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