BUCK电路设计(原创)

μ0 := 0.4π⋅10− 6 H m
ρ := 1.69 × 10− 8Ω⋅m
The magnetic permeablity of vacuum The copper resistance coefficient
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0.12
0.11 ΔB(Vin) 0.1
0.09 0.08
8
10
12
14
Vin
PVe := 200 kW m3
Pcore_max := PVe⋅Ve1
Pcore_max = 0.072 W
refer to core spec ,power loss coefficient @ΔΒ/2
PL1(Vin , Io) := PL1cond(Vin , Io) + Pcore_max Total power loss on L1
PWM(Vin , t) := (10V) if 0 ≤ mod(t , Ts) ≤ Dnm(Vin)⋅Ts 0 if Dnm(Vin)⋅Ts < mod(t , Ts) ≤ Ts
15
15
10.3 VL( VINmax, t)
5.6 VL( VINmin, t)
0.9
11.25 PWM(VINmax , t)
2.2 Q1 SELECT
DC/DC _ BUCK TOPLOGY PAPERDESIGN REV:01
WINK FANG 2007-04-10
Vin Cin Q1 Driver
L1 D1
Voset Co
GND
MODEL:3.3V 3A DC/DC CONVERTER
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Out put power max In put power max Duty max
12
14
Vin
2. Power component selection
2.1 BUCK inductor
VL -voltage wave form
Voltage on inductor
VL(Vin , t) := Vin − Voset if 0 ≤ mod(t , Ts) ≤ Dnm(Vin)⋅Ts −Vo if Dnm(Vin)⋅Ts < mod(t , Ts) ≤ Ts
Voset := 3.3V
Out put voltage
Voripple := 0.1V
Out put max ripple voltage
Vo := Voset + 0.5V
Compensate the out put diode VF
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L inductance (H)
3.2×10− 5
2.9×10− 5
Lmin(Vin) 2.6×10− 5
2.3×10− 5
2×10− 5 8
L1 := 30μH
10
12
14
Vin
Choosing a standard value to meet design
ΔIL(Vin) := Vin − Voset ⋅Dnm(Vin)⋅Ts L1
ID1 := 10mm
Ht := 5mm
Ae1 := 2 × 10− 5m2 Ve1 := 3.6 × 10− 7m3
ID
OD
AL1 := 180n⋅H
Bs1 := 1.5T⋅80%
-----------------------------------------------------------------NL_1 := L1 AL1
t
2
( Iomax) ⋅
t
1
0
ILrms(Vin , Io) := (Io)2 + (ΔIL(Vin))2 12
3.014 3.012 3.011 ILrms(Vin , Iomax) 3.009 3.008 3.006
8
ILpeak(Vin , Io) := Io + ΔIL(Vin) 2
t IL rms approximate formula : IL rms= Io,,in buck topology
10
12
14
Vin
Peak current through inductor
3.5
3.45 ILpeak( Vin, Iomax)
3.4
3.35 8
10
12
14
Vin
Choosing a ring-metal powder core to meet the design requirement
OD1 := 18mm
1.3 design variables define
Tmax := 105
Kt := 1 + 0.005(Tmax − 25)
Kac := 1.2
fs := 100k⋅Hz
ω := 2π fs
Ts := 1 fs
1.4 Basic variables calculation
Po(Io) := Voset Io
,
Ts)⎞⎟⎠
if 0 ≤ mod(t , Ts) ≤ Dnm(Vin)⋅Ts
⎡⎢⎣Io +
ΔIL(Vin) 2
−
Vo ⋅mod[(t L1
−
Dnm(Vin)⋅Ts) , Ts]⎤⎥⎦
if Dnm(Vin)⋅Ts < mod(t , Ts) ≤ Ts
4
IL(VINmin, Iomax , t) 3
IL(VINmax, Iomax , t)
1. variables explain
1.1 variables unit define
k := 103
u := 10− 6
n := 10− 9
p := 10− 12
mil := 2.54⋅10− 5m
ounce := 0.035mm
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7.5 PWM(VINmin , t)
3.75
− 3.8 t
Calculate inductance value ΔI := 0.3Iomax
Lmin(Vin) := Dnm(Vin)⋅Ts⋅(Vin − Vo) ΔI
t
Define the ΔI through inductance,recommend:(0.3-0.4) Io,for tradeoff inductor volume and ripple current. Minimum L value to meet the ΔI requirement
Po(Iomax) = 9.9 W
Pin(Io) := Vo⋅Io ηset
Dnm(Vin) :=
Vo
Vin + (Vo − Voset)
0.5
0.425 Dnm(Vin) 0.35
0.275
0.2
8
10
Max operation temperature of product Thermal coefficient Ac resistance coefficient Minmum PWM frequency,must higher than audio frequency 20k pwm frequency angle frequency pwm period
NL := round(NL_1)
Choosing a whole number
NL = 13
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1.2 spec define
VINmin := 8V
minmum in put voltage
VINnom := 12V
normal in put voltage
VINmax := 15V
maxmum in put voltage
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ηset := 95%
Target efficiency
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CRITICAL CURRENT (A)
1.5 1.125 ICR(Vin) 0.75 0.375
8
10
12
14
Vin
IL wave form
Current through inductor
IL(Vin , Io, t) :=
⎛⎜⎝Io −
ΔIL(Vin) 2
+
Vin
− Voset L1
⋅mod(t
actual ΔI
1
CURRENT A
0.925 ΔIL(Vin) 0.85
0.775
0.7
81012源自14VinCalculate critical current value ICR(Vin) := ΔIL(Vin) 2
Critical Io current
ICR(VINmax) = 0.478 A
Wire_Length := NL⋅(OD1 − ID1 + 2Ht)⋅1.1
The copper wire length
Wire_Length = 0.257 m
ΦL1 := 0.75mm
Choosing a wire diameter
DCR := Wire_Length⋅ρ
π⋅⎛⎜⎝
ΦL1 2
⎞⎟⎠
turns to meet the inductance value
NL_1 = 12.91 NLmin1 := ILpeak(VINmin, Iomax)⋅L1
Bs1⋅Ae1
Saftey turns to avoid the core saturation
NLmin1 = 4.188
IF the NL_1>NLmin1, the design is acceptable
2
DCR = 9.847 × 10− 3 Ω
PL1cond(Vin , Io) := DCR⋅(ILrms(Vin , Io))2⋅Kac
the wire dc resistance conduction loss result from copper wire
ΔB(Vin) := (Vin − Voset)⋅Dnm(Vin)⋅Ts NL⋅Ae1
Iomax := 3A
out put current max
Ioocp := 3.5A
output current ocp limit
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