UNIT 5: POWER ELECTRONICS - EXAM-FOCUSED SHORT NOTES

I. POWER SEMICONDUCTOR DEVICES & CHARACTERISTICS

A. Thyristor (SCR)

• Structure & Symbol: 4-layer (PNPN), 3-terminal (Anode A, Cathode K, Gate G) device. Symbol: diode symbol with gate lead.

• VI Characteristics:

  • Forward Blocking: $$\displaystyle V_{AK} < V_{BO} $$ (Breakover voltage), $$\displaystyle I_G = 0 $$, small leakage current ($$\displaystyle I_{FS} $$).

  • Forward Conduction: Once triggered (by $$\displaystyle I_G $$ or $$\displaystyle V_{AK} > V_{BO} $$), SCR latches on. $$\displaystyle V_{AK} $$ drops to $$\displaystyle V_T $$ (~1-2V).

  • Reverse Blocking: Acts as a diode, blocks until $$\displaystyle V_{AK} = -V_{BR} $$.

• Turn-On Methods:

  1. Gate Triggering: Most common. Apply positive $$\displaystyle I_G $$ to A-K.

  2. dv/dt Triggering: High $$\displaystyle \frac{dv}{dt} $$ charges junction capacitance $$\displaystyle C_{j2} $$, causing $$\displaystyle i_c = C_{j2}\frac{dv}{dt} $$ to trigger.

  3. di/dt Triggering: Excessive $$\displaystyle \frac{di}{dt} $$ locally heats junction, triggers.

  4. Thermal Triggering: High junction temperature increases leakage current.

  5. Light Triggering (LASCR): Light photons generate carriers in junction.

• Two-Transistor Analogy: Equivalent to PNP ($$\displaystyle Q_1 $$) + NPN ($$\displaystyle Q_2 $$) transistors coupled. $$\displaystyle I_A = I_{C1} + I_{C2} $$, $$\displaystyle I_K = I_{E1} + I_{B2} $$, $$\displaystyle I_G = I_{B1} $$. Regenerative action: $$\displaystyle I_{C2} = \beta_2 I_{B2} = \beta_2 I_{E1} = \beta_2 \beta_1 I_{B1} $$.

• Firing/Triggering Circuits:

  • R-Firing: Simple, limited to $\alpha$ range (90°-180° for half-wave).

  • RC-Firing: Widely used. Provides variable $\alpha$ (0°-180°). $$\displaystyle V_G = V_s \left(1 - e^{-t/RC}\right) $$. Adjust $R$ to change $\alpha$.

  • UJT Firing: Relaxation oscillator. $$\displaystyle V_{peak} = \eta V_{BB} + V_D $$, where $\eta$ is intrinsic stand-off ratio. Generates sharp pulse.

  [!TIP] Exam Focus: UJT firing circuit diagram and pulse generation mechanism are frequently asked.

• Protection:

  • Overcurrent: Semiconductor fuses (fast acting), circuit breakers.

  • Overvoltage: Snubber Circuits (RC, RLC) across SCR to limit $$\displaystyle \frac{dv}{dt} $$ and absorb transient energy.

• Series/Parallel Operation:

  • Need for Derating: Non-identical V-I characteristics cause voltage/current imbalance.

  • Static Equalizing: Shunt resistor R across each SCR to equalize steady-state voltage. $$\displaystyle R \ll R_{SCR} $$.

  • Dynamic Equalizing: Shunt capacitor C across each SCR to equalize switching transients. Derivation: For two SCRs $$\displaystyle V_1 = V_{D1} + \frac{1}{C}\int i_1 dt $$, $$\displaystyle V_2 = V_{D2} + \frac{1}{C}\int i_2 dt $$. To force $$\displaystyle V_1 = V_2 $$, need $C$ large enough so that $$\displaystyle \frac{1}{C}\int \Delta i dt \gg \Delta V_D $$.

  • Derating Factor: $$\displaystyle k = 0.1 $$ to $0.2$ typically. Number in series: $$\displaystyle n_s = \frac{V_{system}}{V_{rated} \times (1-k)} $$. Parallel: $$\displaystyle n_p = \frac{I_{system}}{I_{rated} \times (1-k)} $$.

• Ratings: Exceeding $$\displaystyle \frac{dv}{dt} $$ causes false triggering; exceeding $$\displaystyle \frac{di}{dt} $$ causes local hot-spots and damage.

B. Other Power Semiconductor Devices

• GTO (Gate Turn-Off Thyristor):

  • Structure: Similar to SCR but with highly doped P+ layer near gate for efficient turn-off.

  • VI: Latch-up not as strong as SCR. Requires negative gate pulse ($$\displaystyle -I_G $$) to turn off.

  • Applications: High-power inverters, choppers, motor drives (replaces SCR in forced commutation).

• IGBT (Insulated Gate Bipolar Transistor):

  • Structure: MOS gate controlling a bipolar PNP transistor. Combines MOSFET input (high input impedance) with BJT output (low saturation voltage).

  • Conduction: Gate voltage $$\displaystyle V_{GE} > V_{th} $$ forms inversion layer, injectes electrons into P+ region, turns on PNP transistor.

  • Characteristics: High switching speed (MOS), low $$\displaystyle V_{CE(sat)} $$ (BJT). Trade-off: Tail current during turn-off.

• Power MOSFET (n-channel):

  • Structure: Vertical, with source, gate, drain. Gate oxide insulation.

  • Conduction: $$\displaystyle V_{GS} > V_{th} $$ creates channel, electrons flow drain-source.

  • Characteristics: Voltage-controlled, very high input impedance, fast switching, positive temperature coefficient (easy parallel), higher $$\displaystyle R_{DS(on)} $$ than IGBT at high voltages.

• TRIAC (Triode for Alternating Current):

  • Structure: Two SCRs anti-parallel with common gate. 5-layer (PNPNP).

  • Modes of Operation:

    • Mode I (+ve MT1 to MT2): Gate +ve w.r.t MT2. Equivalent to SCR with MT1 as A, MT2 as K.

    • Mode II (-ve MT1 to MT2): Gate -ve w.r.t MT2. Equivalent to SCR with MT2 as A, MT1 as K.

    • Mode III & IV: Less sensitive, require higher $$\displaystyle I_G $$.

  • Applications: Light dimmers, fan speed control, AC motor control.

• DIAC (Diode for Alternating Current):

  • Symbol: Two diodes back-to-back.

  • V-I: Blocks until breakover voltage $$\displaystyle V_{BO} $$ (symmetrical), then conducts with negative resistance region.

  • Application: Triggering TRIAC in phase-control circuits (e.g., in RC network with potentiometer).

• Comparison:

  | Feature | IGBT | MOSFET | GTO | | :--- | :--- | :--- | :--- | | Voltage Rating | Medium-High (600V-6.5kV) | Low-Medium (<200V) | Very High (up to 6kV) | | Current Rating | High (up to 1.5kA) | Medium (up to 100A) | Very High (up to 3kA) | | Switching Speed | Medium (~10kHz) | Very High (>100kHz) | Medium (~1kHz) | | Drive Complexity | Simple (voltage) | Very Simple | Complex (high -ve $$\displaystyle I_G $$) | | Applications | Motor drives, UPS | Switch-mode PSU, low-power | High-power inverters, HVDC |

II. AC-DC CONVERTERS (RECTIFIERS)

A. Single-Phase Converters

• Half-Wave Controlled: Single SCR. $$\displaystyle V_o = \frac{V_m}{2\pi}(1 + \cos\alpha) $$ for R load. Poor performance.

• Full-Wave Uncontrolled (Bridge): 4 diodes. $$\displaystyle V_o = \frac{2V_m}{\pi} $$.

• Fully Controlled Bridge (4 SCRs):

  • Operation with R, RL, RLE loads:

    • R Load: $$\displaystyle v_o = v_s $$ for $\omega t \in [\alpha, \pi+\alpha]$. $$\displaystyle V_{o(avg)} = \frac{2V_m}{\pi}\cos\alpha $$. $\alpha$ range: 0°-180°.

    • RL Load (Inductive): Current continuous if $L$ large. $$\displaystyle v_o $$ follows $$\displaystyle v_s $$ but shifted. $$\displaystyle V_{o(avg)} = \frac{2V_m}{\pi}\cos\alpha $$. $\alpha$ max limited by load time constant.

    • RLE Load (with E): $$\displaystyle V_{o(avg)} = \frac{2V_m}{\pi}\cos\alpha - E $$. For continuous current, $$\displaystyle \alpha \ge \sin^{-1}\left(\frac{E\pi}{2V_m}\right) $$.

  • Waveforms: Key: Load voltage is full-wave rectified sine segments starting at $\alpha$.

  • Modes:

    • Rectification Mode: $$\displaystyle \alpha < 90^\circ $$, $$\displaystyle V_{o(avg)} > 0 $$, power flows AC→DC.

    • Inversion Mode: $$\displaystyle \alpha > 90^\circ $$, $$\displaystyle V_{o(avg)} < 0 $$, power flows DC→AC (requires DC source E).

  [!TIP] Exam Focus: Distinguish waveforms for R vs. RL vs. RLE. Inversion mode requires $$\displaystyle \alpha > 90^\circ $$ and DC source E.

  • Effect of Source Impedance ($$\displaystyle L_s $$): Causes overlap angle $\mu$. During overlap, two SCRs conduct, output voltage is shorted. $$\displaystyle V_o = v_{ab} - 2L_s\frac{di}{dt} $$. $$\displaystyle V_{o(avg)} = \frac{2V_m}{\pi}\cos(\alpha+\frac{\mu}{2}) $$. $\mu$ increases with $$\displaystyle I_d $$ and $$\displaystyle L_s $$.

  • Freewheeling Diode (FWD): Connected across RL load.

    • Purpose: Provides path for load current when SCR off, prevents negative voltage across load, improves PF by making $$\displaystyle i_o $$ unidirectional and more sinusoidal.

    • Operation: When SCR off, FWD conducts, $$\displaystyle v_o = 0 $$, load energy circulates in R-L loop.

• Performance Parameters:

  • $$\displaystyle V_{o(avg)} = \frac{2V_m}{\pi}\cos\alpha $$ (no $$\displaystyle L_s $$)

  • $$\displaystyle I_{o(avg)} = V_{o(avg)}/R $$ (R load)

  • Input Power $$\displaystyle P_{in} = V_{o(avg)} I_{o(avg)} $$ (R load, no FWD)

  • Power Factor $$\displaystyle PF = \frac{P_{in}}{V_{s(rms)} I_{s(rms)}} $$. With FWD, $$\displaystyle I_s $$ more sinusoidal, PF improves.

  • THD: High due to non-sinusoidal $$\displaystyle i_s $$.

B. Three-Phase Converters

• Fully Controlled Bridge (6 SCRs):

  • Operation: Each SCR conducts for 120°. Output is 6-pulse.

  • Waveforms: $$\displaystyle v_o $$ is series of segments from line-to-line voltages ($$\displaystyle v_{ab}, v_{bc}, v_{ca} $$).

  • No Source Inductance: $$\displaystyle V_{o(avg)} = \frac{3\sqrt{6}}{\pi}V_{LL(rms)}\cos\alpha = 1.654 V_{LL}\cos\alpha $$.

  • With Source Inductance $$\displaystyle L_s $$: Overlap angle $\mu$. Each segment reduced by $$\displaystyle 3L_s\frac{di}{dt} $$. $$\displaystyle V_{o(avg)} = \frac{3\sqrt{6}}{\pi}V_{LL}\cos(\alpha+\frac{\mu}{2}) $$.

  • Problem Solving: Given $$\displaystyle V_s $$ (line), $\alpha$, $$\displaystyle I_d $$, $$\displaystyle V_d $$ → Find $$\displaystyle L_s $$, $R$, $\mu$.

    • Use $$\displaystyle V_d = \frac{3\sqrt{6}}{\pi}V_{LL}\cos(\alpha+\frac{\mu}{2}) $$.

    • $\mu$ from $$\displaystyle V_d $$ equation.

    • $$\displaystyle L_s $$ from $$\displaystyle L_s = \frac{V_{LL}\sqrt{6}}{\pi I_d}\sin\mu $$ (approx for 3-phase).

    [!EXAMPLE] (From Nov 2023) Given $$\displaystyle V_{LL}=400V $$, $$\displaystyle \alpha=\pi/4 $$, $$\displaystyle I_d=10A $$, $$\displaystyle V_d=360V $$:

    1. $$\displaystyle 360 = 1.654 \times 400 \times \cos(45^\circ + \mu/2) \Rightarrow \cos(45^\circ+\mu/2) = 0.543 \Rightarrow 45^\circ+\mu/2 = 57^\circ \Rightarrow \mu = 24^\circ $$.

    2. $$\displaystyle L_s = \frac{400\sqrt{6}}{\pi \times 10}\sin24^\circ \approx 2.37 \text{ mH} $$.

    3. Load $R$ not directly from given; if E present, $$\displaystyle V_d = \frac{3\sqrt{6}}{\pi}V_{LL}\cos(\alpha+\mu/2) - E $$.

C. AC Voltage Controllers (Phase-Angle Control)

• Single-Phase Half-Wave: 1 SCR. $$\displaystyle V_{o(rms)} = V_s\sqrt{\frac{1}{2\pi}\int_\alpha^\pi \sin^2\omega t d\omega t} = V_s\sqrt{\frac{1}{2}\left(1 + \frac{\sin2\alpha}{2\alpha}\right)} $$.

• Single-Phase Full-Wave (Bridge): 2 SCRs anti-parallel or 4 SCRs bridge.

  • Circuit:

    DiagramSEARCH: single phase full wave ac voltage controller bridge

  • R Load: $$\displaystyle V_{o(rms)} = V_s\sqrt{\frac{1}{\pi}\int_\alpha^{\pi+\alpha} \sin^2\omega t d\omega t} = V_s\sqrt{\frac{1}{2}\left(1 - \frac{\sin2\alpha}{2\alpha}\right)} $$? Correction: Actually $$\displaystyle V_{o(rms)} = V_s\sqrt{\frac{1}{\pi}\int_\alpha^{\pi+\alpha} \sin^2\omega t d\omega t} = V_s\sqrt{\frac{1}{2}\left(1 + \frac{\sin2\alpha}{\pi}\right)} $$? Let's derive properly:

$$\frac{V_{o(rms)}^2}{V_s^2} = \frac{1}{2\pi}\int_\alpha^{\pi+\alpha} \sin^2\omega t d(\omega t) = \frac{1}{2\pi}\left[ \frac{\omega t}{2} - \frac{\sin2\omega t}{4} \right]_\alpha^{\pi+\alpha}$$

$$= \frac{1}{2\pi}\left( \frac{\pi}{2} - \frac{\sin(2\pi+2\alpha) - \sin2\alpha}{4} \right) = \frac{1}{2\pi}\left( \frac{\pi}{2} - \frac{\sin2\alpha - \sin2\alpha}{4} \right) = \frac{1}{4}$$

    That's wrong because limits are $\alpha$ to $\pi+\alpha$, length $\pi$. Correct:

$$\frac{1}{\pi}\int_\alpha^{\pi+\alpha} \sin^2 x dx = \frac{1}{\pi}\int_\alpha^{\pi+\alpha} \frac{1-\cos2x}{2}dx = \frac{1}{2\pi}\left[ x - \frac{\sin2x}{2} \right]_\alpha^{\pi+\alpha}$$

$$= \frac{1}{2\pi}\left( (\pi+\alpha - \frac{\sin(2\pi+2\alpha)}{2}) - (\alpha - \frac{\sin2\alpha}{2}) \right) = \frac{1}{2\pi}\left( \pi + \frac{\sin2\alpha}{2} - \frac{\sin2\alpha}{2} \right) = \frac{1}{2}$$

    So $$\displaystyle V_{o(rms)} = V_s/\sqrt{2} $$? That's for full-wave uncontrolled. For controlled, the integral is only over conduction period. Actually for full-wave controlled bridge with R load, output is half-wave rectified sine in each half-cycle. Each thyristor conducts from $\alpha$ to $\pi$ in its half-cycle. So:

$$V_{o(rms)}^2 = \frac{1}{2\pi}\left( \int_\alpha^\pi V_m^2\sin^2\omega t d\omega t + \int_{\pi+\alpha}^{2\pi} V_m^2\sin^2\omega t d\omega t \right) = \frac{V_m^2}{\pi}\int_\alpha^\pi \sin^2\omega t d\omega t$$

$$\int_\alpha^\pi \sin^2 x dx = \frac{1}{2}(\pi - \alpha) - \frac{1}{4}(\sin2\pi - \sin2\alpha) = \frac{\pi-\alpha}{2} + \frac{\sin2\alpha}{4}$$

    So 

$$V_{o(rms)} = V_m\sqrt{\frac{1}{\pi}\left( \frac{\pi-\alpha}{2} + \frac{\sin2\alpha}{4} \right)} = V_s\sqrt{\frac{1}{2\pi}(\pi-\alpha) + \frac{\sin2\alpha}{4\pi}}$$

    But $$\displaystyle V_s = V_m/\sqrt{2} $$, so $$\displaystyle V_m = \sqrt{2}V_s $$. Then:

$$V_{o(rms)} = \sqrt{2}V_s\sqrt{\frac{\pi-\alpha}{2\pi} + \frac{\sin2\alpha}{4\pi}} = V_s\sqrt{ \frac{\pi-\alpha}{\pi} + \frac{\sin2\alpha}{2\pi} }$$

    This is the standard expression: 

$$V_{o(rms)} = V_s\sqrt{ \frac{1}{2\pi}(2\pi - 2\alpha + \sin2\alpha) } = V_s\sqrt{ \frac{1}{2\pi}(2(\pi-\alpha) + \sin2\alpha) }$$

    Actually common formula: 

$$V_{o(rms)} = V_s\sqrt{ \frac{1}{2\pi}(2\pi - 2\alpha + \sin2\alpha) } = V_s\sqrt{1 - \frac{\alpha}{\pi} + \frac{\sin2\alpha}{2\pi}}$$

    Yes, that's correct.

*   **RL Load:** Current may be continuous or discontinuous. For continuous, $$\displaystyle V_{o(rms)} $$ expression same as R load? Not exactly because current phase shift. But for PF calc, need $$\displaystyle I_s $$ waveform.

*   **Firing Angle from Power:** For R load, $$\displaystyle P_o = \frac{V_{o(rms)}^2}{R} $$. Given $$\displaystyle P_o $$, $R$, $$\displaystyle V_s $$, solve for $\alpha$: 

$$P_o = \frac{V_s^2}{R}\left(1 - \frac{\alpha}{\pi} + \frac{\sin2\alpha}{2\pi}\right)$$

.

*   **Input Power Factor:** $$\displaystyle PF = \frac{P_o}{V_s I_{s(rms)}} $$. For R load, $$\displaystyle I_{s(rms)} = I_{o(rms)} $$ (same RMS). So $$\displaystyle PF = \frac{V_{o(rms)}}{V_s} $$? No, $$\displaystyle P_o = V_{o(rms)} I_{o(rms)} \cos\phi_o $$, but for R load $$\displaystyle \cos\phi_o=1 $$, so $$\displaystyle P_o = V_{o(rms)} I_{o(rms)} $$. And $$\displaystyle I_{s(rms)} = I_{o(rms)} $$. So $$\displaystyle PF = \frac{V_{o(rms)}}{V_s} $$. That's displacement PF? Actually for non-sinusoidal $$\displaystyle i_s $$, $$\displaystyle PF = \frac{P}{V_s I_{s(rms)}} = \frac{V_{o(rms)}^2/R}{V_s \cdot V_{o(rms)}/R} = \frac{V_{o(rms)}}{V_s} $$. So yes, for R load, PF = $$\displaystyle V_{o(rms)}/V_s $$.

> [!EXAMPLE] (From Jun 2025, Dec 2024) $$\displaystyle V_s=230V $$, $$\displaystyle R=5\Omega $$, $$\displaystyle P_o=5kW $$. $$\displaystyle V_{o(rms)} = \sqrt{P_o R} = \sqrt{5000\times5} = 158.11V $$. Then $$\displaystyle 158.11 = 230\sqrt{1 - \frac{\alpha}{\pi} + \frac{\sin2\alpha}{2\pi}} $$. Solve numerically: $$\displaystyle \alpha \approx 114.5^\circ $$. PF = $$\displaystyle 158.11/230 = 0.688 $$.

• Two-Stage Sequence Control (for RL load):

  • Purpose: Improve PF and reduce harmonics by dividing full cycle into two stages with different $$\displaystyle \alpha_1 $$ and $$\displaystyle \alpha_2 $$.

  • Operation: In first stage (0 to $\pi$), SCR1 triggered at $$\displaystyle \alpha_1 $$. In second stage ($\pi$ to $2\pi$), SCR2 triggered at $$\displaystyle \pi+\alpha_2 $$. By choosing $$\displaystyle \alpha_2 > \alpha_1 $$, load current becomes more continuous, PF improves.

  • Waveforms:

    DiagramCANVAS: two stage sequence control waveforms showing two firing angles per cycle

• Applications: Light dimming, heater control, motor speed control.

III. DC-AC CONVERTERS (INVERTERS)

A. Voltage Source Inverters (VSI)

• Basic Principle: DC input voltage (constant or with small ripple). Switches turned on/off to produce AC output. Commutation: Forced (using auxiliary circuit or load commutation for RL).

• Single-Phase Bridge Inverter:

  • Circuit: 4 switches (SCRs or transistors) in H-bridge, DC source $$\displaystyle V_{dc} $$.

  • 180° Conduction Mode: Each switch conducts 180°. Output $$\displaystyle v_o = +V_{dc} $$ when T1,T2 on; $$\displaystyle -V_{dc} $$ when T3,T4 on; 0 during switching transitions (if any dead-time). Square wave.

    • R Load: $$\displaystyle i_o = v_o/R $$, in phase with $$\displaystyle v_o $$.

    • RL Load: $$\displaystyle i_o $$ lags $$\displaystyle v_o $$. For continuous current, $$\displaystyle i_o = \frac{2V_{dc}}{\pi R}\left( \sin\omega t - \frac{\omega L}{R}\cos\omega t + \frac{L}{R}e^{-\omega L t/\text{?}}\right) $$? Actually steady-state: $$\displaystyle i_o = \frac{2V_{dc}}{\pi}\sum_{n=1,3,5...}^\infty \frac{1}{n^2X_L}\sin n\omega t $$? Better: For square wave $$\displaystyle v_o = V_{dc} \text{sgn}(\sin\omega t) $$, fundamental component $$\displaystyle V_{o1} = \frac{4V_{dc}}{\pi} $$. For RL load, $$\displaystyle I_{o1} = \frac{V_{o1}}{Z} = \frac{4V_{dc}}{\pi\sqrt{R^2+(\omega L)^2}} $$.

    • RMS Load Current: $$\displaystyle I_{o(rms)} = \frac{V_{dc}}{R} $$ for R load? Actually for square wave, $$\displaystyle V_{o(rms)} = V_{dc} $$. So $$\displaystyle I_{o(rms)} = V_{dc}/R $$.

    • Load Power: $$\displaystyle P_o = I_{o(rms)}^2 R $$ (R load) or $$\displaystyle I_{o1}^2 R $$ (RL, fundamental approximation).

  • 120° Conduction Mode: Only 2 switches conduct at a time, each for 120°. Output is 3-level waveform.

    • Output voltage: $$\displaystyle v_o = V_{dc} $$ (T1,T2), $0$ (T1,T3 or T2,T4), $$\displaystyle -V_{dc} $$ (T3,T4).

    • Line Voltage (between two legs): 6-step waveform, same as 3-phase inverter phase voltage.

    • RMS Output: $$\displaystyle V_{o(rms)} = \sqrt{\frac{2}{3}}V_{dc} $$? Actually for 120° mode, conduction ratio 2/3, so $$\displaystyle V_{o(rms)} = V_{dc}\sqrt{\frac{2}{3}} \approx 0.816V_{dc} $$.

• Three-Phase Bridge Inverter (6 switches):

  • 180° Conduction Mode: Each switch 180°, 120° apart. Output phase voltages are 120° shifted square waves. Line voltages are 6-step (quasi-square).

    • Star Load: $$\displaystyle V_{AN} = \frac{2}{3}V_{dc} $$ for 120°, then 0 for 60°? Actually: For 180° mode, each phase voltage is square wave of amplitude $$\displaystyle V_{dc}/2 $$? Let's derive: With DC bus $$\displaystyle V_{dc} $$, midpoint O. When T1 on, T2 off, T3 on? Standard 6-step: T1,T2,T3 conduct in pairs. $$\displaystyle V_{AO} = V_{dc}/2 $$ when T1 on, $$\displaystyle -V_{dc}/2 $$ when T4 on? Actually if DC source $$\displaystyle V_{dc} $$ with center tap O, then each phase to O is $$\displaystyle V_{dc}/2 $$. But usually without center tap, using 6 switches and DC bus $$\displaystyle V_{dc} $$. Then phase voltage $$\displaystyle V_{AN} $$ is square wave between $$\displaystyle +V_{dc}/3 $$ and $$\displaystyle -2V_{dc}/3 $$? This is messy.

    Better: For 3-phase bridge with DC source $$\displaystyle V_{dc} $$ and no center tap, the output line-to-line voltage $$\displaystyle V_{ab} $$ is 6-step: $$\displaystyle +V_{dc} $$, 0, $$\displaystyle -V_{dc} $$ etc. Phase voltage $$\displaystyle V_{AN} $$ depends on load connection. For star load with isolated neutral, $$\displaystyle V_{AN} = \frac{1}{3}(2V_{AO} - V_{BO}) $$? Actually standard result: For 180° mode, phase voltage RMS $$\displaystyle V_{ph(rms)} = \frac{V_{dc}}{\sqrt{6}} $$? Let's recall: For 6-step inverter, line voltage RMS $$\displaystyle V_{LL(rms)} = \sqrt{\frac{2}{3}}V_{dc} $$. Then for star load, phase voltage RMS $$\displaystyle V_{ph(rms)} = V_{LL(rms)}/\sqrt{3} = \frac{V_{dc}}{\sqrt{6}} \approx 0.408V_{dc} $$. For delta load, phase voltage = line voltage.

  • 120° Conduction Mode: Only 2 switches conduct at a time. Each switch 120°. Output phase voltages are not independent. For star load, $$\displaystyle V_{AN} $$ waveform: $$\displaystyle +V_{dc}/2 $$ for 60°, 0 for 60°, $$\displaystyle -V_{dc}/2 $$ for 60°, etc? Actually: With 120° mode, at any time two switches on (one from upper, one from lower leg). The output phase voltage is either $$\displaystyle +V_{dc}/2 $$, $$\displaystyle -V_{dc}/2 $$, or 0. The RMS value: $$\displaystyle V_{ph(rms)} = V_{dc}\sqrt{\frac{1}{3}} $$? Let's compute: Conduction pattern per phase: conducts 120° out of 360°, but when conducting, voltage is either $$\displaystyle +V_{dc}/2 $$ or $$\displaystyle -V_{dc}/2 $$? Actually in 120° mode, each phase is connected to positive rail for 120° and negative rail for 120°, and floating for 120°. So waveform: $$\displaystyle +V_{dc}/2 $$ for 120°, $$\displaystyle -V_{dc}/2 $$ for 120°, 0 for 120°. RMS: $$\displaystyle V_{ph(rms)} = \sqrt{\frac{1}{2\pi}\left( \int_0^{2\pi/3} (V_{dc}/2)^2 d\theta + \int_{2\pi/3}^{4\pi/3} 0^2 d\theta + \int_{4\pi/3}^{2\pi} (-V_{dc}/2)^2 d\theta \right)} = \sqrt{\frac{1}{2\pi}\left( 2 \times \frac{2\pi}{3} \times \frac{V_{dc}^2}{4} \right)} = \sqrt{\frac{V_{dc}^2}{6}} = \frac{V_{dc}}{\sqrt{6}} $$? That's same as 180°? That can't be. Let's recalc: For 120° mode, each phase is connected to positive for 120°, negative for 120°, and floating for 120°. But when floating, voltage is not zero; it's determined by other phases. Actually in 120° mode, the load is connected in a way that at any time two phases are connected across the DC source. So the phase voltages are not independent. For star load with isolated neutral, the phase voltage waveform is not a simple square wave. Typically, for 120° mode, the line-to-line voltage is same as 180° mode (6-step), but phase voltages differ. Actually many texts say for 3-phase inverter, 120° mode gives same line voltages as 180° mode? No, 180° mode gives 6-step line voltages. 120° mode also gives 6-step line voltages? Let's think: In 180° mode, each switch 180°, so at any time three switches on (one per leg upper or lower). That gives line voltage directly between two legs: either $$\displaystyle +V_{dc} $$, 0, or $$\displaystyle -V_{dc} $$. In 120° mode, only two switches on at a time, so line voltage is still between two legs, but one leg might be floating? Actually if only two switches on, say T1 (A+) and T6 (C-), then line voltage $$\displaystyle V_{AB} = V_{AO} - V_{BO} $$. But V_BO is floating because no switch on B leg. So V_BO is determined by load. So line voltage is not simply $$\displaystyle V_{dc} $$ or 0. This is complicated. Usually for 3-phase bridge, 180° mode is standard. 120° mode is used for some special applications. For exam, focus on 180° mode.

  [!TIP] Exam Focus: 3-phase inverter 180° mode is very high frequency. Remember: For star load, $$\displaystyle V_{ph(rms)} = \frac{V_{dc}}{\sqrt{6}} $$, $$\displaystyle I_{ph(rms)} = \frac{V_{ph(rms)}}{R} $$ for R load. Power $$\displaystyle P = 3 I_{ph(rms)}^2 R $$.

  [!EXAMPLE] (From Nov 2023) $$\displaystyle V_{dc}=200V $$, star R=10Ω/phase. $$\displaystyle V_{ph(rms)} = 200/\sqrt{6} = 81.65V $$. $$\displaystyle I_{ph(rms)} = 8.165A $$. $$\displaystyle P = 3 \times 8.165^2 \times 10 = 2000W $$.

• Pulse Width Modulation (PWM):

  • Single-Pulse Modulation: One pulse per half-cycle. Width $2\delta$. Fundamental RMS: $$\displaystyle V_{o1} = \frac{4V_{dc}}{\pi}\sin\delta $$. Can eliminate $n$th harmonic if $$\displaystyle \delta = n\pi/2 $$? Actually for single pulse, harmonics at $k\pi/\delta$. To eliminate $n$th harmonic, set $$\displaystyle \delta = n\pi/2 $$? Not exactly. For single pulse, Fourier: $$\displaystyle b_n = \frac{4V_{dc}}{n\pi}\sin(n\delta) $$. So to eliminate $n$th harmonic, $$\displaystyle \sin(n\delta)=0 \Rightarrow n\delta = m\pi \Rightarrow \delta = m\pi/n $$. Choose m such that $$\displaystyle \delta < \pi/2 $$.

  • Multiple Pulse & Sinusoidal PWM: Concepts only. Sinusoidal PWM compares triangular carrier with sinusoidal reference. Fundamental amplitude controlled by modulation index $$\displaystyle m_a $$.

B. Current Source Inverter (CSI)

• Principle: Constant input current $$\displaystyle I_{dc} $$ (from current source, large inductor). Output voltage depends on load.

• Circuit: 4 SCRs in bridge, with large inductor $$\displaystyle L_{dc} $$ and sometimes capacitor $$\displaystyle C_{dc} $$ across source to absorb voltage spikes.

• Operation: SCRs turned on/off to commutate current. Load current approximately constant (due to $$\displaystyle L_{dc} $$). Output voltage is quasi-square.

• Comparison with VSI:

  | CSI | VSI | | :--- | :--- | | Input current constant | Input voltage constant | | Requires load commutation or auxiliary circuit | Self-commutating (with transistors) | | More suitable for high-power, low-speed drives | More common, higher switching freq | | Output current waveform is square | Output voltage waveform is square |

C. Special Inverters

• McMurray-Bedford Inverter: Uses auxiliary thyristor and capacitor for forced commutation. Auxiliary thyristor conducts, capacitor discharges through main thyristor to turn it off.

• Three-Phase Series Inverter: Two inverters in series, output is sum. Used for high voltage.

IV. AC-AC CONVERTERS (CYCLOCONVERTERS)

A. Single-Phase Cycloconverters

• Principle: Phase-controlled, frequency step-down ($$\displaystyle f_o = f_{in}/n $$). No energy storage; direct conversion.

• Mid-Point Configuration: Two single-phase converters (positive and negative group) connected to separate secondary windings with common midpoint. Each group conducts for half cycle.

  • Circuit:
    DiagramSEARCH: single phase mid-point cycloconverter

• Bridge Configuration: Two full-wave bridges (positive/negative group). More common.

  • Circuit:
    DiagramSEARCH: single phase bridge cycloconverter

• Operation (Step-Down, Resistive Load):

  • For output frequency $$\displaystyle f_o = f_{in}/2 $$, each SCR conducts for one input cycle.

  • Waveforms: Input: 50Hz sine. Output: 25Hz sine-like but with steps. Each half-cycle of output is composed of segments from input cycles.

  • Firing: Positive group SCRs fired during positive half-cycles of output, negative group during negative half-cycles. Firing angle $\alpha$ varies to approximate sine output.

• Step-Up Cycloconverter: Rare, uses forced commutation. Output frequency > input frequency.

B. Three-Phase to Single-Phase Cycloconverter

• Principle: Uses three-phase supply to reduce harmonic content. Three single-phase cycloconverters (one per phase) connected in series/parallel to form single-phase output.

• Circuit Concept: Each phase of 3-phase supply feeds a single-phase converter (mid-point or bridge). Their outputs are combined.

• Advantage: Lower harmonic distortion due to 3-phase input.

V. DC-DC CONVERTERS (CHOPPERS)

A. Classification & Basic Topologies

• Step-Down (Buck): $$\displaystyle V_o = \alpha V_s $$, $$\displaystyle \alpha = T_{on}/T $$. Switch in series with source. Inductor smooths current.

  • Continuous Conduction (CCM): $$\displaystyle i_{min} > 0 $$. $$\displaystyle V_o = \alpha V_s $$.

  • Discontinuous Conduction (DCM): $$\displaystyle i_{min}=0 $$. $$\displaystyle V_o = \frac{1}{2}\alpha(1+\alpha)V_s $$? Actually: $$\displaystyle V_o = V_s \frac{\alpha}{1 + \frac{R}{2Lf}(1-\alpha)} $$? Better derive: Energy balance: $$\displaystyle V_s T_{on} - V_o T = L(i_{max}^2 - i_{min}^2)/(2R) $$. With $$\displaystyle i_{min}=0 $$, $$\displaystyle i_{max} = (V_s - V_o)T_{on}/L $$, and $$\displaystyle V_o = i_{avg}R $$, $$\displaystyle i_{avg} = i_{max}/2 $$. Solve: $$\displaystyle V_o = V_s \frac{\alpha}{1 + \frac{R\alpha}{2Lf}(1-\alpha)} $$? Actually standard: $$\displaystyle V_o = V_s \frac{\alpha}{1 + \frac{R}{2Lf}(1-\alpha)} $$? Let's not overcomplicate. For exam, CCM formula $$\displaystyle V_o = \alpha V_s $$ is primary.

• Step-Up (Boost): $$\displaystyle V_o = \frac{V_s}{1-\alpha} $$. Switch in series with inductor, diode to output.

  • Operation: Switch on: $L$ stores energy, $$\displaystyle i_L $$ ramps up. Switch off: $L$ releases energy via diode to load. $$\displaystyle V_o = V_s + L\frac{di}{dt} $$.

• Step-Up/Step-Down (Buck-Boost): $$\displaystyle V_o = -\frac{\alpha}{1-\alpha}V_s $$. Polarity reversed. Inductor stores energy when switch on, releases to load when off via diode.

B. Types Based on Quadrant Operation

• Type-A (First Quadrant):

  • Circuit: Switch S, diode D, load R-L-E (E can be battery back-emf).

  • Operation: Motoring. S on: $$\displaystyle v_L = V_s - E $$, $i$ increases. S off: D conducts, $$\displaystyle v_L = -E $$, $i$ decreases.

  • Waveforms:

    DiagramCANVAS: Type-A chopper waveforms showing i_L, v_L, v_s

  • Problem Solving (Given $$\displaystyle V_s, T, T_{on}, R, L, E $$):

    1. Continuity: Check if $$\displaystyle i_{min} > 0 $$. Average voltage across L = 0 over T: $$\displaystyle (V_s - E)T_{on} + (-E)(T-T_{on}) = 0 \Rightarrow V_s T_{on} = E T \Rightarrow \alpha = E/V_s $$. If $$\displaystyle \alpha_{actual} > \alpha_{min} = E/V_s $$, continuous. If $$\displaystyle \alpha_{actual} < E/V_s $$, discontinuous.

    2. Average Output Current $$\displaystyle I_{avg} $$: For CCM, $$\displaystyle I_{avg} = \frac{V_s - E}{R} \cdot \frac{T_{on}}{T} $$? Actually from volt-sec balance: $$\displaystyle V_s T_{on} - E T = 0 $$? That gives $$\displaystyle V_s \alpha = E $$, which is only for boundary. For CCM, average voltage across L = 0: $$\displaystyle (V_s - E)T_{on} + (-E)(T-T_{on}) = 0 \Rightarrow V_s T_{on} = E T \Rightarrow \alpha = E/V_s $$. That seems to say for CCM, $\alpha$ must equal $$\displaystyle E/V_s $$? That's not right. Let's derive properly:

    For Type-A chopper with R-L-E load, in CCM, the average voltage across inductor over one period is zero:

$$(V_s - E - I_{avg}R)T_{on} + (-E - I_{avg}R)(T-T_{on}) = 0$$

    Solve for $$\displaystyle I_{avg} $$:

$$(V_s - E - I_{avg}R)T_{on} - (E + I_{avg}R)(T-T_{on}) = 0$$

$$(V_s - E)T_{on} - I_{avg}R T_{on} - E(T-T_{on}) - I_{avg}R(T-T_{on}) = 0$$

$$(V_s - E)T_{on} - E(T-T_{on}) - I_{avg}R T = 0$$

$$V_s T_{on} - E T - I_{avg}R T = 0$$

$$\Rightarrow I_{avg} = \frac{V_s T_{on} - E T}{R T} = \frac{V_s \alpha - E}{R}$$

    So $$\displaystyle I_{avg} = \frac{V_s \alpha - E}{R} $$. For CCM, need $$\displaystyle i_{min} > 0 $$.

    3. **Maximum & Minimum Current:** Ripple: $$\displaystyle \Delta I = \frac{(V_s - E)T_{on}}{L} = \frac{(V_s - E)\alpha T}{L} $$? Actually during on-time: $$\displaystyle di/dt = (V_s - E - iR)/L \approx (V_s - E)/L $$ if R small. More precisely: $$\displaystyle \Delta I = \frac{(V_s - E - I_{avg}R)T_{on}}{L} $$? But $$\displaystyle I_{avg}R $$ is average voltage drop. Actually ripple is approximately: $$\displaystyle \Delta I \approx \frac{(V_s - E)T_{on}}{L} $$ during on, and $$\displaystyle \frac{E T_{off}}{L} $$ during off, but since $$\displaystyle V_s T_{on} = E T $$ for boundary? Let's derive exactly:

    During on: $$\displaystyle L\frac{di}{dt} = V_s - E - iR $$. This is linear if we ignore iR term? For small R, approximate: $$\displaystyle i_{max} - i_{min} \approx \frac{(V_s - E)T_{on}}{L} $$.

    During off: $$\displaystyle L\frac{di}{dt} = -E - iR $$, so $$\displaystyle i_{min} - i_{max} \approx \frac{E T_{off}}{L} $$.

    But from volt-sec balance: $$\displaystyle (V_s - E)T_{on} = (E + I_{avg}R)(T-T_{on}) $$? Actually from zero average: $$\displaystyle (V_s - E - I_{avg}R)T_{on} = (E + I_{avg}R)(T-T_{on}) $$. So $$\displaystyle (V_s - E)T_{on} - I_{avg}R T_{on} = E(T-T_{on}) + I_{avg}R(T-T_{on}) $$. Rearr: $$\displaystyle (V_s - E)T_{on} - E(T-T_{on}) = I_{avg}R T $$. That's same as before.

    For ripple, we can solve differential equations exactly if needed, but for exam, approximate formulas are often used:

$$I_{max} = I_{avg} + \frac{\Delta I}{2}, \quad I_{min} = I_{avg} - \frac{\Delta I}{2}$$

    And $$\displaystyle \Delta I = \frac{(V_s - E)T_{on}}{L} $$ (if R small) or more accurately $$\displaystyle \Delta I = \frac{(V_s - E - I_{avg}R)T_{on}}{L} = \frac{(V_s - E)T_{on}}{L} - \frac{I_{avg}R T_{on}}{L} $$. But since $$\displaystyle I_{avg}R $$ is small, often ignored.

    > [!EXAMPLE] (From Jun 2025, Dec 2024) $$\displaystyle V_s=220V $$, $$\displaystyle T=2000\mu s $$, $$\displaystyle T_{on}=600\mu s $$, $$\displaystyle R=1\Omega $$, $$\displaystyle L=5mH $$, $$\displaystyle E=24V $$.

    > 1. **Continuity:** $$\displaystyle \alpha = 0.3 $$, $$\displaystyle E/V_s = 24/220 = 0.109 $$. Since $$\displaystyle \alpha > E/V_s $$, continuous (CCM).

    > 2. $$\displaystyle I_{avg} = \frac{V_s \alpha - E}{R} = \frac{220\times0.3 - 24}{1} = 66 - 24 = 42A $$.

    > 3. Ripple: $$\displaystyle \Delta I \approx \frac{(V_s - E)T_{on}}{L} = \frac{(220-24)\times600\times10^{-6}}{5\times10^{-3}} = \frac{196\times0.6}{5} = 23.52A $$. Then $$\displaystyle I_{max} = 42 + 11.76 = 53.76A $$, $$\displaystyle I_{min} = 42 - 11.76 = 30.24A $$. (More accurate: $$\displaystyle \Delta I = \frac{(V_s - E - I_{avg}R)T_{on}}{L} = \frac{(220-24-42)\times0.6}{5} = \frac{154\times0.6}{5} = 18.48A $$, so $$\displaystyle I_{max}=51.24A $$, $$\displaystyle I_{min}=32.76A $$). Check boundary: $$\displaystyle \alpha_{min}=E/V_s=0.109 $$, so CCM valid.

• Type-B (Second Quadrant): Regenerative braking. Load is inductive, energy fed back to source. Only one switch and diode.

• Type-C (Two-Quadrant): Combines Type-A and Type-B. Two switches (S1,S2) and two diodes (D1,D2). Can handle motoring and regenerative braking.

C. Special Choppers

• Morgan Chopper: Uses auxiliary thyristor TA for commutation. TA turns on, capacitor C discharges through main thyristor TM to turn it off.

  • Circuit:

    DiagramSEARCH: Morgan chopper circuit

  • Waveforms:

    DiagramCANVAS: Morgan chopper waveforms showing TA, TM, capacitor voltage

• Jones Chopper: Uses two auxiliary thyristors and a capacitor. Similar principle.

D. Control Techniques & Analysis

• Current Limit Control (CLC): Keeps load current within $$\displaystyle I_{max} $$ and $$\displaystyle I_{min} $$. Switch on until $$\displaystyle I_{max} $$, off until $$\displaystyle I_{min} $$. Frequency varies or fixed with variable on/off times.

• Step-Down Chopper with Switch Drop $$\displaystyle V_{on} $$:

  • $$\displaystyle V_{avg} = \alpha (V_s - V_{on}) + V_{on} $$? Actually when switch on, output = $$\displaystyle V_s - V_{on} $$ (drop across switch). When off, output = 0 (diode conducts). So average: $$\displaystyle V_{avg} = \alpha (V_s - V_{on}) $$.

  • $$\displaystyle V_{rms} = \sqrt{\alpha} (V_s - V_{on}) $$.

  • Efficiency: $$\displaystyle P_{in} = V_s I_{avg} $$? Actually input power from source: $$\displaystyle P_{in} = V_s I_{s(avg)} $$. $$\displaystyle I_{s(avg)} = I_{avg} $$ (since inductor makes source current continuous and equal to load current in CCM). So $$\displaystyle P_{in} = V_s I_{avg} $$. $$\displaystyle P_{out} = V_{avg} I_{avg} $$. So $$\displaystyle \eta = \frac{V_{avg}}{V_s} = \alpha \frac{V_s - V_{on}}{V_s} $$.

  [!EXAMPLE] (From May 2024) $$\displaystyle V_s=220V $$, $$\displaystyle V_{on}=1.5V $$, $$\displaystyle f=10kHz $$, $$\displaystyle D=0.8 $$, $$\displaystyle R=20\Omega $$.

  1. $$\displaystyle V_{avg} = 0.8 \times (220-1.5) = 0.8 \times 218.5 = 174.8V $$.

  2. $$\displaystyle V_{rms} = \sqrt{0.8} \times 218.5 = 0.8944 \times 218.5 = 195.4V $$.

  3. $$\displaystyle I_{avg} = V_{avg}/R = 8.74A $$. $$\displaystyle P_{out} = I_{avg}^2 R = 8.74^2 \times 20 = 1527W $$. $$\displaystyle P_{in} = V_s I_{avg} = 220 \times 8.74 = 1923W $$. $$\displaystyle \eta = 1527/1923 = 79.4\% $$.

VI. ANALYSIS TECHNIQUES & MISCELLANEOUS

A. Waveform Analysis

• Key Skill: For each converter/inverter/chopper, be able to draw:

  • Source voltage $$\displaystyle v_s $$

  • Gate signals

  • Switch/device currents & voltages

  • Load voltage & current

• Focus on: Firing angle $\alpha$, overlap angle $\mu$, conduction intervals, freewheeling periods.

B. Performance Calculations - Problem Solving Guide

1. Firing Angle from Power (AC Voltage Controller): Use $$\displaystyle P_o = \frac{V_s^2}{R}\left(1 - \frac{\alpha}{\pi} + \frac{\sin2\alpha}{2\pi}\right) $$ for single-phase full-wave R load. Solve numerically.

2. Power Factor: $$\displaystyle PF = \frac{P}{V_s I_{s(rms)}} $$. For non-sinusoidal $$\displaystyle i_s $$, $$\displaystyle I_{s(rms)} $$ from Fourier or RMS of waveform.

3. Average/RMS/Max/Min Currents:

   • Choppers: Use $$\displaystyle I_{avg} = \frac{V_s \alpha - E}{R} $$ (Type-A CCM), $$\displaystyle I_{max/min} = I_{avg} \pm \Delta I/2 $$, $$\displaystyle \Delta I = \frac{(V_s - E)T_{on}}{L} $$ (approx).

   • Inverters: For square wave output, $$\displaystyle I_{rms} = V_{dc}/R $$ (R load). For RL, use fundamental approximation: $$\displaystyle I_1 = \frac{4V_{dc}}{\pi\sqrt{R^2+(\omega L)^2}} $$.

4. Overlap Angle $\mu$ (Rectifiers): For single-phase: $$\displaystyle \mu = \frac{2\omega L_s I_d}{V_m(1-\cos\mu)} \approx \frac{2\omega L_s I_d}{V_m} $$ for small $\mu$. For three-phase: $$\displaystyle \mu = \frac{2\omega L_s I_d}{3V_{LL}/\pi} \sin\mu $$? Actually: $$\displaystyle V_d = \frac{3\sqrt{6}}{\pi}V_{LL}\cos(\alpha+\mu/2) $$, and $\mu$ from $$\displaystyle L_s \frac{di}{dt} = \frac{V_{LL}\sqrt{6}}{\pi}(\cos\alpha - \cos(\alpha+\mu)) $$. Solve iteratively.

5. Device Ratings for Series/Parallel: $$\displaystyle n_s = \frac{V_{sys}}{V_{rated}(1-k)} $$, $$\displaystyle n_p = \frac{I_{sys}}{I_{rated}(1-k)} $$, with $$\displaystyle k=0.1-0.2 $$.

C. Harmonics & Reduction

• Sources: Non-linear switching (rectifiers, inverters, AC controllers).

• Harmonics in Rectifiers: 6-pulse (3-phase) has $6k\pm1$ harmonics. Single-phase has all odd harmonics.

• Reduction Methods:

  • PWM: Spreads spectrum, reduces low-order harmonics.

  • Multi-pulse: 12-pulse (two 6-pulse with 30° phase shift) cancels 5th,7th,11th,13th.

  • Input Filters: LC filters tuned to harmonic frequencies.

  • Active Filters: Inject compensating currents.

D. Applications

• SCRs: HVDC, motor speed control, welding, battery charging.

• IGBT/MOSFET: Switch-mode power supplies, motor drives, UPS.

• Inverters: UPS, AC motor drives, solar inverters, induction heating.

• Choppers: DC motor drives, regenerative braking, battery chargers.

• Cycloconverters: High-power, low-speed AC motor drives (cement mills, ship propulsion).

• AC Voltage Controllers: Light dimmers, furnace control, fan speed control.

Final Exam Strategy:

1. Draw First: Always sketch circuit and key waveforms for explanation questions.

2. Formulas: Memorize core equations: $$\displaystyle V_{o(avg)} $$ for converters, $$\displaystyle V_o $$ for choppers, $$\displaystyle V_{o(rms)} $$ for AC controllers.

3. Numericals: For Type-A chopper, first check continuity ($$\displaystyle \alpha > E/V_s $$). For 3-phase converter, use $$\displaystyle V_d = 1.654 V_{LL}\cos(\alpha+\mu/2) $$.

4. Comparisons: Be clear on VSI vs. CSI, IGBT vs. MOSFET.

5. Protection: Snubber design (RC values), series/parallel derating.

All the best! Focus on past paper patterns – they repeat.