71128 datasheet(2/7 Pages) VISHAY

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AN731

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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)

(Vr

* Vx)

and

(Vx

* Vr2)

(Vr2

* Vc)

(1)

Let:

+ R2

and

+ R3

(2)

Solve for VX,

+ (1 ) m1)Vr * m1Vo

and

+ (1 ) m2)Vr2 * m2Vc

(3)

Equate the above two equations and solve for VO:

+ 1

m1 )

1 Vr

* 1 ) m2

Vr2

) m2

+ b ) aVc

(4)

Where:

+ m2

and

+ 1

m1 )

1 Vr

* 1 ) m2

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:

+ Vo2 * Vo1

Vc2

* Vc1 +

(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

* Vr2

Vo2

) a(Vr2 * Vc2) * Vr

(8)

Note:

+ R2

(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:

+ 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

) Vc2 + 1.3 V–3.4 V

1.2

) 2.7 + 0.95V

(11)

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