DC-AC Converter (Inverter) with RL Load

§ 01

Introduction & Theory

A power inverter converts DC to AC. The single-phase full-bridge Voltage Source Inverter (VSI) uses four switches (IGBTs/MOSFETs) to produce an AC waveform from a DC source. The output voltage waveform shape and harmonic content depend on the modulation strategy.

≈ Voltage Source Inverter (VSI)

Input is a stiff DC voltage source. Directly controls AC output voltage. Uses feedback diodes to allow current flow when switches are OFF (essential for RL loads).

∿ Current Source Inverter (CSI)

Input is a current source. Directly controls AC output current. Less common in modern low-voltage applications.

Single-phase full-bridge VSI circuit — Fig. 1 — Single-phase full-bridge VSI: four IGBT switches with anti-parallel feedback diodes.

Feedback (freewheeling) diodes are essential for RL loads because the output current I_o is not in phase with voltage V_o. When a switch turns OFF, the inductor drives current through the corresponding feedback diode back to the DC source.

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Modulation Strategies

Square Wave vs. PWM

Square wave produces high THD (~48%) but simple gate drive. Unipolar PWM with sinusoidal reference reduces THD significantly. Higher modulation index (MI) ⇒ larger fundamental output voltage.

§ 02

Square Wave Modulation

Setup: Square Wave

R = 12.5 Ω, L = 6 mH, V_dc = 100 V. Switches S1&S4 ON for first half-cycle, S2&S3 ON for second half-cycle. Output = ±100 V square wave at 50 Hz.

Square wave inverter Simulink model — Fig. 2 — Simulink model: single-phase bridge inverter with square wave gating.

Square wave pulse generator waveforms — Fig. 3 — Complementary pulse generators for square wave modulation (50% duty cycle each).

Square Wave Output — Key Metrics

\[V_{o,fundamental} = \frac{4V_{dc}}{\pi} \approx 127.3\text{ V (peak)},\quad \text{THD} \approx 48.3\%\]

Square wave contains all odd harmonics: 3rd (33%), 5th (20%), 7th (14%), etc. Harmonic amplitudes fall as 1/n.

§ 03

Unipolar PWM Modulation

Setup: Unipolar PWM

Same RL load and V_dc. Sinusoidal reference (50 Hz) compared with high-frequency triangular carrier to generate gate pulses. Test MI = 1.0, 0.6, and 0.2.

Unipolar PWM inverter Simulink model — Fig. 4 — Simulink model: unipolar sinusoidal PWM inverter with RL load.

Fig. 5 — Sine PWM block parameters: set amplitude ratio (MI) and carrier frequency.

Unipolar PWM Output Voltage

\[V_{o,fundamental,peak} = M_I \cdot V_{dc},\quad M_I = \frac{V_{ref,peak}}{V_{carrier,peak}}\]

MI = Modulation Index. For MI = 1.0 and V_dc = 100 V: V_o,fund,peak = 100 V. For MI = 0.6: V_o,fund,peak = 60 V.

Advantage of Unipolar PWM: Harmonic energy is pushed to higher frequencies (near carrier frequency and sidebands), far from the fundamental. This makes filtering much easier than for square wave inverters.

§ 04

Observation Table

S.No	Input Voltage / Mode	V_o,fund RMS (V)	V_o,THD (%)	V_o RMS at 150 Hz (V)	I_o,fund RMS (A)	I_o RMS at 150 Hz (A)	I_o,THD (%)
1	Square Wave — 100 V
2	Unipolar PWM, MI = 1.0, 100 V
3	Unipolar PWM, MI = 0.6, 100 V
4	Unipolar PWM, MI = 0.2, 100 V

§ 05

Results

I. Square Wave Modulation

Attach the MATLAB Simulink circuit diagram of square wave modulation inverter.
Attach waveforms: (a) Output Voltage (b) Output Current
Attach the FFT plot of output voltage and output current. Record THD, V_fundamental, 3rd harmonic, 5th harmonic.

II. Unipolar PWM Modulation (MI = 0.2)

Attach the MATLAB Simulink circuit diagram of unipolar PWM inverter.
Attach waveforms at MI = 0.2: (a) Output Voltage (b) Output Current
Attach FFT plots at MI = 0.2. Compare THD with square wave case.

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Expected Finding

PWM Dramatically Reduces THD vs. Square Wave

Square wave THD ≈ 48%. Unipolar PWM (MI = 1.0) achieves THD below 5% for the output current due to the RL load filtering action. This justifies the additional complexity of PWM control.