寄生参数计算
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S = 20mm L = 8mm
PCB Lp = 12.5nH
Two parallel cylindrical conductors of radius rw
j i
L
s
2 2 µ0L L rw r L w Lpii = Lpjj = ln + + 1 + − + 1 2π rw rw L L
1 0.8
[Ω / m]
S = PCB trace cross section
rAC rAC max
Wide flat conductors have less AC resistance than round or square cross section conductors 6
0.6 0.4 0.2 0 0 2 4 6 8
I Infinite length rW s
µ0 −1 s l e = cosh π 2rW
s I
[H / m]
8
Equivalent Circuit
le L le L (le/2)L (le/2)L
L is wire length
The external inductance can be associated to any of the two conductors Is there a right way to do that?
Average depth of current penetration
T W
δ
AC resistance:
πµ 0 1 rAC ≅ f [Ω / m] 2( W + T ) σ
5
AC Resistance of a Rectangular Conductor
T W
δ
rAC
πµ 0 ≅ f S σ + 2 W W 1
I2
1 2
L
I2 I3
3
Lp22 Lp23 Lp24 Lp44 I4 I3 Lp33
I1
4
s
Lp11 I1
ห้องสมุดไป่ตู้
Lp21
I4
I1=I2=I3=I4=I, Lp11=Lp33=Lpt, Lp22=Lp44=Lpl Lp21=Lp23=Lp41=Lp43=0, Lp24=Lp42=Lml, Lp13=Lp31=Lmt
15
LLoop = 2(Lpl − Lml ) + 2(Lpt − Lmt )
Total Inductance of a Rectangular Loop
Lpl- Lml Lpt- Lmt Lpl- Lml
µ0L s s Lpl − Lml ≈ ln − 2π rw L
Above some tens of kHz inductance is of major concern!
[dB] 40 20 0
z(f ) rDC
r (f ) rDC
23
0.001
0.01
0.1
1
10 [MHz]
Resistors
1) Wire wound 2) Film type 3) Composition
9
Yes, use Partial Inductance Concept
I2
1 2
I2 I3
3
Lp22 Lp23 Lp24 Lp44 I4 I3 Lp33
Lp11 I1
Lp21
I1
4
I4
Voltage across segment 2:
10
dI3 dI2 dI1 dI4 V2 = Lp22 − Lp21 + Lp23 − Lp24 dt dt dt dt
17
Formula Comparison
l eL ε = 1− LLoop
0.25 0.2 0.15 0.1 0.05 0 0
s =8 rw
s =4 rw
L 200 rw
s =6 rw
50 100
150
Example: AWG 20 rw = 16mil (406.4µm), L = 1cm LLoop = 10.88nH
W S
Conductor Internal Inductance
rW
µ0 l i _ DC = 8π
[H/ m]
µ0 1 1 li _ AC ≅ 4πrW πσ f
[H / m]
The internal inductance plays no role at high frequency
7
External Inductance
Series impedance of two conductors AWG20 (26x34) at 2 mm distance (copper) External inductance: DC Resistance: le ≈ 620 nH/m rDC ≈ 33 mΩ/m
22
Impedance per Unit Length
2 2 L L µ0L s s Lpij = ln + + 1 + − + 1 L L 2π s s
13
Partial Self and Mutual Inductance
If s/L <<1
Partial Self Inductance
Ii Bii
i
Segment i
Lpii
si ∞
11
∫ s =
i
r r Biidsi Ii
Partial Mutual Inductance
Ij
j i
Segment j
Bij
Lpij
si ∞
12
∫ s =
i
r r Bij dsi Ij
Partial Self and Mutual Inductance
Relevant EMI “Components”
• Only one third of the components affecting EMI are on the schematics • Another third are parasitic elements within components • The final third are created by PC board trace routing, and component mounting, placement, and even orientation • Any peace of wire and PCB trace has a nonzero impedance
Rideal
Lp1 > Lp2 > Lp3
R
Lp
CP
24
R Z ( jω) = jωLp + 1+ jωRCp
Parasitic Inductance
Except for wire wound resistors, parasitic inductance is mainly due to lead length and separation
1m [µH]
1,616 1,640 1,663 1,687 1,708 1,733 1,756 1,778 1,801 1,825 1,849 1,872 1,893 1,916
1,269 1,292 1,315 1,338 1,361 1,385 1,408 1,431 1,454 1,477 1,500 1,524 1,547 1,570 1,593
2
Resistance of a Cylindrical Conductor
Uniform current distribution inside the conductor
DW
DC: rDC
4 = 2 σπD W
[Ω / m]
3
(σ = conductivity)
Resistance of a Cylindrical Conductor
δ
DW
Non uniform current distribution inside the conductor due to skin effect
Skin depth (σ = conductivity)
1 AC: δ = µ 0 σπf
4
1 µ0 rAC ≅ f DW πσ
[Ω / m]
Resistance of a Rectangular Conductor
µ0L 2L rw Lpii = Lpjj ≈ − 1+ ln 2π rw L
µ0L 2L s Lpij ≈ − 1+ ln L 2π s
14
Total Inductance of a Rectangular Loop
L
L
25 26 27 28 29 30 31 32 33 34 35 36 37 38 .0179 .0159 .0142 .0126 .0113 .0100 .0089 .0080 .0071 .0063 .0056 .0050 .0045 .0040
1cm 10cm [nH]
115,64 118,00 120,26 122,65 124,82 127,26 129,59 131,72 134,11 136,49 138,85 141,11 143,22 145,57 6,999 7,231 7,453 7,688 7,903 8,144 8,374 8,585 8,822 9,059 9,292 9,518 9,727 9,961
External Inductance and Capacitance
rW s If s > 5rW
µ0 s l e ≅ ln [H/ m] π rW
21
+ + + + + +
s
- - - - - -
π ε0 c≅ s ln rW
[F / m]
Example
Parasitic Components
Outline
• Conductor parasitic elements
– resistance – inductance – capacitance
• • • • • •
1
Resistors Capacitors Inductors Ferrite beads Parasitic components in circuit layout Transformers
LLoop ≈ 2(Lpl − Lml )
16
Lpt- Lmt
s << 1 L
Total Inductance of a Rectangular Loop
LLoop ≈ 2(Lpl − Lml )
Because the partial mutual inductance Lm between two conductors increases as their separation is reduced, whereas their partial self inductance Lp does not change, the net partial inductance Lp-Lm decreases as the conductors are brought closer together.
18
leL = 9.105nH
Partial Self Inductance
1cm 10cm 1m AWG dw [in] [nH] [nH] [µH]
10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 .1019 .0907 .0808 .0720 .0641 .0571 .0508 .0453 .0403 .0359 .0320 .0285 .0253 .0226 .0201 3,726 3,932 4,140 4,349 4,562 4,777 4,995 5,211 5,432 5,652 5,873 6,096 6,326 6,545 6,773 81,07 83,37 85,65 87,94 90,24 92,54 94,86 97,14 99,46 101,77 104,06 106,36 108,74 110,99 113,33
AWG dw [in] [nH]
19
Capacitance
Infinite length
+ + + + + +
s
- - - - - -
Homogeneous medium: l ec = µ ε
µ 0ε0 c= = le
20
π ε0 −1 s cosh 2rW
[F / m]
PCB Lp = 12.5nH
Two parallel cylindrical conductors of radius rw
j i
L
s
2 2 µ0L L rw r L w Lpii = Lpjj = ln + + 1 + − + 1 2π rw rw L L
1 0.8
[Ω / m]
S = PCB trace cross section
rAC rAC max
Wide flat conductors have less AC resistance than round or square cross section conductors 6
0.6 0.4 0.2 0 0 2 4 6 8
I Infinite length rW s
µ0 −1 s l e = cosh π 2rW
s I
[H / m]
8
Equivalent Circuit
le L le L (le/2)L (le/2)L
L is wire length
The external inductance can be associated to any of the two conductors Is there a right way to do that?
Average depth of current penetration
T W
δ
AC resistance:
πµ 0 1 rAC ≅ f [Ω / m] 2( W + T ) σ
5
AC Resistance of a Rectangular Conductor
T W
δ
rAC
πµ 0 ≅ f S σ + 2 W W 1
I2
1 2
L
I2 I3
3
Lp22 Lp23 Lp24 Lp44 I4 I3 Lp33
I1
4
s
Lp11 I1
ห้องสมุดไป่ตู้
Lp21
I4
I1=I2=I3=I4=I, Lp11=Lp33=Lpt, Lp22=Lp44=Lpl Lp21=Lp23=Lp41=Lp43=0, Lp24=Lp42=Lml, Lp13=Lp31=Lmt
15
LLoop = 2(Lpl − Lml ) + 2(Lpt − Lmt )
Total Inductance of a Rectangular Loop
Lpl- Lml Lpt- Lmt Lpl- Lml
µ0L s s Lpl − Lml ≈ ln − 2π rw L
Above some tens of kHz inductance is of major concern!
[dB] 40 20 0
z(f ) rDC
r (f ) rDC
23
0.001
0.01
0.1
1
10 [MHz]
Resistors
1) Wire wound 2) Film type 3) Composition
9
Yes, use Partial Inductance Concept
I2
1 2
I2 I3
3
Lp22 Lp23 Lp24 Lp44 I4 I3 Lp33
Lp11 I1
Lp21
I1
4
I4
Voltage across segment 2:
10
dI3 dI2 dI1 dI4 V2 = Lp22 − Lp21 + Lp23 − Lp24 dt dt dt dt
17
Formula Comparison
l eL ε = 1− LLoop
0.25 0.2 0.15 0.1 0.05 0 0
s =8 rw
s =4 rw
L 200 rw
s =6 rw
50 100
150
Example: AWG 20 rw = 16mil (406.4µm), L = 1cm LLoop = 10.88nH
W S
Conductor Internal Inductance
rW
µ0 l i _ DC = 8π
[H/ m]
µ0 1 1 li _ AC ≅ 4πrW πσ f
[H / m]
The internal inductance plays no role at high frequency
7
External Inductance
Series impedance of two conductors AWG20 (26x34) at 2 mm distance (copper) External inductance: DC Resistance: le ≈ 620 nH/m rDC ≈ 33 mΩ/m
22
Impedance per Unit Length
2 2 L L µ0L s s Lpij = ln + + 1 + − + 1 L L 2π s s
13
Partial Self and Mutual Inductance
If s/L <<1
Partial Self Inductance
Ii Bii
i
Segment i
Lpii
si ∞
11
∫ s =
i
r r Biidsi Ii
Partial Mutual Inductance
Ij
j i
Segment j
Bij
Lpij
si ∞
12
∫ s =
i
r r Bij dsi Ij
Partial Self and Mutual Inductance
Relevant EMI “Components”
• Only one third of the components affecting EMI are on the schematics • Another third are parasitic elements within components • The final third are created by PC board trace routing, and component mounting, placement, and even orientation • Any peace of wire and PCB trace has a nonzero impedance
Rideal
Lp1 > Lp2 > Lp3
R
Lp
CP
24
R Z ( jω) = jωLp + 1+ jωRCp
Parasitic Inductance
Except for wire wound resistors, parasitic inductance is mainly due to lead length and separation
1m [µH]
1,616 1,640 1,663 1,687 1,708 1,733 1,756 1,778 1,801 1,825 1,849 1,872 1,893 1,916
1,269 1,292 1,315 1,338 1,361 1,385 1,408 1,431 1,454 1,477 1,500 1,524 1,547 1,570 1,593
2
Resistance of a Cylindrical Conductor
Uniform current distribution inside the conductor
DW
DC: rDC
4 = 2 σπD W
[Ω / m]
3
(σ = conductivity)
Resistance of a Cylindrical Conductor
δ
DW
Non uniform current distribution inside the conductor due to skin effect
Skin depth (σ = conductivity)
1 AC: δ = µ 0 σπf
4
1 µ0 rAC ≅ f DW πσ
[Ω / m]
Resistance of a Rectangular Conductor
µ0L 2L rw Lpii = Lpjj ≈ − 1+ ln 2π rw L
µ0L 2L s Lpij ≈ − 1+ ln L 2π s
14
Total Inductance of a Rectangular Loop
L
L
25 26 27 28 29 30 31 32 33 34 35 36 37 38 .0179 .0159 .0142 .0126 .0113 .0100 .0089 .0080 .0071 .0063 .0056 .0050 .0045 .0040
1cm 10cm [nH]
115,64 118,00 120,26 122,65 124,82 127,26 129,59 131,72 134,11 136,49 138,85 141,11 143,22 145,57 6,999 7,231 7,453 7,688 7,903 8,144 8,374 8,585 8,822 9,059 9,292 9,518 9,727 9,961
External Inductance and Capacitance
rW s If s > 5rW
µ0 s l e ≅ ln [H/ m] π rW
21
+ + + + + +
s
- - - - - -
π ε0 c≅ s ln rW
[F / m]
Example
Parasitic Components
Outline
• Conductor parasitic elements
– resistance – inductance – capacitance
• • • • • •
1
Resistors Capacitors Inductors Ferrite beads Parasitic components in circuit layout Transformers
LLoop ≈ 2(Lpl − Lml )
16
Lpt- Lmt
s << 1 L
Total Inductance of a Rectangular Loop
LLoop ≈ 2(Lpl − Lml )
Because the partial mutual inductance Lm between two conductors increases as their separation is reduced, whereas their partial self inductance Lp does not change, the net partial inductance Lp-Lm decreases as the conductors are brought closer together.
18
leL = 9.105nH
Partial Self Inductance
1cm 10cm 1m AWG dw [in] [nH] [nH] [µH]
10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 .1019 .0907 .0808 .0720 .0641 .0571 .0508 .0453 .0403 .0359 .0320 .0285 .0253 .0226 .0201 3,726 3,932 4,140 4,349 4,562 4,777 4,995 5,211 5,432 5,652 5,873 6,096 6,326 6,545 6,773 81,07 83,37 85,65 87,94 90,24 92,54 94,86 97,14 99,46 101,77 104,06 106,36 108,74 110,99 113,33
AWG dw [in] [nH]
19
Capacitance
Infinite length
+ + + + + +
s
- - - - - -
Homogeneous medium: l ec = µ ε
µ 0ε0 c= = le
20
π ε0 −1 s cosh 2rW
[F / m]