Temperature Measurement System

Fiesta Problem

Instead of using KCL and KVL to find the voltages with respect to currents or vice versa, we used nodal analysis. This way, we end up with a system of 2 equations with w unknowns instead of a system of 3 equations with 3 unknowns. The reference node was given at the bottom. We then used KCL at nodes 1 and 2. We end up with two equations which, if the resistances and currents are given, can be solved for both voltages 1 and 2.

Temperature Measurement System

In this lab, we wire the circuit diagrammed above using a 5V source via WaveForms with Analog Discovery, a resistor, R, of a resistance to be determined, and a thermistor whose resistance varies with temperature according to the graph below. We measure the voltage across the resistor R (V_out) with a digital multimeter (DMM).

We are asked to find a resistance for R that will vary the output voltage (V_out) by a minimum of 0.5 V over a temperature range from 25°C to 37°C. Also, the output voltage must increase as the temperature increases.

Looking at the graph above, we can estimate that 25°C and 37°C correspond to resistances of about 7 kΩ and 11 kΩ, respectively.

Using voltage division we find the output voltages when the temperature is 25°C and when it is 37°C according to their corresponding thermistor resistances. We want the change in voltage between these two temperatures correspond to at least a 0.5 V change so we set the equation up as shown above. Then we simply solve for R, obtaining the two results 4.3668 kΩ and 17.6332 kΩ.

With the information obtained, we wired the circuit using a 4.7 kΩ (measured) resistor. A resistance higher than the calculated value 4.37 kΩ will only increase the output voltage (we need an output voltage change equal or greater than 0.5 V). See the graph below.

This graph shows the curve for V_out with the constraint line V = 0.5 V. The lines cross twice at the values calculated earlier, 4.3668 kΩ and 17.6332 kΩ. This means that at those values for R, the change in voltage, V_out should be exactly 0.5 V. However, according to our criteria, we need 0.5 V or greater. Looking at the graph, we can see that any R value between 4.3668 kΩ and 17.6332 kΩ can be chosen and we will still meet the requirements.

We measured the resistance of the thermistor at room temperature and at body temperature (by pressing it between two fingers). The measured resistances were 11.15 kΩ and 7.09 kΩ, respectively. These values are very close to the values we obtained from the resistance vs. temperature graph for the thermistor (11 kΩ and 7 kΩ).

Solving for output voltage using the equation shown earlier for V_out, we can plug in our value for resistance. We will use 11 kΩ and 7 kΩ for the thermistor resistances. We calculate a value of 1.5 V output voltage for room temperature and 2.0 V for body temperature. This gives us a rise of 0.5 V, like was required.

We now measured the output voltages corresponding to the thermistor at room temperature and at body temperature. We obtained the values 1.51 V and 2.04 V, respectively. This is a rise in 0.53 V which is greater than the 0.5 V minimum required by 30 mV which gives us a 6% experimental error.

Below is a video showing the increase in temperature as the thermistor goes from room temperature to body temperature.

5V source is wired in series with a thermisor and 4.7 kΩ resistor. Resistance of the thermistor goes from 11.5 kΩ to 7.09 kΩ as it goes from room temperature to body temperature (roughly 25°C to 37°C). Thus, the voltage through the 4.7 kΩ resistor increases from 1.51 V to 2.04 V.

Next, we design a temperature measurement system where the output sensitivity is at least 0.1 V/°C.

We attack this problem the same way we did the first time around except that we set the change in V_out equation equal to 1.2 V instead of 0.5 V. This is because from 25°C to 37°C there is temperature change of 12C°. If we multiply these 12C° by 0.1 V/°C, we get 1.2 V.

Solving through, we get a quadratic equation. Solving for the roots of this equation gives us no real solutions. We can then reference the Voltage vs. Resistance graph shown earlier. The V_out curve peaks at about 0.6 V and will never hit 1.2 V. Therefore, it is impossible! You cannot have the voltage reading go up by 0.1 V/°C using any resistor R for the given input voltage and thermistor.