Question

An ideal instrumentation amplifier is used as a signal conditioning circuit. The temperature co-efficient of RTD is \(0.005/^0 C.\) Initially at \(T= 100^0 C\), RTD resistance is \(1k\Omega.\) All the resistance values are given in the figure.


If the temperature is changed by \(50^0 C\) then the magnitude of % change in gain is ________.
vz e—— S00K. Vi o—— - W ;SK /g/'nm §5K W - 500K

• A. 11.88
• B. 18.18
• C. 1.818
• D. 0.1818

Answer 1

The correct option is B 18.18

$\text{RTD resistance at}{0}^{o}C\left(R_0\right)=1k\Omega$

$At{50}^{0}C,$

$R=R_0(1+\alpha \Delta T)$

$R_{{50}^{0}C}=1000(1+0.005\times (50-0\left)\right)$

$R_{{50}^{0}C}=1.25k\Omega$

$\text{Gain of instrumentation of amplifier at}{0}^{0}C\text{,}$

$A_1=\frac{V_0}{V_1-V_2}=\frac{R_2}{R_1}(1+\frac{2R}{R_0})$

$A_1=\frac{500k}{100k}(1+\frac{2\times 5k}{1k})$

$A_1=5\times 11=55$

$\text{Gain of instrumentation of amplifier at}{50}^{0}C\text{,}$

$A_2=\frac{V_0}{V_1-V_2}=\frac{R_2}{R_1}(1+\frac{2R}{R_{{50}^{o}C}})$

$A_2=\frac{500k}{100k}(1+\frac{2\times 5k}{1.25k})$

$A_2=5\times 9=45$

$\text{\% Change in gain =}|\frac{A_2-A_1}{A_1}|\times 100$

$=|\frac{45-55}{55}|\times 100$

$=18.18\%$