Resolucao capitulo 3 - Eletrônica de Potência - Daniel W. Hart Cap 3 PDF

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Resolucão do livro de Eletronica de potência do Daniel W. Hart capitulo 3...

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Solucionário Cap 3 - Power Electronics - D. Hart Engenharia Elétrica Universidade Federal de Ouro Preto (UFOP) 32 pag.

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CHAPTER 3 SOLUTIONS 2/20/10

3-1) a ) I0  b) I rms

V0 Vm 170 /     3.60 A. R R 15 V V 170  rms  m   5.66 A. 2 R 2(15) R

c ) P  I 2 R  5.662 (15)  480 W .  170  d) S  Vrms I rms   (5.66)  679 VA.  2 P 480 W e) pf    0.707  70.7% S 679 VA

3-2) a ) I 0  12 A.; I 0  Vo 

Vm



V0  V0  I 0 R  (12)(20)  240 V . R

; Vm  V o  240   754 V .

754  533 V . 2 N1 240   0.45 N2 533

Vrms 

N 12 b) I o  I o 2   26.7 A. N1 0.45

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3-3) a ) pf 

V P P V V  ; I rms  s , rms ; VR , rms  m ; V s, rms  m 2 S Vs , rms I rms R 2

 Vm   /R 2 1  2  pf     Vs , rms I rms  Vm   Vm  2 2   2 / R  2  b) Displacement pf  cos(1  )  cos(0) 1 1 V 1 Vm I1  1  0; pf  cos(1  1)DF ;  DF  R R 2 2 VR2, rms / R

3-4) Using Eq. 3-15, a) i( t ) 

Vm V sin( t   )  m (sin  ) et /  Z Z

2 2 2 2 Z  R  ( L)  12  (377(0.012))  12.8 

 L 1  377(0.012)    tan    0.361 rad 12  R     L 377(0.012)     0.377 R 12  i( t) 13.2sin( t  0.361)  4.67 e t /0.377 :   3.50 rad  201  b) I avg  4.36 A. ( numerical integration)

  tan1 

c) I rms  6.70 A. ( numerical integration) P  I 2rmsR  (6.70) 2 (12)  538 W. P 538  0.67 d) pf   S (120)(6.70)

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3-5) Using Eq. 3-15, a) i( t ) 

Vm V sin( t   )  m (sin  ) et /  Z Z

Z  R2  ( L)2  102  (377(0.015))2  11.5   L 1  377(0.015)    tan    0.515 rad R 10      L 377(0.015)   0.565   R 10 i( t) 14.8sin( t  0.515)  7.27 et /0.565 :   3.657 rad  209.5 b) I avg  5.05 A. ( numerical integration)

  tan1 

c) I rms  7.65 A. ( numerical integration) P  I 2rmsR  (7.65) 2 (10)  584 W. 584 P d) pf    0.637  63.7% S (120)(7.65)

3-6) Using Eq. 3-15,

a) i( t ) 

Vm V  sin( t   )  m (sin  ) e t /  Z Z

Z  R2  ( L)2  152  (377(0.08))2  33.7     tan1 

 L

1  377(0.08)    tan    1.11 rad 15  R     L 377(0.08)   2.01   15 R  i( t) 10.1sin( t 1.11)  9.02 e t /2.01:   4.35 rad  250  b) I avg  4. 87 A. (numerical integration) 2 2 c) I rms  6.84 A. ( numerical integration) P  I rmsR  (6.84) (15)  701 W. P 701 d) pf    0.427  42.7% S (240)(6.84)

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3-7) Using an ideal diode model, R = 48 Ω for an average current of 2 A.

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8.0A

Current Iavg = 2 A for R = 48 ohms 4.0A (16.700m,2.0030)

Average Current

0A 0s

5ms I(R1)

10ms AVG(I(L1)) Time

15ms

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20ms

3-8) Using Eqs. 3-22 and 3-23, Vm V sin( t   )  dc  Ae t /  Z R V   V A    m sin(   )  dc e / R   Z

a) i( t) 

Z  R 2  ( L )2  102  (377(.075) 2  30.0   L   377(.075)   tan 1     1.23 rad 10  R     L 377(0.075)     2.83 R 10 V 100   sin 1 dc   0.299 rad  17. 1 Vm 240 2

  tan 1 

i( t) 11.3sin( t 1.23) 10  21.2 e  t /2.83;   3.94 rad  226  I avg  3.13 A. (numerical integration), Pdc  V dcI avg  (100)(3.13)  313 W . 2 2 b) I rms  4.81 A. ( numerical integration) PR  I rms R  (4.81) (10)  231 W. P 313  231 c) pf    0.472  47.2% S (240)(4.81)

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3-9) Using Eqs. 3-22 and 3-23, Vm V sin( t  )  dc  Ae  t / Z R V   V A    m sin(   )  dc  e / Z R   a) i( t) 

Z

R 2  ( L) 2  12 2  (377(0.12) 2  46.8 

 L    377(0.12)   tan 1     1.31 rad 12  R     L 377(0.12)     3.77 R 12 V 48   sin 1 dc   0.287 rad  16.4 Vm 120 2

  tan 1 

i( t)  3.63sin( t  1.31)  4.0  7.66 e  t/3.77 ;   4.06 rad  233  I avg  1.124 A. ( numerical integration), Pdc  V dcI avg  (48)(1.124)  54.0 W. 2 b) Irms  1.70 A. ( numerical integration) PR  Irms R  (1.70) 2(12)  34.5 W.

c) pf 

P 54.0  34.5  0.435  43.5%  S (120)(1.70)

3-10) Using Eq. 3-33, Vm V (cos  cos  t)  dc (   t) L L V   48    sin 1  dc   sin 1    0.287 rad .  120 2   Vm  i( t) 

i( t)  4.68  4.50cos( t)  1.23 t A.;   4.483 rad  257 1  Io  i( t) d ( t)  2.00 A.; Pdc  I oVdc  2.00(48)  96 W. 2 

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3-11)

300W

200W L = 0.25 H

100W

0W 0s

5ms AVG(W(Vdc))

10ms

15ms

Time

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20ms

3-12) L ≈ 0.14 H for 50 W (51 W).

100W

(16.670m,51.156) 50W

L = 0.14 H

0W 0s

5ms AVG(W(Vdc))

10ms

15ms

Time

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20ms

3-13) Using Eq. 3-34, a) V0 

Vm





120 2



 54.0 V .; I 0 

V0 54   4.50 A . R 12

b) n Vn Zn In 0 54.02 12.00 4.50 1 84.85

25.6

3.31

2 36.01

46.8

0.77

4

91.3

0.08

7.20

The terms beyond n = 1 are insignificant.

3-14)

Run a transient response long enough to achieve steady-state results (e.g., 1000ms). The peak-topeak load current is approximately 1.48 A, somewhat larger than the 1.35 A obtained using only the first harmonic. (The inductance should be slightly larger, about 0.7 H, to compensate for the approximation of the calculation.)

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3-15) a) Vm 50   3.98 A.  R 4 V Vm / 2 I1  1   2 Z1 R  ( L)2 I0 

25 R  ( L) 2

R2  ( L)2  9  ( L) 2  L

2

 0.05 I0  0.199 A.

25  125    L 0.199

125  0.33 H 2 60

b) A PSpice simulation using an ideal diode model gives 0.443 A p-p in the steady state. This compares with 2(I1)=2(0.199)=0.398 A p-p.

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3-16) a) V0 

Vm





170



 54.1 V

V0  Vdc 54.1  24   3.01 A. R 10 io  1 A.  2 I1  I1  0.5 A.

I0 

Vm  2 V Z1  1  I1

V1 

170  85 V 2 85  170   0.5

2 2 R  ( L)   L

170  450 mH . 377 b) Pdc  Iavg Vdc  (3.01)(24)  72.2 W. L

2 c) PR  Irms R; Irms 

I

2 n, rms

 (3.01)2  (0.5 / 2) 2  3.12 A.

PR  (3.12) 2 (10)  97.4 W .

3-17) a) τ = RC = 10310-3=1 s; τ/T = 60. With τ >> T, the exponential decay is very small and the output voltage has little variation. b) Exact equations:

   tan1 ( RC )     tan 1 (377)    1.5573 rad  90.15 Vm sin   200sin(90.15 )  199.9993 sin   sin  e (2     )/ RC  0    1.391 rad  79.72 Vo  Vm (1  sin  )  3.21 V .

c) Approximation of Eq. 3-51: Vo 

Vm 200   3.33 V . fRC (60)(10 3 )(10 3)

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3-18) a) R = 100 Ω: τ = RC (100)10-3 = 0.1 s; τ/T = 6.

   tan 1 ( RC )     tan 1 (37.7)    1.5973 rad  91.52 Vm sin   200 sin(91.52 )  199.93 sin   sin  e (2    )/ RC )  0    1.0338 rad  59.23 Vo  Vm (1  sin  )  28.16 V . (exact ) Vo 

Vm 200   33.3 V . (approximation ) 3 fRC (60)(100)(10  )

b) R = 10 Ω: τ = RC (10)10-3 = 0.01 s; τ/T = .6.

   tan 1 ( RC )     tan 1 (3.77)   1.830 rad  104.9 Vm sin   200sin(104.9)  193.3 sin   sin e (2    )/ RC )  0    0.2883 rad  16.5 Vo  Vm (1  sin  )  143.2 V . (exact ) Vo 

Vm 200   333V . (approximation ) fRC (60)(10)(10 3 )

In (a) with τ/T=6, the approximation is much more reasonable than (b) where τ/T=0.6.

3-19) a) With C = 4000 µF, RC = 4 s., and the approximation of Eq. 3-51 should be reasonable. Vo 

Vm 120 2   0.707 V . fRC (60)(4)

b) With C = 20 µF, RC = 0.02, which is on the order of one source period. Therefore, the approximation will not be reasonable and exact equations must be used.

   tan 1 ( RC )     tan 1 ((377)(1000)(20(10) 6 )    1.703 rad  97.6 )   0.5324 rad  30.5 ( numerically from Eq. 3  43) Vo  Vm Vm sin  83.6 V .

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3-20) a) With C = 4000 µF, RC = 2 s., and the approximation of Eq. 3-51 should be reasonable. Vo 

120 2 Vm   1.41 V . fRC (60)(2.0)

b) With C = 20 µF, RC = 0.01, which is on the order of one source period. Therefore, the approximation will not be reasonable and exact equations must be used.

   tan 1 ( RC )     tan 1 ((377)(500)(20(10) 6 )    1.83 rad  104.9)   0.2883 rad  16.5 ( numerically from Eq. 3  43) Vo  Vm Vm sin  121 V . 3-21) From Eq. 3-51 C

Vm 120 2   1,886  F fR Vo 60(750)(2) 

  sin 1  1 

Vo  2  1   sin  1   1.417 rad  81.2 Vm   120 2 

sin    I D, peak  Vm  C cos    18.7 A . R   V I D, avg  m  0.226 A. R

3-22) Assuming Vo is constant and equal to Vm, P

Vo2 Vm2 V 2 (120 2) 2   R m   576  R R P 50

From Eq. 3-51 C

Vm 120 2   3, 270  F fR Vo 60(576)(1.5) 

  sin 1  1 

Vo  1.5  1   sin  1   1.438 rad  82.4 Vm  120 2 

sin   I D, peak  Vm  C cos   28.1 A. R   V I D, avg  m  0.295 A. R

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3-23) Using the definition of power factor and Vrms from Eq. 3-53, pf 



Vm 2

2 2 P Vrms Vrms V /R /R    rms S (Vs ,rms )( Is ,rms ) (Vs, rms )( Vrms / R) Vs, rms

 sin 2   sin 2 1 1  sin 2 2      1  2 2 2 4   Vm / 2 2

1

3-24) Vm 120 2 (1  cos )  (1  cos 45 )  46.1 V. 2 2 V2 V  sin 2 b) P  rms ; Vrms  m 1   2 2  R a) Vo 

120 2 0.785 sin(2(0.785))   80.9 V . 1 2 2  80.92 P  65.5 W . 100 P 65.5 80.9  c) S  Vs, rmsI rms  (120)   0.674  67.4%   97.1 VA; pf  S  97.1  100 



3-25) Vm (1  cos  ) 2  2Vo   2 (75)   1   cos 1   1   65.5 or 1.143 rad   cos 1   240 2   Vm  V2 b) P  o ,rms R V 1.143 sin(2(1.143))  sin 2 240 2 Vo ,rms  m 1      147.6 V . 1   2 2 2 2 147.62 P  726 W . 30 P 726  147.6  c) S  V s, rmsI rms  (240)   0.615  61.5%  1181 VA; pf  S  1181  30  a ) vo  I o R  (2.5)(30)  75 V 

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3-26) a) i( t)  5.42sin( t 0.646) 1.33 et /0.754 A.   25   0.524 rad,   3.79 rad  217  ( numerically) 

b) I o 

1 i( t) d( t) 1.80 A. 2 

c) I rms 

1 2



 i ( t) d ( t)  2.80 2

2 A.; Po  PR  I rms R  (2.80) 225 193 W.



3-27) a) i( t )  3.46sin( t  0.615)  6.38 e t/0.707 A.   60  1.047 rad,   3.748 rad  215 ( numerically) 

b) I o 

1 i( t) d( t)  0.893 A. 2  

c) I rms 

1 2 i ( t) d ( t) 1.50 A.; Po  PR  I 2rms R  (1.50) 240  90.3 W. 2 

3-28) α ≈ 46°. Do a parametric sweep for alpha. Use the default (Dbreak) diode, and use Ron = 0.01 for the switch. Alpha of 46 degrees results in approximately 2 A in the load.

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3-29) α ≈ 60.5°. Do a parametric sweep for alpha. Use the default (Dbreak) diode, and use Ron = 0.01 for the switch. Alpha of 60.5 degrees results in approximately 1.8 A in the load.

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3-30) From Eq. 3-61, a) i( t)  4.29 sin( t  1.263)  4.0  7.43 e  t/3.142 A., 0.873   t  3.95 rad 

Io 

1 i (t )d (t ) 1.04 A., Pdc  I oV dc  (1.04)(48)  50.1 W. 2 

b) I rms  c) pf 

1 2



 i ( t) d( t) 1.67 2

2 A.; PR  I rms R  (1.67) 212  33.5 W.

P 50.1  33.5  0.417  41.7%  S (120)(1.67)

3-31) From Eq. 3-61, a) i( t)  2.95sin( t  0.515)  0.96  3.44 e  t/0.565 A., 1.047   t  3.32 rad 

Io 

1 i(t) d (t) 0.454 A., Pdc  I oV dc (0.454)(96) 43.6 W. 2 

b) I rms  c) pf 

1 2



 i ( t) d( t)  0.830 2

2 A.; PR  Irms R  (0.830) 2100  69.0 W.

P 43.6  69.0  0.565  56.5%  S (240)(0.830)

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3-32) α ≈ 75°. Alpha = 75 degrees gives 35 W in the dc voltage source. An Ron = 0.01 for the switch and n = 0.001 for the diode (ideal model).

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3-33) From Eq. 3-61, a) i( t )  5.99sin( t  1.50)  24.0  29.3 e  t /14.1 A., 0.873   t  4.24 rad 

Io 

1 i (t )d (t )  1.91 A., Pdc  I oVdc  (1.91)(48)  91.6 W. 2  1 2

b) I rms 



 i ( t) d( t)  2.93 2

2 A.; PR  I rms R  (2.93) 22 17.1 W.

3-34) α ≈ 81° 3-35) di(t )  Vm sin t Vdc dt di (t ) 1  [Vm sin t Vdc ] or dt L di (t ) 1  [V m sin t V dc] d (t ) L L

t

i ( t ) 

1 (Vm sin t V dc ) d ( t )  L 

Vm V (cos  cos  t )  dc (   t) L L i ( t )  4.34  7.58 cos t 1.82 t A., 1.309   t  4.249





Io 

1 i (t )d ( t ) 1.9 1 A. 2 

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3-36) v0 = vs when S1 on, v0=0 when D2 on

Io 

Vo V 1  V sin(t) d(  t)  m (1 cos ) , Vo  R 2  m 2

 Io 

Vm (1  cos ) 2 R

3-37) I X     u  cos 1 1  L s  ; X s   Ls  377(1.5)(10) 3  0.566  V  m  5(0.452)    u  cos 1 1    10.47 120 2   Vm  X L X s  120 2  5(.566)   1   53.57 V .  1  2Vm      2 2(120)  V (compared to m  54.0 V .)

Vo 



PSpice: Use a current source for the constant load current:

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D1 to D2

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D2 to D1

3-38) u = 20°. Run the simulation long enough for steady-state results. From the Probe output, the commutation angle from D1 to D2 is about 20 degrees, and from D2 to D1 is about 18 degrees. Note that the time axis is changed to angle in degrees here.

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3-39) Run the simulation long enough for steady-state results. From the Probe output, the commutation angle from D1 to D2 is about 16.5 degrees, and from D2 to D1 is about 14.7 degrees. Note that the time axis is changed to angle in degrees here.

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3-40) At ωt = π, D2 turns on, D1 is on because of the current in LS (see Fig. 3-17). KVL; v LS  Vm sin  t  Ls

diD1 did 1   LS dt d (t )

diD1 V t  m  sin( t) d( t)  i D1( ) d t   Ls  Vm [ 1  cos(  u)]  IL Ls V V cos(  u )   cos u  0  m (1  cos u)  I L   m cos u  I L  Ls  Ls at  t    u, iD1  0 

 I X   u  cos 1 1  L s  Vm  

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3-41) At ωt = α, t

V 1 Vm sin(  t) d(  t) 0  m [cos  cos  t]   Ls   Ls V iD 2 ( wt )  IL  is  IL  m [cos   cos  t]  Ls V iD 2 (  u)  0  IL  m [cos  cos(  u)] L s IL  L s   cos  cos(  u) Vm is ( t) 

 I X  u  cos 1 cos   L s    Vm  

3-42) A good solution is to use a controlled half-wave rectifier with an inductor in series with the 48-V source and resistance (Fig. 3-15). The switch will change the delay angle of the SCR to produce the two required power levels. The values of the delay angle depend on the value selected for the inductor. This solution avoids adding resistance, thereby avoiding introducing power losses. 3-43) Several circuit can accomplish this objective, including the half-wave rectifier of Fig. 3-2a and half-wave rectifier with a freewheeling diode of Fig. 3-7, each with resistance added. Another solution is to use the controlled half-wave rectifier of Fig. 3-14a but with no resistance. The analysis of that circuit is like that of Fig. 3-6 but without Vdc. The resulting value of α is 75°, obtain from a PSpice simulation. That solution is good because no resistance is needed, and losses are not introduced. 3-44 and 3-45) The controlled half-wave rectifier o...