Datasheet

1
Q =
= 0.33
S[(3.36)(0.44) - 0.5]
1
2S(100 x 10
-6
)(0.01)
f
ESR
=
= 159kHz
f
p1
=
(500 x 10
3
)(3.3 x 10
-6
)(100 x 10
-6
)
+
1
(100 x 10
-6
)(0.83)
1
2S
1
(3.36)(0.44) - 0.5
>
@
= 2.86 kHz
A
DC
=
0.83
1.8 x 0.02
[(3.36(0.44) - 0.5]
(500 x 10
3
)(3.3 x 10
-6
)
1+
1
0.83
= 15.5
H = feedback gain =
1.27
2.5
= 0.508
F
C
(s) =
(sC
C1
R
C
+ 1)
s
2
C
C1
C
C2
R
C
R
GM
+ s(C
C2
R
GM
+ C
C1
(R
GM
+ R
C
)) + 1
, C
C2
used
F
C
(s) =
(sC
C1
R
C
+ 1)
sC
C1
R
GM
(R
GM
+ R
C
) + 1
, C
C2
not used
F
h
(s) =
+
1
s
2
1
Sf
s
2
s
1
Sf
s
Q
+ 1
F
P(S)
=
1 +
s
Sf
ESR
1 +
s
Sf
p1
R
GM
+ R
C
Sf
ESR
R
GM
R
C
C
C2
=
(F)
LM3477
www.ti.com SNVS141K –OCTOBER 2000–REVISED MARCH 2013
(47)
PLOTTING THE OPEN LOOP RESPONSE
The open loop response is expressed as:
T = A
DC
x A
CM
x H x F
p
(s) x F
c
(s)
Where A
DC
and H are given above and
A
CM
= GM x R
GM
(48)
(49)
(50)
(51)
One can plot the magnitude and phase of the open loop response to analyze the frequency response.
EXAMPLE: COMPENSATION DESIGN
4.5V ≤ V
IN
≤ 5.5V
V
OUT
= 2.5V
I
OUT
= 3A (R = 0.83Ω)
R
SN
= 0.02Ω
L = 3.3µH
R
SL
= 0Ω
C
OUT
= 100µF
R
ESR
= 0.01Ω
First, calculate the power stage parameters using V
IN(MIN)
and R
(MAX)
:
(52)
(53)
(54)
(55)
(56)
In this example, a crossover frequency of 20kHz is chosen, so: f
C
= 20000. R
C
is now calculated using the power
stage information and the target crossover frequency f
C
:
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