Question

An RTD with temperature co-efficient $-0.004{/}^{\circ }C$ is connected to the wheat stone bridge. Initially at $0^{\circ }C$ the RTD resistance is $200\Omega$. The output of the bridge is further connected to an amplifier having gain $100$ and input-output phase relationship is ${180}^{\circ }.$ The output of the amplifier is has to be converted to digital form for further proceeding.



If the temperature increases by ${40}^{\circ }C$ then the digital output is

Answer 1

Given,
$\alpha =-0.004{/}^{\circ }C$

$\Delta T=(40-0)={40}^{\circ }C$

$\therefore$ Resistance of RTD at ${40}^{\circ }C,$

$R_{{40}^{\circ }C}=R_0[1+\alpha \Delta T]$

$\Rightarrow R_{{40}^{\circ }C}=200[1-0.004\times 40]=168\Omega$

$\text{Hence bridge output,}{V}_B=1V[\frac{168}{168+200}-\frac{200}{200+200}]$

$\Rightarrow V_B=-0.043478V$

Now, the gain of amplifier is $100$ and it shifts the phase of ${180}^{\circ },$

$\therefore \text{Output of amplifier}=V_B=0.043475\times 100$

$=4.3475V$

$\text{Now, digital output}=\frac{2^N\times \text{Analog input voltage}}{Referencevoltage}$

$=\frac{2^8\times 4.3478}{10}=111.30368$

$=(1101111.01001{)}_2$