Electronic Components Datasheet Search
 Selected language English  ▼

## 71128 Datasheet(PDF) 2 Page - Vishay Siliconix

 Part No. 71128 Description Simple Solution for Dynamically Programming the Output Voltage of DC-DC Converters Download 7 Pages Scroll/Zoom 100% Maker VISHAY [Vishay Siliconix] Homepage http://www.vishay.com Logo

 2 page AN731Vishay Siliconixwww.vishay.comS FaxBack 408-970-56002Document Number: 7112828-Jan-00CIRCUIT ANALYSISIn a closed-loop power supply, node A will servo to attain avoltage equal to VR. Node B will servo to attain a voltage equalto Vr2. Assuming an ideal op-amp and using Kirchkorf’s currentlaw at node A and B, we have:(Vo* Vr)R1+(Vr* Vx)R2and(Vx* Vr2)R3+(Vr2* Vc)R4(1)Let:M1+ R2R1andM2+ R3R4(2)Solve for VX,Vx+ (1 ) m1)Vr * m1VoandVx+ (1 ) m2)Vr2 * m2Vc(3)Equate the above two equations and solve for VO:Vo+ 1m1 )1 Vr* 1 ) m2m1Vr2) m2m1Vc+ b ) aVc(4)Where:a+ m2m1andb+ 1m1 )1 Vr* 1 ) m2m1Vr2(5)So, VO is a linear function with respect to VC. The function hasslope a and y intercept b.A curve-fitting technique is used to force the equation (4) tofollow the requirement. This is done in two steps:Matching the slope:a+ Vo2 * Vo1Vc2* Vc1 +m2m1(6)Matching one point: Pick point B→VO2 = b + aVC2³ b + Vo2 * aVc2 + 1m1 )1 Vr* 1m1 )a Vr2(7)Equate (4) and (5) and solve for m1m1+Vr* Vr2Vo2) a(Vr2 * Vc2) * Vr(8)Note:m1+ R2R1(9)Since m1 is the ratio of 2 real resistors, it must be a positivenumber. Furthermore, m1 should not be too small or too largeto have realistic resistor values for R1 and R2. There are twovalid scenarios:1. Vr – Vr2 > 0andVo2 + a(Vr2 – Vc2)–Vr > 0 or,2. Vr – Vr2 < 0andVo2 + a(Vr2 – Vc2)–Vr < 0Both of these present a restricted range of values for Vr2 to givea meaningful value of m1. Once Vr2 is chosen correctly, m1and the rest of the parameter values can be determined.DESIGN PROCEDURE AND EXAMPLEGiven: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 * Vo1Vc2* Vc1 +3.4* 0.42.7* 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 usedhere.Vr–Vr2u 0 å Vr2 t Vr + 1.3 Vand,Vo2) a(Vr2–Vc2)–Vr u 0 å Vr2 uVr–Vo2a) Vc2 + 1.3 V–3.4 V1.2) 2.7 + 0.95V(11)