參數(shù)資料
型號: LT1374IS8-5
廠商: LINEAR TECHNOLOGY CORP
元件分類: 穩(wěn)壓器
英文描述: Octal D-Type Transparent Latches With 3-State Outputs 20-SSOP 0 to 70
中文描述: 8.5 A SWITCHING REGULATOR, 560 kHz SWITCHING FREQ-MAX, PDSO8
封裝: 0.150 INCH, PLASTIC, SO-8
文件頁數(shù): 21/28頁
文件大?。?/td> 261K
代理商: LT1374IS8-5
21
LT1374
APPLICATIO
S I
FOR
ATIO
U
Analog experts will note that around 4.4kHz, phase dips
very close to the zero phase margin line. This is typical of
switching regulators, especially those that operate over a
wide range of loads. This region of low phase is not a
problem as long as it does not occur near unity-gain. In
practice, the variability of output capacitor ESR tends to
dominate all other effects with respect to loop response.
Variations in ESR willcause unity-gain to move around,
but at the same time phase moves with it so that adequate
phase margin is maintained over a very wide range of ESR
(
±
3:1).
W
U
U
What About a Resistor in the Compensation Network
It is common practice in switching regulator design to add
a “zero” to the error amplifier compensation to increase
loop phase margin. This zero is created in the external
network in the form of a resistor (R
C
) in series with the
compensation capacitor. Increasing the size of this resis-
tor generally creates better and better loop stability, but
there are two limitations on its value. First, the combina-
tion of output capacitor ESR and a large value for R
C
may
cause loop gain to stop rolling off altogether, creating a
gain margin problem. An approximate formula for R
C
where gain margin falls to zero is:
R Loop
V
G
G
ESR
OUT
MP
MA
Gain =1
(
)
=
(
)(
)(
)
2 42
G
MP
= Transconductance of power stage = 5.3A/V
G
MA
= Error amplifier transconductance = 2(10
–3
)
ESR = Output capacitor ESR
2.42 = Reference voltage
With V
OUT
= 5V and ESR = 0.03
, a value of 6.5k for R
C
would yield zero gain margin, so this represents an upper
limit. There is a second limitation however which has
nothing to do with theoretical small signal dynamics. This
resistor sets high frequency gain of the error amplifier,
including the gain at the switching frequency. If switching
frequency gain is high enough, output ripple voltage will
appear at the V
C
pin with enough amplitude to muck up
proper operation of the regulator. In the marginal case,
subharmonic switching occurs, as evidenced by alternat-
ing pulse widths seen at the switch node. In more severe
cases, the regulator squeals or hisses audibly even though
the output voltage is still roughly correct. None of this will
show on a theoretical Bode plot because Bode is an
amplitude insensitive analysis. Tests have shown that if
ripple voltage on the V
C
is held to less than 100mV
P-P
, the
LT1374 will be well behaved The formula below will give
an estimate of V
C
ripple voltage when R
C
is added to the
loop, assuming that R
C
is large compared to the reactance
of C
C
at 500kHz.
)
=
( )(
( )( )( )
V
R
G
V
V
ESR
)(
V
L f
C RIPPLE
C
MA
IN
OUT
IN
)
(
)( )
2 4
.
G
MA
= Error amplifier transconductance (2000
μ
Mho)
If a computer simulation of the LT1374 showed that a
series compensation resistor of 3k gave best overall loop
response, with adequate gain margin, the resulting V
C
pin
ripple voltage with V
IN
= 10V, V
OUT
= 5V, ESR = 0.1
,
L = 10
μ
H, would be:
=
( )
( )
V
k
V
C RIPPLE
)
(
)( )( )
500 10
=
3
2 10
10 5 0 1 2 4
0 144
.
3
6
3
.
.
This ripple voltage is high enough to possibly create
subharmonic switching. In most situations a compromise
value (<2k in this case) for the resistor gives acceptable
phase margin and no subharmonic problems. In other
cases, the resistor may have to be larger to get acceptable
phase response, and some means must be used to control
ripple voltage at the V
C
pin. The suggested way to do this
is to add a capacitor (C
F
) in parallel with the R
C
/C
C
network
on the V
C
pin. Pole frequency for this capacitor is typically
set at one-fifth of switching frequency so that it provides
significant attenuation of switching ripple, but does not
add unacceptable phase shift at loop unity-gain frequency.
With R
C
= 3k,
C
f R
k
pF
F
C
=
( )( )( )
=
( )
=
5
2
5
2
500 10
3
531
3
π
π
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