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參數(shù)資料
| 型號(hào)： | 71128 |
| 廠商： | Vishay Intertechnology,Inc. |
| 元件分類： | DC/DC變換器 |
| 英文描述： | Simple Solution for Dynamically Programming the Output Voltage of DC-DC Converters |
| 中文描述： | 簡(jiǎn)單的解決方案動(dòng)態(tài)編程的直流輸出電壓- DC轉(zhuǎn)換器 |
| 文件頁數(shù)： | 2/7頁 |
| 文件大?。?/td> | 77K |
| 代理商： | 71128 |
AN731
Vishay Siliconix
www.vishay.com FaxBack 408-970-5600
2
Document Number: 71128
28-Jan-00
In a closed-loop power supply, node A will servo to attain a
voltage equal to V
R
. Node B will servo to attain a voltage equal
to V
r2
. Assuming an ideal op-amp and using Kirchkorf’s current
law at node A and B, we have:
(Vo
Vr)
R1
(Vr
Vx)
R2
and
(Vx
Vr2)
R3
(Vr2
Vc)
R4
(1)
Let:
M1
R2
R1
and
M2
R3
R4
(2)
Solve for V
X
,
Vx
(1
m1)Vr
m1Vo
and
Vx
(1
m2)Vr2
m2Vc
(3)
Equate the above two equations and solve for V
O
:
Vo
1
m1
1 Vr
1
m2
m1
Vr2
m2
b
aVc
(4)
Where:
a
m2
m1
and
b
1
m1
1 Vr
1
m2
m1
Vr2
(5)
So, V
O
is a linear function with respect to V
C
. The function has
slope a and y intercept b.
A curve-fitting technique is used to force the equation (4) to
follow the requirement. This is done in two steps:
Matching the slope:
a
Vo2
Vc2
Vo1
Vc1
m2
m1
(6)
Matching one point: Pick point B
→
V
O2
= b + aV
C2
b
Vo2
aVc2
1
m1
1 Vr
1
m1
a Vr2
(7)
Equate (4) and (5) and solve for m1
m1
Vr
Vr2
Vo2
a(Vr2
Vc2)
Vr
(8)
Note:
m1
R2
R1
(9)
Since m1 is the ratio of 2 real resistors, it must be a positive
number. Furthermore, m1 should not be too small or too large
to have realistic resistor values for R1 and R2. There are two
valid scenarios:
1. Vr – Vr2 > 0
and
Vo2 + a(Vr2 – Vc2)–Vr > 0 or,
2. Vr – Vr2 < 0
and
Vo2 + a(Vr2 – Vc2)–Vr < 0
Both of these present a restricted range of values for V
r2
to give
a meaningful value of m1. Once V
r2
is chosen correctly, m1
and the rest of the parameter values can be determined.
Given:
A = (V
C1
, V
O1
) = (0.2 V, 0.4 V), B = (V
C2
, V
O2
) = (2.7 V, 3.4 V)
V
r
= 1.3 V, R1 = 22.1 k . Also, 1 V < V
X
< 3 V.
Calculate the slope, a:
a
Vo2
Vc2
Vo1
Vc1
3.4
2.7
0.4
0.2
1.2
(10)
Determine V
r2
:
Choose a sensible value of V
r2
to satisfy either (1) or (2) above.
Since it is easier to derive a value for V
r2
that is smaller than V
r
(by using a simple resistor voltage divider), scenario (1) is used
here.
Vr–Vr2
and,
0
Vr2
Vr
1.3 V
Vo2
Vr–Vo2
a
a(Vr2–Vc2)–Vr
0
Vr2
Vc2
1.3 V–3.4 V
1.2
2.7
0.95V
(11)
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