Inverter

A three-phase sinusoidal PWM (SPWM) inverter is a six-switch bridge whose switching instants are fixed by comparing three sinusoidal reference waves (120° apart) with a common high-frequency triangular carrier. The output line voltages are pulse trains whose fundamental is sinusoidal and adjustable in both magnitude and frequency.

Circuit

      +----+--------+--------+
      |    |        |        |
 Vdc/2   S1 D1    S3 D3    S5 D5
  ---  |    |        |        |
  O    +-a  +--b     +--c     |
  ---  |    |        |        |
 Vdc/2   S4 D4    S6 D6    S2 D2
      |    |        |        |
      +----+--------+--------+
        a, b, c -> 3-phase load

Each leg (S1–S4, S3–S6, S5–S2) is switched in a complementary way (with a small dead time). Diodes carry reactive current back to the source.

Principle of operation

1. Three references $v_{ra}, v_{rb}, v_{rc}$ of amplitude $A_r$ and frequency $f_r$ (desired output frequency), displaced by 120°.
2. One triangular carrier $v_c$ of amplitude $A_c$ and frequency $f_c \gg f_r$.
3. Leg a: if $v_{ra} > v_c$, S1 is ON and $v_{aO} = +V_{dc}/2$; if $v_{ra} < v_c$, S4 is ON and $v_{aO} = -V_{dc}/2$. Legs b and c work the same way with $v_{rb}$, $v_{rc}$.
4. Line voltage $v_{ab} = v_{aO} - v_{bO}$ takes values $+V_{dc}$, 0, $-V_{dc}$.

 vc /\/\/\/\/\/\/\/\/\/\/\/\/   carrier
 vra  .-''-.          .-''-
     /      \        /       ref a
 ---'--------\------/--------
               '-..-'
 vaO +Vdc/2 |-| |--| |---| |--|
     -Vdc/2 | |_|  |_|   |_|
          (wide pulses near
           +peak of vra)
 vab  +Vdc  |||| ||||
         0 -------------------
      -Vdc         |||| ||||

Key quantities

• Amplitude modulation index $m_a = A_r/A_c$. For $m_a \le 1$ (linear range):

$$\begin{aligned} \hat V_{aO1} &= m_a\frac{V_{dc}}{2} \\ V_{L1,rms} &= \frac{\sqrt3}{2\sqrt2}\,m_a V_{dc} = 0.612\,m_a V_{dc} \end{aligned}$$

• Frequency modulation ratio $m_f = f_c/f_r$. It is chosen odd and a multiple of 3, so that the three phases have identical waveforms and triplen carrier harmonics cancel in the line voltages.
• Output frequency is changed by changing $f_r$; output voltage by changing $m_a$.

Harmonics

• Harmonics appear as side-bands around $m_f$, $2m_f$, $3m_f$, … (e.g. $m_f \pm 2$, $2m_f \pm 1$).
• Low-order harmonics are almost absent, so a small filter (or the load inductance itself) gives near-sinusoidal current.
• Overmodulation ($m_a > 1$): output rises less than linearly and low-order harmonics (5th, 7th) appear; at very large $m_a$ the inverter becomes a six-step (square-wave) inverter.

Advantages and uses

• Voltage and frequency control within the inverter (no variable dc link needed).
• Low harmonic distortion, small filter.
• Higher switching losses than six-step operation.
• Used in variable-speed induction motor drives (V/f control), UPS and grid-tie inverters.

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

     +-------+-----------+
     |       |           |
     |   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 $-I_0$.

 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 as $1/(n\cdot nX_L)$ because the motor inductance filters them.
• Feedback diodes are essential with a motor load; without them the inductive current could not reverse and large voltage spikes would occur.

In the 180° conduction mode of a three-phase bridge inverter each switch conducts for 180°, three switches are ON at any time (one from each leg), and switches are gated in the order S1, S2, …, S6 at 60° intervals. The output phase voltage is a six-step wave and the line voltage is a 120° quasi-square wave.

Circuit

      +----+--------+--------+
      |    |        |        |
 Vdc/2   S1 D1    S3 D3    S5 D5
  ---  |    |        |        |
  O    +-a  +--b     +--c     |
  ---  |    |        |        |
 Vdc/2   S4 D4    S6 D6    S2 D2
      |    |        |        |
      +----+--------+--------+
        a, b, c -> 3-phase load

Switches S1, S3, S5 connect phases a, b, c to the positive rail; S4, S6, S2 to the negative rail. S1 and S4 (same leg) are complementary.

Conduction table and phase voltages (star resistive load, each phase $R$)

($V = V_{dc}$.) Example, mode 1: phases a and c are on the positive rail, b on the negative. The equivalent circuit is $R/2$ (a, c in parallel) in series with $R$: current $i = \dfrac{V}{1.5R}$, so $v_{bn} = -iR = -2V/3$ and $v_{an} = v_{cn} = iR/2 = V/3$.

Waveforms

 gates: S1 |######......|   (0-180)
        S2  ..|######...|   (60-240)
        S3  ....|######.|   (120-300)
 van 2V/3      __
      V/3   __|  |__
        0 -|--------|-------------
     -V/3            |__    __|
    -2V/3               |__|
 vab   +V  |_____|
        0 -|     |__|      |__|--
       -V           |_____|
           0  60 120 180 240 300 360

• Phase voltage $v_{an}$: $V/3$, $2V/3$, $V/3$, $-V/3$, $-2V/3$, $-V/3$ in successive 60° steps. $v_{bn}$, $v_{cn}$ are the same wave shifted by 120° and 240°.
• Line voltage $v_{ab} = v_{an} - v_{bn}$: $+V$ for 0–120°, 0 for 120–180°, $-V$ for 180–300°, 0 for 300–360°. $v_{bc}$, $v_{ca}$ are shifted by 120°, 240°.

Important values

$$\begin{aligned} V_{ph,rms} &= \frac{\sqrt2}{3}V_{dc} = 0.4714\,V_{dc}, \qquad V_{L,rms} = \sqrt{\frac23}\,V_{dc} = 0.8165\,V_{dc} \\ v_{an} &= \sum_{n=6k\pm1}\frac{2V_{dc}}{n\pi}\sin n\omega t \end{aligned}$$

Triplen harmonics are absent. Because two switches of a leg must not conduct together, a small dead time is given between turning S1 off and S4 on.

A three-phase sinusoidal PWM inverter is a six-switch bridge in which each leg is switched by comparing its own sinusoidal reference (120° apart from the others) with a common triangular carrier. The switching instants are the crossing points of the reference and the carrier.

Circuit

      +----+--------+--------+
      |    |        |        |
 Vdc/2   S1 D1    S3 D3    S5 D5
  ---  |    |        |        |
  O    +-a  +--b     +--c     |
  ---  |    |        |        |
 Vdc/2   S4 D4    S6 D6    S2 D2
      |    |        |        |
      +----+--------+--------+
        a, b, c -> 3-phase load

Operation

• References: $v_{ra} = A_r\sin\omega t$, $v_{rb} = A_r\sin(\omega t - 120^\circ)$, $v_{rc} = A_r\sin(\omega t - 240^\circ)$.
• Carrier: triangle of peak $A_c$, frequency $f_c = m_f f_r$ ($m_f$ odd multiple of 3).
• Leg a: $v_{ra} > v_c \Rightarrow$ S1 ON, $v_{aO} = +V_{dc}/2$; $v_{ra} < v_c \Rightarrow$ S4 ON, $v_{aO} = -V_{dc}/2$.
• Line voltage $v_{ab} = v_{aO} - v_{bO}$ is a train of pulses of $\pm V_{dc}$ and 0 whose widths vary sinusoidally.
• Fundamental (linear range $m_a = A_r/A_c \le 1$): $\hat V_{aO1} = m_a V_{dc}/2$, $V_{L1,rms} = 0.612\,m_a V_{dc}$.

 vc  /\  /\  /\  /\  /\  /\
    /  \/  \/  \/  \/  \/  \
 vra    .--''''--.
 ------'----------'----------
 vaO   _  ___ _____ ___  _
 +   _| ||   |     |   || |_
 -           (S1 ON when vra>vc)

How the switching instants of one leg are found

1. Natural sampling (analogue comparison). In one carrier half-period the carrier is a straight line. For a rising slope starting at $t_k$:

$$v_c(t) = -A_c + \frac{4A_c}{T_c}(t - t_k), \qquad t_k \le t \le t_k + \frac{T_c}{2}$$

The switching instant $t_s$ is the root of

$$A_r\sin\omega t_s = -A_c + \frac{4A_c}{T_c}(t_s - t_k)$$

which is transcendental and is solved numerically (or by an analogue comparator in hardware). At $t_s$ the leg changes state: S1 OFF → S4 ON on the rising slope (as the carrier rises above the reference), and S4 OFF → S1 ON on the falling slope.

2. Regular (uniform) sampling (used in microcontrollers/DSP). The reference is sampled once per carrier period at $t_k$ and held. Then the crossing is found from a straight-line equation:

$$\begin{aligned} \text{duty of S1: } d_k &= \frac{1}{2}\left(1 + m_a\sin\omega t_k\right) \\ t_{on,S1} &= d_k T_c, \qquad t_{on,S4} = (1-d_k)T_c \end{aligned}$$

The ON pulse of S1 is centred in the carrier period (symmetric sampling). For example, $m_a = 0.8$ at $\omega t_k = 90^\circ$ gives $d = 0.9$: S1 is ON for 90% of that carrier period.

3. Complementary gating with dead time. The pair in a leg (S1–S4) receives complementary signals; a short dead time $t_d$ (both OFF) is inserted at every transition to avoid a shoot-through short circuit. During $t_d$ the load current flows through a diode.

The same procedure applies to legs b and c using $v_{rb}$, $v_{rc}$, which gives three balanced pole voltages.

The given wave (steps of 200 V and 400 V, each 60° wide) is the per-phase voltage of a 180°-conduction three-phase inverter with $V_{dc}/3 = 200$ V, i.e. $V_{dc} = 600$ V.

RMS value

Each 60° block occupies $\pi/3$ rad. Using the half cycle (the negative half is a mirror image):

$$\begin{aligned} V_{rms} &= \left[\frac{1}{\pi}\left(200^2\cdot\frac{\pi}{3} + 400^2\cdot\frac{\pi}{3} + 200^2\cdot\frac{\pi}{3}\right)\right]^{1/2} \\ &= \left[\frac{40000 + 160000 + 40000}{3}\right]^{1/2} = \sqrt{80000} \\ &= 282.84\ \text{V} \end{aligned}$$

(Check: $\dfrac{\sqrt2}{3}V_{dc} = 0.4714\times600 = 282.84$ V.)

Peak value of the fundamental

The wave has half-wave and quarter-wave (odd) symmetry, so $a_n = 0$ and

$$\begin{aligned} b_n &= \frac{2}{\pi}\int_0^{\pi}v(\theta)\sin n\theta\,d\theta \\ &= \frac{2}{n\pi}\Big[200\big(1 - \cos\tfrac{n\pi}{3}\big) + 400\big(\cos\tfrac{n\pi}{3} - \cos\tfrac{2n\pi}{3}\big) + 200\big(\cos\tfrac{2n\pi}{3} - \cos n\pi\big)\Big] \end{aligned}$$

For $n = 1$: $\cos 60^\circ = 0.5$, $\cos 120^\circ = -0.5$, $\cos 180^\circ = -1$:

$$\begin{aligned} b_1 &= \frac{2}{\pi}\big[200(0.5) + 400(1) + 200(0.5)\big] = \frac{2}{\pi}\times600 \\ &= 381.97\ \text{V} \end{aligned}$$

(Check: $\dfrac{2V_{dc}}{\pi} = \dfrac{2\times600}{\pi} = 381.97$ V.) RMS of fundamental $= 381.97/\sqrt2 = 270.09$ V.

The same formula gives $b_3 = 0$ (no triplen harmonics), $b_5 = 76.39$ V, $b_7 = 54.57$ V.

Answer: $V_{rms} = 282.84$ V; peak fundamental $V_{1m} = 381.97$ V (270.09 V rms).

A single-phase inverter converts dc into ac by alternately connecting the load to the dc source in opposite directions. With a resistive load and square-wave switching, the output is a square wave of amplitude $V_s$ (full bridge) or $V_s/2$ (half bridge), and the current has the same shape.

Circuit (full bridge)

     +-------+-----------+
     |       |           |
     |      S1          S3
    Vs       |           |
     |       A---[ R ]---B
     |       |    v0     |
     |      S4          S2
     |       |           |
     +-------+-----------+

Operation

• $0 < t < T/2$: S1 and S2 ON. Point A is at $+V_s$ relative to B: $v_0 = +V_s$, $i_0 = V_s/R$.
• $T/2 < t < T$: S3 and S4 ON. $v_0 = -V_s$, $i_0 = -V_s/R$.
• S1–S4 (and S3–S2) of a leg must never be ON together. Output frequency $f = 1/T$ is set by the gating rate.
• With a pure R load the current reverses exactly with the voltage, so the feedback diodes do not conduct.
• Half bridge: two switches and a centre-tapped source give $v_0 = \pm V_s/2$.

 v0  +Vs |________
         |        |
       0 |--------+--------- t
         |        |________
     -Vs  S1,S2 ON  S3,S4 ON
         0       T/2       T

RMS value

$$V_{0,rms} = \left[\frac{2}{T}\int_0^{T/2}V_s^2\,dt\right]^{1/2} = V_s$$

Fundamental component (Fourier analysis)

The wave is odd with half-wave symmetry, so $a_0 = 0$, $a_n = 0$ and only odd sine terms remain:

$$\begin{aligned} b_n &= \frac{2}{\pi}\int_0^{\pi}V_s\sin n\theta\,d\theta = \frac{2V_s}{n\pi}\big[-\cos n\theta\big]_0^{\pi} = \frac{2V_s}{n\pi}(1 - \cos n\pi) \\ b_n &= \frac{4V_s}{n\pi}\ (n\ \text{odd}), \qquad b_n = 0\ (n\ \text{even}) \\ v_0(t) &= \sum_{n=1,3,5,\dots}\frac{4V_s}{n\pi}\sin n\omega t \end{aligned}$$

For $n = 1$:

$$\begin{aligned} V_{1m} &= \frac{4V_s}{\pi} = 1.273\,V_s \\ V_{1,rms} &= \frac{4V_s}{\sqrt2\,\pi} = 0.9003\,V_s \end{aligned}$$

For a half bridge, replace $V_s$ by $V_s/2$: $V_{1m} = 2V_s/\pi$, $V_{1,rms} = 0.45\,V_s$.

Distortion: $THD = \sqrt{V_{rms}^2 - V_1^2}/V_1 = \sqrt{1 - 0.9003^2}/0.9003 = 48.3\%$. Example: $V_s = 220$ V gives $V_{1,rms} = 198.1$ V.

For a star-connected resistive load, the 180° mode gives a six-step phase voltage ($V/3$, $2V/3$) and the 120° mode gives a quasi-square phase voltage ($\pm V/2$ for 120°, zero for 60°).

Given: $V_{dc} = 450$ V, $R = 10\ \Omega$ per phase, star connection.

(a) 180° conduction mode

Phase voltage is $V_{dc}/3 = 150$ V and $2V_{dc}/3 = 300$ V in 60° steps.

(i) RMS load (phase) current

$$\begin{aligned} V_{ph,rms} &= \left[\frac{1}{\pi}\left(\Big(\tfrac{V}{3}\Big)^2\tfrac{\pi}{3}\cdot2 + \Big(\tfrac{2V}{3}\Big)^2\tfrac{\pi}{3}\right)\right]^{1/2} = \frac{\sqrt2}{3}V_{dc} \\ &= 0.4714\times450 = 212.13\ \text{V} \\ I_{L,rms} &= \frac{212.13}{10} = 21.21\ \text{A} \end{aligned}$$

(ii) RMS thyristor current — each thyristor carries the phase current for 180° of every 360°:

$$I_{T,rms} = \frac{I_{L,rms}}{\sqrt2} = \frac{21.21}{\sqrt2} = 15.0\ \text{A}$$

(iii) Load power

$$P = 3I_{L,rms}^2R = 3\times21.21^2\times10 = 13500\ \text{W} = 13.5\ \text{kW}$$

(b) 120° conduction mode

Only two thyristors conduct at a time; two phases are in series across $V_{dc}$, the third is open. So during conduction:

$$i = \frac{V_{dc}}{2R} = \frac{450}{20} = 22.5\ \text{A}, \qquad v_{ph} = \pm\frac{V_{dc}}{2} = \pm225\ \text{V}$$

Each phase conducts for 120° out of every 180°.

(i) RMS load current

$$\begin{aligned} V_{ph,rms} &= \frac{V_{dc}}{2}\sqrt{\frac{2}{3}} = \frac{V_{dc}}{\sqrt6} = 183.71\ \text{V} \\ I_{L,rms} &= 22.5\sqrt{\tfrac{2}{3}} = 18.37\ \text{A} \end{aligned}$$

(ii) RMS thyristor current — each thyristor carries 22.5 A for 120° of 360°:

$$I_{T,rms} = 22.5\sqrt{\tfrac{120}{360}} = \frac{22.5}{\sqrt3} = 12.99\ \text{A}$$

(iii) Load power

$$P = 3\times18.37^2\times10 = 10125\ \text{W} = 10.125\ \text{kW}$$

The 180° mode delivers more power (higher switch utilisation), while the 120° mode has an idle 60° in each leg that reduces the risk of shoot-through.

For a square-wave full-bridge inverter, the output voltage contains only odd harmonics, $v_0 = \sum \frac{4V_s}{n\pi}\sin n\omega t$. Each harmonic current is found by dividing by the load impedance at that harmonic, and the currents are added (superposition).

Given: $R = 4\ \Omega$, $L = 35$ mH, $C = 155\ \mu$F, $V_s = 230$ V, $f = 50$ Hz, $\omega = 314.16$ rad/s.

i) Load current up to the fifth harmonic

$$\begin{aligned} V_{nm} &= \frac{4V_s}{n\pi} = \frac{292.85}{n}\ \text{V} \\ Z_n &= \sqrt{R^2 + \Big(n\omega L - \frac{1}{n\omega C}\Big)^2}, \qquad \theta_n = \tan^{-1}\frac{n\omega L - 1/(n\omega C)}{R} \end{aligned}$$

The current is $i_n = (V_{nm}/Z_n)\sin(n\omega t - \theta_n)$. The fundamental sees a capacitive load, so it leads:

$$i_0(t) = 28.31\sin(314.16t + 67.25^\circ) + 3.69\sin(942.48t - 81.30^\circ) + 1.15\sin(1570.8t - 85.50^\circ)\ \text{A}$$

ii) RMS value of fundamental load current

$$I_{1} = \frac{28.31}{\sqrt2} = 20.02\ \text{A}$$

iii) Power absorbed by the load and fundamental power

RMS load current (harmonics up to the 5th):

$$\begin{aligned} I_3 &= \frac{3.691}{\sqrt2} = 2.61\ \text{A}, \qquad I_5 = \frac{1.148}{\sqrt2} = 0.81\ \text{A} \\ I_0 &= \sqrt{20.02^2 + 2.61^2 + 0.81^2} = 20.20\ \text{A} \end{aligned}$$ $$\begin{aligned} P_0 &= I_0^2R = 20.20^2\times4 = 1632.5\ \text{W} \\ P_1 &= I_1^2R = 20.02^2\times4 = 1602.6\ \text{W} \end{aligned}$$

(Including all higher harmonics changes $P_0$ only slightly, to about 1633.6 W.)

Answer: $I_{1,rms} = 20.02$ A; load power $\approx 1632.5$ W; fundamental power $= 1602.6$ W.

A three-phase bridge inverter converts dc into three-phase ac using six thyristors (with commutation circuits) or self-commutated switches and six feedback diodes. In the 120° conduction mode each thyristor conducts for 120°, only two thyristors (one upper, one lower, in different legs) conduct at any time, and a new thyristor is fired every 60° in the order T1, T2, T3, T4, T5, T6.

Circuit

     +-----+--------+--------+
     |     |        |        |
     |    T1 D1    T3 D3    T5 D5
    Vdc    |        |        |
     |     +--a     +--b     +--c
     |     |        |        |
     |    T4 D4    T6 D6    T2 D2
     |     |        |        |
     +-----+--------+--------+
      a, b, c -> star load (R per phase)

T1, T3, T5 connect phases a, b, c to the positive bus; T4, T6, T2 to the negative bus. The numbering gives the firing sequence T1 → T2 → … → T6 at 60° intervals.

Principle of working

• In each 60° interval, one phase is connected to $+V_{dc}$, one to $-V_{dc}$ (0 V bus), and the third is left open (floating).
• The two connected phases are in series across $V_{dc}$, so each carries $V_{dc}/(2R)$, and the phase voltages are $+V_{dc}/2$ and $-V_{dc}/2$; the open phase has 0 V (resistive load).
• There is a 60° gap between the turn-off of an upper switch and the turn-on of the lower switch of the same leg, so the risk of shoot-through is low.

($V = V_{dc}$.)

Firing sequence and waveforms

 wt:   0   60  120 180 240 300 360
 T1    |#######|               |
 T2        |#######|
 T3            |#######|
 T4                |#######|
 T5                    |#######|
 T6 ###|                   |####
 van V/2 _______
       0 |       |___         ___
    -V/2             |_______|
 vbn V/2         _______
       0 ___ ___|       |___
    -V/2 ___|               |___
 vab   V ___
     V/2    |___             ___
    -V/2        |___     ___|
      -V            |___|

• Phase voltage $v_{an}$: $+V/2$ for 120°, 0 for 60°, $-V/2$ for 120°, 0 for 60° (quasi-square). $v_{bn}$, $v_{cn}$ lag by 120° and 240°.
• Line voltage $v_{ab} = v_{an} - v_{bn}$: six-step wave with levels $V$, $V/2$, $-V/2$, $-V$, $-V/2$, $V/2$.

Main results

$$\begin{aligned} V_{ph,rms} &= \frac{V_{dc}}{\sqrt6} = 0.408\,V_{dc}, \qquad V_{L,rms} = \frac{V_{dc}}{\sqrt2} = 0.707\,V_{dc} \\ \hat V_{an1} &= \frac{4}{\pi}\cdot\frac{V_{dc}}{2}\cos30^\circ = \frac{\sqrt3\,V_{dc}}{\pi} = 0.551\,V_{dc} \end{aligned}$$

Only harmonics of order $6k\pm1$ are present. Compared with the 180° mode, switch utilisation and output voltage are lower, and with an inductive load the open phase voltage is not zero (it depends on the load), which is a drawback.

With the centre point of the dc source grounded, the single-phase (half-bridge) inverter gives a square wave of $\pm V_s/2$ across the load. Its Fourier series contains only odd harmonics, each of which drives a current through the RL impedance at that frequency.

Given: $V_s = 200$ V (so $\pm100$ V), $R = 5\ \Omega$, $L = 20$ mH, $f = 50$ Hz, $\omega = 314.16$ rad/s.

Fourier series of output voltage

$$v_0 = \sum_{n=1,3,5,\dots}\frac{4(V_s/2)}{n\pi}\sin n\omega t = \sum_{n\ \text{odd}}\frac{2V_s}{n\pi}\sin n\omega t = \sum_{n\ \text{odd}}\frac{127.32}{n}\sin n\omega t$$

Harmonic currents

$$Z_n = \sqrt{R^2 + (n\omega L)^2}, \qquad \theta_n = \tan^{-1}\frac{n\omega L}{R}, \qquad I_{nm} = \frac{V_{nm}}{Z_n}$$

Load current up to the 3rd order:

$$i_0(t) = 15.86\sin(314.16t - 51.49^\circ) + 2.18\sin(942.48t - 75.14^\circ)\ \text{A}$$

Harmonic factor and THD of output voltage

$$\begin{aligned} V_{0,rms} &= \frac{V_s}{2} = 100\ \text{V} \\ V_{1,rms} &= \frac{127.32}{\sqrt2} = 90.03\ \text{V} \\ HF_n &= \frac{V_n}{V_1} = \frac{1}{n} \;\Rightarrow\; HF_3 = \frac{1}{3} = 33.33\%,\ HF_5 = 20\% \end{aligned}$$ $$\begin{aligned} THD &= \frac{\sqrt{V_{0,rms}^2 - V_{1,rms}^2}}{V_{1,rms}} = \frac{\sqrt{100^2 - 90.03^2}}{90.03} \\ &= \frac{43.52}{90.03} = 0.4834 = 48.34\% \end{aligned}$$

Answer: $i_0 = 15.86\sin(\omega t - 51.49^\circ) + 2.18\sin(3\omega t - 75.14^\circ)$ A; $HF_3 = 33.33\%$ (lowest-order harmonic); $THD = 48.34\%$.

The given phase voltage (200 V, 400 V, 200 V steps of 60°) is the six-step wave of a 180° conduction inverter with $V_{dc}/3 = 200$ V, so $V_{dc} = 600$ V. It is odd with half-wave symmetry, so only odd sine terms exist.

Fourier coefficient

$$\begin{aligned} b_n &= \frac{2}{\pi}\int_0^{\pi}v_{an}\sin n\theta\,d\theta \\ &= \frac{2}{n\pi}\Big[200\big(1-\cos\tfrac{n\pi}{3}\big) + 400\big(\cos\tfrac{n\pi}{3}-\cos\tfrac{2n\pi}{3}\big) + 200\big(\cos\tfrac{2n\pi}{3}-\cos n\pi\big)\Big] \\ &= \frac{400}{n\pi}\Big[2 + \cos\tfrac{n\pi}{3} - \cos\tfrac{2n\pi}{3}\Big] \quad (n\ \text{odd}) \end{aligned}$$

Fundamental ($n = 1$)

$$\begin{aligned} b_1 &= \frac{400}{\pi}\big[2 + 0.5 - (-0.5)\big] = \frac{1200}{\pi} = 381.97\ \text{V (peak)} \\ V_{1,rms} &= \frac{381.97}{\sqrt2} = 270.09\ \text{V} \end{aligned}$$

Check with the standard result: $\hat V_{1} = \dfrac{2V_{dc}}{\pi} = \dfrac{2\times600}{\pi} = 381.97$ V.

Third harmonic ($n = 3$)

$\cos\pi = -1$, $\cos 2\pi = 1$:

$$b_3 = \frac{400}{3\pi}\big[2 + (-1) - 1\big] = 0, \qquad V_{3,rms} = 0$$

The third harmonic (and every triplen harmonic) is absent from the phase voltage of a 180° inverter with a balanced star load, because triplen components of the three pole voltages are in phase and appear only in the neutral voltage $v_{nO}$.

(For reference: overall rms $= \sqrt{(200^2+400^2+200^2)/3} = 282.84$ V; next harmonic $b_5 = 76.39$ V peak.)

Answer: $V_{1,rms} = 270.09$ V; $V_{3,rms} = 0$.

In single-pulse width modulation (SPWM with one pulse per half cycle), the inverter output has one pulse of width $\delta$ (centred at $\pi/2$ and $3\pi/2$) per half cycle. The pulse width, and so the output voltage, is set by the modulation index.

Generation and effect of modulation index

 carrier (2f)  /\      /\
 ref Ar  -----/--\----/--\---- (dc level)
             /    \  /    \
 v0  +Vs      |====|
         ----'    '----.    .--
     -Vs               |====|
             |<-d->|

• A rectangular reference of amplitude $A_r$ is compared with a triangular carrier of amplitude $A_c$ at twice the output frequency. The switch is ON while the reference exceeds the carrier.
• Modulation index $M = A_r/A_c$ ($0 \le M \le 1$); the pulse width is $\delta = M\pi$.
• RMS output voltage:

$$V_{0,rms} = \left[\frac{2}{2\pi}\int_{(\pi-\delta)/2}^{(\pi+\delta)/2}V_s^2\,d(\omega t)\right]^{1/2} = V_s\sqrt{\frac{\delta}{\pi}} = V_s\sqrt{M}$$

So increasing $M$ widens the pulse and raises the rms output from 0 (at $M = 0$) to $V_s$ (at $M = 1$, square wave). Example: $M = 0.5$ gives $V_{0,rms} = 0.707V_s$.

Fourier series

The wave is odd and has half-wave symmetry, so $a_0 = a_n = 0$ and only odd $n$ exist:

$$\begin{aligned} b_n &= \frac{2}{\pi}\int_{(\pi-\delta)/2}^{(\pi+\delta)/2}V_s\sin n\theta\,d\theta = \frac{2V_s}{n\pi}\Big[\cos n\tfrac{\pi-\delta}{2} - \cos n\tfrac{\pi+\delta}{2}\Big] \\ &= \frac{4V_s}{n\pi}\sin\frac{n\pi}{2}\sin\frac{n\delta}{2} \end{aligned}$$ $$v_0(t) = \sum_{n=1,3,5,\dots}\frac{4V_s}{n\pi}\sin\frac{n\pi}{2}\sin\frac{n\delta}{2}\sin n\omega t$$

(The factor $\sin(n\pi/2) = \pm1$ only fixes the sign; many texts write $v_0 = \sum\frac{4V_s}{n\pi}\sin\frac{n\delta}{2}\sin n\omega t$.)

Fundamental value

$$\begin{aligned} V_{1m} &= \frac{4V_s}{\pi}\sin\frac{\delta}{2}, \qquad V_{1,rms} = \frac{4V_s}{\sqrt2\,\pi}\sin\frac{\delta}{2} = 0.9V_s\sin\frac{\delta}{2} \end{aligned}$$

• At $\delta = \pi$ (square wave), $V_{1,rms} = 0.9V_s$.
• A harmonic $n$ is eliminated when $\sin(n\delta/2) = 0$, i.e. $\delta = 2\pi/n$; e.g. $\delta = 120^\circ$ removes the 3rd harmonic.
• Drawback: at small pulse widths (low output voltage) the harmonic content is high, which led to multiple-pulse and sinusoidal PWM.

In a three-phase VSI with 180° conduction, each switch conducts for half a cycle, the two switches of a leg are complementary, and three switches (one per leg) are ON at any time. Gate signals are given in the order S1, S2, …, S6 every 60°, giving six switch states (modes) per cycle.

Circuit

     +-----+--------+--------+
     |     |        |        |
     |    S1 D1    S3 D3    S5 D5
    Vdc    |        |        |
     |     +--a     +--b     +--c
     |     |        |        |
     |    S4 D4    S6 D6    S2 D2
     |     |        |        |
     +-----+--------+--------+
      a, b, c -> star load (R per phase)

Switch states

Let a switching function be 1 when the upper switch of a leg is ON (phase on $+V_{dc}$) and 0 when the lower one is ON.

($V = V_{dc}$.) The states 000 and 111 (zero states) are not used in six-step operation.

Phase voltage in a mode (star resistive load)

Mode 1: a and c connected to $+V$, b to the negative bus. Equivalent circuit: $R\parallel R = R/2$ in series with $R$:

$$i_b = \frac{V}{R/2 + R} = \frac{2V}{3R}, \quad v_{bn} = -i_bR = -\frac{2V}{3}, \quad v_{an} = v_{cn} = \frac{V}{3}$$

Mode 2: only a on the positive bus; b, c in parallel on the negative bus, giving $v_{an} = 2V/3$, $v_{bn} = v_{cn} = -V/3$. The other modes follow the same way. In general, using the pole voltages $v_{aO}$, $v_{bO}$, $v_{cO}$ ($\pm V/2$):

$$v_{an} = \frac{2v_{aO} - v_{bO} - v_{cO}}{3}$$

Phase voltage waveforms

 mode    1   2   3   4   5   6
 van 2V/3    ___
      V/3 __|   |__
        0 ----------+---------- wt
     -V/3           |__     __
    -2V/3              |___|
 vbn 2V/3            ___
      V/3        ___|   |___
        0 ------+----------+---
     -V/3    ___|           |
    -2V/3 __|
 vcn 2V/3                    ___
      V/3 ___             __|
        0    |__       __|
     -V/3       |__ __|
    -2V/3          |
         0  60 120 180 240 300 360

Each phase voltage is a six-step wave ($V/3$, $2V/3$, $V/3$, $-V/3$, $-2V/3$, $-V/3$), and $v_{bn}$, $v_{cn}$ lag $v_{an}$ by 120° and 240°.

Results: $V_{ph,rms} = \frac{\sqrt2}{3}V_{dc} = 0.471V_{dc}$; fundamental peak $= \frac{2V_{dc}}{\pi}$; no triplen harmonics.

A single-phase full-wave (full-bridge) inverter with square-wave switching gives $v_0 = \pm V_s$ across the load; with a resistive load the current is a square wave of $\pm V_s/R$.

Given: $R = 5\ \Omega$, $V_s = 30$ V (unit taken as volts).

(i) RMS output voltage

$$V_{0,rms} = \left[\frac{2}{T}\int_0^{T/2}V_s^2\,dt\right]^{1/2} = V_s = 30\ \text{V}$$

(ii) RMS of fundamental component

From $v_0 = \sum_{n\ \text{odd}}\frac{4V_s}{n\pi}\sin n\omega t$:

$$V_{1,rms} = \frac{4V_s}{\sqrt2\,\pi} = 0.9003\times30 = 27.01\ \text{V}$$

(iii) Output power

$$P_0 = \frac{V_{0,rms}^2}{R} = \frac{30^2}{5} = 180\ \text{W}$$

(Fundamental power alone is $27.01^2/5 = 145.9$ W.)

(iv) Peak current in each thyristor

Each pair (T1, T2 or T3, T4) carries the full load current for its half cycle:

$$I_{T,peak} = \frac{V_s}{R} = \frac{30}{5} = 6\ \text{A}$$

Also, average thyristor current $= 6\times0.5 = 3$ A and rms thyristor current $= 6/\sqrt2 = 4.24$ A.

Answer: $V_{0,rms} = 30$ V; $V_{1,rms} = 27.01$ V; $P_0 = 180$ W; $I_{T,peak} = 6$ A.