It can be shown that the closed-loop converter input impedance is
given by:
T(s)
1 = 1
1
+ 1
Z i(s) Z N (s) 1 + T(s) Z D(s) 1 + T(s)
where T(s) is the converter loop gain.
At frequencies below the loop crossover frequency, the input
impedance is approximately equal to ZN, which is a negative
resistance.
When an undamped or lightly damped input filter is connected to the
regulator input port, the input filter can interact with ZN to form a
negative resistance oscillator.
Input filter
Lf
Buck converter
with input filter
Converter
L
1
330 μH
Vg
+
–
30 V
i
+
100 μH
2
Cf
470 μF
C
R
100 μF
3Ω
v
–
D = 0.5
Small-signal model
Input filter
Lf
Converter model
1:D
+
–
330 μH
g +
–
Vg Cf
I
470 μF
Zo(s) Zi(s)
L
+
100 μH
C
R
100 μF
3Ω
–
L
1:D
+
Z D(s) = 12 sL + R || 1
sC
D
C
R
ZD(s)
–
40 dBΩ
30 dBΩ
1
2π LC
1.59 kHz
1
2πRC
530 Hz
fo =
f1 =
R = 12 Ω
D2
ωL
D2
|| ZD ||
|| ZN ||
20 dBΩ
1
ωD 2C
10 dBΩ
R0 /D 2
C = fo = 3 → 9.5 dB
L
f1
Q=R
0 dBΩ
100 Hz
1 kHz
10 kHz
f
test(s)
test(s)
0
1:D
0
test
+
–
Z N (s) =
+
test
Vg ZN(s)
L
+
+
s 0
I
C
–
–
R
0
–
Hence,
Solution:
test(s) = I (s)
–
Z N (s) =
Vg (s)
test(s) = –
D
Vg (s)
D
I(s)
= – R2
D
40 dBΩ
Qf → ∞
30 dBΩ
Lf
20 dBΩ
Cf
|| Zo ||
10 dBΩ
Zo(s)
0 dBΩ
Z o(s) = sL f || 1
sC f
R0 f =
ωL f
– 10 dBΩ
ff =
– 20 dBΩ
Lf
= 0.84 Ω
Cf
1
= 400 Hz
2π L f C f
100 Hz
1
ωC f
1 kHz
f
No resistance, hence poles are undamped (infinite Q-factor).
In practice, losses limit Q-factor; nonetheless, Qf may be very large.
Z o Z N , and
Zo ZD
40 dBΩ
Qf → ∞
30 dBΩ
f1 = 530 Hz
fo = 1.59 kHz
12 Ω
ωL
D2
|| ZN ||
20 dBΩ
10 dBΩ
|| Zo ||
1
ωD 2C
0 dBΩ
– 10 dBΩ
|| ZD ||
R0 /D 2
Q=3
R0f
ωL f
1
ωC f
ff = 400 Hz
– 20 dBΩ
100 Hz
1 kHz
10 kHz
f
Can meet
inequalities
everywhere
except at
resonant
frequency ff.
Need to damp
input filter!
Z
1+ o
ZN
Z
1+ o
ZD
10 dB
0 dB
– 10 dB
0˚
Zo
ZN
– 180˚ ∠
Z
1+ o
ZD
1+
– 360˚
10 kHz
1 kHz
100 Hz
f
|| Gvd ||
40 dB
30 dB
∠ Gvd
|| Gvd ||
Dashed lines: no
input filter
20 dB
10 dB
0 dB
∠ Gvd
0˚
– 10 dB
Solid lines:
including effect of
input filter
– 180˚
– 360˚
100 Hz
– 540˚
10 kHz
1 kHz
f
Undamped filter:
Two possible approaches:
Lf
Lf
Cf
Cf
Rf
Zo(s)
Rf
Lf
Cf
To meet the requirement Rf || ZN ||:
Lf
R f R2
D
Cf
Rf
The power loss in Rf is Vg2/ Rf,
which is larger than the load power!
Lf
A solution: add dc blocking
capacitor Cb.
Rf
Cf
Choose Cb so that its impedance is
sufficiently smaller than Rf at the
filter resonant frequency.
Cb
|| Zo ||, with large Cb
Lf
Rf
Rf
R0f
Cf
ωL f
Cb
ff
1
ωC f
40 dBΩ
f1 = 530 Hz
30 dBΩ
fo = 1.59 kHz
12 Ω
ωL
D2
|| ZN ||
20 dBΩ
1
ωD 2C
10 dBΩ
– 10 dBΩ
R0 /D 2
Q=3
|| Zo ||
Rf = 1 Ω
0 dBΩ
ωL f
– 20 dBΩ
100 Hz
|| ZD ||
R0f
1
ωC f
ff = 400 Hz
1 kHz
10 kHz
f
|| Gvd ||
40 dB
30 dB
∠ Gvd
|| Gvd ||
Dashed lines: no
input filter
20 dB
10 dB
0 dB
∠ Gvd
0˚
Solid lines:
including effect of
input filter
– 10 dB
– 90˚
– 180˚
10 kHz
1 kHz
100 Hz
f
Simulation of control-to-output responses
Undamped vs. damped input filters
Plotting the filter output impedances
Simulink files for previous slide
!
! ! ! !
! ! !
! !" ! ! #" ! Resulting control-to-output transfer function
Rf –Cb Parallel Damping
Rf –Lb Series Damping
Lf
v1
+
Lb
v2
Rf
Rf
+
–
Cf
Cb
v1
–
+
–
Lf
+
Cf
v2
–
Rf –Lb Parallel Damping
Lb
Rf
Lf
v1
+
–
Cf
+
• Size of Cb or Lb can become
very large
v2
• Need to optimize design
–