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ELE754 Power Electronics · Lesson 3 of 5

Chopper Analysis

Take the buck's switch and diode, drop the capacitor, and drive an R-L load or a dc motor directly. Nothing holds the load voltage steady now, so the current bends into exponentials with time constant $\tau = L/R$. The chopper still fixes the average. $\tau$ decides the rest: the ripple, whether the current runs out, and how much of each harmonic gets through.

Sept 22 lecture, slides 37 to 61 Mohan 3-2-4-1, 13-7-3 Midterm Oct 20, 8 AM
How to do this lesson · about 75 minutes, 80 at most
  1. 5 min Pre-questions on page 1 of your guided notes. Guess if you have to.
  2. 35 min Lecture mode: 13 minutes of narration, plus pauses to fill in boxes and answer checks.
  3. 30 min F2025 midterm Q2 and one harmonics problem, formula sheet open. The rest is extra.
  4. 5 min Last page of your notes, then "quiz me on the chopper" in chat.
Download guided notes

Open the notes in GoodNotes beside this page. When the narrator says "write that in B6", the matching answer on this page carries a B6 tag; F2 tags mark sketches.

τ/Ts = 0.5
D = 0.40

$V_d = 100$ V, $R = 10\ \Omega$, $T_s = 1$ ms. Slide the time constant: the bend and the ripple change, the average $I_{o,avg} = DV_d/R$ (cyan) does not. The dotted midpoint $(I_{\max} + I_{\min})/2$ only lands on it when $\tau \gg T_s$.

01 · slides 37 to 38

The chopper meets an R-L load

Slide 37 is the front half of the buck: $T_1$ (an IGBT) and the freewheeling diode $D_1$. Dr. Xu's margin note calls it a voltage chopper: its output $v_o$ is $V_d$ while $T_1$ is on and 0 while $D_1$ freewheels. There's no capacitor this time. The load is an inductor in series with a resistor, and he labelled it a first-order filter.

One-quadrant means $I_o > 0$ and $V_o > 0$ only, because $T_1$ and $D_1$ each conduct one way. The two-quadrant chopper on slide 62 adds negative current ($I_o = \pm$); that's next lecture.

  1. $T_1$ on, so $v_o = V_d$ (class note). KVL around the load: $V_d = L\frac{di_o}{dt} + Ri_o$. Starting at $I_{\min}$ and heading for $V_d/R$: $$i_o(t) = \frac{V_d}{R}\left(1 - e^{-t/\tau}\right) + I_{\min}e^{-t/\tau}, \quad 0 \le t < t_{on} \quad (1)$$
  2. $D_1$ on, so $v_o = 0$: $0 = L\frac{di_o}{dt} + Ri_o$. Starting at $I_{\max}$, heading for 0, clock restarted at $t_{on}$: $$i_o(t) = I_{\max}e^{-(t - t_{on})/\tau}, \quad t_{on} \le t < T_s \quad (2)$$
  3. Both come from one template, slide 38's note (Dr. Xu wrote "ELE202" beside it): $$\tau\frac{dx}{dt} = -x + b \;\Rightarrow\; x(t) = x_0e^{-t/\tau} + b\left(1 - e^{-t/\tau}\right), \qquad \tau = \frac{L}{R}$$ $x_0 = x(0)$ is where you start, $b = x(\infty)$ is where you're heading. Divide the on-time KVL by R and you get exactly this form with $b = V_d/R$. Class note: $I_{\min} > 0$. With an R-L load the current decays toward zero but never reaches it, so this load is always in continuous current mode.
Why not straight lines this time?

In the buck, C held $V_o$ constant, so $v_L$ was constant and $i_L$ was a ramp. Here $v_L = v_o - Ri_o$ moves as $i_o$ moves, so the slope keeps changing. Slide 38: "$i_o$ does not increase or decrease linearly!" When $\tau \gg T_s$ each interval only uses the first, nearly straight bit of the exponential. The hero demo at the top shows both cases.

Predict first

$T_1$ has just turned off and $D_1$ picks up the current. What does $i_o$ do?

02 · slide 39

$I_{\max}$ and $I_{\min}$ in steady state

Dr. Xu wrote "in steady state" twice on slide 39: every period starts where the last one ended. That gives two conditions for two unknowns.

  1. End of the on-time, from (1): $$i_o(t_{on}) = I_{\max} = \frac{V_d}{R}\left(1 - e^{-t_{on}/\tau}\right) + I_{\min}e^{-t_{on}/\tau} \quad (3)$$
  2. End of the period, from (2): $$i_o(T_s) = I_{\min} = I_{\max}e^{-(T_s - t_{on})/\tau} \quad (4)$$
  3. The step the slide skips: put (4) into (3). The two exponentials multiply to $e^{-T_s/\tau}$: $$I_{\max} = \frac{V_d}{R}\left(1 - e^{-t_{on}/\tau}\right) + I_{\max}e^{-T_s/\tau} \;\Rightarrow\; I_{\max}\left(1 - e^{-T_s/\tau}\right) = \frac{V_d}{R}\left(1 - e^{-t_{on}/\tau}\right)$$
  4. $$\boxed{I_{\max} = \frac{V_d}{R}\,\frac{1 - e^{-t_{on}/\tau}}{1 - e^{-T_s/\tau}}}$$
  5. Back into (4), then multiply top and bottom by $e^{T_s/\tau}$: $$I_{\min} = \frac{V_d}{R}\,\frac{\left(1 - e^{-t_{on}/\tau}\right)e^{-(T_s - t_{on})/\tau}}{1 - e^{-T_s/\tau}} \;\Rightarrow\; \boxed{I_{\min} = \frac{V_d}{R}\,\frac{e^{t_{on}/\tau} - 1}{e^{T_s/\tau} - 1}}$$ Both boxes are on your formula sheet, row "One quadrant chopper with RL load". Watch the signs: $I_{\max}$ has negative exponents, $I_{\min}$ positive ones.
03 · slides 40 to 42

Example II: an R-L load at 75%

A one-quadrant chopper with an R-L load: $V_d = 120$ V, $f_{sw} = 1$ kHz, $L = 4.5$ mH, $R = 1.5\ \Omega$, $D = 0.75$. Find (1) the current ripple $\Delta I$ and (2) the averages $I_o$ and $V_o$.

Try it on paper first, then reveal one step at a time.

  1. $\tau = \frac{L}{R} = \frac{4.5\text{ mH}}{1.5\ \Omega} = 3$ ms, $\;T_s = \frac{1}{f_{sw}} = 1$ ms, $\;t_{on} = DT_s = 0.75$ ms.
  2. $$I_{\max} = \frac{120}{1.5}\cdot\frac{1 - e^{-0.75/3}}{1 - e^{-1/3}} = 80\times\frac{0.2212}{0.2835} = 62.4\text{ A}$$
  3. $$I_{\min} = 80\times\frac{e^{0.25} - 1}{e^{1/3} - 1} = 80\times\frac{0.2840}{0.3956} = 57.4\text{ A}$$
  4. ΔI = I_max − I_min = 4.99 A (the slide prints 4.9 A; see the note below).
  5. $V_{o,avg} = DV_d = 0.75\times120 = $ 90 V, and $I_{o,avg} = \frac{V_{o,avg}}{R} = \frac{90}{1.5} = $ 60 A, because the inductor has no dc voltage drop.
Slide 41 rounds before it subtracts

Unrounded, $I_{\max} = 62.43$ A and $I_{\min} = 57.44$ A. The slide rounds $I_{\min}$ up to 57.5 A, so its $\Delta I = 62.4 - 57.5 = 4.9$ A; the true ripple is 4.99 A. A ripple is a small difference of two big numbers, so on the exam keep four figures until you subtract.

Two class notes on this example

Slide 41: the buck formula is crossed out

Next to step 5 Dr. Xu wrote $\frac{V_d - V_o}{L}t_{on}$ and put an X through it. That formula needs a constant voltage on the far side of L: the buck's capacitor, or a battery ("$V_{bat}$ = const"). Here the far side is $Ri_o$, which moves. Because $\tau = 3$ ms is three times $T_s$, the curves are nearly straight and the buck-style estimate $\frac{120 - 90}{4.5\text{ mH}}\times0.75$ ms $= 5.0$ A lands close, but it's an approximation. Use $I_{\max} - I_{\min}$.

Slide 42: which average is exact?

He drew the dc equivalent circuit: averaged over a period, $V_L = 0$, so L is a short for dc and $I_{o,avg} = V_{o,avg}/R$ exactly. The slide's second route, $I_{o,avg} \approx \frac{I_{\max} + I_{\min}}{2}$, he circled and marked "not equal": exponential segments aren't symmetric, so the midpoint is only close when R is very small ($\tau \gg T_s$). Here it's 59.93 A.

Predict first

A one-quadrant chopper drives an R-L load: $V_d = 200$ V, $D = 0.6$, $R = 2\ \Omega$, $L = 5$ mH, $f_{sw} = 2$ kHz. You find $I_{\max} = 62.38$ A and $I_{\min} = 57.59$ A. What is the average load current?

04 · slides 43 to 45

Add a back emf: the dc motor load

Slide 43 swaps the load for a dc motor's equivalent circuit: armature inductance L, armature resistance R and back emf E. Dr. Xu sketched where you'd meet it: a 12 V PC fan, chopped by a MOSFET with a freewheeling diode across the motor. He labelled E a constant voltage source. The motor's speed changes slowly compared with $T_s$, so over one period E doesn't move.

Every KVL loop now has an extra E (the red boxes Dr. Xu drew on slide 44): $v_o = L\frac{di_o}{dt} + Ri_o + E$.

  1. $T_1$ on: $V_d = L\frac{di_o}{dt} + Ri_o + E$, so the current heads for $\frac{V_d - E}{R}$: $$i_o(t) = \frac{V_d - E}{R}\left(1 - e^{-t/\tau}\right) + I_{\min}e^{-t/\tau} \quad (1)$$
  2. $D_1$ on: $0 = L\frac{di_o}{dt} + Ri_o + E$, so it heads for $-\frac{E}{R}$, which is below zero: $$i_o(t) = -\frac{E}{R}\left(1 - e^{-(t - t_{on})/\tau}\right) + I_{\max}e^{-(t - t_{on})/\tau} \quad (2)$$
  3. Same steady-state move as slide 39, and E comes out as one shift: $$I_{\max} = \frac{V_d}{R}\,\frac{1 - e^{-t_{on}/\tau}}{1 - e^{-T_s/\tau}} - \frac{E}{R}, \qquad I_{\min} = \frac{V_d}{R}\,\frac{e^{t_{on}/\tau} - 1}{e^{T_s/\tau} - 1} - \frac{E}{R}$$ Why a plain shift: the circuit is linear, so the response to the pulse train and the response to E add. A constant E on its own drives a constant $-E/R$ through the load.
  4. Averages (slide 44): $V_{o,avg} = \frac{\text{Area }A_1}{T_s} = DV_d$ still, and the dc circuit gives $$I_{o,avg} = \frac{V_{o,avg} - E}{R}$$ These four are the CCM column of the "RLE load" row on your formula sheet. Its $I_{\min}$ line ends in "$> 0$": that's the condition for using the column at all.

Example II's numbers with $E = 30$ V (magenta) against $E = 0$ (blue): the whole waveform drops by $E/R = 20$ A. The ripple stays 4.99 A and the average falls from 60 A to 40 A.

Predict first

Example II (120 V, 75%, 4.5 mH, 1.5 Ω) now drives a motor with $E = 30$ V. What happens to $I_{\max}$, $I_{\min}$ and $\Delta I$?

05 · slides 46 to 51

When the current runs out: DCM

Raise E (a lightly loaded motor spins faster) and $I_{\min} = (\text{R-L value}) - E/R$ keeps falling. The CCM formula happily returns a negative number, but the current can't follow: $T_1$ and $D_1$ only conduct one way. So the current reaches zero at some instant $t_x$ and stays there until $T_1$ turns on again. That's discontinuous current mode, with three intervals (slide 46):

Example III. Dashed: the curve the CCM formulas assume, which would dip to −84 A. The real current stops at zero at $t_x$, and from there the output sits at E.

Average output (slide 46): $$V_{o,avg} = \frac{\text{Area }A_1 + \text{Area }A_2}{T_s} = \frac{t_{on}}{T_s}V_d + \frac{T_s - t_x}{T_s}E \;\ne\; DV_d$$ Dr. Xu marked $A_1$ "due to $V_d$", $A_2$ "due to E", and $t_x$ "unknown". So DCM needs one more formula.

  1. Each period now starts at zero, so put $I_{\min} = 0$ in (1): $$I_{\max} = \frac{V_d - E}{R}\left(1 - e^{-t_{on}/\tau}\right) \quad (3)$$ Not the CCM $I_{\max}$. Once you know it's DCM, recompute it.
  2. At $t_x$ the current in (2) reaches zero: $$0 = -\frac{E}{R}\left(1 - e^{-(t_x - t_{on})/\tau}\right) + I_{\max}e^{-(t_x - t_{on})/\tau} \quad (4)$$
  3. The skipped algebra: collect the exponential terms. $$\left(I_{\max} + \frac{E}{R}\right)e^{-(t_x - t_{on})/\tau} = \frac{E}{R} \;\Rightarrow\; t_x - t_{on} = \tau\ln\left(1 + \frac{RI_{\max}}{E}\right)$$
  4. Use (3) for $RI_{\max}$ and fold $t_{on}$ into the log as $\tau\ln e^{t_{on}/\tau}$: $$\boxed{t_x = \tau\ln\left\{e^{t_{on}/\tau}\left[1 + \frac{V_d - E}{E}\left(1 - e^{-t_{on}/\tau}\right)\right]\right\}}$$ $t_x$ is measured from the start of the period, not from turn-off. The DCM column of your formula sheet has (3), $I_{\min} \equiv 0$, this $t_x$ and $V_{o,avg}$.
Mode check: two ways, both from class (slide 50)
  1. Assume CCM and compute $I_{\min}$ with the CCM formula. $I_{\min} > 0$: CCM. $I_{\min} < 0$: DCM.
  2. Or assume DCM and compute $t_x$. $t_x > T_s$: CCM, the current never ran out. $t_x < T_s$: DCM.
  3. If it's DCM, throw away the CCM $I_{\max}$ and $V_{o,avg} = DV_d$. Use $I_{\min} \equiv 0$, $I_{\max} = \frac{V_d - E}{R}\left(1 - e^{-t_{on}/\tau}\right)$, $t_x$, then $V_{o,avg}$ and $I_{o,avg} = \frac{V_{o,avg} - E}{R}$.

The two checks always agree; the bench computes both. With $E = 0$ the log blows up ($t_x \to \infty$): an R-L load is always CCM.

Example III (slides 49 to 51)

A one-quadrant chopper drives a dc motor: $V_d = 110$ V, $f_{sw} = 400$ Hz, $L = 0.2$ mH, $R = 0.25\ \Omega$, $D = 0.5$, $E = 40$ V. Find (1) the mode, (2) $I_{\max}$ and $I_{\min}$, (3) the averages $I_o$ and $V_o$.

  1. $\tau = 0.8$ ms, $T_s = 2.5$ ms, $t_{on} = 1.25$ ms. Assume CCM: $$I_{\min} = 440\times\frac{e^{1.5625} - 1}{e^{3.125} - 1} - 160 = 76.25 - 160 = -84\text{ A} < 0$$ so DCM.
  2. $I_{\max} = \frac{110 - 40}{0.25}\left(1 - e^{-1.5625}\right) = 280\times0.7904 = $ 221.3 A and $I_{\min} = 0$ (the slide prints 221.2 A).
  3. $t_x = 0.8\text{ ms}\times\ln\left\{e^{1.5625}\left[1 + \frac{70}{40}(0.7904)\right]\right\} = 0.8\text{ ms}\times\ln 11.37 = $ 1.945 ms, less than $T_s = 2.5$ ms: DCM confirmed the second way.
  4. $V_{o,avg} = \frac{1.25}{2.5}\times110 + \frac{2.5 - 1.945}{2.5}\times40 = 55 + 8.88 = $ 63.9 V (slide: 64 V).
  5. $I_{o,avg} = \frac{63.9 - 40}{0.25} = $ 95.5 A. The slide's rounded 64 V gives 96 A: dividing a small difference by a small R magnifies rounding, so carry the digits.
Predict first

You assume CCM for a chopper driving a motor, and the CCM formula gives $I_{\min} = -84$ A. What does that tell you?

Predict first

In DCM, between $t_x$ and $T_s$, what is $v_o$?

06 · the bench

Run the chopper yourself

Everything here comes from the slide formulas, so it doubles as an answer checker. Load a preset, then push it around: raise E until the badge flips to DCM, shrink L and watch the ripple grow, or watch which harmonics the load lets through.

Presets from the course
CCM

Harmonic spectrum, peak values (slides 56 to 61)

Show the numbers
h|Voh| (V)Zh (Ω)|Ioh| (A)rms (A)
07 · slides 52 to 54

Fourier series in one page

Any periodic waveform is a dc value plus sinusoids at whole multiples of its frequency. Slide 52's definitions, with $\omega = 2\pi/T$: $$f(t) = f_0 + \sum_{n=1}^{\infty}a_n\cos(n\omega t) + \sum_{n=1}^{\infty}b_n\sin(n\omega t)$$ $$f_0 = \frac{1}{T}\int_0^T f(t)\,dt, \qquad a_n = \frac{2}{T}\int_0^T f(t)\cos(n\omega t)\,dt, \qquad b_n = \frac{2}{T}\int_0^T f(t)\sin(n\omega t)\,dt$$ Each pair combines into one sinusoid, $c_n\sin(n\omega t + \theta_n)$, with $c_n = \sqrt{a_n^2 + b_n^2}$ (Dr. Xu: "magnitude") and $\theta_n = \tan^{-1}(a_n/b_n)$ ("phase angle"). n = 1 is the fundamental; the rest are harmonics.

3

Slide 53's square wave swings $\pm V_m$: $$v_o = \frac{4V_m}{\pi}\left(\sin\omega t + \frac{1}{3}\sin 3\omega t + \frac{1}{5}\sin 5\omega t + \dots\right)$$ Only odd harmonics. Class note: the wave is half-wave symmetric, $f(t + T/2) = -f(t)$, and that cancels every even harmonic. The amplitudes fall as $1/n$, so three terms already look square (slide 54).

Extra if you have time: push the builder to 25 terms. The overshoot at each edge never shrinks below about 9% of the jump (the Gibbs effect); it just gets narrower.

08 · slides 55 to 57

The chopper's harmonics

Slide 55 applies this to the chopper output: $v_o = V_d$ for $0 \le t < DT_s$ and 0 for the rest of the period, with D anywhere from 0 to 1 (class note). Step through slide 56, with the algebra it skips.

  1. dc: $$V_{o,dc} = \frac{1}{T_s}\int_0^{DT_s}V_d\,dt = DV_d$$
  2. $a_n$: only the pulse contributes. $$a_n = \frac{2}{T_s}\int_0^{DT_s}V_d\cos(n\omega t)\,dt = \frac{2V_d}{n\omega T_s}\Big[\sin(n\omega t)\Big]_0^{DT_s} = \frac{2V_d}{n\omega T_s}\sin(n\omega DT_s)$$ The slide's middle step writes $\frac{1}{n\omega}\int\dots d(\omega t)$; with $\frac{1}{n\omega}$ in front the variable should be $d(n\omega t)$. Its next line is right.
  3. The key substitution: $\omega = 2\pi f_{sw} = 2\pi/T_s$, so $\omega T_s = 2\pi$. Then $n\omega DT_s = 2\pi nD$ and $\frac{2V_d}{n\omega T_s} = \frac{V_d}{n\pi}$: $$\boxed{a_n = \frac{V_d}{n\pi}\sin(2\pi nD)}$$
  4. $b_n$ the same way: $$b_n = \frac{2V_d}{n\omega T_s}\Big[-\cos(n\omega t)\Big]_0^{DT_s} \;\Rightarrow\; \boxed{b_n = \frac{V_d}{n\pi}\left[1 - \cos(2\pi nD)\right]}$$ Class note: both are functions of D (times $V_d$), not of $T_s$. Together they're the $v_o(t)$ row on your formula sheet: $a_n$ is the cosine coefficient, $b_n$ the sine one.
  5. Not on the sheet, and it saves time: $$V_{on} = \sqrt{a_n^2 + b_n^2} = \frac{2V_d}{n\pi}\left|\sin(n\pi D)\right|$$ With $x = n\pi D$: $\sin^2 2x + (1 - \cos 2x)^2 = 2 - 2\cos 2x = 4\sin^2 x$. So $V_{on} = 0$ whenever nD is a whole number. In Example IV ($D = 0.4$) the 5th and 10th vanish.
Predict first

A chopper runs at $D = 0.25$. Which harmonic of $v_o$ is zero?

Each harmonic sees its own impedance (slide 57)

The load is linear, so treat each harmonic as its own sinusoidal source and add the responses (superposition). Slide 57 draws three equivalent circuits:

dc L = shortR dc: I = V/R V₁ ω₁LR fundamental Vh hω₁LR h-th harmonic

Dr. Xu's note on the dc circuit: no voltage drop on the inductor.

Not on the formula sheet

The sheet gives the $v_o(t)$ series only (and writes $\frac{V_d}{n\pi}$ outside the sum, as slide 56 does; the n belongs inside). You bring: $V_{on} = \sqrt{a_n^2 + b_n^2}$, $\omega = 2\pi f_{sw}$, $Z_h$, $I_{oh} = V_{oh}/Z_h$, and rms = peak/√2. $Z_h$ grows with h, so the R-L load passes low harmonics and crushes high ones. That's the "first-order filter" from slide 37, written again on slide 61.

Predict first

Example IV: $R = 10\ \Omega$, $L = 1$ mH, $f_{sw} = 1$ kHz. What impedance does the 3rd harmonic see?

09 · slides 58 to 61

Example IV: harmonics of an R-L load

A one-quadrant chopper with an R-L load: $V_d = 100$ V, $D = 0.4$, $R = 10\ \Omega$, $L = 1$ mH, $f_{sw} = 1$ kHz. Find the harmonic components of (1) $v_o$ and (2) $i_o$, up to the 5th. Try it, then reveal.

  1. dc: $V_{o,dc} = DV_d = 40$ V and $I_{o,dc} = 40/10 = 4$ A.
  2. n = 1, with $\frac{V_d}{\pi} = 31.83$: $a_1 = 31.83\sin(0.8\pi) = 31.83\times0.5878 = 18.71$ and $b_1 = 31.83\left[1 - \cos(0.8\pi)\right] = 31.83\times1.809 = 57.58$, so $V_{o1} = \sqrt{18.71^2 + 57.58^2} = $ 60.55 V peak.
  3. The rest, computed the same way (the shortcut $\frac{2V_d}{n\pi}|\sin(n\pi D)|$ gives the same $V_{on}$ column):
    nanbnVon peakZn (Ω)Ion peak
  4. n = 1 current: $Z_1 = \sqrt{10^2 + (2\pi\times10^3\times10^{-3})^2} = \sqrt{100 + 39.48} = 11.81\ \Omega$ and $I_{o1} = \frac{60.55}{11.81} = $ 5.13 A peak. The same move gives $I_{o2} = 1.17$, $I_{o3} = 0.58$, $I_{o4} = 0.56$ and $I_{o5} = 0$ A (the last column).
  5. rms (slide 60, class note: for a sine wave), rms = peak/√2: $V_{o1} = 42.81$ V rms and $I_{o1} = 3.625$ A rms. For the whole $v_o$, go back to the definition Dr. Xu wrote, $V_{rms} = \sqrt{\frac{1}{T_s}\int_0^{T_s}v_o^2\,dt} = V_d\sqrt{D} = 63.25$ V.
Slide 61, recomputed: every bar labelled
Reading slide 61

On the current spectrum, the "0.58A" label sits over both h = 3 and h = 4, and "0.26A" sits over h = 6. Slide 60 has the values: $I_{o3} = 0.58$ A and $I_{o4} = 0.56$ A. The 0.26 A is the 6th harmonic: $V_{o6} = 10.09$ V (the "10.1V" label) over $Z_6 = 39.0\ \Omega$. Nothing on the slide is wrong, but don't read 0.26 A as $I_{o4}$.

Compare the two spectra: the 2nd harmonic is 31% of the fundamental in $v_o$ but only 23% in $i_o$. That's the first-order filter Dr. Xu wrote above the chart.

10 · past exam, about 30 min

F2025 midterm, Question 2 (15%)

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

An IGBT-based one-quadrant chopper powers an RLE load. $R_a = 0.47\ \Omega$, $L_a = 1.0$ mH, $V_d = 200$ V, $D = 0.5$, $E_a = 90$ V, $f_{sw} = 1$ kHz.

2.1 Determine the chopper operating mode5%

Try it first. Hint: assume CCM and look at the sign of $I_{\min}$.

  1. $T_s = 1$ ms, $\tau = \frac{L_a}{R_a} = \frac{1.0\text{ mH}}{0.47\ \Omega} = 2.128$ ms, $t_{on} = DT_s = 0.5$ ms. So $t_{on}/\tau = 0.235$ and $T_s/\tau = 0.470$.
  2. $$I_{\min} = \frac{V_d}{R_a}\,\frac{e^{t_{on}/\tau} - 1}{e^{T_s/\tau} - 1} - \frac{E_a}{R_a} = 425.5\times0.4415 - 191.5 = -3.61\text{ A}$$
  3. I_min < 0, so DCM. Second way: $t_x = 0.985$ ms is less than $T_s$.
2.2 Find the ripple current $\Delta I_a$5%
  1. DCM, so $I_{\min} = 0$ and $\Delta I_a = I_{\max} - 0$, using the DCM $I_{\max}$. (The CCM one would give 46.16 A, which is wrong here.)
  2. $$\Delta I_a = \frac{V_d - E_a}{R_a}\left(1 - e^{-t_{on}/\tau}\right) = 234.0\times0.2094 = $$ 49.02 A
2.3 Find the average output voltage $V_{a,avg}$5%
  1. $$t_x = \tau\ln\left\{e^{0.235}\left[1 + \frac{110}{90}(0.2094)\right]\right\} = 2.128\text{ ms}\times(0.235 + 0.2279) = 0.9849\text{ ms}$$
  2. $$V_{a,avg} = \frac{t_{on}}{T_s}V_d + \frac{T_s - t_x}{T_s}E_a = 100 + 1.36 = $$ 101.36 V, not $DV_d = 100$ V.
  3. Not asked, but one more line: $I_{a,avg} = \frac{101.36 - 90}{0.47} = 24.17$ A.
Two slips in the posted solution

It writes $t_x = 9.4891\times10^{-4}$ s $= 0.9849$ ms. Those two don't match: the first has two digits swapped. The correct value is $9.8491\times10^{-4}$ s $= 0.9849$ ms, and that's what gives 101.36 V.

It also writes $\tau = 2.1$ ms, but its −3.61 A comes from the unrounded 2.128 ms. With 2.1 ms you'd get −3.96 A. Same conclusion, DCM, but keep $\tau$ to four figures: $I_{\min}$ is a small difference of two big numbers.

Harmonics of a chopper

Same method as Example IV with new numbers (from the grill's Fourier set). About 12 minutes.

A one-quadrant chopper feeds an R-L load: $V_d = 120$ V, $D = 0.25$, $R = 5\ \Omega$, $L = 2$ mH, $f_{sw} = 500$ Hz. Find (1) the dc, 1st, 2nd, 3rd and 4th harmonic components of $v_o$ (peak), and (2) the matching components of $i_o$, peak and rms.

(1) Harmonics of $v_o$
  1. $V_{o,dc} = DV_d = $ 30 V. Keep $\frac{V_d}{\pi} = 38.20$ handy; $2\pi nD = 90°, 180°, 270°, 360°$.
  2. n = 1: $a_1 = 38.20\sin 90° = 38.20$, $b_1 = 38.20(1 - \cos 90°) = 38.20$, so $V_{o1} = $ 54.02 V.
  3. n = 2: $a_2 = 19.10\sin 180° = 0$, $b_2 = 19.10(1 - \cos 180°) = 38.20$, so $V_{o2} = $ 38.20 V.
  4. n = 3: $a_3 = 12.73\sin 270° = -12.73$, $b_3 = 12.73(1 - \cos 270°) = 12.73$, so $V_{o3} = $ 18.01 V.
  5. n = 4: $4D = 1$, so $a_4 = b_4 = 0$ and V_o4 = 0. Check with $\frac{2V_d}{n\pi}|\sin(n\pi D)|$: 54.02, 38.20, 18.01, 0.
(2) Harmonics of $i_o$
  1. dc: L is a short, so $I_{o,dc} = 30/5 = $ 6 A.
  2. $\omega_1L = 2\pi\times500\times0.002 = 6.283\ \Omega$. So $Z_1 = \sqrt{25 + 6.283^2} = 8.030\ \Omega$, $Z_2 = 13.52\ \Omega$, $Z_3 = 19.50\ \Omega$.
  3. $I_{o1} = 54.02/8.030 = $ 6.727 A peak, 4.757 A rms
  4. $I_{o2} = 38.20/13.52 = $ 2.824 A peak, 1.997 A rms; $\;I_{o3} = 18.01/19.50 = $ 0.923 A peak, 0.653 A rms; $\;I_{o4} = 0$.

Extra if you have time

Skip these to stay under 80 minutes; the grill has more of the same.

Formula-sheet drill (grill fs-09)

$V_d = 100$ V, $R = 10\ \Omega$, $L = 10$ mH, $f_s = 1$ kHz, $D = 0.5$. Find $I_{\max}$, $I_{\min}$, $\Delta I$ and the average current.

  1. $\tau = 1$ ms, $T_s = 1$ ms, $t_{on} = 0.5$ ms.
  2. $I_{\max} = 10\times\frac{1 - e^{-0.5}}{1 - e^{-1}} = 10\times\frac{0.3935}{0.6321} = $ 6.225 A
  3. $I_{\min} = 10\times\frac{e^{0.5} - 1}{e - 1} = 10\times\frac{0.6487}{1.7183} = $ 3.775 A
  4. $\Delta I = $ 2.449 A; average $= DV_d/R = $ 5 A. At $D = 0.5$ the two exponentials mirror each other, so the midpoint is exactly 5 A too.

Same question, CCM this time (grill)

$R_a = 0.5\ \Omega$, $L_a = 2$ mH, $V_d = 240$ V, $E_a = 100$ V, $f_{sw} = 1$ kHz, $D = 0.5$. Find the mode, $\Delta I_a$, $V_{a,avg}$ and $I_{a,avg}$.

  1. $\tau = 4$ ms, $t_{on}/\tau = 0.125$, $T_s/\tau = 0.25$. $I_{\min} = 480\times\frac{e^{0.125} - 1}{e^{0.25} - 1} - 200 = 480\times0.4688 - 200 = $ 25.02 A > 0: CCM
  2. $I_{\max} = 480\times\frac{1 - e^{-0.125}}{1 - e^{-0.25}} - 200 = 480\times0.5312 - 200 = 54.98$ A, so $\Delta I_a = $ 29.96 A
  3. CCM: $V_{a,avg} = DV_d = $ 120 V and $I_{a,avg} = \frac{120 - 100}{0.5} = $ 40 A.
11 · before the exam

Traps that cost marks

  1. Using the buck's $\frac{V_d - V_o}{L}t_{on}$ for the ripple. No capacitor holds the load voltage; use $I_{\max} - I_{\min}$ from the sheet.
  2. Calling $\frac{I_{\max} + I_{\min}}{2}$ the exact average. The exact one is $\frac{V_{o,avg} - E}{R}$, because L has no dc drop.
  3. Mixing exponent signs. $I_{\max}$: $e^{-t_{on}/\tau}$ and $e^{-T_s/\tau}$. $I_{\min}$: $e^{+t_{on}/\tau}$ and $e^{+T_s/\tau}$.
  4. Skipping the mode check with a motor load. Compute the CCM $I_{\min}$ first; negative means DCM.
  5. Keeping the CCM $I_{\max}$ or $V_{o,avg} = DV_d$ after finding DCM. In DCM, $I_{\max} = \frac{V_d - E}{R}\left(1 - e^{-t_{on}/\tau}\right)$ and $V_{o,avg} > DV_d$.
  6. Measuring $t_x$ from turn-off. The sheet's $t_x$ counts from the start of the period.
  7. Rounding $\tau$ or $I_{\min}$ early. They're small differences of big numbers: keep four figures.
  8. Adding R and $h\omega_1L$ arithmetically, or using $\omega_1L$ for every harmonic. $Z_h = \sqrt{R^2 + (h\omega_1L)^2}$.
  9. Mixing peak and rms. The Fourier coefficients are peak values; rms = peak/√2 for each harmonic.
  10. No units, no marks: the midterm cover says "Marks will be lost if there is no appropriate units."