Instrumentation Measurement Lab Midterm 1: Temperature Senso ✓ Solved

Instrumentation Measurement Labmidterm1 A Temperature Sensor Is Exp

Instrumentation Measurement & Lab Midterm 1. A temperature sensor is exposed to a sudden change of 20 o C to 80 o C. The sensor outputs 0.02 volts for every oC of temperature and has a 2.3 second time constant. a. What is the sensor output voltage at 1.5 seconds? b. At what time, t, does the sensor output become 1.0 volt? 2. A sensor, R, that changes resistance is used in a bridge circuit as shown below. a. What sensor resistance, R, will null the bridge? b. What is the off-null voltage if R changes by 0.5 Ω? 3. A measurement system has noise above 20 kHz and the data signal frequency is between 100 and 500 Hz. Design an RC filter that reduces the noise by 95%, i.e., only 5% is left. Show the schematic and the component values. 4. Given a Type J TC with a 25 o C reference. a. What is the temperature if the TC voltage is 22.87 mV? b. What voltage would result from a temperature of 225 o C? 5. A 10 - bit ADC has a 5.00 - volt reference. a. What binary output is produced by an input of 3.04 volts? b. Suppose the output is found to be 1F4h. What is the possible input voltage? 6. A 12 bit bipolar DAC has a 10 - volt reference. a. If the hex input is 5D7h what is the DAC output voltage? b. What input is required to get a zero volt output?

Paper For Above Instructions

This paper analyzes various components in an instrumentation measurement lab midterm examination, focusing on temperature sensors, bridge circuits, RC filters, thermocouples, analog-to-digital converters (ADCs), and digital-to-analog converters (DACs). It addresses the specific problems outlined in the prompt with calculated answers and circuit designs as required.

1. Temperature Sensor Analysis

The given temperature sensor experiences a sudden temperature change from 20°C to 80°C. The sensor's output is a function of temperature change. Specifically, the sensor produces 0.02 volts for each degree Celcius. To find the output voltage at a specific time, we use the following relationship:

Voltage Output = Initial Voltage + (Change in Temperature x Sensitivity) x (1 - e^(-t/τ)),

where τ (tau) is the time constant (2.3 seconds) and t is the time in seconds after the temperature change.

Initially, at 20°C, the voltage output is:

V_initial = 20°C * 0.02 V/°C = 0.4 V.

After a change to 80°C:

For t = 1.5 seconds:

V_output = 0.4 V + (60°C 0.02 V/°C) (1 - e^(-1.5/2.3))

Calculating e^(-1.5/2.3) yields approximately 0.594, leading to:

V_output = 0.4 V + 1.2 V (1 - 0.594) = 0.4 V + 1.2 V 0.406 = 0.4 V + 0.4872 V = 0.8872 V.

Therefore, the sensor output voltage at 1.5 seconds is approximately 0.89 V.

For time t when the output reaches 1.0 V:

Setting the output voltage equation equal to 1.0 V:

1.0 V = 0.4 V + (1.2 V)(1 - e^(-t/2.3))

Solving for e^(-t/2.3):

0.6 V = (1.2 V)(1 - e^(-t/2.3))

e^(-t/2.3) = 1 - 0.6/1.2 = 0.5.

Taking the natural log:

-t/2.3 = ln(0.5), t = -2.3 ln(0.5).

Calculating gives t ≈ 1.609 seconds.

2. Bridge Circuit Analysis

For a bridge circuit containing a varying resistance R, this requires balancing the bridge for null voltage:

The bridge balances when the ratio of resistances are equal: R1/R2 = R3/R (where R1, R2, and R3 are the known resistances in the bridge).

Assuming R1 = R2 and the bridge needs to be null, choose R values accordingly to find out sensor resistance. An off-null voltage can be calculated based on the change in resistance (ΔR) and its effect on the circuit’s output.

3. RC Filter Design

To design an RC filter that reduces noise above 20 kHz while allowing data signal frequencies (100 Hz to 500 Hz) to pass, we utilize a low-pass filter characterized by:

The cutoff frequency f_c is defined as:

f_c = 1 / (2πRC).

To reduce the noise by 95%, set the cutoff frequency slightly above data signal frequencies, at around 500 Hz:

Choosing R = 10 kΩ, then solving for C gives:

C = 1 / (2π 500 10,000) = 3.18 µF.

Schematic:

Input -> R -> C -> Ground (Output taken across C).

4. Type J Thermocouple Analysis

The Type J Thermocouple reference temperature is 25°C. Given a known TC voltage:

(a) To find temperature from TC voltage of 22.87 mV:

Using the conversion tables or equations for Type J, we find the corresponding temperature is:

Temp = TC_voltage / Sensitivity = 22.87 mV / (50 µV/°C) = 457.4°C.

(b) To find voltage at 225°C:

Voltage = Temperature x Sensitivity ≈ 225°C * (50 µV/°C) = 11,250 µV or 11.25 mV.

5. ADC Analysis

The 10-bit ADC with a 5.00-volt reference yields outputs based on:

Binary Output = (Input Voltage / V_ref) * (2^n - 1).

(a) For an input of 3.04 volts, the output is:

Output = (3.04 / 5.00) * 1023 ≈ 624.

(b) Given an output of 1F4h, convert hex to decimal: 1F4 = 508; find possible input voltage:

Input Voltage = (508 / 1023) * 5.00 ≈ 2.48 volts.

6. DAC Analysis

For a 12-bit bipolar DAC with a 10-volt reference:

(a) For a hex input of 5D7h, convert to decimal (1487):

DAC Output = (Input / (2^n - 1)) V_ref = (1487 / 4095) 10 ≈ 3.63 volts.

(b) A zero-volt output occurs when the hex input is all zeros or effectively zero volts.

Conclusion

This laboratory midterm covers essential concepts in instrumentation, emphasizing the calculation and design of measuring systems.

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