Elements of Power Electronics (second edition)

As an author, one always hopes to find and correct all errors and have a perfect result. Although there have been many layers of checks and corrections to this book, some errors came out only in printing and others slipped through. The information below provides corrections to the texts. We appreciate your understanding.

FIRST PRINTING (2015) U.S. Second Edition errata

Figs. 15.5 and 15.6 somehow did not reproduce properly in the printing process. Fig. 8.25 is missing the photgraph that should appear as d). The axis did not reproduce correctly in Fig. 11.13d. Fig. 12.42 has a symbol out of position.
The original of Fig. 15.5 is linked here.
The original of Fig. 15.6 is linked here.
The original of Fig. 8.25d is linked here.
The original of Fig. 11.13 is linked here. Thanks to Yue Cao for pointing this out.
The original of Fig. 12.42 is linked here.

Fig. 2.29: The bottom horizontal line that defines the current source was cropped out in printing.
Fig. 2.50: The horizontal line that defines the bottom connections was cropped out in printing.
Fig. 2.51: The horizontal line that defines the bottom connections was cropped out in printing.

p. 102, Example 3.4.1: in the second to last line on the page, the current ramp is 5000 A/s. The 3 should be a superscript to reflect this.
p. 119, below eq. (3.41): "boost converter must have positive output larger than" rather than "input". Thanks to Jessica Bardo for pointing this out.
p. 128, Example 3.5.3: The output side inductance is 259.2 mH. (It must be higher than the input side inductance since there are more turns on the output side.) Thanks to Johan Koster for pointing this out.
p. 167, problem 30: R = 10 Ω, not 10 V.
Fig. 4.71: The inductor symbol is missing next to L_in.
p. 192, eqs. (4.37) and (4.38) are correct as written only for m=2. The broader result for alpha+pi/m <= pi/2 is the same as in (4.31). For alpha+pi/m > pi/2, the result is [m V₀/(2 pi)] x [1-sin(alpha-pi/m)]. In (4.38), this is multiplied by I_out. Thanks to Johan Koster for pointing this out.
p. 193 eq. 4.40, the upper integral limit should be min(alpha+pi/m,pi/2). Thanks to Johan Koster for pointing this out.
p. 236 eq. (4.100) should match eq. (4.37), per error on p. 192 given above.
Fig. 5.40: The horizontal line that defines the bottom connections was cropped out in printing.
p. 308, near the bottom, the triangle waveform peak-to-peak value is 3.00 V, as shown in eq. (6.13).
p. 309, eq. (6.14), the peak-to-peak voltage in the numerator should be shown as 3.00 V. The end result, however, is still 2.6 A and the losses are shown correctly. Thanks to Xichen Jiang for pointing these out.
p. 329, center, the reference current is I_ref = 0.5 A.
p. 385, the first full line of text: the reluctance is equal to l/(μA). The equal sign somehow was dropped. Thanks to Jessica Bardo for pointing this out.
p. 401, Fig. 8.25: The photo that should appear as part (d) did not print. See above for the original.
p. 403, last paragraph: Table 7.1, not Table 11.1. Thanks to Jessica Bardo for pointing this out.
pp. 470-471, Example 9.13.1: In the last line on p. 470, the total on-state loss has been written incorrectly. It is 5.6 W, not 8.1 W. On p. 471, without the snubber, the sum of switching and on-state loss is 7.4 W, and the efficiency is 77%. With the snubber, the loss is 6.7 W, and the efficiency is 79%. The thermal computations in eqs. (9.46) and (9.47) are correct as printed. Thanks to Yue Cao for pointing this out.
p. 555, eq. (11.33): The factor of D₁ in the numerator should not be present, as that term in (11.32) has been replaced with k_p.
p. 566, eq. (11.45): The factor of D₁ in the numerator should not be present, so in the lower part of the equation, the numerator value should be 12 V instead of 3.3 V. The effect in Fig. 11.39 is that the magnitude is shown low by a factor of 12/3.3, a small vertical shift upward by 11.2 dB in the magnitude would correct this. The angle is unchanged.
p. 606, third line below eq. (12.59), “ratio ω_p/ω_z = 57.7” not “ratio ω_p/ω_p = 57.7”
p. 619, Fig. 12.42: The symbol i associated with the load current was printed about 12 mm to the left of the drawn location, so incorrectly shows the industor current as ii. Thanks to Jessica Bardo for pointing this out.
p. 664, problem 13.5 part b, q_2,2 is delayed from q_1,1 by 1/4 of a switching period...
p. 665, problem 13.2, the modulating function is cos^-1[k cos(ω_outt)].
Fig. 14.21: The horizontal line that defines the bottom connections was cropped out in printing.
Fig. 14.24: The equations and results are consistent if the arrow direction shown for i_d is reversed.
p. 695, Example 14.4.2: there is a numerical error that alters some of the outcome, beginning in the second line. The second integral yields -3.78 μVs. This coefficient should appear in place of -4.76 in eq. (14.51), and the result is V_out = 48 - f_switch(13.7 x 10^-6) V. To produce 12 V, a switching frequency of 2.64 MHz is needed. This gives a total period of 380 ns. The (minimum) total time spent in (0,0), (0,1), and (1,1) is still 304 ns. The result in eq. (14.52) is T > 304 ns, f_switch < 3.29 MHz. When f_switch = 3.29 MHz, the output voltage is 48 - 3.29(13.7) = 3.03 V. Thanks to Jim Sebek and to Kyle Rasmussen for pointing this out.
p. 730: second paragraph, the boundary is the line k_v(v_C - V_out) + k_i(i_L - V_out/R_load) = 0.

FIRST PRINTING (2016) International Second Edition errata

p. 102, Example 3.4.1: in the second to last line on the page, the current ramp is 5000 A/s. The 3 should be a superscript to reflect this.
p. 119, below eq. (3.41): "boost converter must have positive output larger than" rather than "input". Thanks to Jessica Bardo for pointing this out.
p. 128, Example 3.5.3: The output side inductance is 259.2 mH. (It must be higher than the input side inductance since there are more turns on the output side.) Thanks to Johan Koster for pointing this out.
p. 192, eqs. (4.37) and (4.38) are correct as written only for m=2. The broader result for alpha+pi/m <= pi/2 is the same as in (4.31). For alpha+pi/m > pi/2, the result is [m V₀/(2 pi)] x [1-sin(alpha-pi/m)]. In (4.38), this is multiplied by I_out. Thanks to Johan Koster for pointing this out.
p. 193 eq. 4.40, the upper integral limit should be min(alpha+pi/m,pi/2). Thanks to Johan Koster for pointing this out.
p. 236 eq. (4.100) should match eq. (4.37), per error on p. 192 given above.
p. 258 Figure 5.15, The horizontal line that defines the bottom switch and connections was cropped out in printing.
p. 308, near the bottom, the triangle waveform peak-to-peak value is 3.00 V, as shown in eq. (6.13).
p. 309, eq. (6.14), the peak-to-peak voltage in the numerator should be shown as 3.00 V. The end result, however, is still 2.6 A and the losses are shown correctly. Thanks to Xichen Jiang for pointing these out.
p. 337, international edition, problem 6.22 should read: “22. A VSI delivers 50 kW in the form of a square wave into a 5 Ω load at 400 Hz. It is desired to add tuned traps to remove 1200 Hz and 2000 Hz unwanted components. Find L and C values that will accomplish this without altering the output power (based on the 400 Hz component) by more than 2%.”
p. 385, the first full line of text: the reluctance is equal to l/(μA). The equal sign somehow was dropped. Thanks to Jessica Bardo for pointing this out.
p. 401, Fig. 8.25: The photo that should appear as part (d) did not print. See above for the original.
p. 403, last paragraph: Table 7.1, not Table 11.1. Thanks to Jessica Bardo for pointing this out.
pp. 470-471, Example 9.13.1: In the last line on p. 470, the total on-state loss has been written incorrectly. It is 5.6 W, not 8.1 W. On p. 471, without the snubber, the sum of switching and on-state loss is 7.4 W, and the efficiency is 77%. With the snubber, the loss is 6.7 W, and the efficiency is 79%. The thermal computations in eqs. (9.46) and (9.47) are correct as printed. Thanks to Yue Cao for pointing this out.
p. 491, international edition, problem 3, second line, add “is” to read “modulation depth is such that”.
p. 555 eq. (11.33), remove D₁ in numerator.
p. 566 eq. (11.45), remove D₁ in numerator, change (3.3 V) to (12 V) in numerator.
p. 606, third line below eq. (12.59), “ratio ω_p/ω_z = 57.7” not “ratio ω_p/ω_p = 57.7”
p. 619, Fig. 12.42: The symbol i associated with the load current was printed about 12 mm to the left of the drawn location, so incorrectly shows the industor current as ii. Thanks to Jessica Bardo for pointing this out.
Fig. 14.24: The equations and results are consistent if the arrow direction shown for i_d is reversed.
p. 695, Example 14.4.2: there is a numerical error that alters some of the outcome, beginning in the second line. The second integral yields -3.78 μVs. This coefficient should appear in place of -4.76 in eq. (14.51), and the result is V_out = 48 - f_switch(13.7 x 10^-6) V. To produce 12 V, a switching frequency of 2.64 MHz is needed. This gives a total period of 380 ns. The (minimum) total time spent in (0,0), (0,1), and (1,1) is still 304 ns. The result in eq. (14.52) is T > 304 ns, f_switch < 3.29 MHz. When f_switch = 3.29 MHz, the output voltage is 48 - 3.29(13.7) = 3.03 V. Thanks to Jim Sebek and to Kyle Rasmussen for pointing this out.
p. 730: second paragraph, the boundary is the line k_v(v_C - V_out) + k_i(i_L - V_out/R_load) = 0.

Errata will be updated as issues are found.