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LT1576

LT1576首页预览图
型号: LT1576
PDF文件:
  • LT1576 PDF文件
  • LT1576 PDF在线浏览
功能描述: 1.5A, 200kHz Step-Down Switching Regulator
PDF文件大小: 290.41 Kbytes
PDF页数: 共28页
制造商: LINER[Linear Technology]
制造商LOGO: LINER[Linear Technology] LOGO
制造商网址: http://www.linear.com
捡单宝LT1576
PDF页面索引
120%
21
LT1576/LT1576-5
APPLICATIONS INFORMATION
WUU
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Analog experts will note that around 7kHz, phase dips
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
will
cause 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).
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
C
OUT
MP MA
Gain =1
()
=
()()()()
121.
G
MP
= Transconductance of power stage = 1.5A/V
G
MA
= Error amplifier transconductance = 1(10
–3
)
ESR = Output capacitor ESR
1.21 = Reference voltage
With V
OUT
= 5V and ESR = 0.1, a value of 27.5k 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
LT1576 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 200kHz.
V
R G V V ESR
VLf
C RIPPLE
C MA IN OUT
IN
()
=
()( )
()()()
()()()
121.
G
MA
= Error amplifier transconductance (1000µMho)
If a computer simulation of the LT1576 showed that a
series compensation resistor of 15k 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 = 30µH, would be:
V
k
V
C RIPPLE
()
=
()
()
()()()
()
()()
=
15 1 10 10 5 01 121
10 30 10 200 10
0151
3
63
•..
••
.
This ripple voltage is high enough to possibly create
subharmonic switching. In most situations a compromise
value (<10k 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
= 15k,
C
fR
k
pF
F
C
=
()()()
=
()
()
=
5
2
5
2 200 10 15
265
3
π
π
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