Question

Problem 3 For the given measurement system, which measures temperature using an RTD, of type PT100, that is 100 Ohms at 0 C. If R2-R3 1000 Ohms, R1 110, At what temperature the bridge will be balanced, knowing that RTD is approximated with the given 1. equation Rold(1+0.004(change in temperature)) Rew= 2. If the amplifier has a gain of 10; determine accepted RA values of Rf and Rin RS N If the output of the amplifire is limited from -2.5 V 3. to 3 V, what are the temperature limits of the No device. 4. Assume that this sensor was calibrated to measure RE temperatures between -50 and 0 degrees using stnadrad current signal. What would be the reading of the sensor at 2 mA, 9 mA, 22 mA

Answer 1

Brdge co1 be balanced oWhen CDe knasthat Ra R2eRg-1000 R 10n yp= Rod C1 0.004 AT at o'c ) 4T-T-O T Rold= 100 110 1000 (rm0.001 T) RTD: 100 100(1+ 0.004T) 100 (1+0.004T)110. 10 0.25x10- 25 O.004 T O.1 boidge coll be balanced At T:25°c RF ww o Vo -RF Vos 13FXV2 14 RF Rin12F4Rn + Rin BF V2 Vo Rn RF Vo Goin RE (V2-V) Vo= Rim Rin Main = 10 Noco Fven All Re& Rm that gahs by ths Oonditien col caonk fine 10 Rin RF = 2

cWe know that Wo AV Rin Vo 9anges buem to BV -2.5 When Vo= 3V,then AV= 0.3 le, V2-V, = 0.3 5x RTD V. 2.5 R+RTD Re+ Ra V2 = 2.8 5 x 100 O.004 T 2.8 110+ 100(1+o,004T) 110 (2,8) + 2go + 280 (0.004T 500 + 2T 2T- 280(0.0047)-1N0(2 8) + 280-500 0.88 T 88 temp Max T 100 0c nM Min temperature, AVE -025 Vo= -2.5 -0.25 +9.5=8.85 225 + 225(a oo4 T) 500+ 81- 9.25(110) LMin 21.5 T2 -25%C ,IT temp

Max temp- 100'c Kin temp: -25"c The RTD gengoo is 100 (1+0.004 B AT RT= 6tween the Caliboatren s done to temperateres 1tea sure -50°C to Oc RT (1) As T 1) the At O'c CDe knoo RT-100 2 don knouo we At -50%c 50) 1+0.004x- 100 80n CDOuld be At 2mA, reading Coould be I00 22MA, reading (H-2MA) mA 0.001 (-2mA) 00-80 (n-2mA) 4-20 20MA $4-80 4=804(M-2) At И- 80+ (1-2) 872 Pordiny

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