Diode Rectifiers

One diode, resistive load

Chapter 2 is rectifiers: ac of fixed magnitude and frequency (50, 60 or 400 Hz) in, dc out, with solid-state devices and no rotating machine (slide 2). A diode rectifier gives a fixed dc output, because nobody tells a diode when to turn on. An SCR rectifier gives an adjustable one, because you choose when each thyristor fires. That comes later.

Average output voltage (slide 3)

1. Average = area over one period. In angle, $T = 2\pi$, and $v_o = v_s$ only from 0 to $\pi$: $$V_{o,avg} = \frac{\text{Area } A_1}{T} = \frac{1}{2\pi}\int_0^{\pi}\sqrt{2}V_s\sin(\omega t)\,d\omega t$$
2. $$= \frac{\sqrt{2}V_s}{2\pi}\Big[-\cos(\omega t)\Big]_0^{\pi} = \frac{\sqrt{2}V_s}{2\pi}\big(1 - (-1)\big) = \frac{\sqrt{2}V_s}{2\pi}\cdot 2$$ The step the slide skips: $-\cos\pi = +1$ and $+\cos 0 = +1$, so the bracket is 2.
3. $$\boxed{V_{o,avg} = \frac{\sqrt{2}}{\pi}V_s = 0.45V_s}$$ Dr. Xu circled the 0.45 and labelled $V_{o,avg}$ "DC side". Not on your formula sheet.

Add an inductor: the diode stays on past $\pi$

Put L in series with R (Dr. Xu: "RL load"). An inductor's current can't jump. When $v_s$ crosses zero at $\pi$, current is still flowing, so the diode stays on and $v_o$ follows $v_s$ below zero. It lets go only when L has given back all its energy and $i_s$ reaches zero, at an angle $\theta > 180^\circ$. Step through slide 5's four periods.

Average output voltage (slide 4)

1. $v_o = v_s$ for the whole conduction interval, 0 to $\theta$, including the negative piece from $\pi$ to $\theta$: $$V_{o,avg} = \frac{\text{Area } A_1 - \text{Area } A_2}{T} = \frac{1}{2\pi}\int_0^{\theta}\sqrt{2}V_s\sin(\omega t)\,d\omega t$$ Dr. Xu circled the upper limit: $\theta$, "not $\pi$". One integral to $\theta$ subtracts $A_2$ by itself, because the sine is negative past $\pi$.
2. $$= \frac{\sqrt{2}V_s}{2\pi}\Big[-\cos(\omega t)\Big]_0^{\theta} \;\Rightarrow\; \boxed{V_{o,avg} = \frac{\sqrt{2}}{2\pi}V_s(1 - \cos\theta)}$$ Check: $\theta = \pi$ gives $0.45V_s$ again. Any $\theta > 180^\circ$ makes $1 - \cos\theta < 2$, so the inductor lowers the average. In the example, $\theta = 225.8^\circ$ gives 45.84 V instead of 54.02 V. Not on your formula sheet.
3. And $\theta$ itself? Dr. Xu's red note: "$\theta$: no formula". It is where the RL current returns to zero, an equation with a sine and an exponential in it. You get it by trial and error or simulation, then use the formula above. An exam would give you $\theta$, or ask which way it moves.

Harmonics, THD and power factor

Zoom out. A sinusoidal ac source, in steady state (both are Dr. Xu's margin notes), feeds a nonlinear circuit such as a rectifier. The voltage stays a sine wave; the current doesn't. Slide 6 sketches the typical culprit: narrow pulses near the voltage peaks. Fourier splits that current into sines:

Each $I_{sk}$ is an rms value, which is why every term carries $\sqrt{2}$ (circled in class). Different frequencies are orthogonal, so their cross terms average to zero and rms values add in squares: $I_s = \sqrt{I_{s1}^2 + I_{s2}^2 + I_{s3}^2 + \dots}$. Total harmonic distortion compares everything that isn't the fundamental with the fundamental: $$\text{THD} = \frac{\sqrt{I_s^2 - I_{s1}^2}}{I_{s1}} = \frac{\sqrt{I_{s2}^2 + I_{s3}^2 + \dots}}{I_{s1}}$$

Which current does the work? (slide 7)

1. Average power with $v_s = \sqrt{2}V_s\sin\omega t$: $$P = \frac{1}{2\pi}\int_0^{2\pi}\sqrt{2}V_s\sin(\omega t)\left[\sum_{k=1}^{\infty}\sqrt{2}I_{sk}\sin(k\omega t + \varphi_k)\right]d\omega t$$
2. The step the slide skips. Product to sum: $$\sin(\omega t)\sin(k\omega t + \varphi_k) = \tfrac{1}{2}\big[\cos((k-1)\omega t + \varphi_k) - \cos((k+1)\omega t + \varphi_k)\big]$$ For $k \ge 2$ both cosines run through whole cycles in $2\pi$, so they integrate to zero.
3. Only $k = 1$ survives, and its first cosine is the constant $\cos\varphi_1$: $$P = \frac{1}{2\pi}\cdot 2V_sI_{s1}\cdot\tfrac{1}{2}\cos\varphi_1\cdot 2\pi$$
4. $$\boxed{P = V_sI_{s1}\cos\varphi_1}\qquad\boxed{S = V_sI_s}\ \text{(rms values)}$$ Class notes: P is "real power" in W, kW, MW, and only $I_{s1}$ contributes to it. S is the "total capacity" in VA, kVA, MVA. "Harmonic currents occupy capacity" and cause extra losses in lines and transformers. Neither P nor S is on your formula sheet.

Power factor (slide 8)

1. $$\text{PF} = \frac{P}{S} = \frac{V_sI_{s1}\cos\varphi_1}{V_sI_s} = \frac{I_{s1}}{I_s}\cos\varphi_1 = \text{DF}\cdot\text{DPF}$$ DPF $= \cos\varphi_1$, the displacement power factor, uses the angle between $v_s$ and the fundamental current (Dr. Xu wrote "fundamental" over $I_{s1}$). DF $= I_{s1}/I_s \le 1$ is the distortion factor.
2. Square the THD definition: $$\text{THD}^2 = \frac{I_s^2}{I_{s1}^2} - 1 \;\Rightarrow\; \frac{I_{s1}}{I_s} = \frac{1}{\sqrt{1 + \text{THD}^2}}$$ Class note: THD down means $I_{s1}/I_s$ up, so DF up.
3. $$\boxed{\text{PF} = \frac{1}{\sqrt{1 + \text{THD}^2}}\cos\varphi_1}$$ DF, THD and both PF forms are on page 2 of your formula sheet, under "Power Factor, THD and distortion factor".

Single-phase bridge: a square-wave line current

Four diodes let both half cycles reach the load the same way round. Dr. Xu's margin: the load can be R, RL or capacitive (RC). Flip the source polarity and watch which pair conducts; switch the load between slide 9's R and slide 10's large L.

Dr. Xu's reasoning from class: for $v_s \ge 0$, $v_{D1}, v_{D2} \ge 0$, so $D_1$ and $D_2$ turn on, while $v_{D3}, v_{D4} < 0$ keep $D_3$ and $D_4$ off. For $v_s < 0$ they swap. Either way the load sees $v_d = |v_s|$, and each diode conducts $180^\circ$.

1. $v_d = |v_s|$ repeats every $\pi$, so average one hump: $$V_d = \frac{\text{Area } A_1 + \text{Area } A_2}{T} = \frac{1}{\pi}\int_0^{\pi}\sqrt{2}V_s\sin(\omega t)\,d\omega t$$
2. $$= \frac{\sqrt{2}V_s}{\pi}\cdot 2 = \boxed{\frac{2\sqrt{2}}{\pi}V_s = 0.90V_s}$$ Twice the single diode's value. On your formula sheet (single-phase, inductive load), and it holds for an R load too.

With a pure R the line current is $i_s = v_s/R$ in both halves: a sine in phase with $v_s$, PF = 1. The distortion comes from the load. Slide 10's second assumption: L is very large, so $i_d$ is constant ("constant current", in red). Replace the load by a current source $I_d$. Dr. Xu sketched the real $i_d$ with a small ripple; the formulas use the flat one.

1. The diodes still switch at the zero crossings, so $v_d = |v_s|$ and $V_d = 0.90V_s$ is unchanged.
2. $i_s = +I_d$ while $D_1$, $D_2$ conduct and $-I_d$ while $D_3$, $D_4$ do: a square wave in phase with $v_s$. Its Fourier series (slide 11): $$i_s = \frac{4}{\pi}I_d\left(\sin\omega t + \frac{1}{3}\sin 3\omega t + \frac{1}{5}\sin 5\omega t + \dots\right)$$ The skipped step: $b_k = \frac{2}{\pi}\int_0^{\pi}I_d\sin k\omega t\,d\omega t = \frac{2I_d}{k\pi}(1 - \cos k\pi)$, which is $\frac{4I_d}{k\pi}$ for odd $k$ and 0 for even $k$. Class note: "3rd, 5th, 7th: odd order harmonics. AC: no even order."
3. The fundamental's peak is $I_{s1,peak} = \frac{4}{\pi}I_d$, so its rms is $$I_{s1} = \frac{4}{\sqrt{2}\,\pi}I_d = \frac{2\sqrt{2}}{\pi}I_d = 0.9I_d$$ and the total rms is $I_s = \sqrt{\frac{1}{T}\int_0^T I_d^2\,dt} = I_d$, because $i_s^2 = I_d^2$ at every instant (slide 12).
4. $$\text{THD} = \frac{\sqrt{I_d^2 - \frac{8}{\pi^2}I_d^2}}{\frac{2\sqrt{2}}{\pi}I_d} = \sqrt{\frac{\pi^2}{8} - 1} = 0.4834$$ $I_d$ cancels: every square wave has 48.34% THD, whatever its height.
5. $i_{s1}$ is in phase with $v_s$, so $\varphi_1 = 0$ and DPF $= \cos 0^\circ = 1$. DF $= \frac{I_{s1}}{I_s} = \frac{2\sqrt{2}}{\pi} = 0.9$, so $$\boxed{\text{PF} = \text{DF}\cdot\text{DPF} = 0.9}$$

Three-phase bridge: a six-step line current

Six diodes: $D_1$, $D_3$, $D_5$ on top and $D_4$, $D_6$, $D_2$ below. With $I_d$ flowing, the top diode whose anode sees the highest phase voltage conducts, and the bottom diode whose cathode sees the lowest. So $v_d$ is always the largest line-to-line voltage: six humps per cycle, a "six-pulse" rectifier.

1. Each hump of $v_d$ is the slice of a line-to-line sine, peak $\sqrt{2}V_{LL}$, from $60^\circ$ to $120^\circ$, and the pattern repeats every $\pi/3$.
2. $$V_{d,avg} = \frac{1}{\pi/3}\int_{\pi/3}^{2\pi/3}\sqrt{2}V_{LL}\sin\theta\,d\theta = \frac{3}{\pi}\sqrt{2}V_{LL}\left(\tfrac{1}{2} + \tfrac{1}{2}\right) = \boxed{\frac{3\sqrt{2}}{\pi}V_{LL} = 1.35V_{LL}}$$ $V_{LL}$ is the rms line-to-line voltage. On the sheet.
3. Each diode conducts $120^\circ$. Phase a: $D_1$ carries $+I_d$ while $v_{an}$ is highest ($30^\circ$ to $150^\circ$), $D_4$ carries $-I_d$ while it is lowest ($210^\circ$ to $330^\circ$), and $i_a = 0$ in the two $60^\circ$ gaps. The rms sees $I_d^2$ for $240^\circ$ of $360^\circ$: $$I_{a,rms} = \sqrt{\tfrac{2}{3}}\,I_d = 0.816I_d\quad\text{(on the sheet)}$$
4. From the sheet's series, $$i_a = \frac{2\sqrt{3}}{\pi}I_d\left[\sin\omega t - \frac{1}{5}\sin 5\omega t - \frac{1}{7}\sin 7\omega t + \dots\right]$$ the fundamental's peak is $\frac{2\sqrt{3}}{\pi}I_d$, so its rms is $$I_{a1} = \frac{2\sqrt{3}}{\sqrt{2}\,\pi}I_d = \frac{\sqrt{6}}{\pi}I_d = 0.780I_d$$ Not on the sheet. Same move as single phase: peak off the series, divide by $\sqrt{2}$.
5. DF $= \frac{I_{a1}}{I_{a,rms}} = \frac{\sqrt{6}/\pi}{\sqrt{2/3}} = \frac{3}{\pi} = 0.955$. $i_{a1}$ is centred on the peak of $v_{an}$, so DPF = 1 and $$\boxed{\text{PF} = \frac{3}{\pi} = 0.955},\qquad \text{THD} = \sqrt{\frac{\pi^2}{9} - 1} = 31.1\%$$
6. Harmonics: no even ones (half-wave symmetry) and no multiples of 3 (the $60^\circ$ gaps cancel them). Only $h = 6n \pm 1$: 5, 7, 11, 13, ..., each $I_{ah} = I_{a1}/h$. Better than single phase on every count: PF 0.955 against 0.9, THD 31% against 48%, and the dc ripple at six times line frequency instead of two.

Capacitive load Preview

Phone chargers, PCs and LED drivers put a big capacitor right after the bridge. In the exam's Figure 3a the source feeds the bridge through $L_s$ (line plus transformer inductance), and C holds the dc side at a nearly constant $V_d$.

1. Start. The capacitor holds the bridge output at $V_d$, so $D_1$, $D_2$ stay reverse biased until $v_s$ climbs to $V_d$: $$\sqrt{2}V_s\sin\theta_b = V_d \;\Rightarrow\; \theta_b = \arcsin\frac{V_d}{\sqrt{2}V_s}$$
2. While they conduct, KVL around source, $L_s$ and capacitor gives $L_s\frac{di_d}{dt} = \sqrt{2}V_s\sin\omega t - V_d$. With $\theta = \omega t$, $dt = d\theta/\omega$: $$\omega L_s\frac{di_d}{d\theta} = \sqrt{2}V_s\sin\theta - V_d$$
3. Integrate from $\theta_b$, where $i_d = 0$: $$\boxed{i_d = \frac{1}{\omega L_s}\Big[\sqrt{2}V_s(-\cos\theta + \cos\theta_b) - V_d(\theta - \theta_b)\Big]}$$ The formula-sheet line. $\theta - \theta_b$ came from integrating $d\theta$, so it must be in radians.
4. Peak. $di_d/d\theta = 0$ where $v_s$ falls back to $V_d$. By the symmetry of the sine, $$\theta_p = \pi - \theta_b$$ The current peaks after the voltage peak: it keeps rising as long as $v_s - V_d > 0$.
5. End. After $\theta_p$ the current falls, but the diodes stay on until it reaches zero at $\theta_f$: $$i_d(\theta_f) = 0 \iff \int_{\theta_b}^{\theta_f}\big(\sqrt{2}V_s\sin\theta - V_d\big)\,d\theta = 0$$ The area where $v_s > V_d$ equals the area where $v_s < V_d$ (red and yellow on the bench below). $\theta_f$ sits inside a cosine and on its own, so there is no closed form: trial and error.

Why the line current is so bad

The load draws current all the time, but the capacitor can only be refilled between $\theta_b$ and $\theta_f$: $46^\circ$ of every $180^\circ$ in the exam problem. The same charge in a much shorter time comes as a tall, narrow pulse, full of 3rd, 5th and 7th harmonics. For the exam values: THD about 132%, PF about 0.59, crest factor about 3.

Fixing it (the F2025 solution's three methods)

1. Passive or active power filter. Add L or LC on the ac side of the rectifier. More inductance makes the pulse lower and wider; an active filter supplies the harmonic currents itself so the line doesn't have to.
2. Zig-zag transformer, or a multi-phase rectifier. Phase-shifted windings make the harmonics of different rectifiers cancel, and a zig-zag winding gives the triplen currents a local path that keeps them out of the upstream transformer and neutral.
3. Power factor correction (PFC) circuit. An active front end, usually a boost converter right after the bridge (Dr. Xu sketched one on slide 10), forces $i_s$ to follow the shape of $v_s$: THD near 0, PF near 1.

Your sheet: what's there, what isn't

Open it with the button at the top. Everything this lesson takes from the sheet sits in three boxes on page 2. The first list is what you can copy; the second is what you must carry in your head.

F2025 midterm, Question 3 (25%)

Work it like the real thing: Open the formula sheet (this year's, the one you'll get on Oct 20), try each part on paper in your notes, then reveal.

F2025 final, Question 3, parts 2 and 3 (10%)

Traps that cost marks

1. Using the peak where the formula wants rms. $V_s$ and $V_{LL}$ are rms; $0.45V_s$, $0.90V_s$ and $1.35V_{LL}$ already contain the $\sqrt{2}$.
2. Assuming $\theta = 180^\circ$ with an RL load. It's always more, so $V_{o,avg} = \frac{\sqrt{2}}{2\pi}V_s(1 - \cos\theta)$ is below $0.45V_s$.
3. Taking the Fourier coefficient as rms. $\frac{4}{\pi}I_d$ and $\frac{2\sqrt{3}}{\pi}I_d$ are peaks: $I_{s1} = 0.9I_d$ and $I_{a1} = 0.780I_d$.
4. THD divided by $I_s$. It's divided by $I_{s1}$.
5. PF $= \cos\varphi_1$ for a distorted current. That's only DPF; PF $=$ DF $\cdot$ DPF.
6. Counting harmonics as power. $P = V_sI_{s1}\cos\varphi_1$ uses only the fundamental; $S = V_sI_s$ uses everything.
7. Three-phase with single-phase numbers (0.9, 48%), or $I_{a,rms} = 0.816I_d$ used as the fundamental.
8. Degrees in $V_d(\theta - \theta_b)$. That term needs radians; only the cosines care about your calculator's mode.
9. Putting $\theta_p$ at $90^\circ$, or ending the pulse at $\theta_p$. The peak is at $\pi - \theta_b$; the pulse ends at $\theta_f$, where the areas balance.
10. No units, no marks: the midterm cover says "Marks will be lost if there is no appropriate units." Give angles in rad and degrees.