Power Electronics (EX-502) - Unit 4 Short Notes

UNIT 4: POWER ELECTRONICS

Short Notes Based on RGPV Past Papers (2023-2025)

I. POWER SEMICONDUCTOR DEVICES

A. Thyristor (SCR)

1. Structure & Symbol: Four-layer (PNPN), three terminals (Anode, Cathode, Gate). Symbol: diode with gate line.

2. Two-Transistor Analogy: Equivalent to two complementary transistors (PNP & NPN) with positive feedback. Latch-up occurs when $$\displaystyle I_G $$ triggers, maintaining conduction even after $$\displaystyle I_G $$ removal.

3. Turn-On Methods:

   • Gate Triggering: Most common. Apply positive $$\displaystyle I_G $$ to Anode.

   • dv/dt Triggering: High $$\displaystyle \frac{dV}{dt} $$ charges junction capacitance, causing false turn-on. Avoid with snubber.

   • Thermal Triggering: High temperature increases leakage current → thermal runaway.

   • Light Triggering (LASCR): Light (fiber optic) generates carriers in gate region.

4. Key Ratings & Protection:

   • dv/dt rating: Max allowable $$\displaystyle \frac{dV}{dt} $$ without gate signal. Exceeding causes unintended turn-on.

   • di/dt rating: Max allowable $$\displaystyle \frac{dI}{dt} $$ during turn-on. Exceeding damages local hotspots.

   • Protection:

     • Overcurrent: Fuses, circuit breakers.

     • Overvoltage: Snubber circuits (RC, RCD), varistors.

5. Series/Parallel Operation:

   • Series: Voltage sharing. Need static equalization (shunt resistors) for steady-state voltage imbalance; dynamic equalization (shunt capacitors) for transient imbalance.

   • Parallel: Current sharing. Use small stray inductance or ballast resistors.

   • Derivation: For series SCRs, resistor $$\displaystyle R = \frac{V_{max} - V_{min}}{I_{diff}} $$, capacitor $C$ chosen to equalize voltage during transients.

6. Firing Circuits:

   • UJT Firing Circuit: UJT generates a pulse. $$\displaystyle V_{BB} $$ charges capacitor $C$ via $$\displaystyle R_1 $$ until $$\displaystyle V_{PEAK} $$, UJT fires, discharging $C$ through pulse transformer to SCR gate.

   • Other: R, RC, RLC firing circuits for phase control.

[!TIP]

Common Pitfall: Confusing dv/dt rating (turn-on risk) with di/dt rating (turn-on damage). Both require snubbers.

B. Gate Turn-Off Thyristor (GTO)

1. Structure: Similar to SCR but with highly doped p+ layer near cathode for efficient hole extraction.

2. VI Characteristics: Similar to SCR but with reverse blocking capability in off-state. Requires negative gate pulse for turn-off.

3. Turn-On/Turn-Off:

   • Turn-On: Positive gate current pulse ($$\displaystyle I_{GM} $$).

   • Turn-Off: High-current-density negative gate pulse ($$\displaystyle I_{GM-} $$). Requires commutation circuit.

4. Advantages over SCR: Can be turned off by gate signal → simpler inverters, no forced commutation needed. Disadvantage: Lower $dv/dt$ rating, higher gate drive power.

5. Applications: AC motor drives, high-power inverters, choppers.

C. Power MOSFET

1. n-channel Enhancement Type:

   • Structure: Source, Drain, Gate. Gate oxide insulation.

   • Operation: $$\displaystyle V_{GS} > V_{th} $$ creates inversion layer (channel) → conduction.

2. Conduction: Majority carriers (electrons) → unipolar → fast switching, low switching loss.

3. Switching Characteristics: Fast turn-on/off (ns), no minority carrier storage. Limitations: High on-resistance ($$\displaystyle R_{DS(on)} $$) at high voltages.

4. Applications: Low-voltage (<200V), high-frequency (kHz-MHz) DC-DC converters, switch-mode power supplies.

D. Insulated Gate Bipolar Transistor (IGBT)

1. Structure: MOSFET gate controlling a BJT. Combines MOSFET input (voltage-driven) with BJT output (low saturation voltage).

2. VI & Switching:

   • VI: Similar to BJT but controlled by $$\displaystyle V_{GE} $$.

   • Switching: Faster than BJT, slower than MOSFET. Tail current during turn-off due to minority carriers.

3. Advantages: High input impedance (MOSFET), low conduction loss (BJT). Good for medium power (600V-6.5kV, up to ~500A).

4. Applications: AC motor drives, UPS, inverters, traction.

E. Other Devices

1. TRIAC:

   • Structure: Two SCRs inverse-parallel with common gate.

   • Modes:

     • Mode I (+ve MT1 to MT2): Gate +ve, MT2 +ve.

     • Mode II (-ve MT1 to MT2): Gate -ve, MT2 -ve.

     • Mode III & IV: Less sensitive, require higher gate current.

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

2. DIAC:

   • Structure: Two-terminal, bidirectional trigger diode. No gate.

   • Triggering: Conducts when breakover voltage ($$\displaystyle V_{BO} $$) exceeded in either direction. Used to trigger TRIACs.

3. Device Comparison:

   | Feature | MOSFET | IGBT | GTO | |---------|--------|------|-----| | Drive | Voltage | Voltage | Current | | Switching Speed | Very High | Medium | Low | | Voltage Rating | Low (<200V) | Medium-High | High | | On-State Loss | High $$\displaystyle R_{DS(on)} $$ | Low $$\displaystyle V_{CE(sat)} $$ | Low $$\displaystyle V_{TM} $$ | | Turn-Off | Easy | Easy | Requires negative $$\displaystyle I_G $$ |

II. AC-DC CONVERTERS (RECTIFIERS)

A. Single-Phase Converters

1. Half-Wave Controlled Rectifier:

   • R Load: $$\displaystyle V_{avg} = \frac{V_m}{\pi}(1 + \cos\alpha) $$, $$\displaystyle I_{avg} = V_{avg}/R $$.

   • RL Load (Continuous): $$\displaystyle V_{avg} = \frac{V_m}{\pi}(1 + \cos\alpha) $$. Extinction angle $$\displaystyle \beta > \pi $$.

   • RLE Load: $$\displaystyle V_{avg} = V_m \cos\alpha - I_{avg}R - E $$. Inversion possible if $$\displaystyle E > V_m \cos\alpha $$.

2. Full-Wave Converters:

   • Half-Bridge: Two capacitors split supply. Output $$\displaystyle v_o = v_{an} - v_{bn} $$.

   • Full Bridge (Fully Controlled):

     • Circuit: Four SCRs. Output $$\displaystyle v_o = |v_s| $$ for $$\displaystyle \alpha=0 $$.

     • Waveforms: For RL load, $$\displaystyle i_o $$ continuous if $L$ large. $\alpha$ determines rectification ($$\displaystyle \alpha < 90° $$) or inversion ($$\displaystyle \alpha > 90° $$).

     • Performance:

       • $$\displaystyle V_{avg} = \frac{2V_m}{\pi}\cos\alpha $$ (R, continuous RL).

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

       • Input $$\displaystyle PF = \frac{V_{avg} I_{avg}}{V_s I_s} = \frac{\cos\alpha}{1 + \frac{2}{\pi}\left(\frac{\omega L}{R_s}\right)} $$ (with source inductance $$\displaystyle L_s $$).

3. Freewheeling Diode (FWD):

   • Operation: During SCR off-period, FWD conducts, storing energy in $L$ → continuous $$\displaystyle i_o $$, improved $PF$ (no negative $$\displaystyle v_o $$).

   • Power Factor Improvement: Eliminates negative half-cycles in $$\displaystyle i_s $$ → higher $PF$ than without FWD.

4. Effect of Source Inductance (Overlap Angle $\mu$):

   • Overlap: When transferring between SCR pairs, both conduct → $$\displaystyle v_o $$ reduces.

   • Output Voltage: $$\displaystyle V_{avg} = \frac{2V_m}{\pi}\cos(\alpha + \mu/2) $$.

   • $\mu$ depends on $$\displaystyle L_s $$ and $$\displaystyle I_{avg} $$: $$\displaystyle \cos\alpha - \cos(\alpha+\mu) = \frac{\omega L_s I_{avg}}{V_m} $$.

B. Three-Phase Converters

1. Full-Wave Fully Controlled Bridge:

   • Circuit: Six SCRs. Each pair conducts for $120°$.

   • Waveforms: For $$\displaystyle \alpha=45° $$, $$\displaystyle v_o $$ is 6-pulse. Each SCR fires $60°$ apart.

   • Output Voltage:

     • No overlap: $$\displaystyle V_{avg} = \frac{3\sqrt{6}}{\pi}V_{LL} \cos\alpha = 1.654 V_{LL} \cos\alpha $$.

     • With overlap $\mu$: $$\displaystyle V_{avg} = \frac{3\sqrt{6}}{\pi}V_{LL} \cos(\alpha + \mu/2) $$.

   • Numerical Example (Nov 2023): Given $$\displaystyle V_s=400V $$ (line), $$\displaystyle \alpha=\pi/4 $$, $$\displaystyle I_{avg}=10A $$, $$\displaystyle V_{load}=360V $$.

     $$\displaystyle V_{avg} = 1.654 \times 400 \times \cos(\pi/4) = 468.5V $$ (no overlap).

     But $$\displaystyle V_{load} = V_{avg} - I_{avg}R - L_s $$ drop.

     Overlap: $$\displaystyle \cos\alpha - \cos(\alpha+\mu) = \frac{\omega L_s I_{avg}}{V_m} $$ where $$\displaystyle V_m = \sqrt{2/3} V_{LL} $$?

     Correction: For 3-phase, $$\displaystyle V_m $$ (phase peak) = $$\displaystyle V_{LL}/\sqrt{3} $$.

     Actually: $$\displaystyle V_{avg} = \frac{3\sqrt{2}}{\pi} V_{LL} \cos(\alpha+\mu/2) = 1.35 V_{LL} \cos(\alpha+\mu/2) $$?

     Standard: $$\displaystyle V_{dc} = \frac{3\sqrt{2}}{\pi} V_{LL} \cos\alpha = 1.35 V_{LL} \cos\alpha $$ for no overlap.

     Given $$\displaystyle V_{load}=360V $$, $$\displaystyle V_{LL}=400V $$, so $$\displaystyle 1.35 \times 400 \times \cos\alpha = 360 \Rightarrow \cos\alpha=0.667 $$, $$\displaystyle \alpha=48.2° $$ (close to 45°).

     Then $\mu$ from $$\displaystyle V_{avg} $$ with $\mu$: $$\displaystyle 360 = 1.35 \times 400 \times \cos(45°+\mu/2) \Rightarrow \cos(45°+\mu/2)=0.6667 \Rightarrow 45°+\mu/2=48.2° \Rightarrow \mu=6.4° $$.

     $$\displaystyle L_s $$ from $$\displaystyle \cos\alpha - \cos(\alpha+\mu) = \frac{\omega L_s I_{avg}}{V_m} $$ with $$\displaystyle V_m = \sqrt{2/3} V_{LL} $$?

     Actually: $$\displaystyle \cos\alpha - \cos(\alpha+\mu) = \frac{2\omega L_s I_{avg}}{\sqrt{6} V_{LL}} $$?

     Use: $$\displaystyle V_{avg} = \frac{3\sqrt{2}}{\pi} V_{LL} \cos\alpha - \frac{3\omega L_s}{\pi} I_{avg} $$.

     So $$\displaystyle \frac{3\omega L_s}{\pi} I_{avg} = \frac{3\sqrt{2}}{\pi} V_{LL} (\cos\alpha - \cos(\alpha+\mu/2)) $$?

     Simpler: Overlap angle formula: $$\displaystyle \cos\alpha - \cos(\alpha+\mu) = \frac{\omega L_s I_{avg}}{V_m} $$ where $$\displaystyle V_m $$ is peak phase voltage = $$\displaystyle V_{LL}/\sqrt{3} $$.

     $$\displaystyle \omega L_s I_{avg} = V_m (\cos\alpha - \cos(\alpha+\mu)) $$.

     $$\displaystyle \cos45°=0.7071 $$, $$\displaystyle \cos(45°+6.4°)=\cos51.4°=0.6235 $$, difference=0.0836.

     $$\displaystyle V_m = 400/\sqrt{3}=230.9V $$, $$\displaystyle \omega=2\pi\times50=314 $$.

     $$\displaystyle L_s = \frac{230.9 \times 0.0836}{314 \times 10} = 6.15\,mH $$.

     $$\displaystyle R = (V_{avg,no\mu} - V_{load})/I_{avg} = (1.35\times400\times0.7071 - 360)/10 = (381.8-360)/10=2.18\,\Omega $$.

III. AC VOLTAGE CONTROLLERS

A. Single-Phase AC Controllers

1. Half-Wave Controller:

   • Circuit: Single thyristor (or anti-parallel) in series with load.

   • R Load: $$\displaystyle V_{rms} = V_s \sqrt{\frac{1}{2\pi}(2\pi-\alpha+\sin2\alpha)} $$.

   • RL Load: $$\displaystyle i_o $$ delayed. Conduction angle $$\displaystyle \gamma > \pi-\alpha $$.

2. Full-Wave Controllers:

   • Mid-Point: Two SCRs across split transformer. Each conducts for $\pi$.

   • Bridge (Anti-Parallel): Four SCRs (or two TRIACs). Output $$\displaystyle v_o = |v_s| $$ for $$\displaystyle \alpha=0 $$.

3. Performance Calculations:

   • Firing Angle from Power (R load, full-wave bridge):

     $$\displaystyle P_o = \frac{V_s^2}{R} \cdot \frac{1}{2\pi}(2\pi-\alpha+\sin2\alpha) $$ → solve for $\alpha$.

   • Input Power Factor: $$\displaystyle PF = \frac{P_o}{V_s I_{s,rms}} $$.

   • Example (Jun 2025, Dec 2024): $$\displaystyle R=5\Omega $$, $$\displaystyle V_s=230V $$, $$\displaystyle P_o=5kW $$.

     $$\displaystyle P_o = \frac{230^2}{5} \cdot \frac{1}{2\pi}(2\pi-\alpha+\sin2\alpha) = 5290 \cdot \frac{1}{2\pi}(2\pi-\alpha+\sin2\alpha) $$.

     $$\displaystyle 5000 = 842.5 \times (2\pi-\alpha+\sin2\alpha) $$ → $$\displaystyle 2\pi-\alpha+\sin2\alpha = 5.935 $$.

     Solve iteratively: $\alpha \approx 65.5°$.

     $$\displaystyle I_{s,rms} = \frac{V_s}{R} \sqrt{\frac{1}{2\pi}(2\pi-\alpha+\sin2\alpha)} = 46 \times \sqrt{0.945} = 44.7A $$.

     $$\displaystyle PF = \frac{5000}{230 \times 44.7} = 0.487 $$ (lagging due to RL? But R load → should be same as displacement? Actually for R load, $$\displaystyle PF = \cos\alpha $$ only for phase control? No, for AC controller with R load, $$\displaystyle PF = \frac{1}{2\pi}(2\pi-\alpha+\sin2\alpha)^{1/2} \cdot \cos\alpha $$? Actually: $$\displaystyle PF = \frac{P_o}{V_s I_{s,rms}} = \frac{\frac{V_s^2}{R} \cdot k}{V_s \cdot \frac{V_s}{R} \sqrt{k}} = \sqrt{k} $$ where $$\displaystyle k = \frac{1}{2\pi}(2\pi-\alpha+\sin2\alpha) $$. So $$\displaystyle PF = \sqrt{k} = \sqrt{0.945}=0.972 $$? That can't be right because phase control distorts current.

     Correction: For R load, $$\displaystyle i_s = v_s/R $$ only when SCR on. So $$\displaystyle I_{s,rms} = \frac{V_m}{R\sqrt{2}} \sqrt{\frac{2\pi-\alpha}{\pi}} $$? Actually: $$\displaystyle i_s = \frac{v_s}{R} $$ for $\omega t \in [\alpha, \alpha+\gamma]$ but for R load $$\displaystyle \gamma=\pi $$.

     $$\displaystyle I_{s,rms}^2 = \frac{1}{2\pi} \int_\alpha^{\alpha+\pi} \left(\frac{V_m \sin\omega t}{R}\right)^2 d\omega t = \frac{V_m^2}{2\pi R^2} \int_\alpha^{\alpha+\pi} \sin^2\omega t d\omega t = \frac{V_m^2}{2\pi R^2} \left[ \frac{\omega t}{2} - \frac{\sin2\omega t}{4} \right]_\alpha^{\alpha+\pi} = \frac{V_m^2}{2\pi R^2} \left( \frac{\pi}{2} - \frac{\sin(2\alpha+2\pi)-\sin2\alpha}{4} \right) = \frac{V_m^2}{2\pi R^2} \left( \frac{\pi}{2} - \frac{\sin2\alpha}{4} + \frac{\sin2\alpha}{4} \right) = \frac{V_m^2}{2\pi R^2} \cdot \frac{\pi}{2} = \frac{V_m^2}{4R^2} $$.

     So $$\displaystyle I_{s,rms} = \frac{V_m}{2R} = \frac{\sqrt{2}V_s}{2R} = \frac{V_s}{\sqrt{2}R} $$. That's independent of $\alpha$? That seems wrong because when $\alpha$ increases, conduction time decreases.

     Mistake: For R load, conduction angle is $\pi$, but integration limits from $\alpha$ to $\alpha+\pi$:

     $$\displaystyle \int_\alpha^{\alpha+\pi} \sin^2\omega t d\omega t = \int_\alpha^{\alpha+\pi} \frac{1-\cos2\omega t}{2} d\omega t = \frac{1}{2}[\omega t]_\alpha^{\alpha+\pi} - \frac{1}{4}[\sin2\omega t]_\alpha^{\alpha+\pi} = \frac{\pi}{2} - \frac{1}{4}(\sin(2\alpha+2\pi)-\sin2\alpha) = \frac{\pi}{2} - \frac{1}{4}(\sin2\alpha - \sin2\alpha) = \frac{\pi}{2} $$.

     Yes, so $$\displaystyle I_{s,rms} = \frac{V_m}{2R} $$ constant? But that can't be because when $$\displaystyle \alpha=180° $$, no conduction, $$\displaystyle I_{s,rms}=0 $$.

     The issue: For $$\displaystyle \alpha > 0 $$, the integral from $\alpha$ to $\alpha+\pi$ is still $\pi/2$ only if $\alpha+\pi \leq 2\pi$. But if $$\displaystyle \alpha=150° $$, then $$\displaystyle \alpha+\pi=330° $$, still within $2\pi$. So indeed for R load, $$\displaystyle I_{s,rms} $$ is constant regardless of $\alpha$? That contradicts known results.

     Reality Check: For single-phase AC controller with R load and full-wave (bridge), the RMS current is:

     $$\displaystyle I_{s,rms} = \frac{V_m}{R} \sqrt{\frac{1}{2\pi}(2\pi-\alpha+\sin2\alpha)} $$? Actually that's for half-wave?

     Let's derive properly for full-wave bridge:

     $$\displaystyle v_o = |v_s| $$ when SCRs on. For R load, $$\displaystyle i_o = v_o/R $$. But input current $$\displaystyle i_s $$: for positive half, current flows through one SCR pair; for negative half, through the other. So $$\displaystyle i_s $$ waveform is same as $$\displaystyle i_o $$ but bidirectional.

     $$\displaystyle i_s^2 $$: For $\omega t \in [\alpha, \pi-\alpha]$? No, for full-wave bridge with R load, each SCR conducts for $\pi$ but shifted. Actually: For $\omega t \in [\alpha, \pi-\alpha]$? That's not right.

     Standard result: For full-wave controlled rectifier with R load,

     $$\displaystyle I_{s,rms} = \frac{V_m}{R} \sqrt{\frac{1}{2\pi}(2\pi-\alpha+\sin2\alpha)} $$? That's for half-wave?

     Actually, for single-phase full-wave bridge (diode or SCR), with R load:

     $$\displaystyle I_{s,rms} = \frac{V_m}{R} \sqrt{\frac{1}{2\pi}(2\pi-\alpha+\sin2\alpha)} $$? Let's check with $$\displaystyle \alpha=0 $$: $$\displaystyle I_{s,rms} = \frac{V_m}{R} \sqrt{\frac{1}{2\pi}(2\pi)} = \frac{V_m}{R} \sqrt{1} = \frac{V_m}{R} $$. But for full-wave bridge with diodes, $$\displaystyle I_{s,rms} = \frac{V_m}{R} \cdot \frac{1}{\sqrt{2}} $$? No, for resistive load, $$\displaystyle I_{rms} = \frac{V_{rms}}{R} = \frac{V_m/\sqrt{2}}{R} $$.

     So formula above gives $$\displaystyle V_m/R $$ which is peak current, not RMS.

     Correct: $$\displaystyle I_{s,rms} = \frac{1}{\sqrt{2\pi}} \sqrt{\int_0^{2\pi} i_s^2 d\omega t} $$.

     For full-wave bridge with R load, $$\displaystyle i_s = \frac{V_m \sin\omega t}{R} $$ for $\omega t \in [\alpha, \alpha+\pi]$ and $$\displaystyle i_s = -\frac{V_m \sin\omega t}{R} $$ for $\omega t \in [\pi+\alpha, 2\pi+\alpha]$? Actually, conduction intervals:

     SCR1,2: $\alpha$ to $\pi+\alpha$? No, each SCR pair conducts for $\pi$ but starting at $\alpha$ and $\pi+\alpha$.

     So $$\displaystyle i_s = \frac{V_m \sin\omega t}{R} $$ for $\omega t \in [\alpha, \pi+\alpha]$? That's $180°$ conduction. But then next pair from $\pi+\alpha$ to $2\pi+\alpha$. So over $0$ to $2\pi$, conduction from $\alpha$ to $\pi+\alpha$ and $\pi+\alpha$ to $2\pi+\alpha$? That covers $2\pi$? Actually $\alpha$ to $\pi+\alpha$ is $\pi$, and $\pi+\alpha$ to $2\pi+\alpha$ is $\pi$, but $2\pi+\alpha$ exceeds $2\pi$. So modulo $2\pi$: conduction from $\alpha$ to $\pi+\alpha$ and from $$\displaystyle \pi+\alpha-2\pi=\alpha-\pi $$ to $\alpha$? That's messy.

     Better: Consider $$\displaystyle i_s^2 $$ is always positive. The conduction pattern repeats every $\pi$. So over $0$ to $2\pi$, there are two conduction intervals of length $\pi$: one from $\alpha$ to $\alpha+\pi$, and another from $\alpha+\pi$ to $\alpha+2\pi$ (which modulo $2\pi$ is $\alpha$ to $\alpha+\pi$ again?). Actually, for full-wave bridge, the output voltage $$\displaystyle v_o = |v_s| $$ when SCRs on. So $$\displaystyle v_o = V_m |\sin\omega t| $$ for $\omega t \in [\alpha, \alpha+\pi]$, and also for $[\pi+\alpha, 2\pi+\alpha]$? But $|\sin\omega t|$ has period $\pi$. So effectively, over $0$ to $2\pi$, $$\displaystyle v_o $$ is $$\displaystyle V_m |\sin\omega t| $$ for $\omega t \in [\alpha, \alpha+\pi]$ and also for $[\alpha+\pi, \alpha+2\pi]$? That's the same as $[\alpha, \alpha+\pi]$ shifted by $\pi$. But $$\displaystyle |\sin(\omega t + \pi)| = |\sin\omega t| $$, so it's the same waveform. So actually, the conduction intervals are:

     First interval: $\alpha$ to $\alpha+\pi$ (covers positive and negative half of $$\displaystyle v_s $$? No, $$\displaystyle v_s $$ changes sign at $\pi$. So during $\alpha$ to $\pi$, $$\displaystyle v_s>0 $$, SCR1,2 conduct; during $\pi$ to $\alpha+\pi$, $$\displaystyle v_s<0 $$, SCR3,4 conduct. So indeed one continuous interval of length $\pi$ starting at $\alpha$. Then next interval starts at $\pi+\alpha$? But $\pi+\alpha$ to $$\displaystyle \pi+\alpha+\pi = 2\pi+\alpha $$, which modulo $2\pi$ is $\alpha$ to $\alpha+\pi$ again? That would mean conduction from $\alpha$ to $\alpha+\pi$ and then from $\alpha$ to $\alpha+\pi$ again? That's overlapping.

     Actually, for full-wave bridge, the SCRs are triggered in pairs: SCR1,2 at $$\displaystyle \omega t = \alpha $$, SCR3,4 at $$\displaystyle \omega t = \pi+\alpha $$. So conduction:

     • SCR1,2: $\alpha$ to $\pi+\alpha$? But when $$\displaystyle \omega t = \pi $$, $$\displaystyle v_s=0 $$, then negative, so SCR1,2 would conduct negative current? SCRs are unidirectional, so SCR1,2 can only conduct positive current. So actually, SCR1,2 conduct from $\alpha$ to $\pi$ (when $$\displaystyle v_s>0 $$), then at $\pi$, $$\displaystyle v_s=0 $$, then SCR3,4 are triggered at $\pi+\alpha$, but between $\pi$ and $\pi+\alpha$, no SCR conducts? That would cause discontinuous current for RL load, but for R load, current becomes zero at $\pi$? For R load, $$\displaystyle i = v_s/R $$, so at $$\displaystyle \omega t=\pi $$, $$\displaystyle v_s=0 $$, $$\displaystyle i=0 $$. So actually for R load, SCR1,2 conduct from $\alpha$ to $\pi$, then current zero until $\pi+\alpha$ when SCR3,4 conduct from $\pi+\alpha$ to $2\pi$. So conduction intervals: $[\alpha, \pi]$ and $[\pi+\alpha, 2\pi]$. Each of length $\pi-\alpha$. So total conduction per cycle: $2(\pi-\alpha)$.

     Then $$\displaystyle I_{s,rms}^2 = \frac{1}{2\pi} \left[ \int_\alpha^\pi \left(\frac{V_m \sin\omega t}{R}\right)^2 d\omega t + \int_{\pi+\alpha}^{2\pi} \left(\frac{-V_m \sin\omega t}{R}\right)^2 d\omega t \right] = \frac{V_m^2}{2\pi R^2} \left[ \int_\alpha^\pi \sin^2\omega t d\omega t + \int_{\pi+\alpha}^{2\pi} \sin^2\omega t d\omega t \right] $$.

     Since $$\displaystyle \sin^2 $$ is periodic with $\pi$, both integrals equal $$\displaystyle \int_\alpha^\pi \sin^2\omega t d\omega t $$.

     $$\displaystyle \int_\alpha^\pi \sin^2\omega t d\omega t = \frac{1}{2} \int_\alpha^\pi (1-\cos2\omega t) d\omega t = \frac{1}{2} [\omega t]_\alpha^\pi - \frac{1}{4} [\sin2\omega t]_\alpha^\pi = \frac{1}{2}(\pi-\alpha) - \frac{1}{4}(\sin2\pi - \sin2\alpha) = \frac{\pi-\alpha}{2} + \frac{\sin2\alpha}{4} $$.

     So total integral = $$\displaystyle 2 \times \left( \frac{\pi-\alpha}{2} + \frac{\sin2\alpha}{4} \right) = (\pi-\alpha) + \frac{\sin2\alpha}{2} $$.

     Thus $$\displaystyle I_{s,rms} = \frac{V_m}{R} \sqrt{\frac{1}{2\pi} \left( \pi-\alpha + \frac{\sin2\alpha}{2} \right)} = \frac{V_m}{R} \sqrt{\frac{1}{2\pi} \left( \pi-\alpha + \frac{\sin2\alpha}{2} \right)} $$.

     But $$\displaystyle \frac{1}{2\pi}(\pi-\alpha + \frac{\sin2\alpha}{2}) = \frac{1}{2} - \frac{\alpha}{2\pi} + \frac{\sin2\alpha}{4\pi} $$.

     For $$\displaystyle \alpha=0 $$: $$\displaystyle I_{s,rms} = \frac{V_m}{R} \sqrt{\frac{1}{2}} = \frac{V_m}{\sqrt{2}R} = \frac{V_s}{R} $$? Since $$\displaystyle V_s = V_m/\sqrt{2} $$, so $$\displaystyle I_{s,rms} = \frac{V_m/\sqrt{2}}{R} = \frac{V_s}{R} $$. That matches: for diode bridge, $$\displaystyle I_{rms}=V_s/R $$ for R load. Good.

     So $$\displaystyle I_{s,rms} = \frac{V_s}{R} \sqrt{1 - \frac{\alpha}{\pi} + \frac{\sin2\alpha}{2\pi}} $$? Since $$\displaystyle V_s = V_m/\sqrt{2} $$, $$\displaystyle V_m = \sqrt{2}V_s $$, so $$\displaystyle I_{s,rms} = \frac{\sqrt{2}V_s}{R} \sqrt{\frac{1}{2\pi}(\pi-\alpha+\frac{\sin2\alpha}{2})} = \frac{V_s}{R} \sqrt{\frac{2}{\pi}(\pi-\alpha+\frac{\sin2\alpha}{2})} = \frac{V_s}{R} \sqrt{2 - \frac{2\alpha}{\pi} + \frac{\sin2\alpha}{\pi}} $$.

     That's messy.

     Standard formula (from textbooks): For single-phase full-wave AC controller (bridge) with R load:

     $$\displaystyle V_{o,rms} = V_s \sqrt{\frac{1}{2\pi}(2\pi-\alpha+\sin2\alpha)} $$? That's for half-wave?

     Actually, output voltage $$\displaystyle v_o $$: For bridge, $$\displaystyle v_o = |v_s| $$ when on. So $$\displaystyle v_o $$ waveform: magnitude $$\displaystyle V_m|\sin\omega t| $$ for $\omega t \in [\alpha, \alpha+\pi]$ and $[\pi+\alpha, 2\pi+\alpha]$? But as above, for R load, conduction only when $$\displaystyle v_s>0 $$ and $$\displaystyle v_s<0 $$ separately. So $$\displaystyle v_o $$ is $$\displaystyle v_s $$ for $\omega t \in [\alpha, \pi]$ and $$\displaystyle -v_s $$ for $[\pi+\alpha, 2\pi]$. So $$\displaystyle v_o $$ is not $$\displaystyle |v_s| $$ continuously; it's $$\displaystyle v_s $$ when positive SCRs conduct, and $$\displaystyle -v_s $$ when negative SCRs conduct. But since $$\displaystyle -v_s $$ is positive when $$\displaystyle v_s $$ negative, effectively $$\displaystyle v_o = |v_s| $$ during conduction intervals. But conduction intervals are $[\alpha, \pi]$ and $[\pi+\alpha, 2\pi]$, so $$\displaystyle v_o = V_m |\sin\omega t| $$ for those intervals, zero elsewhere.

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

     As computed, each integral = $$\displaystyle \frac{\pi-\alpha}{2} + \frac{\sin2\alpha}{4} $$. So total = $$\displaystyle (\pi-\alpha) + \frac{\sin2\alpha}{2} $$.

     Thus $$\displaystyle V_{o,rms} = V_m \sqrt{\frac{1}{2\pi} \left( \pi-\alpha + \frac{\sin2\alpha}{2} \right)} = V_s \sqrt{2 \cdot \frac{1}{2\pi} \left( \pi-\alpha + \frac{\sin2\alpha}{2} \right)} = V_s \sqrt{\frac{1}{\pi} \left( \pi-\alpha + \frac{\sin2\alpha}{2} \right)} $$.

     For $$\displaystyle \alpha=0 $$: $$\displaystyle V_{o,rms} = V_s \sqrt{\frac{1}{\pi}(\pi)} = V_s $$, correct.

     So $$\displaystyle V_{o,rms} = V_s \sqrt{1 - \frac{\alpha}{\pi} + \frac{\sin2\alpha}{2\pi}} $$.

     Then $$\displaystyle P_o = V_{o,rms}^2/R $$, $$\displaystyle I_{s,rms} = I_{o,rms} $$ (since bridge, input current magnitude equals output current).

     So $$\displaystyle PF = \frac{P_o}{V_s I_{s,rms}} = \frac{V_{o,rms}^2/R}{V_s \cdot V_{o,rms}/R} = \frac{V_{o,rms}}{V_s} = \sqrt{1 - \frac{\alpha}{\pi} + \frac{\sin2\alpha}{2\pi}} $$.

     That is displacement factor? Actually for R load, current in phase with voltage during conduction, but waveform not sinusoidal, so $PF$ is not $\cos\alpha$.

     For $$\displaystyle \alpha=65.5° $$, compute:

     $$\displaystyle \alpha=65.5°=1.143 rad $$, $$\displaystyle \sin2\alpha=\sin131°=0.7547 $$.

     $$\displaystyle V_{o,rms}/V_s = \sqrt{1 - \frac{1.143}{\pi} + \frac{0.7547}{2\pi}} = \sqrt{1 - 0.364 + 0.120} = \sqrt{0.756} = 0.869 $$.

     Then $$\displaystyle P_o = V_s^2/R \times (0.869)^2 = 230^2/5 \times 0.756 = 5290 \times 0.756 = 4000W $$? But we need 5000W.

     So my formula gives lower power.

     Wait: The problem states load power is 5kW. With $$\displaystyle R=5\Omega $$, maximum power (when $$\displaystyle \alpha=0 $$) is $$\displaystyle V_s^2/R = 230^2/5 = 10580W $$. So 5kW is possible.

     Solve $$\displaystyle P_o = \frac{V_{o,rms}^2}{R} = \frac{V_s^2}{R} \left(1 - \frac{\alpha}{\pi} + \frac{\sin2\alpha}{2\pi}\right) = 5000 $$.

     So $$\displaystyle 1 - \frac{\alpha}{\pi} + \frac{\sin2\alpha}{2\pi} = \frac{5000 \times 5}{230^2} = \frac{25000}{5290} = 4.726 $$? That's >1, impossible.

     Mistake: $$\displaystyle V_{o,rms}^2 = V_s^2 \left(1 - \frac{\alpha}{\pi} + \frac{\sin2\alpha}{2\pi}\right) $$? For $$\displaystyle \alpha=0 $$, $$\displaystyle V_{o,rms}=V_s $$, so $$\displaystyle V_{o,rms}^2=V_s^2 $$, so the factor should be 1. But $$\displaystyle 1 - 0 + 0 =1 $$, ok.

     But $$\displaystyle \frac{25000}{5290}=4.726 $$, impossible. So maybe it's half-wave?

     The problem says "single phase full wave ac voltage controller". That usually means full-wave bridge. But with $$\displaystyle R=5\Omega $$, $$\displaystyle V_s=230V $$, max power is $$\displaystyle 230^2/5=10580W $$, so 5kW is less, so $$\displaystyle \alpha>0 $$.

     Let's compute factor: $$\displaystyle P_o/P_{max} = V_{o,rms}^2/V_s^2 = 5000/10580=0.4726 $$.

     So $$\displaystyle 1 - \frac{\alpha}{\pi} + \frac{\sin2\alpha}{2\pi} = 0.4726 $$.

     Solve: $$\displaystyle \frac{\alpha}{\pi} - \frac{\sin2\alpha}{2\pi} = 0.5274 $$.

     Multiply $\pi$: $$\displaystyle \alpha - \frac{\sin2\alpha}{2} = 1.657 $$.

     Try $$\displaystyle \alpha=100°=1.745 rad $$: $$\displaystyle \sin200°=-0.342 $$, so LHS=1.745 - (-0.171)=1.916 >1.657.

     $$\displaystyle \alpha=90°=1.571 $$: $$\displaystyle \sin180°=0 $$, LHS=1.571.

     $$\displaystyle \alpha=85°=1.484 $$: $$\displaystyle \sin170°=0.1736 $$, LHS=1.484 - 0.0868=1.397.

     $$\displaystyle \alpha=95°=1.660 $$: $$\displaystyle \sin190°=-0.1736 $$, LHS=1.660 - (-0.0868)=1.747.

     Interpolate: for LHS=1.657, $\alpha \approx 92°$?

     But that gives $$\displaystyle PF = V_{o,rms}/V_s = \sqrt{0.4726}=0.687 $$.

     But the problem asks for firing angle and PF.

     However, the standard formula for full-wave AC controller with R load is:

     $$\displaystyle V_{o,rms} = V_s \sqrt{\frac{1}{\pi} \left( \pi - \alpha + \frac{\sin2\alpha}{2} \right)} $$? That's what I have.

     But some texts give: $$\displaystyle V_{o,rms} = V_s \sqrt{\frac{1}{2\pi}(2\pi-\alpha+\sin2\alpha)} $$ for half-wave?

     Let's check half-wave: For half-wave with R load, $$\displaystyle v_o $$ exists only for $\alpha$ to $\pi$.

     $$\displaystyle V_{o,rms}^2 = \frac{1}{2\pi} \int_\alpha^\pi (V_m \sin\omega t)^2 d\omega t = \frac{V_m^2}{2\pi} \left( \frac{\pi-\alpha}{2} + \frac{\sin2\alpha}{4} \right) = \frac{V_m^2}{4\pi} (\pi-\alpha + \frac{\sin2\alpha}{2}) $$.

     Since $$\displaystyle V_s = V_m/\sqrt{2} $$, $$\displaystyle V_{o,rms} = \frac{V_m}{2\sqrt{\pi}} \sqrt{\pi-\alpha+\frac{\sin2\alpha}{2}} = V_s \sqrt{\frac{1}{\pi}(\pi-\alpha+\frac{\sin2\alpha}{2})} $$.

     That's the same as full-wave? That can't be.

     Ah: For full-wave bridge, the output voltage is applied for both half-cycles, so the effective period is $\pi$? Actually, the RMS calculation over $2\pi$ for full-wave gives:

     $$\displaystyle V_{o,rms}^2 = \frac{1}{2\pi} \times 2 \times \int_\alpha^\pi (V_m \sin\omega t)^2 d\omega t = \frac{1}{\pi} \int_\alpha^\pi V_m^2 \sin^2\omega t d\omega t = \frac{V_m^2}{\pi} \left( \frac{\pi-\alpha}{2} + \frac{\sin2\alpha}{4} \right) = \frac{V_m^2}{2\pi} (\pi-\alpha + \frac{\sin2\alpha}{2}) $$.

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

     That's exactly the same as half-wave? That seems odd.

     But for half-wave, the integral is over $\alpha$ to $\pi$ only once, not twice. So for half-wave:

     $$\displaystyle V_{o,rms}^2 = \frac{1}{2\pi} \int_\alpha^\pi V_m^2 \sin^2\omega t d\omega t = \frac{V_m^2}{2\pi} \left( \frac{\pi-\alpha}{2} + \frac{\sin2\alpha}{4} \right) = \frac{V_m^2}{4\pi} (\pi-\alpha + \frac{\sin2\alpha}{2}) $$.

     So half-wave: $$\displaystyle V_{o,rms} = V_m \sqrt{\frac{1}{4\pi} (\pi-\alpha + \frac{\sin2\alpha}{2})} = V_s \sqrt{\frac{1}{2\pi} (\pi-\alpha + \frac{\sin2\alpha}{2})} $$.

     So full-wave has twice the RMS voltage of half-wave for same $\alpha$.

     So full-wave: $$\displaystyle V_{o,rms} = V_s \sqrt{\frac{2}{\pi} (\pi-\alpha + \frac{\sin2\alpha}{2})} $$? Let's recalc:

     Full-wave: $$\displaystyle V_{o,rms}^2 = \frac{1}{2\pi} \times 2 \times \int_\alpha^\pi V_m^2 \sin^2\omega t d\omega t = \frac{1}{\pi} \int_\alpha^\pi V_m^2 \sin^2\omega t d\omega t = \frac{V_m^2}{\pi} \left( \frac{\pi-\alpha}{2} + \frac{\sin2\alpha}{4} \right) = \frac{V_m^2}{2\pi} (\pi-\alpha + \frac{\sin2\alpha}{2}) $$.

     So $$\displaystyle V_{o,rms} = V_m \sqrt{\frac{1}{2\pi} (\pi-\alpha + \frac{\sin2\alpha}{2})} $$.

     Since $$\displaystyle V_s = V_m/\sqrt{2} $$, $$\displaystyle V_m = \sqrt{2} V_s $$, so

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

     That's what I had.

     For $$\displaystyle \alpha=0 $$: $$\displaystyle V_{o,rms} = V_s \sqrt{\frac{1}{\pi} \pi} = V_s $$, correct.

     For half-wave: $$\displaystyle V_{o,rms} = V_s \sqrt{\frac{1}{2\pi} (\pi-\alpha + \frac{\sin2\alpha}{2})} $$. For $$\displaystyle \alpha=0 $$, $$\displaystyle V_{o,rms} = V_s \sqrt{\frac{1}{2\pi} \pi} = V_s/\sqrt{2} $$, correct.

     So full-wave formula is $$\displaystyle V_{o,rms} = V_s \sqrt{\frac{1}{\pi} (\pi-\alpha + \frac{\sin2\alpha}{2})} $$.

     Then $$\displaystyle P_o = V_{o,rms}^2/R = \frac{V_s^2}{R} \left( \frac{1}{\pi} (\pi-\alpha + \frac{\sin2\alpha}{2}) \right) $$.

     So $$\displaystyle \frac{1}{\pi} (\pi-\alpha + \frac{\sin2\alpha}{2}) = \frac{P_o R}{V_s^2} = \frac{5000 \times 5}{230^2} = \frac{25000}{5290} = 4.726 $$? That's >1, impossible.

     Ah: $$\displaystyle V_s=230V $$ RMS. So $$\displaystyle V_s^2=52900 $$? $$\displaystyle 230^2=52900 $$? Actually $$\displaystyle 230^2=52900 $$? $$\displaystyle 200^2=40000 $$, $$\displaystyle 30^2=900 $$, $$\displaystyle 2*200*30=12000 $$, total 52900. Yes 52900.

     Then $$\displaystyle P_o R = 5000*5=25000 $$.

     So $$\displaystyle \frac{P_o R}{V_s^2} = 25000/52900 = 0.4726 $$.

     So $$\displaystyle \frac{1}{\pi} (\pi-\alpha + \frac{\sin2\alpha}{2}) = 0.4726 $$.

     Multiply $\pi$: $$\displaystyle \pi-\alpha + \frac{\sin2\alpha}{2} = 1.484 $$.

     $$\displaystyle \pi=3.1416 $$, so $$\displaystyle 3.1416 - \alpha + \frac{\sin2\alpha}{2} = 1.484 $$ → $$\displaystyle \alpha - \frac{\sin2\alpha}{2} = 3.1416 - 1.484 = 1.6576 $$.

     Solve: try $$\displaystyle \alpha=100°=1.745 $$ rad: $$\displaystyle \sin200°=-0.342 $$, $$\displaystyle \frac{\sin2\alpha}{2}=-0.171 $$, LHS=1.745 - (-0.171)=1.916 >1.6576.

     $$\displaystyle \alpha=90°=1.571 $$: $$\displaystyle \sin180°=0 $$, LHS=1.571.

     $$\displaystyle \alpha=85°=1.484 $$: $$\displaystyle \sin170°=0.1736 $$, $$\displaystyle \frac{\sin2\alpha}{2}=0.0868 $$, LHS=1.484-0.0868=1.397.

     $$\displaystyle \alpha=95°=1.660 $$: $$\displaystyle \sin190°=-0.1736 $$, $$\displaystyle \frac{\sin2\alpha}{2}=-0.0868 $$, LHS=1.660 - (-0.0868)=1.747.

     So $\alpha \approx 92°$?

     But then $$\displaystyle PF = \frac{P_o}{V_s I_{s,rms}} $$.

     $$\displaystyle I_{s,rms} = I_{o,rms} = V_{o,rms}/R = V_s \sqrt{0.4726}/R = 230 \times 0.6876 /5 = 31.6A $$.

     $$\displaystyle PF = 5000/(230*31.6)=5000/7268=0.688 $$.

     That seems plausible.

     So firing angle $\alpha \approx 92°$, $PF \approx 0.688$.

     But the problem likely expects exact solution.

     Note: The formula I used is for R load. For RL load, it's more complex. But problem says "load of R=5Ω", so pure R.

     So in notes, I'll give the formula and mention solving iteratively.

4. Two-Stage Sequence Control for RL Load:

   • Purpose: Improve power factor and reduce harmonic distortion for inductive loads.

   • Circuit: Two AC controllers in series. First stage (high-power) operates at low frequency (e.g., 1/2 line frequency), second stage (low-power) operates at high frequency (e.g., 1kHz).

   • Operation: First stage selects fundamental voltage magnitude; second stage shapes waveform to approximate sine.

   • Waveforms: Output voltage has two levels of modulation.

   • Advantage: Higher PF, lower harmonics than single-stage for RL loads.

IV. DC-DC CONVERTERS (CHOPPERS)

A. Classification & Basic Operation

1. Step-Down (Buck) Chopper:

   • Circuit: Switch (MOSFET/IGBT) in series with source $$\displaystyle V_s $$, diode across load for freewheeling.

   • Operation: Switch ON ($$\displaystyle T_{on} $$): $$\displaystyle v_o=V_s $$, $$\displaystyle i_L $$ increases. Switch OFF ($$\displaystyle T_{off} $$): diode conducts, $$\displaystyle v_o=0 $$, $$\displaystyle i_L $$ decreases.

   • Average Output Voltage: $$\displaystyle V_o = \alpha V_s $$, where $$\displaystyle \alpha = T_{on}/T $$ (duty cycle).

   • Waveforms: $$\displaystyle v_o $$ is rectangular, $$\displaystyle i_L $$ triangular (with ripple).

2. Step-Up (Boost) Chopper:

   • Circuit: Switch in series with source, inductor in parallel with load via diode.

   • Operation: Switch ON: $$\displaystyle v_L = V_s $$, inductor charges. Switch OFF: $$\displaystyle v_o = V_s + v_L $$, inductor discharges to load.

   • Output Voltage Derivation:

     During ON: $$\displaystyle V_s = L \frac{di}{dt} $$ → $$\displaystyle \Delta i_{on} = \frac{V_s T_{on}}{L} $$.

     During OFF: $$\displaystyle V_o - V_s = L \frac{di}{dt} $$ → $$\displaystyle \Delta i_{off} = \frac{(V_o - V_s) T_{off}}{L} $$.

     Steady state: $$\displaystyle \Delta i_{on} = \Delta i_{off} $$ → $$\displaystyle \frac{V_s T_{on}}{L} = \frac{(V_o - V_s) T_{off}}{L} $$ → $$\displaystyle V_s T_{on} = (V_o - V_s) T_{off} $$ → $$\displaystyle V_s (\alpha T) = (V_o - V_s) ((1-\alpha)T) $$ → $$\displaystyle V_s \alpha = V_o (1-\alpha) - V_s (1-\alpha) $$ → $$\displaystyle V_o (1-\alpha) = V_s (\alpha + 1-\alpha) = V_s $$ → $$\displaystyle V_o = \frac{V_s}{1-\alpha} $$.

   • Note: $$\displaystyle \alpha < 1 $$, $$\displaystyle V_o > V_s $$.

3. Buck-Boost Chopper:

   • Circuit: Switch, inductor, diode, capacitor. Output polarity reversed.

   • Output Voltage: $$\displaystyle V_o = -\frac{\alpha}{1-\alpha} V_s $$.

4. Type-A Chopper (Step-Down with RLE Load):

   • Circuit: Switch, diode, R, L, E (may be motor back-EMF).

   • Operation:

     • ON: $$\displaystyle v_o = V_s $$, $$\displaystyle i_L $$ increases from $$\displaystyle I_{min} $$ to $$\displaystyle I_{max} $$.

     • OFF: $$\displaystyle v_o = 0 $$, $$\displaystyle i_L $$ decreases from $$\displaystyle I_{max} $$ to $$\displaystyle I_{min} $$ via diode.

   • Continuous vs Discontinuous Conduction:

     • Continuous: $$\displaystyle i_L > 0 $$ always. $$\displaystyle I_{min} > 0 $$.

     • Discontinuous: $$\displaystyle i_L $$ reaches zero before next ON. $$\displaystyle I_{min}=0 $$.

     • Criterion: $$\displaystyle I_{min} = I_{avg} - \frac{\Delta i}{2} > 0 $$ for continuous.

   • Analysis for Continuous Current:

     • $$\displaystyle \Delta i = \frac{V_s - E}{L} T_{on} - \frac{-E}{L} T_{off} = \frac{(V_s - E) T_{on} + E T_{off}}{L} = \frac{V_s T_{on} - E T}{L} $$.

     • $$\displaystyle I_{avg} = \frac{V_o - E}{R} = \frac{\alpha V_s - E}{R} $$.

     • $$\displaystyle I_{max} = I_{avg} + \frac{\Delta i}{2} $$, $$\displaystyle I_{min} = I_{avg} - \frac{\Delta i}{2} $$.

     • Condition for continuous: $$\displaystyle I_{min} > 0 \Rightarrow \frac{\alpha V_s - E}{R} > \frac{V_s T_{on} - E T}{2L} $$.

   • Example (Jun 2025, Dec 2024): $$\displaystyle V_s=220V $$, $$\displaystyle T=2000\mu s $$, $$\displaystyle T_{on}=600\mu s $$, $$\displaystyle R=1\Omega $$, $$\displaystyle L=5mH $$, $$\displaystyle E=24V $$.

     • $$\displaystyle \alpha = 600/2000 = 0.3 $$.

     • $$\displaystyle V_o = \alpha V_s = 66V $$.

     • $$\displaystyle \Delta i = \frac{V_s T_{on} - E T}{L} = \frac{220 \times 0.6 \times 10^{-3} - 24 \times 2 \times 10^{-3}}{5 \times 10^{-3}} = \frac{0.132 - 0.048}{0.005} = \frac{0.084}{0.005} = 16.8A $$.

     • $$\displaystyle I_{avg} = \frac{V_o - E}{R} = \frac{66-24}{1} = 42A $$.

     • $$\displaystyle I_{min} = I_{avg} - \frac{\Delta i}{2} = 42 - 8.4 = 33.6A > 0 $$ → Continuous.

     • $$\displaystyle I_{max} = 42 + 8.4 = 50.4A $$.

5. Type-C Chopper (Two-Quadrant):

   • Circuit: Two switches (S1, S2) with antiparallel diodes, common inductive load.

   • Operation:

     • Motoring (First Quadrant): S1 ON → $$\displaystyle v_o=V_s $$, $$\displaystyle i_o>0 $$ (motor forward).

     • Regenerative Braking (Second Quadrant): S2 ON → $$\displaystyle v_o=0 $$, $$\displaystyle i_o<0 $$ (energy returned to source via D2).

   • Waveforms: S1 and S2 never ON together. $$\displaystyle i_o $$ always positive? Actually for regenerative, $$\displaystyle i_o $$ reverses? In two-quadrant, $$\displaystyle v_o \geq 0 $$, $$\displaystyle i_o $$ can be +ve or -ve. With S2 ON, $$\displaystyle v_o=0 $$, but $$\displaystyle i_o $$ can be negative if load is regenerative (e.g., motor acting as generator). So $$\displaystyle v_o \geq 0 $$, $$\displaystyle i_o $$ bidirectional.

6. Morgan Chopper:

   • Circuit: Uses two SCRs and a commutating capacitor. For step-down.

   • Operation:

     • S1 ON: $$\displaystyle v_o=V_s $$, capacitor $C$ charges to $$\displaystyle V_s $$ via D1.

     • S1 OFF, S2 ON: $C$ discharges through S2, reverse-biasing S1, turning it off. $$\displaystyle v_o=0 $$ during commutation.

   • Waveforms: $$\displaystyle v_o $$ has notch during turn-off. Requires $C$ and commutating inductor.

B. Analysis & Control

1. Duty Cycle: $$\displaystyle \alpha = T_{on}/T $$. Controls $$\displaystyle V_o $$ linearly for buck.

2. Current Limit Control (CLC):

   • Principle: Maintain $$\displaystyle i_L $$ between $$\displaystyle I_{min} $$ and $$\displaystyle I_{max} $$ by varying $\alpha$.

   • Advantage: Limits ripple, protects switch.

   • Disadvantage: Variable frequency (unless constant $T$ with variable $$\displaystyle T_{on} $$ and $$\displaystyle T_{off} $$).

3. Performance Parameters:

   • Ripple: $$\displaystyle \Delta i = \frac{V_s - E}{L} T_{on} + \frac{E}{L} T_{off} $$ for Type-A.

   • Efficiency: $$\displaystyle \eta = \frac{P_o}{P_s} = \frac{V_o I_{avg}}{V_s I_{avg} + \text{losses}} $$.

4. Continuous vs Discontinuous:

   • Continuous: $$\displaystyle i_L > 0 $$, smoother, better control.

   • Discontinuous: $$\displaystyle i_L=0 $$ part of cycle, $$\displaystyle V_o $$ equation changes, more ripple.

V. INVERTERS

A. Classification

1. VSI vs CSI:

   | Feature | VSI | CSI | |---------|-----|-----| | Input | Voltage source (capacitor) | Current source (inductor) | | Output | Voltage waveform ≈ square | Current waveform ≈ square | | Commutation | Self-commutated (devices turn off by gate) | Load commutation or forced | | Short-circuit | Risk (low impedance) | Safe (high impedance) | | Applications | General purpose | High-power, motor drives |

2. Single-phase vs Three-phase

3. Self-Commutated vs Line-Commutated: Self-commutated (using MOSFET/IGBT) → independent of load.

B. Single-Phase Inverters

1. Half-Bridge:

   • Circuit: Two switches, two capacitors split DC bus.

   • Operation: S1 ON → $$\displaystyle v_o = +V_s/2 $$; S2 ON → $$\displaystyle v_o = -V_s/2 $$.

2. Full-Bridge:

   • Circuit: Four switches. S1,S4 ON → $$\displaystyle v_o=V_s $$; S2,S3 ON → $$\displaystyle v_o=-V_s $$.

   • Square-Wave (180°): Each switch conducts $180°$. Output: square wave of amplitude $$\displaystyle V_s $$.

   • Waveforms:

     • R Load: $$\displaystyle i_o $$ in phase with $$\displaystyle v_o $$, sinusoidal (due to load inductance? Actually for pure R, $$\displaystyle i_o $$ is square wave too, but usually RL load gives sinusoidal).

     • RL Load: $$\displaystyle i_o $$ lags $$\displaystyle v_o $$, sinusoidal if $L$ large enough.

3. PWM:

   • Single-Pulse Modulation: One pulse per half-cycle. Width $2\delta$ controls fundamental amplitude. Harmonic at $n\omega$ can be eliminated by choosing $\delta$.

   • Sinusoidal PWM: Compare triangular carrier with sinusoidal reference. Fundamental amplitude controlled by modulation index $$\displaystyle m_a $$.

C. Three-Phase Inverters

1. 120° Conduction Mode:

   • Circuit: Six switches (bridge). Each switch conducts $120°$.

   • Sequence: T1, T2, T3, T4, T5, T6 (each $60°$ apart).

   • Waveforms (Star load):

     • Phase voltages: $$\displaystyle v_{AN} $$, $$\displaystyle v_{BN} $$, $$\displaystyle v_{CN} $$ are square waves but with $120°$ shift.

     • Line voltages: $$\displaystyle v_{AB} $$ etc. are 6-step.

   • RMS Load Current (Star, R per phase):

     • $$\displaystyle V_{ph} = \frac{V_s}{\sqrt{2}} $$? Actually for 120° mode, phase voltage RMS: $$\displaystyle V_{ph,rms} = \sqrt{\frac{2}{3}} V_s $$?

       Derivation: $$\displaystyle v_{AN} $$ waveform: $$\displaystyle V_s $$ for $60°$ to $180°$, etc. Over $360°$, $$\displaystyle v_{AN}^2 $$ average:

       $$\displaystyle V_{AN,rms}^2 = \frac{1}{2\pi} \left[ \int_{60°}^{180°} V_s^2 d\omega t + \int_{180°}^{300°} 0 d\omega t + \int_{300°}^{60°+360°} V_s^2 d\omega t \right] = \frac{V_s^2}{2\pi} (120°+120°) = \frac{V_s^2}{2\pi} \times \frac{4\pi}{3} = \frac{2}{3} V_s^2 $$.

       So $$\displaystyle V_{ph,rms} = \sqrt{\frac{2}{3}} V_s $$.

     • $$\displaystyle I_{ph,rms} = V_{ph,rms}/R = \frac{\sqrt{2/3} V_s}{R} $$.

     • Total Power: $$\displaystyle P = 3 I_{ph,rms}^2 R = 3 \times \frac{2}{3} \frac{V_s^2}{R} \times R? $$ Actually $$\displaystyle I_{ph,rms}^2 R = \frac{2}{3} \frac{V_s^2}{R} \times R = \frac{2}{3} V_s^2 $$. So $$\displaystyle P = 3 \times \frac{2}{3} V_s^2 = 2 V_s^2 $$? That can't be right because for star, $$\displaystyle P = \frac{3 V_{ph,rms}^2}{R} = \frac{3 \times (2/3 V_s^2)}{R} = \frac{2 V_s^2}{R} $$? Wait, $$\displaystyle V_{ph,rms}^2 = \frac{2}{3} V_s^2 $$, so $$\displaystyle P = 3 \times \frac{2}{3} V_s^2 / R $$? No: $$\displaystyle P = 3 I_{ph,rms}^2 R = 3 \times (V_{ph,rms}^2/R^2) \times R = 3 V_{ph,rms}^2 / R = 3 \times \frac{2}{3} V_s^2 / R = \frac{2 V_s^2}{R} $$.

       Example (Nov 2023): $$\displaystyle V_s=200V $$, $$\displaystyle R=10\Omega $$/phase star.

       $$\displaystyle V_{ph,rms} = \sqrt{2/3} \times 200 = 163.3V $$.

       $$\displaystyle I_{ph,rms} = 163.3/10 = 16.33A $$.

       $$\displaystyle P = 3 \times 16.33^2 \times 10 = 3 \times 266.7 \times 10 = 8000W $$? Actually $$\displaystyle 16.33^2=266.7 $$, times 3 times 10 = 8000W.

       Or $$\displaystyle P = \frac{2 V_s^2}{R} = \frac{2 \times 40000}{10} = 8000W $$. Correct.

2. 180° Conduction Mode:

   • Operation: Each switch conducts $180°$. Two switches on at any time (one from top, one from bottom).

   • Sequence: T1,T2 → T2,T3 → T3,T4 → T4,T5 → T5,T6 → T6,T1.

   • Waveforms: Phase voltages are square waves with $180°$ width. Line voltages are 6-step.

   • Comparison with 120°: 180° mode has higher fundamental voltage ($$\displaystyle V_{ph} = \frac{\sqrt{6}}{\pi} V_s \approx 0.78 V_s $$ vs $$\displaystyle 0.816 V_s $$ for 120°? Actually for 180°: $$\displaystyle V_{ph,rms} = V_s/\sqrt{2} = 0.707 V_s $$? Let's compute:

     For 180°, $$\displaystyle v_{AN} $$: $$\displaystyle V_s $$ for $180°$, $$\displaystyle -V_s $$ for $180°$. So $$\displaystyle V_{AN,rms} = V_s $$. That's for full-bridge? But in three-phase inverter, with 180° mode, each phase is connected to positive or negative bus for $180°$. So $$\displaystyle v_{AN} $$ is $$\displaystyle V_s $$ or $$\displaystyle -V_s $$ for $180°$ each. So RMS = $$\displaystyle V_s $$. But that's the DC bus voltage? Actually if DC bus is $$\displaystyle V_s $$, then phase voltage amplitude is $$\displaystyle V_s $$? But in full-bridge single-phase, output amplitude is $$\displaystyle V_s $$. For three-phase, with 180° mode, the phase voltage is not exactly $$\displaystyle V_s $$ because of the switching pattern.

     Actually, for 180° mode, each phase terminal is connected to either +V_s or -V_s for $180°$. So the voltage between phase and neutral (if available) is a square wave of amplitude $$\displaystyle V_s $$. But in a three-phase inverter without neutral, the phase voltages are not independent. For star load with neutral, $$\displaystyle v_{AN} $$ is indeed $$\displaystyle V_s $$ for $180°$ and $$\displaystyle -V_s $$ for $180°$, so RMS = $$\displaystyle V_s $$. But then line-to-line voltage $$\displaystyle v_{AB} $$ would be $$\displaystyle 2V_s $$ for $60°$? That seems high.

     Standard result: For 180° mode, fundamental phase voltage $$\displaystyle V_{ph1} = \frac{\sqrt{6}}{\pi} V_s \approx 0.78 V_s $$.

     RMS of square wave is amplitude, but the fundamental is less. So $$\displaystyle V_{ph,rms} $$ of the square wave is $$\displaystyle V_s $$, but the fundamental RMS is $$\displaystyle 0.78 V_s $$.

     So for power calculation with sinusoidal assumption, we use fundamental. But if load is pure R, the current will be square wave too, so RMS current is $$\displaystyle V_s/R $$, and power is $$\displaystyle 3 V_s^2/R $$? That would be huge. Actually, for star-connected R load with 180° mode, the phase current is square wave of amplitude $$\displaystyle V_s/R $$, RMS = $$\displaystyle V_s/R $$, power = $$\displaystyle 3 (V_s/R)^2 R = 3 V_s^2/R $$.

     But for 120° mode, phase current RMS = $$\displaystyle \sqrt{2/3} V_s/R $$, power = $$\displaystyle 2 V_s^2/R $$. So 180° mode gives higher power.

     So in numerical, if $$\displaystyle V_s=200V $$, R=10Ω, star:

     120°: $$\displaystyle P=2 \times 200^2 /10 = 8000W $$.

     180°: $$\displaystyle P=3 \times 200^2 /10 = 12000W $$.

     But the Nov 2023 problem specified 120° mode.

3. McMurray-Bedford Inverter:

   • Type: CSI (Current Source Inverter).

   • Circuit: DC inductor (current source), four SCRs, commutation capacitors.

   • Operation: Uses resonant commutation. Capacitors provide reverse voltage to turn off SCRs.

   • Waveforms: Load current is square wave (from current source), load voltage is sinusoidal (from LC filter).

   • Applications: High-power, high-voltage (e.g., HVDC).

D. Inverter Analysis

1. Load Current for RL Load (Square-wave VSI):

   • For single-phase full-bridge with RL load, $$\displaystyle i_o $$ is triangular if $L$ small, sinusoidal if $L$ large (due to inductance smoothing).

   • Expression: $$\displaystyle i_o(t) = \frac{V_s}{Z} \sin(\omega t - \theta) $$ for steady-state sinusoidal, where $$\displaystyle Z=\sqrt{R^2+(\omega L)^2} $$, $$\displaystyle \theta = \tan^{-1}(\omega L/R) $$.

2. RMS Source Current:

   • For VSI, source current $$\displaystyle i_s $$ is same as switch current. Average $$\displaystyle I_{s,avg} = 0 $$ (AC output). RMS depends on load.

3. Harmonic Content:

   • Square-wave: Odd harmonics: $$\displaystyle 3^{rd}, 5^{th}, 7^{th}, ... $$ with amplitude $$\displaystyle \frac{4V_s}{n\pi} $$ for phase voltage.

   • Reduction:

     • PWM: Shift harmonics to high frequency → easier filtering.

     • Multi-pulse: Use phase-shifted transformers (e.g., 12-pulse).

     • Filters: LC tuned to dominant harmonics.

4. Applications: AC motor drives, UPS, renewable energy (solar inverters), induction heating.

VI. CYCLOCONVERTERS

A. Single-Phase Cycloconverters

1. Mid-Point Configuration:

   • Circuit: Two SCRs per phase, center-tapped transformer.

   • Operation: For step-down ($$\displaystyle f_o < f_i $$), each SCR conducts for less than $180°$ (e.g., $90°$ for $$\displaystyle f_o=f_i/2 $$).

   • Waveforms (Resistive load, $$\displaystyle f_o=f_i/2 $$):

     Input: $$\displaystyle v_s = V_m \sin\omega_i t $$.

     Output: $$\displaystyle v_o $$ follows positive half of $$\displaystyle v_s $$ during first half-cycle, negative half during next, but with phase control. For $$\displaystyle \alpha=0 $$, $$\displaystyle v_o $$ is full-wave rectified $$\displaystyle v_s $$ → output frequency $$\displaystyle f_i $$? Actually for cycloconverter, to get $$\displaystyle f_o=f_i/2 $$, each SCR conducts for $180°$ at output frequency?

     Standard: For single-phase midpoint cycloconverter with resistive load, to get output frequency $$\displaystyle f_o $$, each SCR conducts for $180°$ at output frequency? Actually, the firing angles are varied sinusoidally to synthesize output.

     For simple case $$\displaystyle f_o=f_i/2 $$:

     Positive group: SCR1 fired at $$\displaystyle \omega_i t = 0° $$, SCR2 at $180°$. But then output would be full-wave rectified → DC?

     Actually, cycloconverter works by phase control: For each half-cycle of output, one SCR from positive group and one from negative group conduct. For $$\displaystyle f_o=f_i/2 $$, the firing angle $\alpha$ is varied as $$\displaystyle \alpha = \omega_o t $$?

     Better: The output voltage is generated by connecting the load to the input supply through SCRs with varying firing angle. For sinusoidal output, $\alpha$ is varied to follow a sinusoidal envelope.

     For $$\displaystyle f_o=f_i/2 $$, the output period is twice input period. So over two input cycles, the output goes through one cycle.

     Waveform: Typically, for resistive load, $$\displaystyle v_o $$ is a stepped approximation of a sine wave. For $$\displaystyle f_o=f_i/2 $$, it might be a square wave?

     Common Example: Single-phase midpoint cycloconverter with $$\displaystyle f_o=f_i/2 $$ and $$\displaystyle \alpha=0 $$ gives output:

     Cycle 1 (0 to $$\displaystyle T_i $$): SCR1 conducts → $$\displaystyle v_o = v_s $$ for positive half, SCR2 conducts → $$\displaystyle v_o = -v_s $$ for negative half? That would give $$\displaystyle v_o = |v_s| $$? That's not sinusoidal.

     Actually, for cycloconverter, the firing angles are controlled to produce sinusoidal output. For $$\displaystyle f_o=f_i/2 $$, the firing angle pattern is:

     For positive output: fire SCR1 at $$\displaystyle \alpha = \omega_o t $$, SCR2 at $$\displaystyle \alpha = \omega_o t + \pi $$.

     But then $$\displaystyle v_o = v_s $$ when SCR1 conducts, $$\displaystyle -v_s $$ when SCR2 conducts. So over input cycle, $$\displaystyle v_o $$ is $$\displaystyle v_s $$ for part, $$\displaystyle -v_s $$ for part. To get sinusoidal, the conduction intervals must be varied.

     Simplified: For step-down cycloconverter, the output frequency is lower. The firing pulses are generated from a reference sine wave of output frequency.

     For $$\displaystyle f_o=f_i/2 $$, the reference has half the frequency. So the firing angle $\alpha$ varies slowly.

     Waveform: $$\displaystyle v_o $$ is a series of segments of $$\displaystyle v_s $$ or $$\displaystyle -v_s $$ with widths determined by $\alpha$.

     Note: The problem asks for input waveform $f$ and output $f/2$ for resistive load. So likely they want the basic operation: For each half-cycle of output, one SCR conducts for the entire positive half of input? That would give output DC?

     Actually, to get AC output at $$\displaystyle f_o=f_i/2 $$, you need to reverse polarity every $$\displaystyle T_o = 2 T_i $$. So over two input cycles, output is positive for first input cycle, negative for second. That would be a square wave at $$\displaystyle f_i/2 $$.

     So for resistive load, with $$\displaystyle \alpha=0 $$, the output is a square wave of frequency $$\displaystyle f_i/2 $$ and amplitude $$\displaystyle V_m $$.

     That is the simplest case.

2. Bridge Configuration:

   • Circuit: Four SCRs in bridge.

   • Operation: Similar to midpoint but no center tap. Each SCR conducts for $180°$ at output frequency? Actually, for step-down, each SCR conducts for less than $180°$.

   • Comparison: Bridge requires 4 SCRs per phase vs 2 for midpoint, but no transformer center tap.

3. Principle:

   • Phase-Controlled Operation: Firing angles varied sinusoidally to approximate output sine wave.

   • Frequency Conversion: By controlling conduction intervals, output frequency $$\displaystyle f_o $$ can be any fraction of $$\displaystyle f_i $$ (typically $$\displaystyle f_o < f_i/3 $$ for reliable commutation).

B. Three-Phase Cycloconverters

1. Three-Phase to Single-Phase:

   • Circuit: Six SCRs (three positive, three negative) for single-phase output.

   • Operation: Each phase of supply connected to load through SCRs. Firing signals from three-phase reference.

   • Waveforms: Output is smoother than single-phase due to three-phase input.

2. Applications: High-power, low-speed AC drives (e.g., rolling mills, ship propulsion).

VII. MODULATION TECHNIQUES AND HARMONICS

A. Pulse Modulation

1. Single-Pulse Modulation:

   • Circuit: Inverter with PWM control. Pulse width $2\delta$ varied.

   • Harmonic Elimination: The $$\displaystyle n^{th} $$ harmonic amplitude is $$\displaystyle \frac{4V_s}{n\pi} \cos(n\delta) $$. Set $$\displaystyle \cos(n\delta)=0 $$ to eliminate $$\displaystyle n^{th} $$ harmonic.

   • Example: To eliminate $$\displaystyle 3^{rd} $$ harmonic, choose $$\displaystyle \delta = 30° $$ (since $$\displaystyle \cos(3\delta)=0 $$).

2. Sinusoidal PWM:

   • Principle: Compare sinusoidal reference ($$\displaystyle f_o $$) with high-frequency triangular carrier.

   • Output: Fundamental amplitude proportional to modulation index $$\displaystyle m_a $$ ($$\displaystyle m_a \leq 1 $$). Harmonics at carrier frequency and sidebands.

B. Harmonics

1. Generation: Non-linear switching → non-sinusoidal waveforms → harmonics.

2. Reduction Methods:

   • PWM: Shift harmonics to high frequency → easy filtering.

   • Multi-pulse Converters: Use phase-shifted transformers (e.g., 12-pulse) to cancel low-order harmonics.

   • Passive Filters: LC tuned to specific harmonic frequencies.

   • Active Filters: Inject counter-harmonic currents.

VIII. PROTECTION, COMMUTATION, AND DESIGN

A. Protection

1. Overcurrent: Fuses, semiconductor fuses, circuit breakers, current-limiting reactors.

2. Overvoltage:

   • Snubbers: RC (for dv/dt), RCD (for voltage spike).

   • Varistors: Metal-oxide varistors (MOVs) clamp transient overvoltage.

3. Thermal: Heat sinks, thermal sensors, derating.

B. Commutation

1. Natural (Line-Commutated): AC source provides commutation (e.g., in rectifiers).

2. Forced Commutation:

   • Self-Commutated: Devices like MOSFET, IGBT turn off by gate signal.

   • Auxiliary Forced: Use separate commutating circuit (e.g., in GTO, thyristor inverters).

     • Types:

       • Class A: Load commutation (load provides reverse voltage).

       • Class B: Resonant commutation (LC circuit).

       • Class C: Complementary commutation (auxiliary SCR).

       • Class D: Impulse commutation.

       • Class E: AC line commutation (for inverters).

C. Design Considerations

1. Device Selection:

   • Voltage Rating: $$\displaystyle V_{rating} \geq 1.5 \times V_{peak} $$ (with derating factor 0.1-0.2).

   • Current Rating: $$\displaystyle I_{rating} \geq 1.5 \times I_{peak} $$ (considering $$\displaystyle I^2t $$).

   • Example (Nov 2023): For 6kV, 1kA, derating 0.1 → each SCR rating: $V_{rating} = 6kV/1.1? Actually derating factor 0.1 means use 90% of rating?

     If derating factor $$\displaystyle k=0.1 $$, then number in series $$\displaystyle N_s = \frac{V_{total}}{V_{rating} \times (1-k)} $$.

     Similarly parallel $$\displaystyle N_p = \frac{I_{total}}{I_{rating} \times (1-k)} $$.

     Given $$\displaystyle V_{rating}=1000V $$, $$\displaystyle I_{rating}=200A $$, $$\displaystyle V_{total}=6kV $$, $$\displaystyle I_{total}=1kA $$, $$\displaystyle k=0.1 $$.

     $$\displaystyle N_s = \frac{6000}{1000 \times 0.9} = \frac{6000}{900} = 6.67 \Rightarrow 7 $$.

     $$\displaystyle N_p = \frac{1000}{200 \times 0.9} = \frac{1000}{180} = 5.56 \Rightarrow 6 $$.

     So 7 series, 6 parallel → total 42 SCRs.

2. Heat Sinking: Thermal resistance calculation, forced cooling for high power.

3. Snubber Design:

   • RC Snubber: $C$ limits $dv/dt$, $R$ damps oscillation.

     $$\displaystyle C = \frac{I_{peak}}{dv/dt_{max}} $$, $$\displaystyle R = \sqrt{\frac{L}{C}} $$ (where $L$ is circuit inductance).

   • RCD Snubber: Clamps voltage to $$\displaystyle V_s + V_{D} $$.

IX. SPECIAL TOPICS & SHORT NOTES

1. Jones Chopper:

   • Circuit: Uses two SCRs and a commutating capacitor. For step-down.

   • Operation: Similar to Morgan but with different commutation. Capacitor charges to $$\displaystyle V_s $$ during ON, then discharges to turn off SCR.

   • Waveforms: Notch in output voltage during commutation.

2. Current Limit Control (CLC):

   • Method: Maintain $$\displaystyle i_L $$ between $$\displaystyle I_{min} $$ and $$\displaystyle I_{max} $$ by switching ON/OFF.

   • Implementation: Hysteresis control. When $$\displaystyle i_L $$ reaches $$\displaystyle I_{max} $$, turn OFF; when $$\displaystyle i_L $$ drops to $$\displaystyle I_{min} $$, turn ON.

   • Advantage: Simple, limits ripple.

   • Disadvantage: Variable switching frequency.

3. AC Voltage Controllers:

   • Overview: Phase-angle control of AC power. Used for light dimming, heating, motor speed control.

   • Types: Half-wave, full-wave (mid-point, bridge).

   • Disadvantages: Harmonic generation, low PF for RL loads.

4. Harmonics and Reduction:

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

   • Effects: Heating, torque pulsation, interference.

   • Reduction: PWM, multi-pulse, filters (passive/active).

5. GTO VI Characteristics:

   • Forward: Similar to SCR until latching, then negative resistance region.

   • Reverse: Blocks up to reverse breakdown voltage.

   • Turn-Off: Requires high negative gate current. $$\displaystyle I_{GM-} \approx 1/3 $$ of anode current.

6. Forced Commutation of Thyristor:

   • Need: SCR cannot be turned off by gate → need external circuit.

   • Types:

     • Resonant (LC): Create oscillating current to reverse anode current.

     • Impulse: Use capacitor discharge.

     • Complementary: Use another SCR to divert current.

   • Application: Inverters, choppers.

7. Industrial Applications:

   • SCR: HVDC, motor drives, welding.

   • GTO: Traction drives, high-power inverters.

   • MOSFET: Switch-mode power supplies, DC-DC converters.

   • IGBT: AC motor drives, UPS, induction heating.

   • Converters: Battery charging, electroplating.

   • Inverters: UPS, solar inverters, variable-frequency drives.

   • Cycloconverters: Large cement mills, ship propulsion.

[!IMPORTANT]

Exam Focus:

• SCR: Turn-on methods, dv/dt/di/dt, equalization, UJT firing.

• Rectifiers: Single-phase full-bridge RLE waveforms, source inductance effect, firing angle calculation.

• AC Controllers: Firing angle from power (R load), PF, two-stage control.

• Choppers: Type-A continuous/discontinuous analysis, $$\displaystyle I_{max}/I_{min} $$ calculation.

• Inverters: 120°/180° mode operation, RMS calculations (star load), VSI vs CSI.

• Cycloconverters: Single-phase bridge operation, $f/2$ output.

• Numericals: Always show derivation, use $\boxed{}$ for final answers.