EE3901/EE5901 Sensor Technologies
Week 10 Tutorial

College of Science and Engineering, James Cook University

Last updated 22 April 2022

Questions 1 - 3 require data from the NIST ITS-90 thermocouple database.

Question 1

A type K thermocouple has its reference junction in an ice bath, i.e. the reference temperature is 0 °C. What is the voltage across the thermocouple when the sensing junction is at each of the temperatures below?

(a) 100 °C.
(b) 102 °C.
(c) 609 °C.
(d) 1057 °C.

Answer

Thermocouple voltages are given by

Here, °C. For type K, we find V. Therefore,

Hence the voltages are:
(a) 4.096 mV.
(b) 4.179 mV.
(c) 25.288 mV (note that the 9 is not the last column!)
(d) 43.480 mV.

Question 2

A type R thermocouple is being used to measure temperatures in an industrial setting. Calculate the measured temperature in each of the scenarios below.

(a) °C and mV.
(b) °C and mV.
(c) °C and mV.
(d) °C and mV.

Answer

Rearranging the thermocouple transfer function, we obtain

(a)

(b)

You could also interpolate between the two tabulated voltages, in which case the result is

However thermocouples typically have errors in the range of 1-2 °C so this level of calculation is unlikely to be meaningful.

(c)

(d)

Question 3

You are designing the control system for an industrial furnace that will operate in the range of 500 °C to 1700 °C. You are using a type B thermocouple, an instrumentation amplifier, and a microcontroller. The microcontroller has an analog-to-digital converter with a range of 0 to 3.6 V. Using these components, sketch the circuit layout and calculate the required gain from the instrumentation amplifier.

Aim for a 10% safety margin in the maximum voltage, i.e. design your circuit so the highest voltage it will produce is 90% of the maximum accepted by the microcontroller.

Answer

The temperature of the reference junction is not specified, but we can see in the NIST database that a type B thermocouple’s characteristic values are 0 V plus/minus some noise in the range of 0 to 45 °C. It is reasonable to assume that you have access to an ambient temperature within this range. If the ambient temperature is expected to exceed this value, then only small corrections would be needed, i.e. even if the reference temperature would rise to 100 °C, the characteristic is still only $μ$ . Hence for the purpose of designing this circuit it is reasonable to assume .

The maximum temperature is specified as 1700 °C, which corresponds to 12.433 mV. This value needs to be amplified to a value Hence the required gain is

A suitable interface circuit is then shown in Figure A3.

Question 4

In this question you will consider a circuit design using the Hamamatsu S1223 photodiode. This photodiode produces a current signal that is proportional to the light intensity.

Suppose that you do not want to use a transimpedance amplifier circuit. For simplicity, you will detect the current via a current sense resistor, as shown in Figure Q4. The voltage will be measured using an analog-to-digital converter.

Figure Q4:
Simplified circuit model of a photodiode and current sense resistor. The shunt and series resistances (which are internal to the photodiode) have been neglected for simplicity. The photocurrent is being converted into a voltage by sinking it into a load resistor . There is a DC voltage which ensures that the photodiode operates in reverse bias.
Zoom:

As per the sensor datasheet, the reverse bias current is , but for the purposes of this analysis, you can assume .

The photocurrent varies with time, yet the capacitor and load resistor together form a lowpass filter. Hence the circuit has a limited measurement bandwidth. There is a frequency at which the voltage will drop 3 dB below its DC value.

Your task is to analyse the frequency response of this circuit to predict the measurement bandwidth.

Follow the steps below:

(a) Analyse the AC characteristics of this circuit. Show that the output voltage as function of frequency is

where is the frequency in radians/sec.

(b) Find of the measurement circuit. In other words, find the frequency at which the power in the load resistor drops by half, or equivalently the frequency at which the square of the voltage drops by half. This frequency can be found by solving for in the equation:

(c) From the sensor datasheet, the photocurrent produced under 100 lux of illumination is $μ$ . Suppose that this needs to be converted into an output voltage of V. Calculate the required value of to achieve the voltage target.

(d) Given that the photodiode has a capacitance of , calculate the measurement bandwidth of this circuit.

(e) The S1223 photodiode has an actual sensor bandwidth of 30 MHz. Is this a suitable measurement circuit that can fully utilise the capabilities of the sensor?

Answer

(a) Represent the capacitor with its impedance , then use KCL at the top node to obtain:

Solving for V,

(b) The DC voltage is . Hence the relative voltage is

The cutoff frequency is given by

(d) The -3dB bandwidth of the circuit is

(e) The effective bandwidth of the measurement circuit is far below that of the sensor. Hence, the circuit is not a suitable design since the high sensor bandwidth is inaccessible.

Question 5

The same photodiode from Question 4 is now connected to a transimpedance amplifier.

Design a transimpedance amplifier circuit to achieve a gain of 0.1 V/mA. Use the sensor characteristics as they were given in the question above.

In your design, use the Texas Instruments OPA656 op-amp. You will need to refer to the op-amp’s datasheet to find its gain-bandwidth product and input impedance (specifically, the input capacitance).

Hints:

The design rule for the feedback capacitor is: where is sensor capacitance plus the parasitic capacitance due to the op-amp’s input pins, is the feedback resistor, and is the op-amp’s gain-bandwidth product.
The total parasitic capacitance at the input is the sum of the differential and common-mode input capacitances.

Answer

The circuit design is shown in Figure A5.

Our design task is to choose and .

Analysing the DC characteristics, it is clear that the gain is set via the feedback resistance, i.e. choose .

To choose we must use the design rule from the question. Firstly, we must calculate as the sensor capacitance plus the op-amp’s parasitic input capacitance.

Therefore,

Question 6

There is a gain and bandwidth tradeoff in the transimpedance amplifier design. Assuming a properly specified feedback capacitor, the amplifier bandwidth is given by

where is the op-amp gain-bandwidth product, is the feedback resistor, and is the total input capacitance.

In this question, you will consider the Analog Devices LT6275 op-amp. This device has a gain-bandwidth product of 40 MHz and a total input capacitance of 3.4 pF. Your goal is to achieve a transimpedance gain of 10 V/μA and a circuit bandwidth of 1 MHz. Suppose that your photodiode has 12 pF of capacitance.

(a) Calculate the point if you perform all the amplification in a single stage, i.e. implement the full 10 V/μA in a single op-amp. Comment on the suitability of this circuit.

(b) The design requirement is for a circuit bandwidth of 1 MHz. Starting from this bandwidth limit, calculate the maximum achievable transimpedance gain using the LT6275.

(c) To achieve the high gain and bandwidth simultaneously, you have decided to implement a two-stage amplification circuit with two identical op-amps. The first op-amp will be used as a transimpedance amplifier providing the gain that you calculated in part (b). The second op-amp will be used as a voltage amplifier. Write down how you would use the second op-amp such that your overall circuit achieves both the gain and bandwidth requirements. Does your voltage amplifier fit within the gain-bandwidth envelope of the op-amp?

Answer

(a) The total input capacitance is

To achieve 10 V/μA of gain, we require MΩ. The resulting bandwidth is

This is a quite low bandwidth given that many photodetectors have megahertz bandwidths.

(b) Rearranging the bandwidth equation,

Hence the maximum gain is 0.4134 V/μA.

This could be achieved using a non-inverting amplifier circuit. At 1 MHz, the gain-bandwidth is

This is comfortably inside the 40 MHz capability of the LT6275, hence we would not expect any loss of bandwidth due to the second stage.