Chapter 7-Resonant Inverters
Prob 7-1
Given: Vs = 220 V, R = 4 , L = 40 × 10−6 H, C = 4 × 10−6 F, f0 = 7.5 kHz, toff = 15 µs
ω0 = 2π × 7500 = 47124 rad/s;
ωr = = 61237 /
∴ = 9746.17 Hz
∴ 102.6
α = R/2L = 4/(2×40×10−6) = 50000
a. From Eq. (7.17),
15.36
ω
b. From Eq. (7.18) 7541
c. From Eq. (7.14), Vc = / . 18.33
From Eq. (7.16), Vc1 = 220 + 18.33 = 238.33 V. The peak-to-peak capacitor voltage is
Vpp = 18.33 + 238.33 = 256.66 V.
d. From Eq. (7.7), the peak load current, which is the same as the peak supply current,
occurs at
tm =
Chapter 7-Resonant Inverters
Page # 7-1
Prob 7-2
−6 −6
Vs := 220 C1 := 2⋅ 10 C2 := 2⋅ 10 C := C1 + C2
−6 −6 3 −6
C = 4 × 10 L := 40⋅ 10 R := 1.2 fo := 8.5⋅ 10 tq := 15⋅ 10
4
ω o := 2⋅ π ⋅ fo ω o = 5.341 × 10
Using Eq. (7-4)
2
1 R 4
ω r := − ω r = 7.762 × 10
L⋅ C 4⋅ L2
Chapter 7-Resonant Inverters
Page # 7-2
ωr 1
4 −5
fr := fr = 1.235 × 10 Tr := Tr = 8.095 × 10
2⋅ π fr
Using Eq. (7-6)
R 4
α := α = 1.5 × 10
2⋅ L
α⋅π
z := z = 0.607
ωr
Using Eq. (7-17)
π π
( a) toff := − −5
ωo ωr toff = 1.835 × 10
Using Eq. (7-18)
1
fmax := 3
fmax = 9.013 × 10
⎛
2⋅ ⎜ tq +
π ⎞
ωr ⎟
Using Eq. (7-14) ⎝ ⎠
Vs
Vc := Vc = 263.439
z
e −1
Vc1 := Vs + Vc Vc1 = 483.439
Vpp := Vc + Vc1 Vpp = 746.878
Using Eq. (7-7)
1 ⎛ ωr ⎞ −5
tm := ⋅ atan ⎜ ⎟ tm = 1.778 × 10
ωr ⎝α⎠
Using Eq. (7-5)
V s + V c − α ⋅ tm
Ip :=
ω r⋅ L
⋅e ( )
⋅ sin ω r⋅ tm Ip = 117.093
Ip
Ips := Ips = 58.547
2
Chapter 7-Resonant Inverters
Page # 7-3
Tr
⌠2 2
⎮ ⎛⎜ Vs + Vc − α ⋅ t ⎞
Io := 2⋅ fo⋅ ⎮ ⋅e ⋅ sin ( ω r⋅ t) ⎟ dt I = 68.232
⎮ ⎜ ω r⋅ L ⎟ o
⌡0 ⎝ ⎠
2 3
Po := Io ⋅ R Po = 5.587 × 10
Po
Is := Is = 25.394
Vs
⎡ Tr ⎤
⎢⌠ 2 ⎥
⎢⎮ ⎛⎜ Vs + Vc − α ⋅ t ⎞ ⎥
( b) IA := fo⋅ ⎢ ⎮ ⋅e ⋅ sin ( ω r⋅ t) ⎟ dt ⎥ IA = 25.394
⎜ ω r⋅ L ⎟
⎢⎮ ⎝ ⎠ ⎥
⎣ ⌡0 ⎦
( c) Ipk := IA Ipk = 25.394
:= IR = 48.247
Chapter 7-Resonant Inverters
Page # 7-4
Where
1 10 6
ωr = = = 81650 𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠
√𝐿𝐿𝐿𝐿 �(5×30)
𝜔𝜔 𝑟𝑟 81650
∴ 𝑓𝑓𝑟𝑟 = = = 12995 𝐻𝐻𝐻𝐻 = 12.99 𝑘𝑘𝐻𝐻𝐻𝐻
2𝜋𝜋 2𝜋𝜋
1 𝑡𝑡 𝑟𝑟
∴ 𝑡𝑡𝑟𝑟 = = 76.95 𝜇𝜇𝑠𝑠 ; 𝑡𝑡1 = = 38.47 𝜇𝜇𝑠𝑠
𝑓𝑓𝑟𝑟 2
5
a. Ip = Vs√(C/L) = 220� = 89.81 A.
30
𝜋𝜋
b. IA = fo∫0 𝐼𝐼𝑝𝑝 sin 𝜃𝜃 𝑟𝑟𝜃𝜃 = Ip fo/(𝜋𝜋𝑓𝑓𝑟𝑟 ) = 44.01 A
c. IR = Ip√(fot1/2) = 89.81×√ (20000×38.47× 10−6 )= 55.703 A.
d. The peak-to-peak capacitor voltage Vpp = Vc1 − Vc = 440 V.
e. From Eq. (7.25), f max = 106/(2 × 15) = 33.33 kHz.
f. Because there is no power loss in the circuit, Is = 0.
Prob 7-4
−6 −6
Vs := 220 C1 := 2⋅ 10 C2 := 2⋅ 10 C := C1 + C2
Chapter 7-Resonant Inverters
Page # 7-5
−6 −6 3
C = 4 × 10 L := 20⋅ 10 R := 1.5 fo := 3.5⋅ 10
4
ω o := 2⋅ π ⋅ fo ω o = 2.199 × 10
Using Eq. (7-4)
2
1 R 5
ω r := − ω r = 1.053 × 10
L⋅ C 4⋅ L2
ωr 1
4 −5
fr := fr = 1.676 × 10 Tr := Tr = 5.965 × 10
2⋅ π fr
Tr 1 −4
−5 To := To = 2.857 × 10
t1 := t1 = 2.983 × 10 fo
2
−4
td := To − Tr td = 2.261 × 10
Using Eq. (7-6)
R 4
α := α = 3.75 × 10
2⋅ L
α⋅π
z := z = 1.119
ωr
Using Eq. (7-14)
Vs
Vc := Vc = 106.78
z
e −1
Vc1 := Vs + Vc Vc1 = 326.78
Using Eq. (7-7)
1 ⎛ ωr ⎞ −5
( a) tm := ⋅ atan ⎜ ⎟ tm = 1.167 × 10
ωr ⎝α⎠
Using Eq. (7-5)
V s + V c − α ⋅ tm
Ip :=
ω r⋅ L
⋅e ( )
⋅ sin ω r⋅ tm Ip = 94.357
Chapter 7-Resonant Inverters
Page # 7-6
⎡ Tr ⎤
⎢⌠ 2 ⎥
⎢⎮ ⎛⎜ Vs + Vc − α ⋅ t ⎞ ⎥
( b) IA := fo⋅ ⎢ ⎮ ⋅e ⋅ sin ( ω r⋅ t) ⎟ dt ⎥ IA = 6.07
⎜ ω r⋅ L ⎟
⎢⎮ ⎝ ⎠ ⎥
⎣ ⌡0 ⎦
t
⌠1 2
⎮ ⎛⎜ Vs + Vc − α ⋅ t ⎞
⋅ sin ( ω r⋅ t) ⎟ dt
( c) IR := fo⋅ ⎮ ⋅e IR = 21.098
⎮ ⎜ ω r⋅ L ⎟
⌡0 ⎝ ⎠
( d) Io := 2⋅ IR Io = 42.196
2 3
( e) Po := Io ⋅ R Po = 2.671 × 10
Po
Is := Is = 12.14
Vs
Prob 7-6
−6 −6 3
Vs := 220 C := 2⋅ 10 L := 20⋅ 10 R := 1.2 fo := 3.5⋅ 10
4
ω o := 2⋅ π ⋅ fo ω o = 2.199 × 10
Using Eq. (7-4)
2
1 R 5
ω r := − ω r = 1.552 × 10
L⋅ C 4⋅ L2
ωr 1
4 −5
fr := fr = 2.471 × 10 Tr := Tr = 4.047 × 10
2⋅ π fr
Tr 1 −4
−5 T o := To = 2.857 × 10
t1 := t1 = 2.024 × 10 fo
2
−4
td := To − Tr td = 2.452 × 10
Using Eq. (7-6)
R 4
α := α = 3 × 10
2⋅ L
Chapter 7-Resonant Inverters
Page # 7-7
α⋅π
z := z = 0.607
ωr
Using Eq. (7-15)
(z )
Vs⋅ e + 1
Vc := Vc = 746.878
z
e −1
Vc1 := Vc Vc1 = 746.878
Using Eq. (7-7)
1 ⎛ ωr ⎞ −6
( a) tm := ⋅ atan ⎜ ⎟ tm = 8.889 × 10
ωr ⎝α⎠
Using Eq. (7-5)
V s + V c − α ⋅ tm
Ip :=
ω r⋅ L
⋅e ( )
⋅ sin ω r⋅ tm Ip = 234.186
⎡ ⌠ t1 ⎤
⎢ ⎮ ⎛⎜ Vs + Vc − α ⋅ t ⎞⎟ ⎥
( b) IA := fo⋅ ⎢ ⎮ ⋅e ⋅ sin ( ω r⋅ t) dt ⎥ IA = 10.456
⎜ ω r⋅ L ⎟
⎢⎮⌡0 ⎝ ⎠ ⎥
⎣ ⎦
t
⌠1 2
⎮ ⎛⎜ Vs + Vc − α ⋅ t ⎞
⋅ sin ( ω r⋅ t) ⎟ dt
( c) IR := fo⋅ ⎮ ⋅e IR = 43.783
⎮ ⎜ ω r⋅ L ⎟
⌡0 ⎝ ⎠
( d) Io := 2⋅ IR Io = 87.567
2 3
( e) Po := Io ⋅ R Po = 9.202 × 10
Po
Is := Is = 41.825
Vs
Chapter 7-Resonant Inverters
Page # 7-8
Prob 7-7
a. Because at resonance u = 1 and | ω |max = 1, the peak fundamental load voltage
is Vp = Vi(pk) = 4Vs/π.
PL =Vp 2/2R = 42Vs2/2Rπ2 or 3000 = 42Vs2/2π2 × 7
which gives Vs = 160.9 V.
b. To reduce the load power by (3000/1500 = ) 2, the voltage gain must be reduced by 1.414 at u
= 0.8.
That is, from Eq. (7.35), we get 1 + Qs2(u − 1/u)2 = 2, which gives Qs = 2.22.
c. Qs is defined by
Qs = ω0L/R or 2.22 = 2π × 30 kHz × L/7
which gives L = 82.44 µH.
d. f0 = 1/2π√ LC) or 30 kHz = 1/[2π(√ 82.44 µH × C)], which gives C = 0.341 µF.
Prob 7-8
3 3
PL := 2⋅ 10 Vp := 330 R := 6.5 fo := 25⋅ 10 PRQ := 500
⎛ P ⋅ 2⋅ π 2⋅ R⎞
Vs := ⎝ L ⎠ Vs = 126.642
( a) 4
4⋅ Vs
Vi :=
π Vi = 161.245
Vp
( b) Q :=
Vi Q = 2.047
PL
x := x=4
PRQ
Guess
u := 1.
Given
2
(1 − u )
2
2
+ ⎛⎜ ⎞⎟ = x
u
⎝Q⎠
Chapter 7-Resonant Inverters
Page # 7-9
Uval := Find ( u)
u := Uval u = 1.68
R
( c) L := −5
2⋅ π ⋅ fo⋅ Q L = 2.022 × 10
1
( d) C := −6
C = 2.004 × 10
( 2⋅ π ⋅ fo) ⋅ L
2
Prob 7-9
a. Because at resonance u = 1 and |Z ω |max = 1, the peak fundamental load current
is Ip = Ii(pk) = 4Is/π.
PL =Ip 2R/2 = 42Is2R/(2π2) or 3000 =42Is2×7.5/(2π2)
which gives Is = 22.21 A.
b. To reduce the load power by (3000/1500 = ) 2, the current gain must be reduced by 1.414 at u
= 1.25. That is, from Eq. (7.35), we get 1 + Qp2(u − 1/u)2 = 2, which gives Qp = 2.22.
c. Qs is defined by
Qs = ω0L/R or 2.22 =2π × 30 kHz × L/7.5
which gives L = 88.33 µH.
d. f0 = 1/2π√ LC) or 30 kHz = 1/[2π(√ 88.33µH × C)], which gives C = 0.3 µF.
Prob 7-10
3
Vs := 18 R := 5 fs := 50⋅ 10 k := 0.304 Q := 7
5
ω s := 2⋅ π ⋅ fs ω s = 3.142 × 10
0.4001⋅ R −6
Le := Le = 6.368 × 10
ωs
2.165 −6
Ce := Ce = 1.378 × 10
R⋅ ω s
Chapter 7-Resonant Inverters
Page # 7-10
Q⋅ R −4
L := L = 1.114 × 10
ωs
1 −8
C := C = 9.578 × 10
(
ω s⋅ ω s⋅ L − 0.3533⋅ R )
R C
δ := ⋅ δ = 0.073
2 L
1 4
fo := fo = 4.872 × 10
2⋅ π ⋅ L⋅ C
Prob 7-11
−3 3
Vm := 12 ΔVm := 20⋅ 10 fo := 350⋅ 10
PL := 1.5 Vo := 5
−9 −8
( a) Let C := 10⋅ 10 C = 1 × 10
1
L := −5
L = 2.068 × 10
(
C⋅ 2⋅ π ⋅ fo ) 2
2
Vo
R := R = 16.67
PL
Vo
Io := Io = 0.3
R
Io
Cf := −5
2⋅ fo⋅ ΔVm Cf = 2.14 × 10
Vm
( b) Im := Im⋅ 1000 = 720 mA
R
2
Im
2
IL_rms := Io + IL_rms⋅ 1000 = 590.931 mA
2
IL_dc := Io IL_dc⋅ 1000 = 300 mA
Chapter 7-Resonant Inverters
Page # 7-11
Im
IC_rms :=
2 IC_rms⋅ 1000 = 509.117 mA
IC_dc := 0 IC_dc = 0
Prob 7-12
x = 1.5, Vs = 15 V, f = fmax = 50 kHz, and T = 1/50 kHz = 20 µs. PL = VoIo or 2.5 W = 9Io, which
gives Io = 0.277 A.
The maximum frequency occurs when t5 = 0. Because t1 = t3 = t5 = 0, t2 + t4 = T. Substituting
t4 = 2VsC/Im and using x = (Vs/Io)√(C/L) gives:
2 2
√
which gives C = 0.0902 µF.
Thus, L = (Vs/xIo)2C = 117.56 µH.
Prob 7-13
Vo := 5 Vs := 15 3
PL := 1 f := 40⋅ 10
−6 −6 1
L := 150⋅ 10 C := 0.02⋅ 10 ω o :=
L⋅ C 5
ω o = 5.774 × 10
PL
Io := Io = 0.2
Vo
Using Eq. (7-56)
( a) L
Vp := Io⋅ + Vs Vp = 32.321
C
Chapter 7-Resonant Inverters
Page # 7-12
Ip := Io
Ip = 0.2
C
( b) t1 := Vs⋅ −6
Io t1 = 1.5 × 10
−6
t2 := π ⋅ L⋅ C t2 = 5.441 × 10
Vs C
x := ⋅ x = 0.866
Io L
t3 := L⋅ C⋅ asin ( x) −6
t3 = 1.814 × 10
(
IL3 := −Io⋅ cos ω o⋅ t3 ) IL3 = −0.1
L
(
t4 := Io − IL3 ⋅
Vs
) t4 = 3 × 10
−6
(
t5 := T − t1 + t2 + t3 + t4 ) t5 = 1.324 × 10
−5
Chapter 7-Resonant Inverters
Page # 7-13
