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AN1623 数据表(PDF) 5 Page - STMicroelectronics

部件名 AN1623
功能描述  REGULATION LOOPS PRECISION CALCULATION
PDF  6 Pages
Scroll/Zoom Zoom In 100%  Zoom Out
制造商  STMICROELECTRONICS [STMicroelectronics]
网页  http://www.st.com
标志 STMICROELECTRONICS - STMicroelectronics

AN1623 数据表(HTML) 5 Page - STMicroelectronics

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AN1623 - APPLICATION NOTE
5/6
2.2 Voltage Loop Precision Calculation
2.2.1 Theoretical Calculation
Let’s show some different steps in order to
calculate the voltage precision loop.
According to Figure 4 it is simple to verify that:
Equation 24
where
Equation 25
∆V
lim
tot is the sum of:
Equation 26
where
Equation 27
Equation 28
Equation 29
Equation 30
Equation 31
Substituting
Equation 27 and Equation 31 in
Equation 26 and considering the absolute value of
terms:
Equation 32
Equation 32 is more important in order to
calculate
∆V
lim
tot
and shows many parameters
that characterize voltage variation.
It can be observed that Vio=4mV, Vref=2.5V and
V1<5mV, so the term Vio+V1 can be neglected
with respect to Vref .
2.2.2 Practical Example
In order to better understand the formula, let’s
take a practical example that uses absolute and
relative
tolerances
typical
for
TSM011
applications.
Let’s consider for example:
see
Footnote 1
Assume that Iout will vary between 0 and 5 mA,
therefore:
Vlim = 18V
Gm = 3.5mA/mV
R1 = 30kΩ - R2 = 186kΩ
rewriting
Equation 32
Equation 33
it is possible to obtain a good current variation
()
1
io
ref
1
2
lim
V
V
V
R
R
1
V
−
−


+
=
m
out
1
G
I
V
=
Iout
lim
1
R
lim
2
R
lim
Vio
lim
Vref
lim
tot
lim
I
I
V
V
V
V
∆
+
∆
+
∆
+
∆
+
∆
=
∆
1
2
ref
lim
Vref
lim
R
R
1
V
V
V
+
=
∆
∆
=


+
−
=
∆
∆
=
1
2
io
lim
Vio
lim
R
R
1
V
V
V


+
−
=
∆
∆
=
1
2
m
io
lim
Iout
lim
R
R
1
G
1
I
V
V
()
1
io
ref
1
2
lim
2
R
lim
V
V
V
R
1
R
V
V
−
−
=
∆
∆
=
()
1
io
ref
2
1
2
1
lim
1
R
lim
V
V
V
R
R
R
V
V
−
−
−
=
∆
∆
=
()
()
1
1
io
ref
2
1
2
2
1
io
ref
1
out
1
2
m
io
1
2
ref
1
2
tot
lim
R
V
V
V
R
R
R
V
V
V
R
1
I
R
R
1
G
1
V
R
R
1
V
R
R
1
V
∆
−
−
+
+
∆
−
−
+
+
∆


+
+
+
∆


+
+
+
∆


+
=
%
5
.
0
V
V
ref
ref
±
=
∆
%
1
R
R
1
1
±
=
∆
%
1
R
R
2
2
±
=
∆
mV
2
Vio
±
=
∆
mA
5
.
2
Iout
±
=
∆
1
1
ref
1
2
2
2
ref
1
2
out
1
2
m
io
1
2
ref
ref
1
2
ref
tot
lim
R
R
V
R
R
R
R
V
R
R
I
R
R
1
G
1
V
R
R
1
V
V
R
R
1
V
V
∆
+
∆
+
+
∆


+
+
+
∆


+
+
+
∆


+
=
2.3%
V
V
lim
tot
lim
±
=
∆



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