---
title: "Chopper ON time for separately excited DC motor"
subject: "Utilization of Electrical Energy"
chapter: 3
key: "semester_7/utilization_of_electrical_energy:chopper-on-time-separately-excited-motor"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/chopper-on-time-separately-excited-motor
site: BCT Engineering
---

# Chopper ON time for separately excited DC motor

Utilization of Electrical Energy · Chapter 3: Control of Electric Drive

- Asked in: 2082 Baisakh, 8 marks
- Video: [Speed Control of a separately excited DC motor using an IGBT or MOSFET Chopper](https://www.youtube.com/watch?v=n2XYSspJR_k) — BMSCE Education videos
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/chopper-on-time-separately-excited-motor
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3

## Question

A 2.2 kW, 220 V, 1000 rpm separately excited dc motor has armature resistance of 2 Ω. This motor controlled by a stepdown chopper having a frequency of 250 Hz and input voltage 220 V. The motor is driving a load whose torque is proportional to speed at α = 0.9, motor speed is 1,200 rpm. What should be the ON time of chopper if motor speed is 800 rpm.

## Short answer

For a load where torque is proportional to speed, armature voltage is proportional to speed; thus α2 = α1 × (800/1200) = 0.6. With chopping frequency 250 Hz, ON time Ton = α2 × (1/250) = 2.4 ms.

## Answer

Assumption: field current is constant (rated), so $E = K N$ and $T = K I_a$.

#### Key relations
- Chopper output: $V_a = \alpha V_s$
- Load torque proportional to speed: $T_L \propto N$. Since $T = K I_a$, the armature current is also proportional to speed: $I_a = c\,N$.
- Armature equation:
$$
\begin{aligned} V_a &= E + I_a R_a = K N + R_a c N = (K + R_a c)\,N \end{aligned}
$$
So the armature voltage is **directly proportional to speed** for this load, whatever the values of $K$ and $c$. Hence
$$
\frac{\alpha_2}{\alpha_1} = \frac{V_{a2}}{V_{a1}} = \frac{N_2}{N_1}
$$

#### Duty ratio at 800 rpm
$$
\begin{aligned} \alpha_2 &= \alpha_1 \frac{N_2}{N_1} = 0.9 \times \frac{800}{1200} = 0.6 \end{aligned}
$$

#### ON time
Chopping period:
$$
T = \frac{1}{f} = \frac{1}{250} = 4\ \text{ms}
$$
$$
\begin{aligned} T_{on} &= \alpha_2 T = 0.6 \times 4\ \text{ms} = 2.4\ \text{ms} \\ T_{off} &= 4 - 2.4 = 1.6\ \text{ms} \end{aligned}
$$
Check: $V_{a1} = 0.9 \times 220 = 198$ V at 1200 rpm and $V_{a2} = 0.6 \times 220 = 132$ V at 800 rpm; $132/198 = 800/1200$.

Note: the rated data (2.2 kW, 1000 rpm) are not needed, because both the back emf and the $I_a R_a$ drop scale with speed for a load with $T_L \propto N$.

**Answer: Duty ratio = 0.6, ON time $T_{on}$ = 2.4 ms (OFF time 1.6 ms).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For a separately excited motor with load torque proportional to speed, how is armature voltage related to speed?
   A. Va is proportional to N^2
   B. Va is independent of N
   C. Va is directly proportional to N
   D. Va is inversely proportional to N
   Answer: C. Both E = KN and IaRa (Ia proportional to N) scale with speed, so Va = (K + Ra c)N. (Key point: Armature voltage vs speed)

2. What is the duty ratio at 800 rpm if alpha = 0.9 at 1200 rpm?
   A. 0.9
   B. 0.45
   C. 0.67
   D. 0.6
   Answer: D. alpha2 = 0.9 x 800/1200 = 0.6. (Key point: Duty ratio)

3. With a chopper frequency of 250 Hz and duty ratio 0.6, what is the ON time?
   A. 2.4 ms
   B. 1.6 ms
   C. 4 ms
   D. 0.6 ms
   Answer: A. T = 1/250 = 4 ms, so Ton = 0.6 x 4 = 2.4 ms. (Key point: ON time)

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