PE1 Midterm Guide
Everything for the Power Electronics (1) midterm on Lectures 1–7 in one page: every lecture explained for beginners, both sheets, Section 3 and six past midterms solved step by step, with every number checked by computer simulation.
What to do now
- Read The big picture and the symbol glossary (25 min). Every later page uses those words.
- Learn the four free-mark theory answers that appear in almost every midterm: merits/demerits, ideal vs real switch, the two losses (30 min).
- Go to Lecture 2 and solve the half-wave R-L example with a calculator, without looking. Last year it was worth 7 of 20 marks.
- Then work through Lectures 3–7 in order, solving every card before reading its solution, and finish with the Dec 2025 paper as a timed mock exam.
In plain words explains the idea. Formulas to memorise is the short list for the night before. Waveforms are drawn to scale from real numbers: swipe them sideways on a phone. Question cards state the task, the given data and what to find. Under each card is an on-paper solution written the way you should write it in the exam: formula first, then numbers, lined up on "=", with the answer boxed.
Colour key for all waveforms:
The big picture
The wall socket gives AC: a voltage that swings up and down 50 times a second. Many loads need DC: a battery charger, a DC motor, electroplating, an HVDC line. Power electronics uses electronic switches (diodes, thyristors, transistors) to chop and steer that AC so the load sees a voltage that is, on average, steady and in one direction.
A rectifier does AC → DC. Its output is not a flat line; it is pieces of the sine wave. So we describe it with numbers: its average (the DC part, what a DC meter reads), its rms (what decides heating and power), and how far it is from pure DC (form factor, ripple factor).
| Converter | Input → output | Name | In this midterm? |
|---|---|---|---|
| AC-DC | AC supply → DC load | Rectifier (uncontrolled: diodes; controlled: thyristors/SCRs) | Yes, Lect 2–7 |
| DC-DC | DC → different DC level | Chopper | No (Lect 9–10) |
| DC-AC | DC → AC | Inverter | Only as an idea (HVDC block 2, α > 90°) |
| AC-AC | AC → AC, other level/frequency | AC voltage controller / cycloconverter | Name only |
Lecture → what they ask
| Lecture | Circuit | Typical exam question |
|---|---|---|
| 1 | Devices, ideal switch | Merits/demerits; ideal vs real switch; two losses; SCR holding/latching current |
| 2 | Average/rms; half-wave R; half-wave R-L | Vdc, Irms, PF of half-wave R (with ron); β and Idc, Irms, P, PF of R-L |
| 3 | Half-wave + FWD; + capacitor; R + E | Effect of FWD; draw vo for small/large C; ΔVo, ID,peak, C; battery current, efficiency |
| 4 | Bridge; centre-tap; R and L≫R | Draw vo, io, iD, is; PIV; bridge vs centre-tap |
| 5 | 3-pulse, 6-pulse, 6-phase star; FF, RF, TUF | Vdc = 0.827Vm / 1.654Vm; diode and supply currents; PF; ripple frequency |
| 6 | Controlled half-wave (R, R-L); controlled bridge (R) | Vdc(α), α for a wanted Vdc, Irms, P, PF |
| 7 | Controlled bridge R-L (dis/continuous), L≫R, + E, half-controlled + FWD | "Verify continuous" (α ≤ θ), Vdc = 2Vmcosα/π, Idc, P; rectifier vs inverter |
Glossary of symbols
| Symbol | Meaning | How to get it |
|---|---|---|
| vs, Vs | Supply (source) voltage; vs = Vm sin ωt | "120 V" in a question means 120 V rms |
| Vm | Peak (maximum) of the supply sine | Vm = √2 × Vrms (120 V → 169.7 V) |
| ω, f | Angular frequency (rad/s), frequency (Hz) | ω = 2πf; 50 Hz → 314.16 rad/s |
| ωt | Angle on the waveform (rad); one cycle = 2π | 1 ms at 50 Hz = 2π/20 = π/10 |
| vo, io | Output (load) voltage and current, instantaneous | Read from the waveform |
| Vdc = Vo = Vavg | Average (DC) output voltage. Slides use all three names | Area under vo over one period ÷ period |
| Idc = Io | Average (DC) load current | Idc = Vdc/R (R load and R-L load) |
| Vrms, Irms | Root-mean-square (effective) value | √(area under v² over a period ÷ period) |
| Pdc | DC output power | Vdc Idc |
| Pac | "AC output power" = total (real) power to the load in the slides | Vrms Irms = Irms2R for R |
| S | Apparent power of the supply | Vs,rms × Is,rms |
| PF | Input power factor | P / S (not cos θ!) |
| η | Rectification ratio (efficiency) | Pdc/Pac |
| FF | Form factor | Vrms/Vdc (ideal = 1) |
| RF | Ripple factor | √(FF² − 1) (ideal = 0) |
| TUF | Transformer utilisation factor | Pdc/(VsIs) of the secondary (ideal = 1) |
| CF | Crest factor | Is,peak/Is,rms |
| PIV | Peak inverse voltage: largest reverse voltage across a diode/SCR | Read the most negative vD on the waveform |
| Z, θ, ωτ | Load impedance, its angle, normalised time constant | Z=√(R²+(ωL)²), θ=tan⁻¹(ωL/R), ωτ=ωL/R |
| β | Extinction angle: where the current falls to zero | Solve i(β)=0 with the calculator |
| α | (a) Firing/delay angle of an SCR. (b) In the C-filter and battery circuits: the angle where the diode turns on | (a) given; (b) solved / sin⁻¹(E/Vm) |
| γ | Conduction angle | β − α |
| E | Back-emf: a battery or a DC motor's internal voltage | given |
| ΔVo | Peak-to-peak ripple of the output voltage | Vmax − Vmin |
| ron | Diode forward (ON) resistance | Acts as a resistor in series with R |
| IL, IH | SCR latching and holding current | Lect 1 |
| D (duty cycle) | ton/T of a chopper | Choppers: not in this midterm |
Set the calculator to RADIAN mode for anything with e−ωt/ωτ, β, α in formulas like (β − α) or sin 2β. Degrees only for the final "β = 201°". In the slides, "sin 7" means sin(7 rad).
Lecture 1: Devices and switches
Power electronics controls and converts electric power with semiconductor switches. A switch is either fully ON (passes current, almost no voltage across it) or fully OFF (blocks voltage, no current). Because an ideal switch is never "half on", it wastes almost no power. That is the whole reason power electronics is efficient. Real switches are not perfect, and the imperfections are exactly what the exam asks about.
Merits and demerits (asked in 4 of 6 midterms)
- High efficiency (low losses in the semiconductor devices)
- High reliability
- Long life and less maintenance (no moving parts)
- Flexibility in operation
- Fast dynamic response compared with electromechanical converters
- Small size and less weight → low installation cost
- Generate harmonics in the supply and in the load
- AC-DC and AC-AC converters work at low input power factor under certain conditions
- Regeneration of power (sending it back) is difficult
- Low overload (overhead) capacity
Goal of a power electronic system (Lect 1 slide 9): convert energy from source to load with the highest efficiency, availability and reliability, at the lowest cost, smallest size and least weight.
Ideal switch vs real switch, and the two losses
| Ideal switch | Real (actual) switch | |
|---|---|---|
| ON | Zero voltage across it, can carry infinite current | Small on-state voltage (0.2–3 V for a diode); finite maximum current |
| OFF | Zero current, can block infinite voltage | Small leakage current (µA–mA); finite maximum (breakdown) voltage |
| Switching | Instant (zero time) | Takes time to turn on and off |
| Power loss | Zero; bidirectional current and voltage; four quadrants | Two losses (below) → the device heats up |
- Conduction loss: during the ON and OFF states (on-voltage × current while ON, leakage current × voltage while OFF).
- Switching loss: while the switch is changing state (opening or closing), voltage and current are both non-zero for a short time.
Last year's student wrote "ON losses, OFF losses, switching losses". That is acceptable because the first two together are the conduction loss, but name them as the slide does: conduction loss and switching loss.
Three classes of switches
| Class | Meaning | Examples |
|---|---|---|
| Uncontrolled | No control terminal; the circuit decides ON/OFF | Diode |
| Semi-controlled | Can be turned ON by the gate, but not OFF | Thyristor (SCR), TRIAC |
| Fully controlled | Turned ON and OFF by the control terminal | BJT, MOSFET, IGBT, IGCT, GTO, MCT |
The thyristor (SCR): what to memorise
Other devices (low priority, one line each)
- Power diodes: general purpose (line frequency, trr ≈ 25 µs, up to 6 kA / 8 kV); fast recovery (trr < 1 µs, high-frequency circuits); Schottky (≈0.3 V drop, 50–100 V, low-voltage high-current SMPS).
- TRIAC: two antiparallel thyristors with one gate (MT1, MT2, G); conducts both ways; lamp dimmers, heaters, small motors.
- GTO: turned off by a large negative gate current (≈1/5 of anode current); highest power ratings (<5 kV, <5 kA, <5 kHz).
- IGCT: GTO-like with integrated gate drive, turned off with ≈20 V; <6.5 kV, <4 kA, <1 kHz.
- MCT: MOSFET-controlled thyristor; GTO-like but simpler gate drive.
- BJT: current-controlled (base current); unidirectional current and voltage.
- MOSFET: voltage-controlled (vGS > threshold vT); ON acts as a resistance; fastest.
- IGBT: MOSFET gate + BJT power port: fast switching and low conduction loss; motor drives, PV, wind.
- Comparison: GTO/IGCT 36 MVA, IGBT 6 MVA; IGBT is fast, has simple gate drive and can be paralleled, but has higher on-state loss.
How do actual (real) switches differ from ideal switches?
Ideal: ON → zero voltage, any current; OFF → zero current, any voltage; switches in zero time; zero power loss. Real: ON → small on-voltage and a finite maximum current; OFF → small leakage current and a finite maximum voltage; switching takes time; so it has conduction loss (in the ON and OFF states) and switching loss (during transitions), which heat the device.
Lecture 2: Average, rms and the half-wave rectifier
Average and rms values
Average = area under the curve over one period ÷ the period. Area above the axis counts +, below counts −. A DC meter reads the average. A full sine wave averages to zero, but half a sine does not: its area is 2Vm.
RMS = the DC value that would heat a resistor the same amount. Square the curve, find the area, divide by the period, take the square root. Power in a resistor always uses rms: P = Irms2R.
A bar chart of five test marks: 80 (1 unit wide), 60 (2 units), 95 (1 unit), 75 (1 unit). Find the average.
FIND the average value
Answer: average = 74
A current is +2 A for 3 ms and −1.5 A for 4 ms (period 7 ms). (1) Find its average. (2) Repeat if the negative part is −3 A. (3) What does a DC ammeter read?
FIND Iavg in both cases and the DC-meter reading
Answer: (1) 0 A · (2) −0.857 A · (3) the DC ammeter reads the average: 0 and 0.857 A (reversed)
Two waveforms (one period each): v = 10 V for 2 s, 20 V for 1 s, 30 V for 2 s, 0 V for 1 s (T = 6 s). i rises linearly 0 → 40 mA in 3 s, then −20 mA for 2 s, −40 mA for 2 s, 0 for 1 s (T = 8 s).
FIND Vavg and Iavg
Answer: Vavg = 16.7 V · Iavg = −7.5 mA
One period (T = 10 s) of a voltage: 20 V for 4 s, 30 V for 2 s, −10 V for 2 s, 0 for 2 s.
FIND Vrms
Answer: Vrms = 18.97 ≈ 19 V
The slide writes "(400×4)+(900×2)+(100×2) = 3600" for the numerator, the same thing.
i(t) = 5t for 0 < t < 2 and −10 for 2 < t < 4 (T = 4). It flows in a 2 Ω resistor.
FIND Irms and the average power in the resistor
Answer: Irms = 8.165 A · P = 133.3 W
v(t) = 10 sin t for 0 < t < π, 0 for π < t < 2π (a half-wave rectified sine). It feeds a 10 Ω resistor.
FIND Vrms and the power
Answer: Vrms = 5 V · P = 2.5 W
Shortcut you can quote: half-wave rectified sine → Vrms = Vm/2 = 10/2 = 5 V.
Half-wave rectifier, resistive load
One diode in series with the load. When vs is positive the diode is ON (a closed switch), so the load sees vs. When vs is negative the diode is OFF (an open switch): no current, vo = 0, and the diode itself holds the negative supply voltage. So the output is the positive half-cycles only.
- Vm = √2 × Vrms.
- Vdc = Vm/π, Idc = Vdc/R.
- Vrms = Vm/2, Irms = Vm/(2R).
- Pdc = VdcIdc; Pac = Vrms2/R.
- PF = Pac/(Vs,rms × Irms).
- In S, use the supply rms voltage (120 V), not the output Vrms (84.85 V).
- With a diode resistance ron: the peak current is Vm/(R + ron), and the output peak is that current × R. Don't forget the division (2022 handwritten answer did).
- If the diode is drawn reversed, the output is the negative half-cycles: Vdc = −Vm/π.
A half-wave rectifier with a resistive load.
| Supply | 120 V rms, 50 Hz |
|---|---|
| Load | R = 5 Ω |
| Diode | ideal |
FIND (a) average load current, (b) dc and ac power, (c) power factor
Answer: Io = 10.8 A · Pdc = 583.6 W · Pac = 1440 W · PF = 0.707
From the waveform: vo is the half sine from 0 to π, zero from π to 2π.
Half-wave rectifier, vs = 110 sin ωt V, R = 10 Ω. The diode is ideal except its forward resistance ron = 1 Ω. Sketch vo and find (a) the average output voltage, (b) the diode rms current, (c) the input power factor.
Full solution and the comparison with the handwritten answer are in Dec 2024.
Half-wave rectifier, R-L load (highest priority)
Add an inductor. An inductor hates sudden changes of current: it stores energy while the current rises and gives it back while it falls. So when vs crosses zero at π, the current is still flowing. The inductor's stored energy keeps the diode ON past π, into the negative half-cycle, until the energy is used up and the current reaches zero at the extinction angle β (β > π). During π → β the output voltage follows vs and is negative. That negative bit lowers the average.
- KVL with the diode ON: Vm sin ωt = Ri + L di/dt.
- Forced (steady AC) response: if = (Vm/Z) sin(ωt − θ).
- Natural response: in = A e−t/τ, τ = L/R.
- i(0) = 0 gives A = (Vm/Z) sin θ. Add them: the boxed formula. Set i(β) = 0 to get the β equation; it has no closed form, so use the calculator's SOLVE.
- β must come out between π and 2π. When SOLVE asks for a start value, enter 3.5 (radians).
- θ and ωτ in radians inside the formula.
- PF ≠ cos θ. The slide says so explicitly.
- \(I_{\mathrm{rms}}=V_{\mathrm{rms}}/Z\) is the slides' shortcut. It is not exact (the true rms needs ∫i²). For the lecture numbers the error is under 1%; for a large L it can reach 10%. In the exam use the slide formula. The exact values are shown under each solution so you understand.
A half-wave rectifier feeds an R-L load.
| Load | R = 100 Ω, L = 0.1 H |
|---|---|
| Supply | Vm = 100 V, ω = 377 rad/s |
FIND (a) i(ωt), (b) average current, (c) rms current, (d) power, (e) power factor
Answer: β = 3.502 rad = 201° · Idc = 0.308 A · Irms = 0.470 A · P = 22.1 W · PF = 0.665
Slide 23 prints Irms = 0.457 A, P = 21 W, PF = 0.65. Re-doing its own formula gives 0.470 A, 22.1 W, 0.665. Integrating the real current (and simulating the circuit) gives Irms = 0.474 A, P = 22.4 W, PF = 0.67. The β = 201° and Idc = 0.308 A on the slide are correct.
For the R-L half-wave rectifier, R = 100 Ω, L = 0.1 H, ω = 314 rad/s, Vm = 200 V. Determine (a) an expression for the current, (b) the average current, (c) the rms current, (d) the power absorbed, (e) the power factor.
Lecture 3: Freewheeling diode, capacitor filter, battery load
The lecture quiz repeats the Lect 2 R-L method with new numbers. Do it first as revision.
| Load | R = 50 Ω, L = 0.1 H |
|---|---|
| Supply | Vm = 100 V, ω = 200 rad/s |
FIND i(ωt), Idc, Irms, P, PF
Answer: β = 201.8° · Idc = 0.614 A · Irms = 0.934 A · P = 43.6 W · PF = 0.660
Exact check (∫i²): Irms = 0.942 A, P = 44.4 W, PF = 0.666.
Half-wave with a freewheeling diode (FWD)
Put a second diode D2 across the R-L load, pointing upward. While vs > 0, D1 conducts and the load sees vs. When vs goes negative, the inductor wants to keep its current going. Instead of dragging D1 into the negative half (which made vo negative before), the current now "freewheels" round the loop L → R → D2. So vo = 0 instead of negative. The output is exactly the half-wave rectified sine, so the formulas of the R load come back: \(V_{o}=V_{m}/\pi \).
Half-wave rectifier with FWD and R-L load.
| Load | R = 2 Ω, L = 25 mH |
|---|---|
| Supply | Vm = 100 V, 50 Hz |
FIND average load voltage and current
Answer: Vo = 31.8 V · Io = 15.9 A
From the waveform: vo is the half sine (the FWD clamps the negative half to zero).
Verified by simulating the circuit: average current 15.92 A. L does not change the average; it only smooths the current.
Model answer: with the FWD the negative part of vo disappears, so Vdc and Idc increase (Pdc = VdcIdc increases), the output rms voltage decreases, and the load current becomes continuous and smoother. The power delivered to the load (Irms2R) increases slightly.
Checked by simulation with the Lect 2 load (R = 100 Ω, L = 0.1 H, Vm = 100 V, ω = 377): without FWD Vdc = 30.81 V, Vrms = 50.24 V, P = 22.43 W; with FWD Vdc = 31.83 V, Vrms = 50.00 V, P = 22.54 W. The handwritten answer "Pac will decrease" is therefore wrong.
Half-wave with a capacitor filter
A capacitor across the load works like a small water tank. While the diode conducts, the capacitor charges up to the peak Vm. After the peak, vs falls faster than the capacitor can discharge, so the diode turns off at ωt = θ (just after π/2). The capacitor then feeds the load alone and its voltage decays exponentially (time constant RC). In the next cycle, the rising vs catches up with the capacitor voltage at ωt = 2π + α and the diode turns on again. Result: an almost-flat output with a small saw-tooth ripple ΔVo, but the diode now delivers all its charge in a short, high pulse.
- Vm = √2 Vrms; ωRC = 2πf·R·C.
- θ = π − tan−1(ωRC); Vm sin θ.
- Solve α with SHIFT-SOLVE (start near 0.8 rad).
- ΔVo = Vm(1 − sin α).
- IC,peak = ωCVm cos α; Is,peak = ID,peak = IC,peak + Vm sin α/R.
- For a required ripple use the approximation C = Vm/(fRΔVo).
- The approximation ΔVo ≈ Vm/(fRC) gives 67.9 V for the lecture example, but the exact answer is 49.9 V. Use the exact one when α is asked; use the approximation only to find C.
- Bigger C → smaller ripple but larger peak diode current (shorter conduction).
- For a full-wave rectifier the capacitor discharges for only half a period: use 2f instead of f (Sheet 1 P11). This is an addition: the slides only show the half-wave case.
Half-wave rectifier with a capacitor filter across a resistor.
| Supply | 120 V rms, 50 Hz |
|---|---|
| Load | R = 500 Ω, C = 100 µF |
FIND (a) vo(ωt), (b) ΔVo, (c) iC(ωt), (d) peak diode current, (e) C for ΔVo = 1% of Vm
Answer: θ = 93.64°, α = 45° · ΔVo = 49.86 V · ID,peak = 4.01 A · C = 4000 µF
Check by time-stepping the RC + diode circuit: ΔVo = 49.86 V.
The capacitor current while the diode is off is printed as "−0.339 sin ωt e…". The sin ωt must not be there: \(i_{C}=-0.339\ e^{-(\omega t-1.63)/15.708}\) A. Compare with the general formula on slide 10.
The half-wave rectifier with a capacitor filter has a 220 V rms source at 50 Hz, R = 300 Ω, C = 150 µF. Find (16) the peak-to-peak output variation, (17) the peak diode current, (18) the C that makes ΔVo = 5 V.
For the half-wave rectifier with a capacitor across R, draw vo for a very low, a medium and a very high capacitance.
Key words to write: the capacitor charges to Vm through the diode and discharges through R with time constant RC; larger C (5τ ≫ T) → smaller ripple, shorter diode conduction, higher peak diode current.
Resistive load + back-emf (battery charging)
Charging a battery E through a resistor R. The diode can only conduct when the supply is higher than the battery: from α = sin−1(E/Vm) to π − α. The rest of the time no current flows and the output terminals simply show the battery voltage E. The worst reverse voltage on the diode is at the negative peak: −(Vm + E).
- Vrms: the bracket (½ − α/π + sin2α/2π) is printed squared. It must not be squared. For Vm = 169.7 V, E = 24 V the slide formula gives 62.5 V instead of 86.6 V.
- Irms: the printed formula (with sin α, sin((π−α)/2) and (½ − α/2π)) is not the integral of the current. It happens to be close for small α (13.98 vs 13.98 A here) but is wrong in general (for Vm = 100, E = 80, R = 2 it gives 4.97 A instead of 3.29 A). Use the corrected formula above.
A single-phase source charges a battery through one ideal diode and a 1:1 transformer.
| Supply | 120 V rms, 50 Hz |
|---|---|
| Battery | E = 24 V, internal resistance R = 5 Ω |
FIND average current, rms current, efficiency
Answer: Io = 8.51 A · Irms = 13.98 A · η = 17.29% (MCQ answers c, c, d)
The MCQ key says 17.23 %; the difference is rounding. Choose (d).
Lecture 4: Bridge and centre-tapped full-wave rectifiers
A half-wave rectifier throws away the negative half-cycle. A full-wave rectifier flips it up instead, so the load gets a hump every half-cycle: twice the average voltage and twice the ripple frequency (100 Hz from 50 Hz). Two ways to build it:
- Bridge (4 diodes, no special transformer): D1 + D2 conduct in the positive half, D3 + D4 in the negative half. Each reverse-biased diode sees at most Vm.
- Centre-tapped transformer (2 diodes): each half of the secondary feeds the load for one half-cycle. Only one diode drop in the path, but each diode must block 2Vm.
Resistive load
Highly inductive load (L ≫ R)
A huge inductor keeps the current almost perfectly flat at Io. The voltage across the load is still the full-wave humps, because the diodes still connect the load to |vs|. So: vo unchanged, io flat, each diode carries Io for half a cycle (a rectangle), and the supply current is a square wave ±Io.
- Don't draw vo flat for a highly inductive load (the 2022 handwritten answer did). Only the current is flat.
- Centre-tap: vs1 = vs2 = vs(N2/2N1); Vm is the peak of one half of the secondary. "230-0-230 V" means Vm = 230√2.
- Bridge → better for high voltage (PIV = Vm). Centre-tap → isolation and only one diode drop → better for low-voltage, high-current.
A single-phase full-wave uncontrolled rectifier feeds a resistive load. Repeat for (a) a bridge with a 1:1 isolating transformer and (b) a centre-tapped transformer 1:1:1.
| Supply | 240 V, 50 Hz |
|---|---|
| Load | R = 24 Ω |
FIND the average, peak and rms current, and the peak reverse voltage across each diode
Answer: Io = 9.00 A, Ipeak = 14.14 A, Irms = 10 A in both; per diode Iavg = 4.50 A, Irms = 7.07 A; PIV (a) 339.4 V, (b) 678.8 V
1:1:1 means each half of the secondary gives 240 V, so the load sees the same waveform in (a) and (b).
For the centre-tapped circuit, draw (1) output voltage, (2) output current, (3) diode currents, (4) supply current, for (a) a pure resistive load and (b) a very high inductance.
(a) vo and io full-wave humps; iD1 humps in 0–π, iD2 in π–2π; is (primary) a full sine. (b) Use the L ≫ R figure: vo still humps; io flat at Io; iD1, iD2 rectangles of height Io, alternating every π; is a square wave ±Io (×N2/N1 on the primary side). Label Vm, Io, π, 2π.
A centre-tapped transformer 230-0-230 V feeds a full-wave rectifier whose ideal diodes can carry at most 10 A. Find (12) Vdc, (13) Idc, (14) the RL that gives the maximum dc power, (15) the PIV.
Bridge, Vm = 100 V, 50 Hz, R = 10 Ω, L = 10 mH. Find the diode average and rms current and the power factor. Then the inductor is replaced by a 40 V battery: find the average output current.
The exam key picks 4.77 A and 0.954 for Q20–21: those are the nearest options; exact values are 4.86 A and 0.971 at 50 Hz. The battery formula is the slide's half-wave method applied to the full-wave case (period π), an addition to the slides.
Lecture 5: Three-phase rectifiers and figures of merit
Industry uses three-phase supplies for big loads. Three sine waves are 120° apart, so at every instant one of them is the highest. A diode connected to each phase lets only the highest phase reach the load: the output rides along the tops of the sine waves and never drops low, so it is much closer to pure DC.
- 3-pulse (half-wave): 3 diodes + neutral wire. Output = the highest phase voltage. 3 pulses per cycle (150 Hz), each diode conducts 120°.
- 6-pulse (bridge): 6 diodes. The top group picks the highest phase, the bottom group the lowest, so the output is the highest line-to-line voltage. 6 pulses per cycle (300 Hz), each diode conducts 120°, the pair changes every 60°.
- Six-phase star: a centre-tapped star secondary makes 6 phases; one diode at a time, each for 60°.
Three-phase half-wave (3-pulse)
\(v_{\mathrm{D1}}=v_{a}-v_{o}\). While D1 conducts, vo = va → vD1 = 0. While D2 conducts, vD1 = va − vb: at 180° it is −(√3/2)Vm, at 210° −1.5Vm, at 240° −√3Vm. While D3 conducts, vD1 = va − vc: −√3Vm at 300°, −1.5Vm at 330°, 0 at 390°.
Small slip on slide 4–5: "Vb = 1" at 210° and "Vc = 1" at 330° should read Vm.
Three-phase full-wave bridge (6-pulse)
Six-phase star rectifier
The slide gives ID,rms = 0.396Im = Irms/√3. Its own integral gives √(11.48/75.40) = 0.390 Im, and since each of the 6 diodes carries the load current for 1/6 of the time, ID,rms = Irms/√6 (checked numerically: ratio 0.4082 = 1/√6).
Output evaluation: figures of merit
| Circuit (R load) | Vo/Vm | FF | RF | η | TUF | Ripple f | PIV |
|---|---|---|---|---|---|---|---|
| 1-φ half-wave | 0.318 | 1.571 | 1.211 | 40.5% | 0.287 | f | Vm |
| 1-φ bridge | 0.637 | 1.111 | 0.483 | 81.1% | 0.811 | 2f | Vm |
| 1-φ centre-tap | 0.637 | 1.111 | 0.483 | 81.1% | 0.573 (secondary) | 2f | 2Vm |
| 3-φ 3-pulse | 0.827 | 1.017 | 0.183 | 96.8% | — | 3f | √3Vm |
| 3-φ 6-pulse | 1.654 | 1.0009 | 0.042 | 99.8% | — | 6f | √3Vm |
All values in this table were computed by integrating the waveforms numerically.
(10) In a three-phase full-wave rectifier, the ratio of output ripple frequency to supply frequency is? (23–25) A three-phase bridge has a 480 V rms line-to-line source and a load of 25 Ω in series with 50 mH. Find the diode average current, the diode rms current and the source rms current.
(10) 6.
The 2021 key marks 12.68 A for Q24 and 17.93 A for Q25. Both are impossible: the rms of a current cannot be smaller than its average (25.93 A for the load). Simulating the real R-L circuit gives 14.97 A and 21.17 A (same at 60 Hz). None of the printed options is correct; write the correct value and show the method.
Lecture 6: Controlled half-wave and controlled full-wave (R load)
Replace the diode with a thyristor (SCR). A diode turns on by itself the moment vs becomes positive. An SCR waits: it needs forward bias + a gate pulse. The doctor fires the gate a little later, at the firing (delay) angle α. Everything between 0 and α is cut off, so a bigger α gives a smaller average voltage. That is how we control the DC output. Once on, the SCR stays on until its current falls to zero.
Controlled half-wave, R load
Controlled half-wave rectifier with a resistive load R, α = π/2.
FIND Vdc, Idc, Vrms, Irms in terms of Vm and R
Answer: Vdc = 0.1592Vm · Idc = 0.1592Vm/R · Vrms = 0.3536Vm · Irms = 0.3536Vm/R
Design a controlled half-wave circuit that gives an average of 40 V across a 100 Ω resistor.
| Supply | 120 V rms, 50 Hz |
|---|---|
| Load | R = 100 Ω, wanted Vdc = 40 V |
FIND α, the power absorbed, the power factor
Answer: α = 61.3° · P = 57.2 W · PF = 0.630
The slide writes the rms as "Vm/√2 · √(…/π)"; it is the same as Vm/2 · √(1 − α/π + sin2α/2π). It also gives PF = 0.629 (rounding).
Controlled half-wave, R-L load
Same as the diode R-L case, but the current starts at α instead of 0. The inductor again carries the conduction past π, to β. Conduction angle γ = β − α.
Controlled half-wave rectifier with an R-L load.
| Supply | 120 V rms, 50 Hz |
|---|---|
| Load | R = 20 Ω, L = 0.04 H |
| Firing angle | α = 45° |
FIND (a) i(ωt), (b) Irms, (c) power, (d) power factor
Answer: β = 3.70 rad · Irms = 3.49 A · P = 243.6 W · PF = 0.582
Integrating i² (and simulating the circuit) gives Irms = 3.41 A, P = 232 W, PF = 0.568; also Idc = 2.10 A. The slide answers come from the approximate \(I_{\mathrm{rms}}=V_{\mathrm{rms}}/Z\). Write the lecture formula in the exam.
Controlled full-wave (bridge or centre-tap), R load
Four SCRs in a bridge (or two with a centre-tap). T1 and T2 are fired together at α; T3 and T4 together at π + α. Each half-cycle is trimmed the same way, so the output is two trimmed humps per cycle: double the half-wave value.
A full-wave controlled bridge feeds a resistor.
| Supply | 120 V rms, 50 Hz |
|---|---|
| Load | R = 20 Ω |
| Firing angle | α = 40° |
FIND average load current, power absorbed, source volt-amperes (and PF)
Answer: Io = 4.77 A · P = 672.9 W · S = 696 VA · PF = 0.967
A controlled half-wave rectifier (the switch is drawn reversed, so it conducts in the negative half) has vs = 100 sin 314t and R = 10 Ω. The switch is ideal except that it needs 1 ms to turn on. Find the average output voltage for α = 0° and α = 90°.
Build a controlled half-wave rectifier without a thyristor, keeping the circuit simple.
Use a diode in series with a fully controlled switch (MOSFET/IGBT/transistor) and the load. The diode blocks the negative half-cycle; the switch is turned on at the wanted angle α in the positive half, so the output starts at α exactly like an SCR. Relation between delay time and angle: α = 2π·ton/T (T = 20 ms at 50 Hz). Draw vo = vs from α to π, zero elsewhere.
Lecture 7: Controlled full-wave rectifier with inductive loads
Fully controlled bridge with an R-L load. T1-T2 fire at α, T3-T4 at π + α. After T1-T2 start, the inductor keeps them on past π. Two things can happen:
- Discontinuous: the current dies (at β) before the next pair is fired at π + α. The output is zero for a while.
- Continuous: the current is still flowing when T3-T4 are fired at π + α. Firing them turns T1-T2 off immediately (natural commutation). The current never reaches zero and the output voltage goes negative from π to π + α.
The test is simple: continuous if α ≤ θ = tan−1(ωL/R). Then Vdc = (2Vm/π) cos α, which can even be negative for α > 90° (inverter mode).
Discontinuous current (β < π + α)
The current integrals are printed with 1/(2π): "Idc = (1/2π)∫αβ i d(ωt)" and the same for Irms. For a full-wave rectifier the period is π, so it must be 1/π (the Vdc line on the same slide correctly uses 1/π). With 1/(2π) you would get half the current: 3.65 A instead of 7.29 A in Ex 7.1.
A controlled full-wave bridge with an R-L load.
| Supply | 120 V rms, 50 Hz |
|---|---|
| Load | R = 10 Ω, L = 20 mH |
| Firing angle | α = 60° |
FIND (a) i(ωt), (b) average load current, (c) power, (d) power factor
Answer: discontinuous (α = 60° > θ = 32.1°) · β = 3.70 rad · Idc = 7.29 A · Irms = 9.30 A · P = 866 W · PF = 0.775
Simulating the bridge and integrating i² gives Irms = 8.76 A, P = 768 W, PF = 0.730. The slide's Irms = Vrms/Z overestimates by 6% here. The slide never computes part (b); its answer is Idc = 7.29 A.
Continuous current (α ≤ θ)
Continuous needs i(π + α) ≥ 0. Putting ωt = π + α in the current formula: sin(θ − α)(1 − e−π/ωτ) ≥ 0. The bracket is always positive, so sin(θ − α) ≥ 0 → α ≤ θ.
"Irms = Vrms/Z = Vm/(√2 Z)" is the current of the AC supply connected straight to the R-L load, not of the rectifier. It gives 3.64 A for Ex 7.2, less than the average current 5.40 A, which is impossible. The doctor's own example (slide 25) instead uses Irms ≅ Idc. That is a good approximation only when L ≫ R; with ωL = 31.4 Ω and R = 10 Ω the true Irms is 5.60 A.
A controlled full-wave bridge with an R-L load.
| Supply | 120 V rms, 50 Hz |
|---|---|
| Load | R = 10 Ω, L = 100 mH |
| Firing angle | α = 60° |
FIND (a) show the current is continuous, (b) average load current, (c) power absorbed
Answer: θ = 72.34° > α → continuous · Idc = 5.40 A · P ≈ 291.8 W
Exact (simulation): Irms = 5.60 A, P = 314 W; Idc = 5.40 A agrees.
A controlled full-wave bridge rectifier has a source of 150 V rms at 60 Hz and an R-L load, R = 10 Ω, L = 100 mH, α = 60°. (a) Verify that the load current is continuous. (b) Find the dc component of the current. (c) Find the power absorbed.
Highly inductive load (L ≫ R)
R-L + back-emf (DC motor, battery)
The slides call the emf "vdc" here; don't confuse it with the average output Vo.
Half-controlled bridge with freewheeling diode
Two thyristors + two diodes, plus a freewheeling diode across the load. Whenever the load voltage tries to go negative, the FWD turns on and clamps it to zero. So the output looks exactly like the R-load case, even with an inductive load, and Vdc can never be negative (no inverter mode). During 0 → α the load current freewheels through the FWD and the supply current is zero.
- Which rectifier do you recommend for a controlled high-power industrial application, and why? Draw its output voltage and current.
- Which rectifiers can give an output of double polarity (positive and negative average)?
- What happens to the load power if a freewheeling diode is added to a fully controlled full-wave rectifier with continuous R-L current?
- Three-phase fully controlled bridge (6-pulse): highest average voltage (up to 1.654Vm·cos α), lowest ripple (6f), good TUF and PF, and Vdc controlled by α. Draw vo as segments of the line voltages starting α after each natural crossing, and io almost flat for an inductive load. For a single-phase supply: a controlled bridge with a FWD (continuous current, no negative output).
- Fully controlled converters with a continuous (inductive) load current: single-phase fully controlled bridge or centre-tap, three-phase half-wave controlled, three-phase fully controlled bridge. With α > 90° their average output is negative. Uncontrolled, half-controlled and FWD circuits cannot.
- The FWD removes the negative part of vo: Vdc rises from (2Vm/π)cos α to (Vm/π)(1 + cos α), so Idc and the load power increase, the output rms voltage falls, and the supply current is zero during freewheeling (better PF).
Sheet 1: Single-phase rectifiers (11 problems)
"120 V" in the sheet means rms. Problems that ask to "derive formulas" are answered with the formula first, then the numbers.
| Supply | 120 V, 60 Hz |
|---|---|
| Load | R = 10 Ω, ideal diode |
FIND (a) Vdc (b) Idc (c) PIV (d) average power (e) PF (f) conduction angle
Answer: 54.02 V · 5.40 A · 169.7 V · 720 W · 0.707 · 180°
| Supply | 240 V, 50 Hz |
|---|---|
| Load | R = 10 Ω |
FIND Pdc, Pac, rectification ratio, FF, RF, ripple frequency
Answer: 1167.2 W · 2880 W · 40.53% · 1.571 · 1.211 · 50 Hz
| Supply | 220 V, 50 Hz |
|---|---|
| Load | R = 40 Ω, L = 0.15 H |
FIND waveforms; i(ωt); Idc; Irms; P; input PF
Answer: β = 231.1° · Idc = 2.015 A · Irms = 2.673 A (lecture formula) · P = 285.9 W · PF = 0.486
Waveforms: the same shapes as the R-L figure in Lecture 2, with β = 231°.
∫i² gives Irms = 2.930 A, P = 343.4 W, PF = 0.533 (confirmed by simulation: 2.930 A). The larger ωL/R, the worse Vrms/Z becomes.
| Supply | Vm = 300 V, 50 Hz |
|---|---|
| Load | R = 10 Ω, L = 250 mH, FWD across the load |
FIND waveforms; Vo, Io; Fourier terms of vo up to the 6th harmonic
Answer: Vo = 95.49 V · Io = 9.55 A · vo = 95.49 + 150 sin ωt − 63.66 cos 2ωt − 12.73 cos 4ωt − 5.46 cos 6ωt V
Waveforms: as in the FWD figure; with ωL = 78.5 Ω ≫ R the current is almost flat.
Extra (not in slides): harmonic currents In = Vn/|R + jnωL|: I1 = 1.895 A, I2 = 0.404 A, so Irms = 9.65 A (simulation 9.65 A). Odd harmonics above the 1st are zero (checked numerically).
| Supply | Vm = 100 V, 50 Hz |
|---|---|
| Load | R = 5 Ω, L very large (ripple-free current), FWD |
FIND waveforms; Vo; Io; diode average currents; P; PF
Answer: 31.83 V · 6.366 A · ID1 = ID2 = 3.183 A · 202.6 W · PF = 0.637
Waveforms: vo = half sine; io flat; iD1 = Io for 0–π, iD2 = Io for π–2π; is = iD1. (The sheet says "Figure (2)" but the circuit printed is Figure 1 again.)
| Supply | Vm = 200 V, 50 Hz |
|---|---|
| Load | R = 10 Ω |
FIND waveforms (vo, io, vD, is); formulas and values of Vo, Io, Irms, Is,rms, PF
Answer: 127.32 V · 12.73 A · Irms = Is,rms = 14.14 A · PF = 1
Waveforms: bridge with R.
| Supply | 240 V, 50 Hz |
|---|---|
| Load | R = 10 Ω |
FIND Pdc, Pac, η, FF, RF, ripple frequency
Answer: 4669 W · 5760 W · 81.06% · 1.111 · 0.483 · 100 Hz
| Supply | 220 V, 50 Hz |
|---|---|
| Load | R = 40 Ω, L = 0.15 H |
FIND waveforms; io(ωt); average and rms load and supply currents; P; PF
Answer: Io = 4.952 A · Irms = 5.036 A · Is,avg = 0 · Is,rms = 5.036 A · P = 1014 W · PF = 0.916
The slides only treat L ≫ R for the bridge, so (b) and (d) use an addition: the exact steady-state current (same method as the half-wave R-L, but periodic every π) and a Fourier check.
Integrating the exact current gives the same 5.036 A; the "Vrms/Z" shortcut would give 3.56 A (wrong, it is less than Io).
| Supply | 220 V, 50 Hz |
|---|---|
| Load | R = 40 Ω, L very large |
FIND waveforms; Vo; Io; Irms; Is,avg; Is,rms; P; PF
Answer: 198.07 V · 4.952 A · Irms = 4.952 A · 0 · 4.952 A · 980.8 W · 0.900
Waveforms: L ≫ R figure.
FIND Pdc, Pac, η, FF, RF, ripple frequency (current ripple-free)
Answer: 980.8 W · 1089.4 W · 90.03% · 1.111 · 0.483 · 100 Hz
Pac = VrmsIrms follows the slide definition. The real power in the load is still Io2R = 980.8 W, because the inductor absorbs the ripple voltage.
| Supply | 120 V, 60 Hz, full-wave |
|---|---|
| Load | R = 100 Ω ∥ C = 600 µF |
FIND (a) ΔVo (b) C for ΔVo < 5% of Vm (c) waveforms
Answer: ΔVo ≈ 23.6 V (exact 18.7 V) · C ≥ 1667 µF
Addition to the slides: in a full-wave rectifier the capacitor is recharged every half-period, so the discharge time halves: ΔVo ≈ Vm/(2fRC).
Sheet 2: Three-phase rectifiers (11 problems)
Problems come in pairs: one "derive the relations", the next "compute the numbers" for the same circuit. They are combined below, five cards for eleven problems. Vm = peak phase voltage. Waveforms: 3-pulse and 6-pulse in Lecture 5.
| Supply | balanced 3-φ, Vm = 200 V (phase), 50 Hz, 4-wire |
|---|---|
| Load | R = 10 Ω, ideal diodes |
FIND Vo, Io, Is,avg; Irms, Is,rms; diode avg/rms; conduction angle; ripple frequency; P; S; PF; PIV
Answer: Vo = 165.4 V · Io = 16.54 A · ID,avg = Is,avg = 5.51 A · Irms = 16.81 A · ID,rms = Is,rms = 9.71 A · 120° · 150 Hz · P = 2827 W · S = 4118 VA · PF = 0.686 · PIV = 346.4 V
From the waveform: vo = the highest phase; each diode (and each phase winding) carries the load current for 120° (π/6 → 5π/6 for phase a).
The supply current of a 3-pulse rectifier has a DC component (Io/3 per phase): one reason transformers for it need special design.
| Supply | Vm = 200 V (phase), 50 Hz |
|---|---|
| Load | R = 10 Ω, ripple-free current |
FIND waveforms; Vo, Vrms; Io, Irms; diode avg/rms; PF; ripple frequency; conduction angle
Answer: Vo = 165.4 V · Vrms = 168.1 V · Io = Irms = 16.54 A · ID,avg = 5.51 A · ID,rms = 9.55 A · PF = 0.675 · 150 Hz · 120°
From the waveform: vo unchanged; each diode current is a rectangle of height Io, 120° wide.
| Supply | Vm = 200 V (phase), 50 Hz |
|---|---|
| Load | R = 10 Ω |
FIND waveforms; Vo, Io; Vrms, Irms; Is,rms; diode avg/rms; P; S; PF
Answer: 330.8 V · 33.08 A · 331.1 V · 33.11 A · Is,rms = 27.03 A · ID,avg = 11.03 A · ID,rms = 19.12 A · P = 10962 W · S = 11469 VA · PF = 0.956
| Supply | phase voltage 220 V rms, 50 Hz |
|---|---|
| Load | R = 25 Ω |
FIND averages and rms of vo, io, iD, is; conduction angle; ripple frequency; PIV; P; S; PF
Answer: Vo = 514.6 V · Io = 20.58 A · ID,avg = 6.86 A · Is,avg = 0 · Vrms = 515.1 V · Irms = 20.60 A · ID,rms = 11.89 A · Is,rms = 16.82 A · 120° · 300 Hz · PIV = 538.9 V · P = 10611 W · S = 11102 VA · PF = 0.956
| Supply | phase 220 V rms, 50 Hz |
|---|---|
| Load | R = 25 Ω, ripple-free current |
FIND waveforms; averages/rms of io, iD, is; vo; PF; ripple frequency; conduction angle; PIV
Answer: Io = Irms = 20.58 A · ID,avg = 6.86 A · ID,rms = 11.88 A · Is,avg = 0 · Is,rms = 16.81 A · Vo = 514.6 V · Vrms = 515.1 V · PF = 0.955 · 300 Hz · 120° · PIV = 538.9 V
From the waveform: ia = +Io for 120°, 0 for 60°, −Io for 120°, 0 for 60° (the "L ≫ R" panel of the 6-pulse figure).
Section 3 (whiteboard notes)
Page 1 is the capacitor-filter method for exactly the Lect 3 example (120 V, 50 Hz, R = 500 Ω, C = 100 µF): ωRC → θ = −tan−1(ωRC) + π → α from sin α − sin θ e−(2π+α−θ)/ωRC = 0 with SHIFT-SOLVE → ripple Vm(1 − sin α) → IC,peak = ωCVm cos α and Is,peak = IC,peak + Vm sin α/R. The board also writes ΔVo = Vm/(fRC) and "0.01 × 120√2" for the 1% design. All of it is solved in Ex 3.2.
Page 2 is the battery charger: E = Vm sin α → α = sin−1(E/Vm); vo = vs from α to π − α, E otherwise; Vavg = (Vm/π)cos α + E(½ + α/π); Iavg = (Vavg − E)/R; Pdelivered = Irms2R + EIavg; vD = vs − E → PIV = −(Vm + E). Solved in Ex 3.3. Page 3 is unrelated scribbling.
Past-exam analysis
Six past midterms are covered. They come from three examiners: Dr. Elgohary (last year), Dr. Hamouda (2022, 2024) and Dr. Ghanem (2021). The current lectures are Dr. Elgohary's, so last year's paper carries the most weight. Dr. Hamouda's papers matter too: he co-teaches and Lect 7 is his.
Format and marks
| Exam | Time · marks | Questions | Theory / drawing | Calculation |
|---|---|---|---|---|
| Dec 2025 · Elgohary | 1 h · 20 | Q1: merits/demerits [2], two losses of real switches [2], R-L half-wave principle + waveforms [4]. Q2: R-L half-wave calculation [7], controlled bridge R-L continuous [5] | 8 (40%) | 12 (60%) |
| Dec 2024 · Hamouda | 45 min · 36 | Merits [2], real vs ideal switch [2], bridge + C waveforms [6], recommend rectifier + draw [4], centre-tap R and L≫R waveforms [8], half-wave with ron [6], double-polarity rectifiers [4], 3-phase controlled α=90° drawing [4] | 30 (83%) | 6 (17%) |
| 2023/24 "DU" | (paper missing) | Half-wave with ron + 1 ms turn-on delay, C-filter drawing, effect of FWD, merits, real vs ideal, centre-tap waveforms, HVDC blocks | ≈70% | ≈30% |
| 2023/24 "MU" | (paper missing) | Controlled half-wave without a thyristor, controlled half-wave with turn-on time (α=0°, 90°), FWD in a controlled full-wave, rectifier for industry, centre-tap with D1 burnt | ≈70% | ≈30% |
| Dec 2022 · Hamouda | 45 min · 30 | Merits [2], real vs ideal [2], C-filter drawing [6], centre-tap R and L≫R [8], half-wave with ron [6], HVDC blocks [6] | 24 (80%) | 6 (20%) |
| Dec 2021 · Ghanem | 1 h · 25 MCQ | Half-wave R, battery charging, RF/TUF/η, PF definition, transformer rating, ripple frequency, SCR turn-off, centre-tap, C filter, bridge R-L, bridge + battery, 3-phase bridge R-L | 6 (24%) | 19 (76%) |
How often each topic appeared (6 exams)
| Topic | Lecture | Count | Where |
|---|---|---|---|
| Merits & demerits of power electronics | 1 | 4 | 2022, DU, 2024, 2025 |
| Ideal vs real switch / switch losses | 1 | 4 | 2022, DU, 2024, 2025 |
| Full-wave centre-tap/bridge waveforms, R and L≫R | 4 | 4 | 2022, DU, 2024, MU (D1 burnt) |
| Capacitor filter (draw for small/medium/large C, or ΔV, Ipeak, C) | 3 | 4 | 2021, 2022, DU, 2024 |
| Half-wave R with diode resistance ron (Vavg, ID,rms, PF) | 2 | 3 | 2022, DU, 2024 (+2021 Q1, Q7) |
| Half-wave R-L: β, i(ωt), Idc, Irms, P, PF | 2 | 2 | 2025 (11 of 20 marks), Lect 3 quiz |
| Controlled half-wave / turn-on delay | 6 | 2 | DU, MU |
| Controlled full-wave R-L (continuous test, Vdc, Idc, P) | 7 | 2 | 2025 (5 marks), MU (FWD) |
| Freewheeling diode effect | 3, 7 | 2 | DU, MU |
| HVDC blocks / which rectifier for industry / double polarity | 1, 4, 7 | 4 | 2022, DU, MU, 2024 |
| Battery charging (R + E) | 3 | 1 | 2021 (4 MCQs) |
| Three-phase uncontrolled (3- and 6-pulse) | 5 | 1 | 2021 (4 MCQs); all of Sheet 2 |
| RF / TUF / η / PF definitions | 2, 5 | 1 | 2021 (5 MCQs) |
| SCR holding/latching current | 1 | 1 | 2021 |
Questions that repeat almost word for word
- "Mention three advantages and three disadvantages for power electronic systems." (2022, DU, 2024, 2025)
- "How do actual/real switches differ from ideal switches?" (2022, DU, 2024; 2025 asked it as "the two types of losses")
- "For the circuit shown (centre-tapped), draw output voltage, output current, diode currents, supply current for (a) pure R, (b) very high inductance." (2022, DU, 2024)
- "Draw the expected output voltage when the capacitor is very low / medium / very high." (2022, DU, 2024 with a bridge)
- "Half-wave rectifier, vs = 110 sin ωt, R = …, ron = …: sketch vo; find the average output voltage, the diode rms current, the input power factor." (2022: R=10 kΩ, r=10 Ω; DU: R=100 Ω, r=10 Ω; 2024: R=10 Ω, r=1 Ω)
- "For the R-L half-wave rectifier, R=100 Ω, L=0.1 H, ω=…, Vm=…: (a) expression for i, (b) average current, (c) rms current, (d) power, (e) PF." This is the Lect 2 example with new numbers (2025: ω=314, Vm=200).
- "A controlled full-wave bridge … R=10 Ω, L=100 mH, α=60°: (a) verify continuous current, (b) Idc, (c) power." This is the Lect 7 example with 150 V / 60 Hz (2025).
What the handwritten answers teach us
- Dr. Elgohary copies his lecture examples and changes the numbers. Both calculation questions in 2025 are lecture examples (Lect 2 slide 22, Lect 7 slide 25). Master every lecture example.
- The 2025 student used exactly the slide formulas, including \(I_{\mathrm{rms}}=V_{\mathrm{rms}}/Z\), and found β "by the calculator" (SOLVE). Write the formula, substitute, give the number. No long derivations are needed.
- For "discuss the principle and draw" (2025 Q1.3, 4 marks) the student's note on the answer sheet (in Arabic) says: explain the circuit + draw the waveforms + the equations needed. So: explain the operation, draw vo and io under vs, and add the key equations.
- Drawings are graded on: correct shape, labels (Vm, π, 2π, α, β), and conduction intervals. Always draw vs first and line the other waveforms up under it.
- Theory answers are short bullet lists (6 merits, 4 demerits). See the quick-fire list.
- 2022 Q5(b): wrote the diode rms current as 54.9 A. It is 5.49 mA: they forgot to divide by R + ron.
- 2022 Q4(b): drew vo as a flat line for the highly inductive load. Wrong: vo is still the full-wave rectified sine; only io becomes flat.
- DU and 2024: input PF computed with the load power only (0.643). With the diode resistance counted as part of the circuit, the input PF is 0.707. Both are shown in the solutions below; write the 0.707 method and mention why.
- DU Q3 / MU Q4: "with a FWD, Pac decreases". Wrong: the load power slightly increases. Checked in Lect 3.
- 2025 Q2.2: P = Idc2R = 455.9 W (lecture formula). The exact value is 479.7 W because ωL = 37.7 Ω is not ≫ R. Write the lecture formula in the exam, but know why it is approximate.
Priority map
A grain of salt: three examiners, small sample. Priority combines how often a topic appeared, how many marks it carried, and whether this year's doctor (Elgohary) asked it.
| Topic | Lect | How it appeared | Priority |
|---|---|---|---|
| Half-wave R-L (β, i(ωt), Idc, Irms, P, PF) + principle + waveforms | 2 | 2025: 11/20 marks; lecture example + quiz | Top |
| Merits/demerits, ideal vs real switch, switch losses | 1 | 4 of 6 exams, 4 marks each time | Top |
| Controlled full-wave R-L: continuous test α ≤ θ, Vdc=2Vmcosα/π | 7 | 2025: 5 marks | Top |
| Centre-tap / bridge waveforms (R and L≫R), PIV | 4 | 3 exams × 8 marks (Hamouda) | High |
| Capacitor filter (drawing; ΔVo, ID,peak, C) | 3 | 4 exams | High |
| Half-wave R with ron (Vdc, ID,rms, PF) | 2 | 3 exams × 6 marks | High |
| Controlled half-wave (R, R-L), controlled bridge with R | 6 | MU (2 Qs); 4 lecture examples | Medium-high |
| Freewheeling diode (half-wave, half-controlled bridge) | 3, 7 | DU, MU; Sheet 1 P4–5 | Medium-high |
| Average & rms by area and by integral | 2 | Basis of everything; 6 lecture examples | Medium |
| Battery / back-emf (R + E) | 3, 7 | 2021 MCQs; Section 3 | Medium |
| Three-phase 3-pulse and 6-pulse | 5 | 2021 MCQs; all of Sheet 2 | Medium |
| FF, RF, TUF, η, PF definitions and standard values | 2, 5 | 2021 MCQs; Sheets | Medium |
| HVDC blocks, rectifier for industry, double polarity, rectifier/inverter regions | 1, 7 | Hamouda papers | Medium-low |
| Device details (SCR ratings, TRIAC, GTO, IGBT…) | 1 | 2021 (holding current) | Low |
Past midterms solved
Dec 2025 · Dr. Elgohary · 20 marks · 1 hour (last year)
What are the merits and demerits of power electronic systems?
Merits: high efficiency; high reliability; long life and less maintenance (no moving parts); flexibility in operation; fast dynamic response; small size and weight (low installation cost). Demerits: harmonics in supply and load; low input power factor under some conditions (AC-DC, AC-AC); regeneration of power is difficult; low overload capacity.
Handwritten answer: correct and complete.
Explain the two types of losses which occur in non-ideal switches.
Conduction loss: in the ON state the real switch has a small on-voltage while carrying current, and in the OFF state a small leakage current while blocking voltage, so V × I ≠ 0. Switching loss: the change ON↔OFF takes a finite time during which voltage and current are both significant. Both heat the switch.
Handwritten answer lists "ON losses, OFF losses, switching losses": right idea, but name them conduction (ON + OFF) and switching.
For an R-L load on a single-phase uncontrolled half-wave rectifier, discuss the principle of operation and draw the load current and output voltage.
When vs becomes positive at ωt = 0 the diode is forward-biased and vo = vs. KVL: Vm sin ωt = Ri + L di/dt. The current rises, lagging the voltage; the inductor stores energy. At ωt = π the supply reverses, but the inductor keeps the current (and the diode) going, so vo follows the negative supply until the current reaches zero at the extinction angle β (π < β < 2π). From β to 2π the diode is off, i = 0, vo = 0, and the diode blocks vs.
Handwritten answer was only an outline note. Write the paragraph above plus the drawing and the two equations.
R-L half-wave rectifier.
| Load | R = 100 Ω, L = 0.1 H |
|---|---|
| Supply | ω = 314 rad/s, Vm = 200 V |
FIND (a) i(ωt) (b) Iavg (c) Irms (d) P (e) PF
Answer: β = 3.4458 rad = 197.43° · Idc = 0.622 A · Irms = 0.9569 A · P = 91.6 W · PF = 0.677
Z = 104.82 ✓, θ = 17.4322° ✓, ωτ = 0.314 ✓, β = 3.44585 rad = 197.43° ✓, Vdc = 62.2 V ✓, Idc = 0.622 A ✓, Irms = 0.9568 A ✓, P = 91.549 W ✓, PF = 0.6753 (should be 0.6766; small arithmetic slip). Exact values by integration: Irms = 0.9611 A, P = 92.38 W, PF = 0.6796.
Controlled full-wave bridge, R-L load.
| Supply | 150 V rms, 60 Hz |
|---|---|
| Load | R = 10 Ω, L = 100 mH |
| Firing angle | α = 60° |
FIND (a) verify continuous current (b) Idc (c) P
Answer: θ = 75.14° > 60° → continuous · Idc = 6.752 A · P ≈ 455.9 W
θ = 75.144° ✓, Vdc = 67.524 V ✓, Idc = 6.7524 A ✓, P = 455.949 W ✓ (doctor's method). Exact simulation: Irms = 6.926 A, P = 479.7 W. If you have time, add one line: "Irms ≅ Idc because L is large".
Dec 2021 · Dr. Ghanem · 25 MCQ · 1 hour
Key = the highlighted answer on the paper. Check = independent result (computed).
Worked solutions: Q2–4 in Ex 3.3, Q12–15 and Q19–22 in Lecture 4, Q16–18 in Lecture 3, Q23–25 in Lecture 5.
Dec 2022 · Dr. Hamouda · 30 marks · 45 minutes
Three advantages and three disadvantages of power electronic systems. How do real switches differ from ideal ones?
See Lecture 1 merits and ideal vs real. Handwritten answer: correct (it mixes "I = 0, V = ∞" for the ON state; the ideal ON state is V = 0 and any current).
Half-wave + capacitor: draw vo for very low, medium and very high C.
Handwritten answer drew the decays below the axis toward −Vm: wrong. With a diode-capacitor-R circuit vo never goes negative; it decays toward 0.
Centre-tapped rectifier: draw vo, io, iD1, iD2, is for (a) pure R, (b) very high L.
Handwritten answer: (a) correct; (b) drew vo as a flat line, wrong: vo keeps the full-wave humps, only io is flat.
Half-wave rectifier, the diode is drawn reversed (conducts in the negative half). Diode ideal except ron.
| Supply | vs = 110 sin ωt V |
|---|---|
| Load | R = 10 kΩ, ron = 10 Ω |
FIND sketch vo; (a) Vavg (b) ID,rms (c) input PF
Answer: Vavg = −34.98 V · ID,rms = 5.49451 mA · PF = 0.707
Handwritten: Vdc = −34.979 V ✓; Irms,D = 54.945 A ✗ (forgot ÷R; it is 5.49 mA); PF = 0.706 (load-power version, 0.7064; fine here because ron ≪ R).
HVDC link (power flows left → right): what is block 1 (best choice) and its function? What is block 2? Draw block 1's output voltage and current.
Block 1: a rectifier, best a three-phase fully controlled thyristor bridge (6-pulse; in practice two of them make 12 pulses). It converts the AC power to controllable DC for the line (Vdc set by α). Block 2: an inverter (the same bridge with α > 90°), converting the DC back to AC at the receiving end. Waveforms: the 6-pulse output (humps of the line voltage) and an almost flat current (the line and smoothing reactors are highly inductive).
Handwritten answer: "full-wave bridge rectifier … inverter": accepted; mention three-phase and controlled for full marks.
Mid 2023/2024 "DU" (student-solved paper)
Half-wave, diode reversed, ideal except ron.
| Supply | vs = 110 sin 314t V |
|---|---|
| Load | R = 100 Ω, ron = 10 Ω |
FIND sketch vo; Vo; Irms,D; PF; (d) what happens to Vo if the diode needs 1 ms to turn on?
Answer: Vo = −31.83 V · ID,rms = 0.5 A · PF = 0.707 · (d) Vo = −31.05 V
Student's PF = 0.6428 used only the load power (25 W). The input power is 27.5 W (the diode resistance also absorbs power from the source), so the input PF is 0.707. If unsure, show both lines.
Q2 C-filter, very low / very high C: see 2022 Q3. Q3 effect of FWD on output power: see Lecture 3 (student's "Pac decreases" is wrong). Q4–Q5 merits, real vs ideal: Lecture 1. Q6 centre-tap R and L≫R: Lecture 4 (student's answer correct, including flat io and humped vo). Q7 HVDC with power flowing right → left: now block 2 is the rectifier and block 1 the inverter.
Mid 2023/2024 "MU" (student-solved paper)
Q1 Controlled half-wave without a thyristor: Lecture 6 exam question. Q2 Turn-on time 1 ms, α = 0° and 90°: Vo = −31.05 V and −11.00 V (student ✓; solved in Lecture 6). Q3 = 2022 paper. Q4 FWD on a controlled full-wave: Vdc rises from (2Vm/π)cos α to (Vm/π)(1 + cos α), load power rises (student's "Pac decreases" ✗). Q5 Industry: controlled (adjustable power), full-wave bridge (higher average, less ripple), three-phase for high power, with FWD (continuous current, no negative voltage) (student ✓). Q6 Centre-tap with D1 burnt (open): it becomes a half-wave rectifier using only D2:
Dec 2024 · Dr. Hamouda · 36 marks · 45 minutes
Q1–Q2 [4]: Lecture 1. Q3 [6] bridge + C, three capacitor sizes:
Q4 [4] controlled high-power industrial rectifier: three-phase fully controlled bridge (see Lecture 7 exam question); draw 6-pulse segments starting α late and a flat current. Q5 [8] centre-tap R / L≫R: Lecture 4 (handwritten ✓).
| Supply | vs = 110 sin ωt V |
|---|---|
| Load | R = 10 Ω, ron = 1 Ω |
FIND sketch vo; Vavg; ID,rms; input PF
Answer: 31.83 V · 5 A · PF = 0.707
Handwritten: Io,av = 3.183 A ✓, Vo,av = 31.83 V ✓, Irms = 5 A ✓ (but the formula written, Vm/2√2, is wrong; it is Im/2), PF = 0.643 (load power 250 W only); the input PF counting ron is 0.707.
Q7 [4] double-polarity output: single-phase fully controlled full-wave (bridge/centre-tap), three-phase half-wave controlled, three-phase fully controlled bridge, all with continuous inductive current and α > 90° (handwritten ✓). Q8 [4] three-phase half-wave controlled, α = 90°: this is Lecture 8, outside your scope; answer in the appendix.
Theory quick-fire
Read them aloud the night before.
Merits of power electronic systems (any 3–6)
High efficiency; high reliability; long life, less maintenance (no moving parts); flexibility; fast dynamic response; small size and weight, low installation cost.
Demerits (any 3–4)
Harmonics in supply and load; low input PF under some conditions (AC-DC, AC-AC); regeneration of power is difficult; low overload capacity.
Goal of a power electronic system
Convert energy from source to load with highest efficiency, availability and reliability, lowest cost, smallest size and least weight.
Ideal switch
ON: zero voltage, infinite current capability. OFF: zero current, infinite voltage capability. Switches instantly, zero power loss, bidirectional current and voltage (four quadrants).
Real switch: how it differs
On-state voltage and finite max current; leakage current and finite max voltage when off; finite switching time; so conduction loss + switching loss → heating.
The two losses
Conduction loss (ON and OFF states) and switching loss (during transitions).
Three classes of switches
Uncontrolled (diode); semi-controlled (SCR, TRIAC: on by gate, off by circuit); fully controlled (BJT, MOSFET, IGBT, GTO, IGCT, MCT).
Conditions to turn an SCR on
Forward biased; positive gate current; anode current above the latching current.
Latching vs holding current
Latching: minimum anode current during turn-on to stay on after the gate is removed. Holding: minimum anode current below which the SCR turns off.
How is an SCR turned off?
Not by the gate. Its current must fall below IH: natural (line) commutation when the AC reverses, or forced commutation by an auxiliary circuit.
Converter classes
AC-DC rectifier; DC-AC inverter; DC-DC chopper; AC-AC converter (level and/or frequency).
Applications of rectifiers
Variable-speed DC drives, battery chargers, DC power supplies, electroplating, HVDC transmission.
Bridge vs centre-tap
Bridge: PIV = Vm → high-voltage use, no special transformer. Centre-tap: isolation, one diode drop → low-voltage high-current use; PIV = 2Vm.
Why use a freewheeling diode?
Removes the negative part of vo with inductive loads: higher Vdc, continuous smoother current, main diode/SCR relieved, zero supply current during freewheeling (better PF).
Effect of a capacitor filter, and of increasing C
Smooths vo toward Vm; larger C → less ripple, shorter diode conduction, higher peak diode current.
PF of a rectifier
PF = P/S = mean input power ÷ (Vs,rms Is,rms). Not cos θ of the load.
When is the controlled bridge current continuous?
When α ≤ θ = tan−1(ωL/R). Then Vdc = (2Vm/π) cos α.
Rectifier vs inverter mode
Fully controlled with continuous current: α < 90° → Vdc > 0 (rectifier); α > 90° → Vdc < 0 (inverter, power flows back to AC; needs a DC source like a motor's back-emf).
Ripple frequency of each rectifier (50 Hz)
Half-wave 50 Hz; single-phase full-wave 100 Hz; 3-pulse 150 Hz; 6-pulse 300 Hz.
Standard figures (R load)
HW: Vdc = 0.318Vm, FF 1.57, RF 1.21, η 40.5%, TUF 0.287. FW: 0.637Vm, FF 1.11, RF 0.48, η 81.06%, TUF 0.81. 3-pulse 0.827Vm; 6-pulse 1.654Vm.
TRIAC, GTO, IGBT in one line each
TRIAC: two antiparallel SCRs, one gate, AC dimmers. GTO: SCR turned off by a large negative gate current, highest power. IGBT: MOSFET gate + BJT output, fast and low conduction loss, motor drives.
Mistakes found in the slides, sheets and answers
| # | Where | What it says | Correct |
|---|---|---|---|
| 1 | Lect 2 slide 23 | Irms = 0.457 A, P = 21 W, PF = 0.65 | 0.470 A, 22.1 W, 0.665 by the slide's own formula (exact 0.474 A, 22.4 W, 0.670) |
| 2 | Lect 2 sl. 21, Lect 6 sl. 10, Lect 7 sl. 19 | Irms = Vrms/Z | An approximation (error 1%–9% in the examples); exact needs ∫i². Use it in the exam, know it is approximate |
| 3 | Lect 3 slide 15 | iC = −0.339 sin ωt e… (diode off) | −0.339 e−(ωt−1.63)/15.708 A (no sin ωt) |
| 4 | Lect 3 slide 17 | Vrms with the bracket squared | √[Vm²/2·(½ − α/π + sin2α/2π) + E²(½ + α/π)] |
| 5 | Lect 3 slide 17 | Irms formula with sin α, sin((π−α)/2) | (1/R)√[Vm²(¼ − α/2π + sin2α/4π) − 2VmE cos α/π + E²(½ − α/π)] |
| 6 | Lect 4 slides 11–12 | "bridge rectifier" text and iD3, iD4 labels on the centre-tap slides | Centre-tap has only D1, D2 (copy-paste slip) |
| 7 | Lect 5 slides 4–5 | "Vb = 1", "Vc = 1" | Vm |
| 8 | Lect 5 slide 13 | Six-phase ID,rms = 0.396Im = Irms/√3 | 0.390Im = Irms/√6 |
| 9 | Lect 6 slide 11 | Irms 3.49 A, P 243.6 W, PF 0.581 | Lecture formula OK; exact 3.41 A, 232 W, 0.568 |
| 10 | Lect 7 slide 19 | Idc, Irms integrals with 1/(2π) | 1/π (period π) |
| 11 | Lect 7 slide 20 | Irms 9.31 A, P 866.8 W, PF 0.776 | Lecture formula OK; exact 8.76 A, 768 W, 0.730 |
| 12 | Lect 7 slide 22 | Continuous: Irms = Vm/(√2 Z) | Wrong (smaller than Idc). Use Irms ≅ Idc (as slide 25 does) or the exact value |
| 13 | Lect 7 slide 25 | P = Idc²R = 291.8 W | Approximation; exact 313.6 W (ωL = 31.4 Ω is not ≫ 10 Ω) |
| 14 | Sheet 1 P5 | "Figure (2)" | The figure printed is Figure 1 (same FWD circuit) |
| 15 | Sheet 2 P5, P11 | "3-4", "didoes" | 3-φ, diodes (typos) |
| 16 | 2021 key Q24, Q25 | 12.68 A, 17.93 A | 14.97 A, 21.17 A (no option is correct) |
| 17 | 2021 key Q20, Q21 | 4.77 A, 0.954 | 4.86 A, 0.971 (key = nearest option) |
| 18 | 2022 handwritten Q5(b) | ID,rms = 54.945 A | 5.49 mA |
| 19 | 2022 handwritten Q4(b) | vo flat for L ≫ R | vo = full-wave humps; only io is flat |
| 20 | 2022 / DU handwritten C-filter | vo decays below zero | vo ≥ 0, decays toward 0 |
| 21 | 2022 handwritten Q2 | Ideal ON: I = 0, V = ∞ | Ideal ON: V = 0, any I; OFF: I = 0, any V |
| 22 | DU Q1, 2024 Q6 handwritten | Input PF = 0.643 | 0.707 counting ron loss (0.643 is the load-only ratio) |
| 23 | DU Q3, MU Q4 handwritten | With FWD, Pac decreases | Load power increases slightly; Vrms of vo decreases |
| 24 | 2024 Q6 handwritten | Irms = Vm/2√2 | Im/2 = 5 A (number was right) |
| 25 | 2025 handwritten Q2.1(e) | PF = 0.6753 | 0.6766 (arithmetic) |
Appendix: Lecture 8 question from Dec 2024 (outside your scope)
Q8 [4]: three-phase half-wave controlled rectifier, draw vo for α = 90°. For three-phase converters α is measured from the natural commutation point π/6, so T1 fires at π/6 + 90° = 2π/3.
Built from your course files only. Exam scope: Lectures 1–7. Numbers verified by simulation and numerical integration.
About this guide
Sources used
Every slide and page was read, including slide equations stored as pictures and all scanned or handwritten pages.
| File | What it contains | In scope? |
|---|---|---|
| Lect 1 Introduction (.pptx, 37 slides) | Course info; what power electronics is; goals; advantages/disadvantages; converter types (AC-DC rectifier, DC-AC inverter, DC-DC chopper, AC-AC); applications; ideal vs real switch and the two losses; uncontrolled / semi-controlled / fully-controlled switches; diode types; SCR (turn-on conditions, latching/holding current, ratings, turn-off); TRIAC, GTO, IGCT, MCT, BJT, MOSFET, IGBT; GTO/IGCT/IGBT comparison table. | Yes |
| Lect 2 (3 versions, 24 slides) | Average value and rms value (graphical and by integration, 6 examples); single-phase uncontrolled half-wave rectifier with R load (example 120 V, 5 Ω) and R-L load (derivation of i(ωt), extinction angle β, example R=100 Ω, L=0.1 H). | Yes |
| Lect 3 (.pptx + .pdf, 20 slides) | Half-wave with freewheeling diode (example), with capacitor filter (θ, α, ripple ΔVo, peak diode current, example 120 V / 500 Ω / 100 µF), with resistive load + back-emf E (battery charging), and a quiz (R=50 Ω, L=0.1 H, ω=200). | Yes |
| Lect 4 (.pptx, 16 slides) | Single-phase uncontrolled full-wave: bridge with R load and with highly inductive load; centre-tapped transformer rectifier (R and L≫R); bridge vs centre-tap comparison; a 10-minute in-lecture quiz. | Yes |
| Lect 5 (.pptx, 17 slides) | Three-phase uncontrolled: half-wave (3-pulse) with diode-voltage analysis; full-wave bridge (6-pulse); six-phase star rectifier; output evaluation (FF, RF, TUF, CF); quiz (24 Ω, 240 V, bridge vs centre-tap). | Yes |
| Lect 6-7 (.pptx + .pdf, 35 slides) | Lect 6: single-phase controlled half-wave (R, R-L) and controlled full-wave bridge / centre-tap with R load (4 examples). Lect 7 (originally Dr. Hamouda's 2022-23 slides): controlled full-wave with R-L load, discontinuous and continuous current, highly inductive load, R-L + back-emf, rectifier/inverter regions, half-controlled bridge with FWD. | Yes |
| Lect 8 (.pptx + .pdf, 30 slides) | Three-phase controlled rectifiers (half-wave and bridge, R and R-L, α ranges, rectifier/inverter). | No (after Lect 7) |
| Lect 9-10 Choppers, Lect 11 Choppers commutation ("3elec D") | DC-DC choppers: step-down/up, control strategies, chopper classes A–E; Lect 11 is titled Power Electronics (2), ELE233 by Dr. Elgohary: forced/load commutation of thyristor choppers. | No, see below |
| PE Mid 25-26 Dr Elgohary (2 scanned pages) | Last year's midterm (Dec 2025, 20 marks, 1 h) with a student's handwritten answers. | Key file |
| PE Dr Hamouda (18 pages) | Not one exam but five: p.1–2 Dec 2021 midterm (25 MCQ, Dr. Abdelhady Ghanem, answers highlighted); p.3–6 Dec 2022 midterm, blank (Dr. Hamouda); p.7–8 "Mid 2023/2024 DU" solved by a student; p.9–10 "Mid 2023-2024 MU", another version, solved; p.11–14 Dec 2022 midterm solved by hand; p.15–18 Dec 2024 midterm (Dr. Hamouda, 36 marks) solved by hand. | Key file |
| Sheet 1 (5 pages) | 11 problems: single-phase uncontrolled half-wave (R, R-L, FWD, L≫R), full-wave (R, R-L, L≫R), capacitor filter. | Yes |
| Sheet 2 (4 pages) | 11 problems: three-phase 3-pulse and 6-pulse with R and highly inductive loads. | Yes |
| Section 3 (3 scanned whiteboard pages) | Page 1: capacitor-filter worked method (θ, α, ripple, IC,peak, Is,peak, for 120 V, 500 Ω, 100 µF). Page 2: battery-charger (R + E) formulas and waveform. Page 3: unrelated scribbles (thermodynamics energy equation and a bridge sketch): ignore. | Yes |
The three versions of Lecture 2
All 24 slides of the three files were compared slide by slide (text, equations and number of pictures). They have the same content. The only differences:
- Lect 2.pptx: the editable original, saved 1 Oct 2026 by Dr. Elgohary, title slide says 2026-2027. Its equations are pictures.
- Lect 2.pdf: a PDF export of the same slides made the same day (title slide still says 2025-2026). Equations are real text, so it is the easiest to read.
- Lect 2 (Telegram copy).pdf: last year's export (30 Oct 2025). Identical slides; only the way two equations are stored differs.
So none is "more complete". This guide follows the .pptx (this year's) and uses the PDF to read the equations. The same is true of Lect 3, 6-7 and 8: the .pdf and .pptx versions match.
Are Lectures 9–11 ("3elec D") part of your course?
Not for this midterm. "3elec D" means third-year electrical. Lect 11's title slide says Power Electronics (2), ELE233, a different course. Lect 9-10 (choppers) carry the ELE221 header, and choppers are in your course description (Lect 1, slide 4: "DC choppers + applications"), so they are probably taught after the midterm. The midterm covers Lectures 1–7, so choppers and Lect 8 are left out. One past question needed Lect 8; it is answered briefly in the appendix.
How the numbers were checked
- 356 different numerical results on this page are inserted straight from the verification program's output, so no number was typed by hand twice.
- Every average, rms, FF, RF, η, TUF, PF and PIV was recomputed by integrating the actual waveform numerically (200 000–600 000 points per period), not by re-using the slide formula.
- Every R-L, R-L + FWD, controlled and three-phase R-L answer was confirmed by a time-step simulation of the circuit (diode/SCR switching, 60 cycles to steady state). Simulated and integrated currents agree to better than 0.1%.
- Extinction angles β and capacitor angles α were solved with a bracketed root finder; the capacitor ripple was also confirmed by simulating the RC-diode circuit.
- The three-phase factors (0.827, 0.8407, 1.654, 1.6554, 0.485, 0.552, 0.390) were confirmed by integrating max-of-phases waveforms.
- Every waveform figure is drawn from the same computed data, to scale (angles, β, α, θ and amplitudes are exact).
| Result | Count |
|---|---|
| Question cards solved on this page (lectures, sheets, exams) | 47 + 25 MCQ |
| Slide and sheet mistakes | 15 |
| Answer-key mistakes (2021) | 2 wrong + 2 approximate |
| Handwritten-answer mistakes | 8 |
| Lecture example answers confirmed exactly (β, Idc, Vdc, α, ΔVo, ID,peak, C, all R-load results) | all |
Labelled additions (not in your slides): the exact R-L bridge current (Sheet 1 P8), the Fourier series (Sheet 1 P4), the full-wave ripple 2f (Sheet 1 P11), the full-wave battery formula (2021 Q22). Everything else uses only the doctors' formulas and symbols.