71128 Datasheet, PDF (2/7 Pages) Vishay Siliconix – Simple Solution for Dynamically Programming the Output Voltage of DC-DC Converters

AN731
Vishay Siliconix
CIRCUIT ANALYSIS
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 *
R1
Vr)
+
(Vr
* Vx)
R2
and
(1)
(Vx
* Vr2)
R3
+
(Vr2 *
R4
Vc)
Let:
M1
+
R2
R1
and
(2)
M2
+
R3
R4
Solve for V_X,
Vx + (1 ) m1)Vr * m1Vo
and
(3)
Vx + (1 ) m2)Vr2 * m2Vc
Equate the above two equations and solve for V_O:
ǒ Ǔ ǒ Ǔ Vo +
1
m1
)
1
Vr *
1 ) m2
m1
Vr2
)
m2
m1
Vc
+
b ) aVc
(4)
Where:
a
+
m2
m1
and
(5)
ǒ Ǔ ǒ Ǔ b +
1
m1
)
1
Vr
*
1 ) m2
m1
Vr2
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
ǒ Ǔ ǒ Ǔ ³ b + Vo2 * aVc2 +
1
m1
)
1
Vr *
1
m1
)
a
Vr2
(7)
Equate (4) and (5) and solve for m1
m1
+
Vo2
)
Vr * Vr2
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.
DESIGN PROCEDURE AND EXAMPLE
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 kW. 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 u 0 å Vr2 t Vr + 1.3 V
and,
Vo2 ) a(Vr2–Vc2)–Vr u 0 å Vr2 u
(11)
Vr–Vo2
a
)
Vc2
+
1.3
V–3.4
1.2
V _{) 2.7 +}
0.95V
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Document Number: 71128
28-Jan-00