \(S_1/D_1\):

\(n = 400\,\text{rpm},\; i_0 = +100\,\text{A}\)

(b) Regenerative braking via \(S_2/D_2\):

\(n = 350\,\text{rpm},\; i_0 = -100\,\text{A}\)

Find \(t_{\text{on}}\), \(P_{\text{motor}}\), \(P_{R_a}\), \(P_s\) for each case.

Given:

\(V_s=120\,\text{V}\), \(K\Phi=0.1\,\text{V/rpm}\), \(R_a=0.2\,\Omega\), \(f_c=250\,\text{Hz}\)

Find:

\(t_{\text{on}}\), \(P_{\text{motor}}\), \(P_{R_a}\), \(P_s\) in both motoring and braking modes

Part (a) — Forward Motoring Quadrant I

Step 1 — Back EMF

\[E_a = K\Phi \cdot n = 0.1 \times 400 = 40\,\text{V}\]

Step 2 — Required average voltage

\[V_a = E_a + I_a R_a = 40 + 100\times0.2 = 60\,\text{V}\]

Step 3 — Duty cycle and on-time

\[\alpha = \frac{V_a}{V_s} = \frac{60}{120} = 0.5\] \[t_{\text{on}} = \frac{\alpha}{f_c} = \frac{0.5}{250} = 2\,\text{ms}\]

Step 4 — Power quantities

\[P_{\text{motor}} = E_a\,i_0 = 40\times100 = 4000\,\text{W}\] \[P_{R_a} = i_0^2\,R_a = 100^2\times0.2 = 2000\,\text{W}\] \[P_s = V_s\,(i_s)_{\text{avg}} = 120\times(100\times0.5) = 6000\,\text{W}\]

Supply Current Average

In motoring mode \(S_1\) conducts during \(t_{\text{on}}\), so the average supply current is: \[(i_s)_{\text{avg}} = \alpha\,i_0 = 0.5\times100 = 50\,\text{A}\]

Energy Balance Check

\[P_s = P_{\text{motor}} + P_{R_a} = 4000 + 2000 = 6000\,\text{W}\;\checkmark\]

▶ Results — Part (a)

\(t_{\text{on}} = 2\,\text{ms}\) \(P_{\text{motor}} = 4\,\text{kW}\) \(P_{R_a} = 2\,\text{kW}\) \(P_s = 6\,\text{kW}\)

Part (b) — Regenerative Braking Quadrant II

Step 1 — Back EMF at braking speed

\[E_a = K\Phi \cdot n = 0.1\times350 = 35\,\text{V}\]

Step 2 — Voltage across chopper output

Current is reversed (\(i_0 = -100\,\text{A}\)), so:

\[V_a = E_a - |i_0|\,R_a = 35 - 100\times0.2 = 15\,\text{V}\]

Step 3 — Duty cycle and on-time

In \(S_2/D_2\) mode: \((1-\alpha)V_s = V_a\)

\[\alpha = 1 - \frac{15}{120} = \frac{7}{8} = 0.875\] \[t_{\text{on}} = \frac{0.875}{250} = 3.5\,\text{ms}\]

Step 4 — Power quantities

\[P_{\text{motor}} = E_a\,i_0 = 35\times(-100) = -3500\,\text{W}\] \[P_{R_a} = i_0^2\,R_a = 100^2\times0.2 = 2000\,\text{W}\] \[P_s = V_s\,(i_s)_{\text{avg}} = 120\times\!\left(-100\times\tfrac{1}{8}\right) = -1500\,\text{W}\]

Supply Current in Braking

\(S_2\) conducts for fraction \((1-\alpha) = 1/8\) of the period, returning current to the supply: \[(i_s)_{\text{avg}} = -|i_0|\,(1-\alpha) = -100\times\tfrac{1}{8} = -12.5\,\text{A}\]

Negative \(P_s\) Interpretation

\(P_s = -1500\,\text{W}\) means 1.5 kW is returned to the DC supply. The remaining 2 kW is dissipated in \(R_a\) — the motor acts as a generator.

Power Balance (Braking)

Generator output: \(|P_{\text{motor}}| = 3500\,\text{W}\)
To supply: \(|P_s| = 1500\,\text{W}\)
Lost in \(R_a\): \(P_{R_a} = 2000\,\text{W}\)
\(3500 = 1500 + 2000\;\checkmark\)

▶ Results — Part (b)

\(t_{\text{on}} = 3.5\,\text{ms}\) \(P_{\text{motor}} = -3.5\,\text{kW}\) \(P_{R_a} = 2\,\text{kW}\) \(P_s = -1.5\,\text{kW}\)

3. Regenerative Braking: Speed and Power

Problem 2.2

Problem Statement

Motor & Drive Data

Rated supply voltage	230 V
Rated speed \(N_r\)	500 rpm
Rated armature current \(I_a\)	90 A
Armature resistance \(R_a\)	0.115 Ω
Chopper frequency \(f\)	400 Hz
Field	Constant (separately excited)

Objective

Find motor speed and \(P_{\text{regen}}\) at \(\alpha = 0.5\), rated braking current.
Find maximum safe regenerative braking speed at \(\alpha_{\min} = 0.1\).

Neglect rotational losses.

Given:

\(V_s=230\,\text{V}\), \(N_r=500\,\text{rpm}\), \(I_a=90\,\text{A}\), \(R_a=0.115\,\Omega\), \(f=400\,\text{Hz}\)

Find:

(1) \(N\) and \(P_{\text{regen}}\) at \(\alpha=0.5\) (2) \(N_{\max}\) at \(\alpha_{\min}=0.1\)

Solution

Step 1 — Determine motor constant \(K\)

\[E_{\text{rated}} = V_s - I_a R_a = 230 - 90\times0.115 = 219.65\,\text{V}\] \[\omega_r = \frac{500\times2\pi}{60} = 52.36\,\text{rad/s}\] \[K = \frac{E_{\text{rated}}}{\omega_r} = \frac{219.65}{52.36} = 4.194\,\text{V·s/rad}\]

Part (1) — At \(\alpha = 0.5\) regenerative

In regen mode: \(V_a = \alpha\,V_s\) and \(E_b = V_a + I_a R_a\):

\[V_a = 0.5\times230 = 115\,\text{V}\] \[E_b = 115 + 90\times0.115 = 125.35\,\text{V}\] \[\omega = \frac{E_b}{K} = \frac{125.35}{4.194} = 29.89\,\text{rad/s}\] \[N = 29.89\times\frac{60}{2\pi} = 285.5\,\text{rpm}\] \[P_{\text{regen}} = E_b\,I_a = 125.35\times90 = 11.28\,\text{kW}\]

Part (2) — At \(\alpha_{\min} = 0.1\)

\[V_a = 0.1\times230 = 23\,\text{V}\] \[E_b = 23 + 90\times0.115 = 33.35\,\text{V}\] \[\omega = \frac{33.35}{4.194} = 7.952\,\text{rad/s}\] \[N_{\max} = 7.952\times\frac{60}{2\pi} = 75.9\,\text{rpm}\]

Regen Braking Voltage Equation

In regenerative braking the machine drives current back against the supply. The armature equation reverses sign: \[E_b = V_a + I_a R_a\] Compare with motoring: \(E_b = V_a - I_a R_a\). The back-EMF must be higher than \(V_a\) to push current back.

Speed Control Range

By varying \(\alpha\) from 0.1 to 1.0, the regenerative braking speed can be controlled from 75.9 rpm up to rated speed, maintaining constant rated braking current throughout.

At \(\alpha = 1\) (upper limit)

\[V_a = 230\,\text{V},\quad E_b = 230 + 10.35 = 240.35\,\text{V}\] \[\omega = \frac{240.35}{4.194} = 57.3\,\text{rad/s} = 547\,\text{rpm}\] Slightly above rated speed — upper regenerative limit.

▶ Key Results

Condition	Speed	Regen Power
\(\alpha = 0.5\)	285.5 rpm	11.28 kW
\(\alpha_{\min} = 0.1\)	75.9 rpm	—

\(N = 285.5\,\text{rpm}\) \(P_{\text{regen}} = 11.28\,\text{kW}\)

\(N_{\max} = 75.9\,\text{rpm at}\;\alpha_{\min}\)

4. Regenerative Braking of a DC Series Motor

Problem 2.3

Problem Statement

Motor & Drive Data

Bus voltage \(V_s\)	600 V
Armature resistance \(R_a\)	0.02 Ω
Field resistance \(R_f\)	0.03 Ω
Motor constant \(K_v\)	0.01527 V/(A·rad/s)
Armature current \(I_a\)	250 A
Duty cycle \(k\)	0.60

Objective

A DC–DC converter provides regenerative braking for a series motor. Find:

Chopper output voltage \(V_{\text{ch}}\)
Regenerated power \(P_{\text{regen}}\)
Equivalent resistance \(R_{\text{eq}}\)
Speed range: \(\omega_{\min}\), \(\omega_{\max}\)
Operating speed \(\omega\) at \(k = 0.60\)

Given:

\(V_s=600\,\text{V}\), \(R_a+R_f=0.05\,\Omega\), \(K_v=0.01527\,\text{V/(A·rad/s)}\), \(I_a=250\,\text{A}\), \(k=0.60\)

Find:

\(V_{\text{ch}}\), \(P_{\text{regen}}\), \(R_{\text{eq}}\), \(\omega_{\min}\), \(\omega_{\max}\), \(\omega\)

Solution

Step 1 — Chopper output voltage

\[V_{\text{ch}} = (1-k)\,V_s = 0.4\times600 = 240\,\text{V}\]

Step 2 — Regenerated power

\[P_{\text{regen}} = I_a\,V_s\,(1-k) = 250\times600\times0.4 = 60\,\text{kW}\]

Step 3 — Equivalent resistance

\[R_{\text{eq}} = \frac{V_s}{I_a}(1-k) + (R_a+R_f) = \frac{600}{250}\times0.4 + 0.05 = 1.01\,\Omega\]

Step 4 — Speed range

\[\omega_{\min} = \frac{R_a+R_f}{K_v\,I_a} = \frac{0.05}{0.01527\times250} = 3.274\,\text{rad/s}\] \[\omega_{\max} = \frac{V_s/I_a - (R_a+R_f)}{K_v} = \frac{600/250 - 0.05}{0.01527} = 153.9\,\text{rad/s}\]

Step 5 — Operating speed at \(k = 0.60\)

\[E_b = V_{\text{ch}} + (R_a+R_f)\,I_a = 240 + 0.05\times250 = 252.5\,\text{V}\] \[\omega = \frac{E_b}{K_v\,I_a} = \frac{252.5}{0.01527\times250} = 66.14\,\text{rad/s}\]

Series Motor Difference

In a series motor \(K_v\,I_a\) acts as the combined flux-constant, so speed is: \[\omega = \frac{E_b}{K_v\,I_a}\] Unlike the separately excited machine, flux falls with current, making the series motor suitable for high-torque traction loads.

Speed Range Summary

Condition	Speed (rad/s)
Minimum (\(V_{\text{ch}}=0\))	3.274
At \(k=0.60\)	66.14
Maximum (\(k=1\))	153.9

▶ Key Results

\(V_{\text{ch}} = 240\,\text{V}\) \(P_{\text{regen}} = 60\,\text{kW}\) \(R_{\text{eq}} = 1.01\,\Omega\)

\(\omega_{\min} = 3.274\,\text{rad/s}\) \(\omega_{\max} = 153.9\,\text{rad/s}\) \(\omega = 66.14\,\text{rad/s}\)

5. Four-Quadrant H-Bridge Chopper

Problem 2.4

Problem Statement

Drive Data

Supply voltage \(V_s\)	200 V
Armature resistance \(R_a\)	0.5 Ω
Motor flux constant \(K\Phi\)	0.8 V·s/rad

Bipolar PWM Relation

\[V_a = (2\alpha-1)\,V_s\] \[\alpha = \frac{V_a + V_s}{2\,V_s}\]

Three Operating Points

(a) Forward motoring: \(\omega = +100\,\text{rad/s}\), \(I_a = +20\,\text{A}\)

(b) Forward regen. braking: \(\omega = +80\,\text{rad/s}\), \(I_a = -20\,\text{A}\)

(c) Reverse motoring: \(\omega = -60\,\text{rad/s}\), \(I_a = +20\,\text{A}\)

Find duty cycle \(\alpha\) and identify the operating quadrant for each case.

Given:

\(V_s=200\,\text{V}\), \(R_a=0.5\,\Omega\), \(K\Phi=0.8\,\text{V·s/rad}\)

Find:

Duty cycle \(\alpha\) and quadrant for cases (a), (b), (c)

Solution

Case (a) — Forward Motoring Quadrant I

\[E_b = K\Phi\,\omega = 0.8\times100 = 80\,\text{V}\] \[V_a = E_b + I_a R_a = 80 + 20\times0.5 = 90\,\text{V}\] \[\alpha = \frac{V_a+V_s}{2V_s} = \frac{90+200}{400} = \frac{290}{400} = 0.725\]

Verify

\[V_a = (2\times0.725-1)\times200 = 0.45\times200 = 90\,\text{V}\;\checkmark\] \(V_a>0\), \(I_a>0\) → Quadrant I

Case (b) — Forward Regenerative Braking Quadrant II

\[E_b = 0.8\times80 = 64\,\text{V}\] \[V_a = E_b + I_a R_a = 64 + (-20)\times0.5 = 54\,\text{V}\] \[\alpha = \frac{54+200}{400} = \frac{254}{400} = 0.635\]

Verify

\[V_a = (2\times0.635-1)\times200 = 0.27\times200 = 54\,\text{V}\;\checkmark\] \(V_a>0\), \(I_a<0\) → Quadrant II

Case (c) — Reverse Operation Quadrant IV

\[E_b = 0.8\times(-60) = -48\,\text{V}\] \[V_a = E_b + I_a R_a = -48 + 20\times0.5 = -38\,\text{V}\] \[\alpha = \frac{-38+200}{400} = \frac{162}{400} = 0.405\]

Verify

\[V_a = (2\times0.405-1)\times200 = -0.19\times200 = -38\,\text{V}\;\checkmark\] \(V_a<0\), \(I_a>0\) → Quadrant IV

▶ Summary of All Three Cases

Case	\(\omega\) (rad/s)	\(I_a\) (A)	\(E_b\) (V)	\(V_a\) (V)	Duty Cycle \(\alpha\)	Quadrant
(a) Fwd. motoring	+100	+20	+80	+90	0.725	Q1
(b) Fwd. regen.	+80	−20	+64	+54	0.635	Q2
(c) Reverse	−60	+20	−48	−38	0.405	Q4

6. Rheostatic Braking with DC–DC Converter

Problem 5.1

Problem Statement

Motor & Drive Data

Armature resistance \(R_a\)	0.05 Ω
Braking resistor \(R_b\)	5 Ω
Motor constant \(K_v\)	1.527 V/(A·rad/s)
Armature current \(I_a\)	150 A
Field current \(I_f\)	1.5 A
Duty cycle \(k\)	0.40

Objective

A separately excited DC motor brakes rheostically via a DC–DC converter. Find:

Chopper output voltage \(V_{\text{ch}}\)
Power in braking resistor \(P_b\)
Equivalent circuit resistance \(R_{\text{eq}}\)
Motor speed \(\omega\)
Peak switch voltage \(V_p\)

Given:

\(R_a=0.05\,\Omega\), \(R_b=5\,\Omega\), \(K_v=1.527\,\text{V/(A·rad/s)}\), \(I_a=150\,\text{A}\), \(I_f=1.5\,\text{A}\), \(k=0.40\)

Find:

\(V_{\text{ch}}\), \(P_b\), \(R_{\text{eq}}\), \(\omega\), \(V_p\)

Solution

Step 1 — Chopper output voltage

\[V_{\text{ch}} = R_b\,I_a\,(1-k) = 5\times150\times0.6 = 450\,\text{V}\]

Step 2 — Braking resistor power

\[P_b = I_a^2\,R_b\,(1-k) = 150^2\times5\times0.6 = 67.5\,\text{kW}\]

Step 3 — Equivalent circuit resistance

\[R_{\text{eq}} = R_b(1-k)+R_a = 5\times0.6+0.05 = 3.05\,\Omega\]

Step 4 — Back EMF and motor speed

\[E_b = V_{\text{ch}} + R_a\,I_a = 450 + 0.05\times150 = 457.5\,\text{V}\] \[\omega = \frac{E_b}{K_v\,I_f} = \frac{457.5}{1.527\times1.5} = 199.7\,\text{rad/s}\]

Step 5 — Peak switch voltage

\[V_p = I_a\,R_b = 150\times5 = 750\,\text{V}\]

Rheostatic Braking Circuit

The chopper chops the braking resistor current. During switch-off the current freewheels through a diode. The average voltage across \(R_b\) is: \[V_{\text{ch}} = R_b\,I_a\,(1-k)\] where \((1-k)\) is the fraction of the period the switch is off (current flowing through \(R_b\)).

Peak Voltage Warning

The switch must withstand \(V_p = I_a R_b = 750\,\text{V}\) — the full resistor voltage when the switch opens. Device voltage rating must exceed this with a safety margin.

Speed in rpm

\[N = 199.7\times\frac{60}{2\pi} = 1907\,\text{rpm}\]

▶ Key Results

\(V_{\text{ch}} = 450\,\text{V}\) \(P_b = 67.5\,\text{kW}\) \(R_{\text{eq}} = 3.05\,\Omega\)

\(\omega = 199.7\,\text{rad/s}\) \(V_p = 750\,\text{V}\)

7. Boost Chopper for Regenerative Braking

Problem 8.1

Problem Statement

Motor & Drive Data

Rated motor voltage	200 V
DC bus voltage \(V_{\text{bus}}\)	200 V
Back EMF during braking \(E_b\)	150 V
Armature resistance \(R_a\)	0.2 Ω
Braking current \(I_a\)	40 A
Switching frequency \(f\)	1 kHz

Objective

A boost chopper returns braking energy to a 200 V DC bus. Find:

Duty cycle \(\alpha\)
Regenerated power \(P_{\text{regen}}\)
Minimum inductance \(L_{\min}\) for \(\Delta I \le 5\%\) of \(I_a\)

Why a Boost Chopper?

When \(E_b = 150\,\text{V} < V_{\text{bus}} = 200\,\text{V}\), a step-down (buck) chopper cannot push energy back. A boost chopper steps up the motor voltage to match the bus, enabling regeneration at low speeds.

Given:

\(E_b=150\,\text{V}\), \(R_a=0.2\,\Omega\), \(I_a=40\,\text{A}\), \(V_{\text{bus}}=200\,\text{V}\), \(f=1\,\text{kHz}\)

Find:

\(\alpha\), \(P_{\text{regen}}\), \(L_{\min}\)

Solution

Step 1 — Motor terminal voltage in generator mode

\[V_{\text{mot}} = E_b - I_a R_a = 150 - 40\times0.2 = 142\,\text{V}\]

Step 2 — Boost duty cycle

\[\alpha = 1 - \frac{V_{\text{mot}}}{V_{\text{bus}}} = 1 - \frac{142}{200} = 0.29\]

Step 3 — Verification

\[V_{\text{bus}} = \frac{V_{\text{mot}}}{1-\alpha} = \frac{142}{0.71} = 200\,\text{V}\;\checkmark\]

Step 4 — Regenerated power

\[P_{\text{regen}} = V_{\text{mot}}\,I_a = 142\times40 = 5{,}680\,\text{W}\]

Step 5 — Minimum inductance

\[\Delta I_{\max} = 0.05\times40 = 2\,\text{A}\] \[L_{\min} = \frac{V_{\text{mot}}\,\alpha}{\Delta I_{\max}\,f} = \frac{142\times0.29}{2\times1000} = \frac{41.18}{2000} = 20.6\,\text{mH}\]

Boost vs. Buck Ripple Formula

Buck (step-down):

\[L_{\min} = \frac{V_s\,\alpha\,(1-\alpha)}{2\,\Delta I\,f}\]

Boost (step-up):

\[L_{\min} = \frac{V_{\text{mot}}\,\alpha}{\Delta I\,f}\] The boost inductor stores energy during the switch-on period and releases it to the bus during switch-off.

Bus Power Consistency Check

\[P_{\text{bus}} = V_{\text{bus}}\,I_{\text{bus}}\] \[I_{\text{bus}} = I_a(1-\alpha) = 40\times0.71 = 28.4\,\text{A}\] \[P_{\text{bus}} = 200\times28.4 = 5{,}680\,\text{W}\;\checkmark\]

▶ Key Results

\(\alpha = 0.29\) \(P_{\text{regen}} = 5{,}680\,\text{W}\) \(L_{\min} = 20.6\,\text{mH}\)

8. Boost Chopper Design for a Hoist Drive

Problem 8.2

Problem Statement

Motor Nameplate & Drive Data

Rated voltage	500 V
Rated current \(I_{\text{rated}}\)	100 A
Rated speed \(N_r\)	1200 rpm
Armature resistance \(R_a\)	0.08 Ω
DC bus voltage \(V_{\text{bus}}\)	600 V
Braking conditions
Current \(I_a\)	80 A
Speed during descent \(N\)	900 rpm
Switching frequency \(f\)	800 Hz

Objective

A hoist drive brakes regeneratively during controlled descent via a boost chopper. Find:

Motor flux constant \(K\Phi\)
Duty cycle \(\alpha\) during descent
Regenerated power \(P_{\text{regen}}\)
Minimum inductance \(L_{\min}\) for 8% ripple

Given:

Rated: 500 V, 100 A, 1200 rpm; \(R_a=0.08\,\Omega\); Descent: \(N=900\,\text{rpm}\), \(I_a=80\,\text{A}\), \(V_{\text{bus}}=600\,\text{V}\), \(f=800\,\text{Hz}\)

Find:

\(K\Phi\), \(\alpha\), \(P_{\text{regen}}\), \(L_{\min}\)

Solution

Step 1 — Motor flux constant from rated data

\[E_{\text{rated}} = 500 - 100\times0.08 = 492\,\text{V}\] \[\omega_r = \frac{1200\times2\pi}{60} = 125.66\,\text{rad/s}\] \[K\Phi = \frac{E_{\text{rated}}}{\omega_r} = \frac{492}{125.66} = 3.915\,\text{V·s/rad}\]

Step 2 — Back EMF at descent speed

\[\omega = \frac{900\times2\pi}{60} = 94.25\,\text{rad/s}\] \[E_b = K\Phi\,\omega = 3.915\times94.25 = 368.9\,\text{V}\]

Step 3 — Motor terminal voltage and duty cycle

\[V_{\text{mot}} = E_b - I_a R_a = 368.9 - 80\times0.08 = 362.5\,\text{V}\] \[\alpha = 1 - \frac{V_{\text{mot}}}{V_{\text{bus}}} = 1 - \frac{362.5}{600} = 0.396\]

Step 4 — Regenerated power

\[P_{\text{regen}} = V_{\text{mot}}\,I_a = 362.5\times80 = 29.0\,\text{kW}\]

Step 5 — Minimum inductance

\[\Delta I_{\max} = 0.08\times80 = 6.4\,\text{A}\] \[L_{\min} = \frac{V_{\text{mot}}\,\alpha}{\Delta I_{\max}\,f} = \frac{362.5\times0.396}{6.4\times800} = \frac{143.6}{5120} = 28.0\,\text{mH}\]

Hoist Application Note

During descent, the load drives the motor as a generator. The boost chopper elevates \(V_{\text{mot}} = 362.5\,\text{V}\) to the 600 V bus, recovering potential energy that would otherwise be wasted as heat in a braking resistor.

Energy Recovery Potential

At 29 kW recovered during a 60-second descent: \[\text{Energy} = 29{,}000\times60 = 1{,}740\,\text{kJ} = 483\,\text{Wh}\] Compared to rheostatic braking, this is pure energy saving.

Duty Cycle Increases at Lower Speeds

If descent speed were halved to 450 rpm: \[E_b \approx 184.5\,\text{V},\quad V_{\text{mot}} \approx 178.1\,\text{V}\] \[\alpha = 1 - \frac{178.1}{600} = 0.70\] Higher \(\alpha\) means the switch is ON longer to store more energy in \(L\).

▶ Key Results

\(K\Phi = 3.915\,\text{V·s/rad}\) \(\alpha = 0.396\)

\(P_{\text{regen}} = 29.0\,\text{kW}\) \(L_{\min} = 28.0\,\text{mH}\)