Po ≔ 850 W
Vin_max ≔ 410 V
Vin_min ≔ 335 V
Vin_nom ≔ 385 V
Vo ≔ 12 V
Vf ≔ 0.1 V
η ≔ 95%
Po
= 70.833 A
Io ≔ ――
Vo
fo ≔ 100 KHz
⎛ Po ⎞
⎜――
⎟ ⋅ ((1 - 0.95))
Vo
⎝ 0.95 ⎠
= 0.169 Ω
= 0.632 V
RL ≔ ――
Vloss ≔ ――――――
⎛ Po ⎞
Po
⎜――
⎟
Vin_nom
⎝ Vo ⎠
n_rel ≔ 16
n ≔ ――――= 15.909
2 ⋅ ((Vo + Vf))
Vo_max ≔ Vo ⋅ ((1 + 0.01)) = 12.12 V
Vo_min ≔ Vo ⋅ ((1 - 0.01)) = 11.88 V
2
n ⋅ ((Vo_min + Vf))
= 0.93
Mg_min ≔ ――――――
Vin_max
――――
2
电压的负载调整范围
n ⋅ ((Vo_max + Vf + Vloss))
Mg_max ≔ ―――――――― = 1.221
Vin_min
―――
2
Mg_max1 ≔ Mg_max ⋅ 1.1 = 1.343
⎛⎝Ln ⋅ fn 2 ⎞⎠
Mg ((Ln , Qe , fn)) ≔ ―――――――――――――――
2
2
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
⎛⎝((1 + Ln)) ⋅ fn 2 - 1⎞⎠ + ⎛⎝Qe ⋅ Ln ⋅ fn ⋅ ⎛⎝fn 2 - 1⎞⎠⎞⎠
寻找Qe，Ln
2
1.8
1.6
Mg ((6 , 0.345 , fn))
1.4
1.2
Mg ((6 , 0 , fn))
1
Mg_min
0.8
Mg_max
0.6
0.4
Mg_max1
0.2
0
0
0.15
0.3
0.45
0.6
0.75
0.9
1.05
1.2
1.35
1.5
1.65
fn
Ln ≔ 6
Qe ≔ 0.345
6.0 ⋅ fn 2
Mg1 ((fn)) ≔ Mg ((Ln , Qe , fn)) → ―――――――――――――――
0.5
2
2⎞
⎛
2
2 ⎛
2
⎛
⎞
⎞
⎝⎝7.0 ⋅ fn - 1.0⎠ + 4.2849 ⋅ fn ⋅ ⎝fn - 1.0⎠ ⎠
⎤
⎡
0.0
⎢ -0.84206901035452211362 + 0.90010510273395272836i ⎥
⎢
⎥
solve ⎢ -0.84206901035452211362 - 0.90010510273395272836i ⎥
d
Mg2 ((fn)) ≔ ――
→⎢
-0.44968650806590878811
Mg1 ((fn)) ――
⎥
d fn
⎢
⎥
0.44968650806590878811
⎢ 0.84206901035452211362 - 0.90010510273395272836i ⎥
⎢
⎥
⎣ 0.84206901035452211362 + 0.90010510273395272836i ⎦
解得
fn ≔ 0.45
Mg_ap ≔ Mg ((Ln , Qe , fn)) = 1.426
⎛ n_rel 2 ⎞ Vo
= 35.154 Ω
Re ≔ ⎜8 ⋅ ―――
⎟ ⋅ ――
π 2 ⎠ Io
⎝
⎛ n_rel 2 ⎞
Vo
Re_over ≔ ⎜8 ⋅ ―――
⎟ ⋅ ―――= 31.958 Ω
2
π
⎝
⎠ Io ⋅ 110%
1
= 131.228 nF
Cr ≔ ――――――
2 ⋅ π ⋅ Qe ⋅ fo ⋅ Re
1
Lr ≔ ―――――
= 19.302 μH
2
((2 ⋅ π ⋅ fo)) ⋅ Cr
Cr_rel ≔ 136 nF
Lr_rel ≔ 20 μH Lm ≔ Lr_rel ⋅ Ln = 120 μH
―――
Cr_rel
Qe_max ≔‾‾‾‾‾‾
= 0.379
――――
Lr_rel
Re_over
1
Lr ≔ ―――――
= 19.302 μH
2
((2 ⋅ π ⋅ fo)) ⋅ Cr
‾‾‾‾‾‾
‾L‾r‾_‾
re‾l‾
―――
Cr_rel
Qe_max ≔ ――――
= 0.379
Re_over
n ⋅ ((Vo_min + Vf))
= 0.93
Mg_min ≔ ――――――
Vin_max
――――
2
有上述可知
Mg_max1 ≔ Mg_max ⋅ 1.1 = 1.343
⎛⎝Ln ⋅ fn 2 ⎞⎠
Mg ((Ln , Qe , fn)) ≔ ―――――――――――――――
2
2
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
⎛⎝((1 + Ln)) ⋅ fn 2 - 1⎞⎠ + ⎛⎝Qe ⋅ Ln ⋅ fn ⋅ ⎛⎝fn 2 - 1⎞⎠⎞⎠
⎡ -1.352850749561555632 ⎤ ⎡ -1.353 ⎤
solve ⎢ 1.352850749561555632 ⎥ ⎢ 1.353 ⎥
Mg ((6 , 0 , fn1)) ＝ Mg_min ――
→⎢
⎥=⎢
⎥
⎢ -0.27263432422905386715 ⎥ ⎢ -0.273 ⎥
⎣ 0.27263432422905386715 ⎦ ⎣ 0.273 ⎦
求解fn1，fn2的值
⎡ -0.53618121915164250972 ⎤ ⎡ -0.536 ⎤
solve ⎢ -0.39441204738192898492 ⎥ ⎢ -0.394 ⎥
Mg ((6 , 0.345 , fn2)) ＝ Mg_max1 ――
→⎢
⎥=⎢
⎥
⎢ 0.39441204738192898492 ⎥ ⎢ 0.394 ⎥
⎣ 0.53618121915164250972 ⎦ ⎣ 0.536 ⎦
fmin ≔ fo ⋅ 0.536 = 53.6 KHz
计算等效电路负载电流
fmax ≔ fo ⋅ 1.353 = 135.3 KHz
⎛ 3.14 ⎞ Io ⋅ 1.1
= 5.406 A
Ioe ≔ ⎜―――
⎟ ⋅ ―――
n_rel
‾‾
⎝⎜ 2 ⋅ 2 ⎠⎟
2 ⋅ ‾‾
2
Vo
⋅ n_rel ⋅ ―――――= 4.277 A
Im ≔ ―――
π
2 ⋅ π ⋅ fmin ⋅ Lm
计算等效电路电感电流
Ir ≔ ‾‾‾‾‾‾‾‾‾
Im 2 + Ioe 2 = 6.894 A
选择谐振电容Cr
ICr ≔ 2.6 A
ICr
VCr ≔ ――――――= 56.766 V
2 ⋅ π ⋅ fmin ⋅ Cr_rel
VCr_RMS ≔
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
⎛ Vin_max ⎞ 2
2
⎜――――
⎟ + VCr = 212.714 V
2
⎝
⎠
⎛ Vin_max ⎞
VCr_max ≔ ⎜――――
2 ⋅ VCr = 285.28 V
⎟ + ‾‾
2
⎝
⎠
MOSFET
变压器设计
Vq1_peak ≔ Vin_max = 410 V
Iq_rms ≔ ICr = 2.6 A
窗口利用系数
ΔBmax ≔ 0.3 T
电感比
Ko ≔ 0.35
Kf ≔ 6
⎛
1⎞
A
3
PT ≔ ⎜ ‾‾
2 +―
IJ ≔ 6 ――
电流密度
⎟ ⋅ Po = ⎛⎝2.097 ⋅ 10 ⎞⎠ W
η⎠
⎝
mm 2
PT
= 1.035 cm 4
AP ≔ ―――――――――
Ko ⋅ Kf ⋅ IJ ⋅ ΔBmax ⋅ fmin
‖
⎛ AP ⎞ 1.143
AP ≔ ‖ Ap ← ⎜――
cm 4
⎟
4
‖
⎝ cm ⎠
‖
|
Ap
‖ if ――
1
>
|
‖ cm 4
|
‖ ‖
‖ ‖ Ap ← Ap ||
‖
‖ Ap
|
|
|
|
|
|
|
|
|
|
AP = 1.04 cm 4
Tc ≥ 210 度
N≔1
E ≔ 32 mm
磁芯窗口面积
电感系数
μH
AL ≔ 4.86 ――
N2
B ≔ 14.35 mm
Bsat ≔ 0.4 T
F ≔ 11.2 mm
⎛E-B⎞
AW ≔ ⎜――⎟ ⋅ F = 98.84 mm 2
⎝ 2 ⎠
AW ⋅ Ae = 1.166 cm 4
Recheck_AP ≔ ‖ if AP < 0.95 ⋅ ⎛⎝AW ⋅ Ae⎞⎠| |
‖
||
‖ ‖
“FAIL
”
||
‖
‖
|
|
‖ “PASS”
磁芯有效截面积
Ae ≔ 118 mm 2
Recheck_AP ≔ ‖ if AP < 0.95 ⋅ ⎛⎝AW ⋅ Ae⎞⎠| |
‖
||
‖ ‖‖ “FAIL ”
||
‖
|
|
‖ “PASS”
Recheck_AP = “PASS”
The voltage across the primary winding can be calculated as
Vp ≔ n_rel ⋅ ((Vo_max + Vf + Vloss)) = 205.625 V
The half switching cycle period is around
1
t ≔ ―――= ⎛⎝9.328 ⋅ 10 -6⎞⎠ s
2 ⋅ fmin
According to Faraday's law
ΔB ≔ 2 ⋅ ΔBmax
((Vo_max + Vf + Vloss))
＝ Np ⋅ Ae ⋅ ΔB
n_rel ⋅ ――――――――
2 ⋅ fmin
((Vo_max + Vf + Vloss))
= 27.092
Np_min ≔ n_rel ⋅ ――――――――
2 ⋅ fmin ⋅ Ae ⋅ ΔB
Np_rel
Np_rel ≔ 32
=2
Ns ≔ ―――
n_rel
计算电流密度前，先算路径有效值电流
1 π
⋅ ―⋅ Io = 78.676 A
Irect_rms ≔ ――
‾‾
2 2
((Vo_max + Vf + Vloss))
= 4.284 A
Imag_peak ≔ n_rel ⋅ ――――――――
4 ⋅ fo ⋅ Lm
1
8
⋅ ――
⋅ Imag_peak = 2.455 A
Imag_FHA ≔ ――
2
π
‾‾
2
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
⎛ Irect_rms ⎞ 2
2
2 ⋅ Ir_rms = 7.773 A
Ir_rms ≔ ⎜――――
⎟ + Imag_FHA = 5.496 A Ir_peak ≔ ‾‾
⎝ n_rel ⎠
A
Ir_rms
= 0.55 mm 2
JP ≔ 10 ――
SP ≔ ―――
其中
2
JP
mm
集肤深度
ϕp ≔ 0.1 mm
2 ⋅ 66.1 mm
ε ≔ ――――= 0.418 mm
‾‾‾
fo
――
Hz
Choose 0.1mm twisting multi-wires in primary coil
⎛
⎞
SP
Numberpri ≔ ceil ⎜――――
⎟ = 70
((ϕp)) 2 ⎟ Choose 0.1mm*36 twisting multi-wires in primary coil
⎜
π⋅
⎜⎝ ―――
4 ⎟⎠
Choose 0.1mm twisting multi-wires in secondary coil
A
Js ≔ 6 ――
mm 2
Irect_rms
= 13.113 mm 2
SS ≔ ――――
Js
铜损计算
ρcu ≔ 2 ⋅ 10 -6 Ω ⋅ cm
Lperturn ≔ 105 mm
Lperturn ⋅ Np_rel
Rp ≔ ρcu ⋅ ――――――= 0.122 Ω
SP
⎛
⎞
SS
3
NumberSEC ≔ ceil ⎜――――
⎟ = 1.67 ⋅ 10
2
((ϕp)) ⎟
⎜
π⋅
⎜⎝ ―――
4 ⎟⎠
环境60° C
Lperturn ⋅ Ns ⎛
= ⎝3.203 ⋅ 10 -4⎞⎠ Ω
RS ≔ ρcu ⋅ ―――――
SS
⎛ Irect_rms ⎞ 2
2
Pcu ≔ ⎜――――
⎟ ⋅ 2 ⋅ RS + Ir_rms ⋅ Rp = 5.676 W
‾‾
2
⎜⎝
⎟⎠
铁损计算，先查磁损耗密度，再由公式可算出
Pcv_Lr ≔ Ve_Lr ⋅ Pcv_Lr
输出电容计算
The rectifier’s full-wave output current is expressed as
π
⋅ Io = 78.676 A
Irect ≔ ―――
2 ⋅ ‾‾
2
Then, for the load current (Io), the capacitor’s RMS current rating at about 100 kHz is calculated as
⎛ ‾‾‾‾‾‾
⎞
π2
- 1 ⎟ ⋅ Io = 34.243 A
Ico ≔ ⎜ ――
⎜⎝ 8
⎟⎠
⎛ ‾‾‾‾‾‾
⎞
π2
- 1 ⎟ ⋅ Io = 34.243 A
Ico ≔ ⎜ ――
⎜⎝ 8
⎟⎠
Vo_peak_peak ≔ Vo ⋅ ((0.01 + 0.01)) = 0.24 V
Vo_peak_peak
ESRmax ≔ ―――――= 0.002 Ω
⎛π
⎞
⎜―⋅ Io⎟ ⋅ 2
⎝4
⎠