EE-188 Turn in on paper at the beginning of class (52 points tota Operational am

Question

EE-188

Turn in on paper at the beginning of class (52 points tota Operational amplifiers are often used to amplify a sensor output. This problem will walk you through the design of a simple temperature measuring device based on a platinum wire Sensor. Goal: Build a circuit that will provide a calibrated output of 10mVoC and a second output of 10mVOF. (The final circuit can be seen oat the end of this homework but we will work out each stage in turn.) The platinum wire sensor has a resistance of 1002 at 0oC and 1382 at 1000, or a change of 0.3852/ °C. (The arrow through the resistor in the circuit means that it is a variable resistor.) Assume a 5mA source will excite the sense resistor The first stage of our circuit will be to buffer the output of the sensor so we do not load the sensor circuit. 0.5mA A. (6 points) What is V1 when the temperature is 0°C, 1°C and 100 C? (Use at least three decimal points.) B. (10 points)The output voltage of the resistor changes by I*ART where ART 0.3852/ C. We need amplify this so the output will be 10mV per degree using a non-inverting amplifier OmV T gain for a 10mV change per degree centigrade In What is the required gain for this circuit? Choose values of R1 and R2 to achieve this, the values should be between 1k and 100ks2. (Rin can be assumed to be 1K-10k2.) C. (6 points) What is V2 for a temperature of 0°C and 1 C. What is the difference between the two voltages? (Hint: The difference should be exactly 10mV! The resistance of the platinum wire will be 100.3852 1°C.)

Explanation / Answer

A)
At 0 deg C ---> R =100 Ohm, I =0.5 mA that is V1 =IR =50 mV
At 1 deg C ---> R =100.385 Ohm, I =0.5 mA that is V1 =50.1925 mV
At 100 deg C ---> R =138 Ohm, I =0.5 mA that is V1=69 mV
(Obs. Given the temperature change of 0.385 ohm/deg C the resistance at 100 deg C should be 138.5 Ohm and thus V1 =69.25 mV)

B)
G=Gain = 10 mV/(0.385*0.5) mV =51.948 (=52)
For noninverting amplifier we have
G=Gain = 1+R2/R1 that is R2/R1 =50.948
We choose R1 =1 K and R2 =51 K
Rin = R1 ||R2 = R1*R2/(R1+R2) = 1K
Rin is for canceling In+ and In- polarization currents.

C)
For 0 deg C V1 =50 mV that is V2 =G*V1 =52*0.050 =2.60 V
For 1 deg C V1 =50.1925 mV that is V2 =G*V1 =52*0.0501925 =2.61001 V
Difference between two voltages is
Delta(V2) =(2.61001-2.60) V = 10.01 mV

D)
For the circuit in the figure one has
(http://en.wikipedia.org/wiki/Operational_amplifier_applications#Differential_amplifier_.28difference_amplifier.29)
Vc = V2 -R5/R4*V(off)
Taking V2 =1.30 V and Vc = 0 V one has
0 = 1.3-(R5/R4) *V(off)
1.3 = R5/R4 *V(off)
For R5 = R4 one has V(off) =2.6 V

For 20 deg C , R(T)=100+20*0.385 =107.7 Ohm, that is V1 =107.7*0.5 =53.85 mV,
that is V2 =G*V1 =52*0.05385 =2.8002 V, that is Vc =V2-V(off) =2.8002-2.6 =0.2002 V

For 100 deg C, V1 =69 mV, that is V2 =G*V1 =52*0.069 = 3.588 V that is Vc =3.588-2.8 =0.988 V

E)
If Vc =0 V (0 deg C) and R10=R11 = 1K then VF =V(off) -(-R8/R7)*(Vc) =Voff = 0.32 V
(From F =(9/5)*C +32)

If T = 1 deg C (33.8 F) then Vc =10 mV (we have for Vc, 10 mv/dec C) then
0.338 V = 0.32 +(R8/R7)*0.010
0.01*(R8/R7)=0.018
R8/R7 =1.8
We chose R7 =1 K and R8 =1.8 K

Thus for 100 deg C (212 deg F) one has
V(F) =V(off) +R8/R7*Vc =0.32+(1.8/1)*1 V =2.12 V

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