Question

A J-type thermocouple has an output voltage \( V_\theta = (13650 + 50 \theta_x) \, \mu \text{V} \), where \( \theta_x \) is the junction temperature in Celsius (°C). The thermocouple is used with reference junction compensation, as shown in the figure. The instrumentation amplifier used has a gain \( G = 20 \). If \( \theta_\text{Ref} = 1 \, \degree \text{C} \), for an input \( \theta_x = 100 \, \degree \text{C} \), the output \( V_o \) of the instrumentation amplifier in millivolt is

• A. 98 mV
• B. 99 mV
• C. 100 mV
• D. 101 mV

Hint:

For thermocouples, the output voltage is linearly related to the junction temperature. The final output from the instrumentation amplifier is scaled by the amplifier's gain.

Answer 1

The Correct Option is B

The thermocouple's output voltage \( V_\theta \) is given by: \[ V_\theta = 13650 + 50 \theta_x \, \mu \text{V}. \] For \( \theta_x = 100 \, \degree \text{C} \), we substitute this value into the equation: \[ V_\theta = 13650 + 50 \times 100 = 13650 + 5000 = 18650 \, \mu \text{V}. \] The reference junction temperature \( \theta_\text{Ref} = 1 \, \degree \text{C} \), so: \[ V_{\theta_\text{Ref}} = 13650 + 50 \times 1 = 13700 \, \mu \text{V}. \] The voltage difference is: \[ \Delta V = V_\theta - V_{\theta_\text{Ref}} = 18650 - 13700 = 4950 \, \mu \text{V}. \] Now, the instrumentation amplifier has a gain \( G = 20 \), so the output voltage \( V_o \) is: \[ V_o = G \times \Delta V = 20 \times 4950 = 99000 \, \mu \text{V} = 99 \, \text{mV}. \] Final Answer: 99 mV