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71128 Datasheet(PDF) 2 Page - Vishay Siliconix

Part No. 71128
Description  Simple Solution for Dynamically Programming the Output Voltage of DC-DC Converters
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Maker  VISHAY [Vishay Siliconix]
Homepage  http://www.vishay.com
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AN731
Vishay Siliconix
www.vishay.com
S FaxBack 408-970-5600
2
Document Number: 71128
28-Jan-00
CIRCUIT ANALYSIS
In a closed-loop power supply, node A will servo to attain a
voltage equal to VR. Node B will servo to attain a voltage equal
to Vr2. 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 VX,
Vx
+ (1 ) m1)Vr * m1Vo
and
Vx
+ (1 ) m2)Vr2 * m2Vc
(3)
Equate the above two equations and solve for VO:
Vo
+ 1
m1 )
1 Vr
* 1 ) m2
m1
Vr2
) m2
m1
Vc
+ b ) aVc
(4)
Where:
a
+ m2
m1
and
b
+ 1
m1 )
1 Vr
* 1 ) m2
m1
Vr2
(5)
So, VO is a linear function with respect to VC. 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 * Vo1
Vc2
* Vc1 +
m2
m1
(6)
Matching one point: Pick point B
→VO2 = b + aVC2
³ 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 Vr2 to give
a meaningful value of m1. Once Vr2 is chosen correctly, m1
and the rest of the parameter values can be determined.
DESIGN PROCEDURE AND EXAMPLE
Given:
A = (VC1, VO1) = (0.2 V, 0.4 V), B = (VC2, VO2) = (2.7 V, 3.4 V)
Vr = 1.3 V, R1 = 22.1 kW. Also, 1 V < VX < 3 V.
Calculate the slope, a:
a
+ Vo2 * Vo1
Vc2
* Vc1 +
3.4
* 0.4
2.7
* 0.2 +
1.2
(10)
Determine Vr2:
Choose a sensible value of Vr2 to satisfy either (1) or (2) above.
Since it is easier to derive a value for Vr2 that is smaller than Vr
(by using a simple resistor voltage divider), scenario (1) is used
here.
Vr–Vr2
u 0 å Vr2 t Vr + 1.3 V
and,
Vo2
) a(Vr2–Vc2)–Vr u 0 å Vr2 u
Vr–Vo2
a
) Vc2 + 1.3 V–3.4 V
1.2
) 2.7 + 0.95V
(11)




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