Inverters and PWM techniques

Voltage and current source inverters, half- and full-bridge and six-step outputs with their harmonics, and sinusoidal PWM (modulation index, frequency ratio, V/f drives), with a square-wave THD example and a three-phase SPWM drive example.

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Why it matters

Every variable-frequency drive (VFD), servo amplifier, UPS, solar inverter and electric-vehicle traction drive contains an inverter that turns a DC link into AC of adjustable voltage and frequency. Pulse-width modulation (PWM) is how that inverter controls its output voltage and keeps low-order harmonics out of the motor. Choosing a DC-link voltage, a switching frequency or a modulation index for a drive needs the relations in this topic.

Key ideas

What an inverter does. An inverter converts DC to AC by switching the DC source across the load in alternating directions. Unlike a chopper, the output is AC, so its useful measures are the RMS value, the amplitude and frequency of the fundamental component, and the harmonic content. The "duty cycle × V_dc" rule of a chopper does not give an inverter's AC output.

Voltage source and current source inverters.

  • A voltage source inverter (VSI) is fed from a stiff DC voltage (a large DC-link capacitor). The output voltage is a switched waveform set by the inverter; the load decides the current. Almost all industrial drives, UPS units and solar inverters are VSIs built with IGBTs or MOSFETs.
  • A current source inverter (CSI) is fed from a stiff DC current (a large series inductor). Used in some large medium-voltage drives.
  • Each switch in a VSI has an anti-parallel (freewheeling) diode so that the lagging current of an inductive load (a motor) has a path when the switch turns off.

Single-phase bridges.

  • Half-bridge: two switches and a split DC capacitor; the output swings between +V_dc/2 and −V_dc/2.
  • Full bridge (H-bridge): four switches; the output swings between +V_dc and −V_dc, or can be held at zero.
  • The two switches of one leg must never be on together: that shorts the DC link (shoot-through). A short dead time (typically 0.5–3 µs) is inserted between turning one off and the other on.

Square-wave output and its harmonics. If each pair of switches conducts for 180°, the output is a square wave. Its Fourier series has only odd harmonics, with amplitudes falling as 1/n (3rd = 33 %, 5th = 20 % of the fundamental). The output voltage can only be varied by varying V_dc. These low-order harmonics cause extra heating and torque pulsation in motors.

Three-phase six-step inverter. Three legs, each conducting for 180°, give a six-step line-voltage waveform. Triplen harmonics (3rd, 9th, …) cancel in the line voltages; the 5th, 7th, 11th, 13th remain.

Sinusoidal PWM (SPWM). A sinusoidal reference of the desired output frequency f_r is compared with a high-frequency triangular carrier f_c. The switch is on when the reference is above the carrier.

  • Amplitude modulation index m_a = V_ref(peak) / V_carrier(peak). For m_a ≤ 1 (linear range) the fundamental amplitude is proportional to m_a.
  • Frequency modulation ratio m_f = f_c / f_r. Harmonics appear in groups around m_f, 2m_f, …, far above the fundamental, where the motor or load inductance filters them easily. m_f is chosen as an odd integer (and, for three-phase, a multiple of 3) to avoid even and triplen-related harmonics.
  • For m_a > 1 (overmodulation) the output rises more slowly and low-order harmonics return; in the limit the output becomes a square wave.
  • So one inverter controls both frequency (via f_r) and voltage (via m_a) from a fixed DC link. This is exactly what V/f speed control of an induction motor needs: synchronous speed ∝ f, and V is raised with f to keep the air-gap flux constant.

Other PWM methods. Single-pulse and multiple-pulse (uniform) PWM vary pulse width to control voltage; selective harmonic elimination chooses notch angles to cancel particular harmonics; space-vector PWM (SVPWM) in three-phase drives gives about 15 % more fundamental line voltage than SPWM in the linear range and is the standard in digital motor controllers.

Choosing switching frequency. Higher f_c pushes harmonics higher (smoother current, less audible noise, smaller filters) but increases switching losses and EMI. IGBT drives typically run at 2–16 kHz; MOSFET inverters at tens of kHz or more.

Formulas

V_n = 4·V_dc / (n·π) — peak of the nth harmonic of a full-bridge square wave (n odd). V_1,rms = 4·V_dc / (π·√2) ≈ 0.900·V_dc — fundamental RMS, full-bridge square wave; total RMS of the square wave is V_dc.

  • For a half-bridge, replace V_dc by V_dc/2.

THD = √(V_rms² − V_1,rms²) / V_1,rms — total harmonic distortion (≈ 48.3 % for a square wave).

V_L,rms = √(2/3)·V_dc ≈ 0.816·V_dc; V_L1,rms = (√6 / π)·V_dc ≈ 0.780·V_dc — three-phase six-step (180° conduction) line voltage, total and fundamental.

m_a = V_ref / V_carrier; m_f = f_c / f_r V_1,peak = m_a·V_dc — single-phase full bridge, SPWM, m_a ≤ 1 (half-bridge: m_a·V_dc/2). V_L1,rms = (√3 / (2·√2))·m_a·V_dc ≈ 0.612·m_a·V_dc — three-phase SPWM fundamental line voltage, m_a ≤ 1.

  • V_dc: DC input voltage (V); n: harmonic order; V_ref, V_carrier: peak reference and carrier amplitudes (V); f_r: output (reference) frequency (Hz); f_c: carrier (switching) frequency (Hz).

Worked examples

Example 1 (standard: square-wave full bridge). A single-phase full-bridge inverter runs from V_dc = 220 V with 180° square-wave switching into a resistive load of 10 Ω. Find the RMS output voltage, the RMS fundamental, the power delivered at the fundamental and the THD.

  1. A full-bridge square wave of ±V_dc has V_rms = V_dc = 220 V.
  2. V_1,rms = 4·V_dc/(π·√2) = 0.9003 × 220 = 198.1 V.
  3. Total power P = V_rms²/R = 220²/10 = 4840 W; fundamental power P_1 = 198.1²/10 = 3923 W.
  4. THD = √(220² − 198.1²)/198.1 = √(48 400 − 39 230)/198.1 = 95.8/198.1 = 0.483.

V_rms = 220 V, V_1,rms = 198.1 V, P_1 ≈ 3.92 kW of 4.84 kW total, THD ≈ 48.3 %. Nearly a fifth of the power is in harmonics, which is why square-wave inverters are not used for motors.

Example 2 (GATE level: three-phase SPWM drive). A three-phase VSI with a 600 V DC link uses sinusoidal PWM with a 1950 Hz carrier to drive an induction motor at 50 Hz with m_a = 0.8. Find (a) the RMS fundamental line voltage, (b) m_f and where the dominant harmonics lie, (c) the minimum DC-link voltage that would give 400 V line in the linear range.

  1. V_L1,rms = 0.6124·m_a·V_dc = 0.6124 × 0.8 × 600 = 293.9 V.
  2. m_f = f_c/f_r = 1950/50 = 39 — odd and a multiple of 3, as required. Dominant harmonics are grouped around the 39th, i.e. near 1950 Hz, with sidebands; there are no significant low-order (5th, 7th) harmonics.
  3. Maximum linear output is at m_a = 1: V_dc,min = 400/0.6124 = 653.2 V.
  4. For comparison, six-step operation from 600 V would give 0.7797 × 600 = 467.8 V but with 5th and 7th harmonics present.

(a) 293.9 V (b) m_f = 39, harmonics near 1.95 kHz (c) about 653 V. (A 400 V supply rectified by a diode bridge gives only about 540 V, so a 400 V drive cannot reach full motor voltage with SPWM; SVPWM or slight overmodulation is used.)

Common mistakes

  • Applying the chopper rule V = D·V_dc to an inverter's AC output.
  • Forgetting that a half-bridge swings only ±V_dc/2.
  • Mixing peak and RMS: SPWM gives m_a·V_dc as the peak fundamental for a full bridge.
  • Using the single-phase SPWM formula for a three-phase line voltage (it carries the factor √3/(2√2)).
  • Assuming the output stays proportional to m_a above 1 (overmodulation is non-linear).
  • Choosing an even m_f, or ignoring dead time and shoot-through when explaining a leg.
  • Thinking higher switching frequency is always better: switching loss and EMI rise with it.

For GATE ME

Expect conceptual MCQs on VSI versus CSI, the role of anti-parallel diodes, which harmonics appear in square-wave and six-step outputs, and how m_a and m_f affect output; and short numericals on fundamental and harmonic amplitudes of square-wave inverters, six-step line voltage, and SPWM output voltage for a given m_a and DC link. Practise the Fourier coefficients 4V_dc/(nπ) and the peak/RMS conversions.

Quick check

  1. What is the peak of the fundamental of a full-bridge square wave from 100 V DC?
  2. In the square wave, what is the ratio of the 5th-harmonic amplitude to the fundamental?
  3. A single-phase full-bridge SPWM inverter on 300 V DC has m_a = 0.6. What is the peak fundamental output?
  4. Why does every switch in a VSI have an anti-parallel diode?
  5. Reference 50 Hz, carrier 5 kHz. What is m_f?

Answers: 1. 127.3 V. 2. 1/5 (20 %). 3. 180 V. 4. To carry the reactive (lagging) load current when the switch is off. 5. 100.

Interview questions

All Electrical Circuits and Electronics interview questions

Try answering each one aloud before you open it.

  1. 1.What is an inverter and what is its primary function in electrical circuits?Concept

    An inverter converts DC into AC by switching the DC source across the load in alternating directions with power semiconductors such as IGBTs or MOSFETs. It lets AC loads run from batteries, solar panels or a rectified DC link. In motor drives its real job is to produce AC of adjustable voltage and frequency, which is how a variable-frequency drive controls induction-motor speed.

  2. 2.Explain the concept of Pulse Width Modulation (PWM) and its significance in inverters.Concept

    In PWM the inverter switches at a high carrier frequency and varies the width of each pulse so that the local average of the output follows a desired waveform. In sinusoidal PWM a sine reference at the output frequency is compared with a triangular carrier; the modulation index m_a = V_ref/V_carrier sets the fundamental amplitude (peak m_a·V_dc for a full bridge in the linear range), and the reference frequency sets the output frequency. Its significance is that one inverter on a fixed DC link controls both voltage and frequency, while the harmonics are pushed up to around the carrier frequency where the load inductance filters them easily.

  3. 3.How does a PWM inverter differ from a square wave inverter?Concept

    A square-wave inverter switches each device once per cycle, so its output has large low-order odd harmonics (the 3rd is a third and the 5th a fifth of the fundamental, about 48% THD) and its voltage can only be changed by changing the DC input. A PWM inverter switches many times per cycle with varying pulse widths, so it can vary the output voltage internally through the modulation index and its harmonics sit near the switching frequency, far above the fundamental. The cost is higher switching loss and EMI, but motor current is much closer to sinusoidal, with less heating and torque ripple.

  4. 4.Why is PWM preferred over square-wave or stepped-wave operation in inverter applications?Application

    PWM controls both the output voltage and frequency from a fixed DC link, so no controlled rectifier or separate DC chopper is needed for voltage control. It moves the harmonics from the 3rd, 5th and 7th up to groups around the carrier frequency, where the load inductance or a small filter removes them, giving nearly sinusoidal current. The penalties are more switching events, so higher switching losses and EMI, and slightly lower maximum fundamental voltage than six-step operation.

  5. 5.What happens if the PWM frequency is too low in an inverter application?Application

    The harmonics sit around the switching frequency, so a low carrier frequency brings them closer to the fundamental where the load inductance filters them less. Motor current then has more ripple, causing extra copper and iron loss, torque ripple and audible magnetic noise if the frequency is in the hearing range. A low frequency ratio also makes subharmonics more likely if the carrier is not synchronised to the reference. Switching losses in the inverter itself fall, so the choice is a trade-off.

  6. 6.In what scenarios would a pure sine wave inverter be preferred over a modified sine wave inverter?Application

    A pure sine wave inverter is preferred in scenarios where sensitive electronic devices are used, such as medical equipment, audio/video equipment, and precision instruments. These devices require a clean and stable AC waveform to function correctly, which a pure sine wave inverter provides.

  7. 7.A single-phase full-bridge inverter on a 48 V DC supply uses sinusoidal PWM with a modulation index of 0.5. Find the peak and RMS values of the fundamental output voltage.Numerical

    In the linear range (m_a ≤ 1) a full-bridge SPWM inverter gives a fundamental of peak value V_1,peak = m_a·V_dc. So V_1,peak = 0.5 × 48 = 24 V, and V_1,rms = 24/√2 ≈ 17.0 V. Note that the simple chopper rule V = D·V_dc does not apply to an inverter's AC output.

  8. 8.What are the potential drawbacks of using a high PWM frequency in an inverter?Application

    Using a high PWM frequency can lead to increased switching losses in the inverter's power electronic components, which can reduce overall efficiency. It may also require more advanced and expensive components to handle the higher frequencies, and can lead to electromagnetic interference (EMI) issues.

  9. 9.Explain how PWM can be used to control the speed of an AC motor.Application

    The synchronous speed of an induction motor is N_s = 120·f/P, so speed is varied by changing the supply frequency. A PWM inverter sets the output frequency through the frequency of its sine reference and the output voltage through the modulation index. To keep the air-gap flux constant the voltage is changed in proportion to frequency (constant V/f) up to rated speed; above it the voltage stays at its maximum and the motor runs in field weakening. The high-frequency switching harmonics are filtered by the motor's inductance, so the motor current is nearly sinusoidal.

  10. 10.If an inverter has an efficiency of 90% and the input power is 100 W, what is the output power?Numerical

    The output power of an inverter can be calculated using the formula: Output Power = Input Power × Efficiency. Here, Input Power = 100 W and Efficiency = 90% = 0.9. Therefore, Output Power = 100 W × 0.9 = 90 W.

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