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# 71128

## Simple Solution for Dynamically Programming the Output Voltage of DC-DC Converters

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AN731
Vishay Siliconix
Simple Solution for Dynamically Programming the
Output Voltage of DC-DC Converters
INTRODUCTION
Figure 1 shows a typical dc-to-dc converter configuration. As in
many PWM controllers, the non-inverting input of the voltage
feedback error amplifier is internally connected to the
reference voltage, Vr. The output voltage of the converter is set
by a resistor divider network R1 and R2. This configuration
enables a fixed output voltage that is equal to or greater than
the reference voltage. But in many power conversion designs,
it is useful to vary the output to a value lower than the reference
voltage and to dynamically adjust the output voltage. The
op-amp circuit shown in Figure 3 gives designers a simple way
to do this. Scaling the output voltage with respect to a control
voltage VC is totally flexible. This application note describes
how to use this simple solution and provides a step-by-step
design procedure to assist designers in calculating the specific
parameters required by their circuits.
DESIGN REQUIREMENT
Reference voltage, Vr
Control voltage: VC = VC1 to VC2
Output voltage: VO = VO1 at VC = VC1, VO = VO2 at VC = VC2,
and VO is linear for any value of VC in its range.
These design requirements are shown in Figure 2. The output
voltage is a linear function with respect to the control voltage.
The function crosses two end points A = (VC1, VO1) and B =
(VC2, VO2).
VIN Power Stage
VOUT
PWM
+
EA
PWM Controller
FB
Vr
R1
R2
FIGURE 1. Typical DC/DC Converter Configuration
VO (V)
B (VC2, VO2)
A (VC1, VO1)
VC (V)
FIGURE 2. Output Voltage vs Control Voltage Requirement
VIN
Document Number: 71128
28-Jan-00
Power Stage
VOUT
PWM
+
EA
PWM Controller
FB
Vr
R1
A
R2
VX
R3
LM358
B
+ Vr2
R4
Op-Amp Circuit
FIGURE 3. Op-Amp Circuit Offers Programmable Output Function
VC
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AN731
Vishay Siliconix
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 *
R1
Vr)
+
(Vr
* Vx)
R2
and
(Vx
* Vr2)
R3
+
(Vr2 *
R4
Vc)
(1)
Let:
M1
+
R2
R1
and
M2
+
R3
R4
Solve for VX,
Vx + (1 ) m1)Vr * m1Vo
and
Vx + (1 ) m2)Vr2 * m2Vc
(2)
(3)
Equate the above two equations and solve for VO:
ǒ Ǔ ǒ ǓVo +
1
m1
)
1
Vr *
1 ) m2
m1
Vr2
)
m2
m1
Vc
+
b ) aVc
Where:
(4)
a
+
m2
m1
and
ǒ Ǔ ǒ Ǔb +
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
Vc2
*
*
Vo1
Vc1
+
m2
m1
Matching one point: Pick point B VO2 = b + aVC2
(6)
ǒ Ǔ ǒ Ǔ³ b + Vo2 * aVc2 +
1
m1
)
1
Vr *
1
m1
)
a
Vr2
Equate (4) and (5) and solve for m1
(7)
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 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
Vc2
*
*
Vo1
Vc1
+
3.4
2.7
*
*
0.4
0.2
+
1.2
Determine Vr2:
(10)
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
1.2
V ) 2.7 +
0.95V
(11)
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Document Number: 71128
28-Jan-00

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