---
title: "Single-phase inverter with AC motor load"
subject: "Power Electronics"
chapter: 5
key: "semester_7/power_electronics:single-phase-inverter-ac-motor"
source: https://www.bctengineer.com/q/semester_7/power_electronics/single-phase-inverter-ac-motor
site: BCT Engineering
---

# Single-phase inverter with AC motor load

Power Electronics · Chapter 5: Inverter

- Asked in: 2081 Baishakh, 8 marks; 2080 Baishakh, 8 marks; 2073 Chaitra, 8 marks; 2070 Chaitra, 8 marks (asked 4 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/single-phase-inverter-ac-motor
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5

## Question

Explain the operation of a single phase inverter (square wave output voltage) with ac motor as load. Derive the equation for current drawn by the motor for positive half cycle and negative half cycle.

## Short answer

With square‑wave ±Vs output and motor as series RL, current lags. For 0<t<T/2, i0(t)=Vs/R − (Vs/R + I0)e^(−t/τ); for T/2<t<T, i0(t)=−Vs/R + (Vs/R + I0)e^(−(t−T/2)/τ), where I0 = (Vs/R)·(1−e^(−T/(2τ)))/(1+e^(−T/(2τ))).

## Answer

A single-phase full-bridge inverter with square-wave output applies $+V_s$ for half a period and $-V_s$ for the other half to the motor. An ac motor is modelled per phase as a series R–L load (back EMF neglected or lumped in R–L), so the current is exponential and lags the voltage; feedback diodes carry current when it is opposite to the voltage.

#### Circuit
```text
     +-------+-----------+
     |       |           |
     |   T1 D1       T3 D3
    Vs       |           |
     |       A--[R  L]---B
     |       |  motor    |
     |   T4 D4       T2 D2
     |       |           |
     +-------+-----------+
```

#### Operation (steady state)
- **$0 < t < T/2$:** T1, T2 are gated, $v_0 = +V_s$. At $t = 0$ the current is still negative ($-I_0$), so it first flows through **D1, D2** back to the source. When it reaches zero, T1 and T2 take over and $i_0$ rises to $+I_0$.
- **$T/2 < t < T$:** T3, T4 are gated, $v_0 = -V_s$. The positive current first flows through **D3, D4**, falls to zero, then T3, T4 conduct and $i_0$ goes to $-I_0$.

```text
 v0  +Vs|________
        |        |
      0 |--------+--------- t
        |        |________
     -Vs           T/2     T
 i0     |      ___
     +I0|    /     \
      0 |---/-------\------- t
     -I0|__/         \___/
        D1D2|T1T2|D3D4|T3T4
```

#### Current during the positive half cycle ($0 \le t \le T/2$)
$$
\begin{aligned}
V_s &= Ri_0 + L\frac{di_0}{dt}, \qquad i_0(0) = -I_0 \\
i_0(t) &= \frac{V_s}{R} - \left(\frac{V_s}{R} + I_0\right)e^{-t/\tau}, \qquad \tau = \frac{L}{R}
\end{aligned}
$$

At $t = T/2$, $i_0 = +I_0$ (half-wave symmetry):

$$
\begin{aligned}
I_0 &= \frac{V_s}{R} - \left(\frac{V_s}{R} + I_0\right)e^{-T/2\tau} \\
I_0 &= \frac{V_s}{R}\cdot\frac{1 - e^{-T/2\tau}}{1 + e^{-T/2\tau}}
\end{aligned}
$$

The diodes D1, D2 conduct until $i_0 = 0$, i.e. for

$$
t_1 = \tau\ln\!\left[\frac{2}{1+e^{-T/2\tau}}\right]
$$

#### Current during the negative half cycle ($T/2 \le t \le T$)
$$
\begin{aligned}
-V_s &= Ri_0 + L\frac{di_0}{dt}, \qquad i_0(T/2) = +I_0 \\
i_0(t) &= -\frac{V_s}{R} + \left(\frac{V_s}{R} + I_0\right)e^{-(t - T/2)/\tau}
\end{aligned}
$$

#### Remarks
- Output voltage: $V_{0,rms} = V_s$; fundamental $V_{1,rms} = \dfrac{4V_s}{\sqrt2\,\pi} = 0.9V_s$.
- The harmonic currents (3rd, 5th, …) cause extra copper loss and torque pulsation in the motor; their magnitudes fall roughly as $1/n^2$ (the harmonic voltage falls as $1/n$ and the reactance $nX_L$ rises as $n$), so the motor inductance filters them.
- Feedback diodes are essential with a motor load; without them the inductive current would have no path when the conducting switches turn off, and large voltage spikes would occur.

## Check yourself

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

1. At t = 0 in the positive half cycle (T1, T2 gated), current is first carried by:
   A. D1 and D2, returning energy to source, until it reaches zero
   B. T3 and T4
   C. T1 and T2 immediately
   D. D3 and D4
   Answer: A. The current is still negative from the previous half cycle. (Key point: Start of positive half cycle)

2. The current in the positive half cycle is:
   A. i0 = Vs/R (1 - e^(-t/tau))
   B. i0 = Vs/R - (Vs/R + I0)e^(-t/tau), tau = L/R
   C. i0 = -Vs/R + (Vs/R + I0)e^(-t/tau)
   D. i0 = Vs/R e^(-t/tau)
   Answer: B. With i0(0) = -I0, the solution of Vs = Ri + L di/dt gives this expression. (Key point: Current equation)

3. I0 is found from half-wave symmetry:
   A. I0 = Vs/R
   B. I0 = (Vs/R)(1 + e^(-T/2tau))
   C. I0 = (Vs/R)(1 - e^(-T/2tau))/(1 + e^(-T/2tau))
   D. I0 = Vs/(R e^(-T/2tau))
   Answer: C. Setting i0(T/2) = +I0 gives this. (Key point: Peak current)

4. Why are feedback diodes essential?
   A. They raise the output voltage
   B. They filter harmonics
   C. They set the frequency
   D. The inductive current cannot reverse instantly; diodes give it a path and avoid spikes
   Answer: D. Without them the current would have no path when it opposes the voltage. (Key point: Need for diodes)

## Related in Power Electronics

- [Half-bridge square-wave inverter at 120 Hz](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Asquare-wave-inverter-120hz) — Power Electronics, chapter 5
- [Single-phase current source inverter with AC motor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Acsi-ac-motor) — Power Electronics, chapter 5
- [Square-wave inverter RL load current up to 5th order](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ainverter-square-wave-rl-fifth-order) — Power Electronics, chapter 5, asked 2×
- [Square-wave inverter ±100 V: fundamental load current](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ainverter-100v-rl-fundamental) — Power Electronics, chapter 5, asked 2×
- [Single-phase square-wave inverter rms and fundamental](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Asquare-wave-inverter-rms-fundamental) — Power Electronics, chapter 5
- [Single-phase inverter fundamental output voltage](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Asingle-phase-inverter-r-fundamental) — Power Electronics, chapter 5

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