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BG3105 Biomedical Instrumentation

Tutorial Part I Solution

Lecture 1-3 (Introduction to Bio-instrumentation): Q1. Find the equivalent resistance of the following circuit?

Solution:

6 𝑘Ω

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Q2. What is the output voltage (𝑉0) of the following amplifier circuit?

Solution:

12 V.

As we know that no current flows into the OP-AMP or the positive and the negative terminal are at the same voltage (𝑉1). The diagram below shows the flow of currents.

𝑖1 =5𝑉

5𝑘Ω= 1𝑚𝑚

𝑉1 = 2𝑘Ω × 1𝑚𝑚 = 2𝑉

𝑖2 =1𝑉 − 𝑉1

1𝑘Ω=

1𝑉 − 2𝑉1𝑘Ω

= −1𝑚𝑚

𝑉0 = 𝑉1 − 10𝑘Ω× 𝑖2 = 2𝑉 + 10𝑘Ω × 1𝑚𝑚 = 12𝑉

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Q3. Given a resistor 𝑅 with 5% accuracy, what will be the percent accuracy in 1𝑅

? i) 2.5% ii) 5% iii) -5% iv)

15%.

Solution: ii

𝑐 = 1/𝑅

𝑑𝑐 = −1𝑅2 𝑑𝑅

𝑑𝑐𝑐 = −

𝑑𝑅𝑅

�𝑑𝑐𝑐 � = �

𝑑𝑅𝑅 �

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Q4. A DC voltage across a resistor is measured with an accuracy of 3%. If the resistor is valued with a 5% accuracy, what is the percent accuracy in determining the power dissipated by the resistor from values of the resistor and measured voltage?

Solution:

𝑃 =𝑉2

𝑅 = 𝑉 × 𝑉 ×1𝑅

�𝑑𝑃𝑃 � = �

𝑑𝑉𝑉 � + �

𝑑𝑉𝑉 � + �

𝑑𝑅𝑅 � = 3% + 3% + 5% = 11%

The percent accuracy in power measurement will be 11%.

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Q5. A Wheatstone bridge for measuring resistance is shown below, where 𝑅1, 𝑅2, and 𝑅3 are resistors of the bridge circuit and 𝑅𝑇 is the resistor to be measured. The bridge circuit is balanced by adjusting 𝑅3 such that the voltmeter indicates zero reading. Then the value of 𝑅𝑇 can be represented in terms of the bridge resistors.

(i) Show that the balanced condition for the bridge circuit is 𝑅1

𝑅2= 𝑅3

𝑅𝑇= 𝑐, where 𝑐 is a

constant.

(ii) Find the sensitivity for measurement of 𝑅𝑇 near the balanced point. (iii) If the relative measurement accuracy of resistors 𝑅1, 𝑅2 and 𝑅3 is 1%, what is the relative measurement accuracy of 𝑅𝑇?

Solution: (i) 𝑉 = 𝑅2

𝑅1+𝑅2𝐸 − 𝑅𝑇

𝑅3+𝑅𝑇𝐸. When the bridge is balanced, 𝑉 = 0.Thus, at balance:

𝑅2

𝑅1 + 𝑅2=

𝑅𝑇𝑅3 + 𝑅𝑇

⇒ 𝑅2𝑅3 = 𝑅1𝑅𝑇 ⇒𝑅1𝑅2

=𝑅3𝑅𝑇

= 𝑐,𝑤ℎ𝑒𝑒𝑒 𝑐 𝑖𝑖 𝑐𝑐𝑐𝑖𝑐𝑐𝑐𝑐

(ii)

∆𝑉 = �𝑅2

𝑅1 + 𝑅2−

𝑅𝑇 + ∆𝑅𝑇𝑅3 + 𝑅𝑇 + ∆𝑅𝑇

�𝐸

= �1

𝑅1/𝑅2 + 1 −1 + ∆𝑅𝑇/𝑅𝑇

𝑅3/𝑅𝑇 + 1 + ∆𝑅𝑇/𝑅𝑇�𝐸 = �

11 + 𝑐 −

1 + ∆𝑅𝑇/𝑅𝑇1 + 𝑐 + ∆𝑅𝑇/𝑅𝑇

�𝐸

= �(1 + 𝑐 + ∆𝑅𝑇/𝑅𝑇)− (1 + 𝑐)(1 + ∆𝑅𝑇/𝑅𝑇)

(1 + 𝑐)(1 + 𝑐 + ∆𝑅𝑇/𝑅𝑇) �𝐸

=−𝑐∆𝑅𝑇/𝑅𝑇

(1 + 𝑐)(1 + 𝑐 + ∆𝑅𝑇/𝑅𝑇)𝐸

∆𝑉 ≈ −𝑐

(1 + 𝑐)2∆𝑅𝑇𝑅𝑇

𝐸,𝑓𝑐𝑒 𝑖𝑚𝑐𝑠𝑠 ∆𝑅𝑇

(iii) 𝑅𝑇 = 𝑅2𝑅3𝑅1

⇒ ∆𝑅𝑇𝑅𝑇

= �∆𝑅1𝑅1�+ �∆𝑅2

𝑅2�+ �∆𝑅3

𝑅3� = 3%

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Q6. You are measuring temperature of a solution using a mercury thermometer. You have done measurements 10 times and obtained the following readings. 30.1, 30.2, 29.8, 32.1, 30.5, 29.9, 30.1, 29.5, 31.1, and 30.6. Calculate the mean and the standard deviation of the measured temperatures.

Solution:

Mean:

𝑋� = ∑𝑋𝑖𝑛

= 30.1+30.2+29.8+32.1+30.5+29.9+30.1+29.5+31.1+30.610

= 30.39

Standard Deviation:

𝑆 = �∑(𝑋𝑖−𝑋�)2

𝑛−1= �(30.1−30.39)2+(30.2−30.39)2+(29.8−30.39)2+⋯

10−1= 0.7505

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Q7. Below are some hypothetical results from mammogram for breast cancer screening. What is the specificity of the mammogram?

Breast cancer No breast cancer Total Mammogram positive 70 180 250 Mammogram negative 30 720 750

Total 100 900 1000 What are the sensitivity and specificity and prevalence of the disease? Solution: Sensitivity = 70/100 = 70% Specificity = 720/900 = 80% Prevalence = 100/1000 = 10%

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Lecture 4 (Pressure, Flow, and Temperature): Q8. Consider the potentiometer measurement circuit on the right. (i) What is the expected reading of the voltmeter? (ii) If the internal resistance of the voltmeter is 𝑹𝑽, what is the measured voltage?

Solution: (i) 𝑽 = 𝑹𝒙

𝑹𝑽𝒔

(ii) The equivalent circuit is

The reading is:

𝑽𝒏𝒏𝒏 =𝑹𝒙//𝑹𝒗

(𝑹− 𝑹𝒙) + 𝑹𝒙//𝑹𝒗𝑽𝒔 =

𝑹𝒙𝑹𝒗𝑹𝒙 + 𝑹𝒗

(𝑹 − 𝑹𝒙) + 𝑹𝒙𝑹𝒗𝑹𝒙 + 𝑹𝒗

𝑽𝒔

=𝑹𝒙𝑹𝒗

𝑹𝑹𝒗 + 𝑹𝑹𝒙 − 𝑹𝒙𝟐𝑽𝒔

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Q9. A capacitive pressure measurement scheme is shown below where the movable and fixed metal diaphragms form a capacitor 𝐶. Assume that the common area 𝑚 of the diaphragm, the dielectric constant 𝜀 of the capacitor are known, and the distance 𝑑 between the diaphragm is inversely proportional to the gas pressure.

(i) Show that the gas pressure 𝑝 can be measured by measuring the capacitance value 𝐶.

(ii) If the percent accuracy in measuring 𝐶 is 5%, the known value of the area 𝑚 has a percent accuracy 3% and 𝜀 and 𝑘 have fixed values, what is the percent accuracy in measuring the pressure 𝑝?

Solution:

(i) Since the distance 𝑑 is inversely proportional to the pressure 𝑝, there is a constant 𝑘 such that 𝑑 = 𝑘

𝑝, 𝐶 = 𝜀𝜀

𝑑= 𝜀𝜀

𝑘𝑝, 𝑝 = 𝑘

𝜀𝜀𝐶

(ii) 𝑑𝑝 = 𝑘𝜀𝜀𝑑𝐶 − 𝑘𝑘

𝜀𝜀2𝑑𝑚 = 𝑘𝑘

𝜀𝜀�𝑑𝑘𝑘� − 𝑘𝑘

𝜀𝜀�𝑑𝜀𝜀� = 𝑝 𝑑𝑘

𝑘− 𝑝 𝑑𝜀

𝜀

𝑑𝑝𝑝

= 𝑑𝑘𝑘− 𝑑𝜀

𝜀, �𝑑𝑝

𝑝� = �𝑑𝑘

𝑘� + �𝑑𝜀

𝜀�

Percent accuracy in measuring pressure 𝑝 = 5%+3% = 8%

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Q10. If blood flowing in a blood vessel with a radius of 2 mm cuts a magnetic field with flux density 𝐵 = 0.1 tesla, and the blood volume flow rate is 10 cm3/s, what is the electric potential that can be detected across the blood vessel? Solution:

𝑄 =𝜋𝑈𝑑2

4 =𝜋𝑉𝑑4𝐵

𝑉 =4𝑄𝐵𝜋𝑑 =

4 × 10 × 0.013 × 0.13.14 × 2 × 2 × 10−3 = 0.318 𝑚𝑉

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Q11. Which one is a piezoelectric material? i) PVDF (Polyvinylidene fluoride) ii) PZT (Lead Zirconium Titanate) iii) Quartz iv) All of the above

Solution: iv

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Q12. Explain briefly the working principle of thermocouple temperature sensor? Solution:

When the junctions of two dissimilar metals are placed at two different temperatures, then there is potential difference between the two junctions. By measuring this potential difference and knowing one of the junction temperatures (reference junction temperature), the other junction temperature can be measured. Typically, junction is placed at a location whose temperature needs to be measure. The other junction is placed at a known temperature (typically at zero degree C). Thus by measuring the voltage between the two junctions one can measure the unknown temperature.

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Q13. The resistance of a semiconductor thermistor is a function in the absolute temperature

in Kelvin (K) expressed as 𝑅(𝑇) = 𝑅(𝑇0) 𝑒𝛽(𝑇0−𝑇)/(𝑇𝑇0) = 𝑅(𝑇0) 𝑒�1𝑇−

1𝑇0�𝛽

It is known that 𝛽 = 3150 and 𝑅(300) = 1600 Ω. Find the sensitivity of the thermistor at 𝑇 = 300 𝐾 𝑐𝑐𝑑 𝑇 = 310 𝐾, respectively. Solution:

𝑅(𝑇) = 𝑅(𝑇0) 𝑒�1𝑇−

1𝑇0�𝛽

𝑑𝑅(𝑇)𝑑𝑇

= 𝑅(𝑇0) 𝑑 �𝑒�

1𝑇−

1𝑇0�𝛽�

𝑑𝑇= 𝑅(𝑇0) 𝑒�

1𝑇−

1𝑇0�𝛽𝑑 ��1

𝑇 −1𝑇0�𝛽�

𝑑𝑇

𝑑𝑅(𝑇)𝑑𝑇

= 𝑅(𝑇0) 𝑒�1𝑇−

1𝑇0�𝛽 × 𝛽 ×

𝑑 �1𝑇 −

1𝑇0�

𝑑𝑇= 𝑅(𝑇0) 𝑒�

1𝑇−

1𝑇0�𝛽 × 𝛽 ×

𝑑 �1𝑇�

𝑑𝑇

𝑑𝑅(𝑇)𝑑𝑇

= 𝑅(𝑇0) 𝑒�1𝑇−

1𝑇0�𝛽 × 𝛽 × �−

1𝑇2� = 𝑅(𝑇) 𝛽 �−

1𝑇2�

𝑑𝑅(𝑇)𝑑𝑇

= −𝛽𝑇2

𝑅(𝑇)

𝑅(300) = 1600 𝛺, 𝑅(310) = 1600 × 𝑒�1310−

1300�×3150 = 1140 𝛺

∆𝑅∆𝑇

(300) = −31503002

× 𝑅(300) = −56 𝛺/𝑇

∆𝑅∆𝑇

(310) = −31503102

× 𝑅(310) = −37.4 𝛺/𝑇

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Lecture 5 (Electrical Safety): Q14. What are macroshock and microshock? Solution

macroshock: When a person touches a hot wire and the ground with his limbs.

Microschock: When a small amount of current directly passes through the heart.

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Q15. Explain how the isolation transformer works?

Solution

An isolation transformer is used to transfer electrical power from an AC source to some equipment/device isolating the equipment/device and protect against electrical shock (macroshock). With an isolation transformer there is no “return path” so one could touch the “hot” wire without getting shock.

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Q16. Why cannot Ground-fault interrupter (GFI) be used on life support device circuitry?

Solution:

GFI stops the power supply when there is a fault. This may lead to death of the patient on life support device.

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Q17. Explain how a line isolation monitor (LIM) can be used to detect a neutral line shorted to the ground.

Solution

To detect fault of the neutral line the Line isolation monitor (LIM) needs to be connected between the “hot” and “ground” lines. When there is no fault the current flowing through the LIM is very small due to the very small stray capacitance. Once the neutral get shorted to the ground, the current flowing through the LIM increases. The ammeter connected in the LIM shows the fault.

Equivalent circuit

The current flowing through the LIM is 𝐼𝐿𝐿𝐿 = 𝑉0

�𝑅2+� 1𝜔𝜔�

2. Since the leakage capacitance C is

very small, the current is also very small.

When the neutral gets shorted to the ground the equivalent circuit becomes,

Now the current flowing through the LIM is 𝐼𝐿𝐿𝐿 = 𝑉0𝑅

.

So the LIM is set in such a way that whenever there is a current flow higher than the set limit, it will set an alarm, which indicates that there is a fault in the neutral wire.

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Q18. Calculate the maximal safe capacitance between a liquid-filled catheter and dc-isolated pressure sensor leads for a 230 volts, 50 Hz fault in the sensor leads if its equivalent circuit is as shown in Figure below.

Solution:

As this is a microshock hazard, so the maximal safe current is 10 µA.

From the circuit, we have 𝑍 = 50 𝑘Ω + 300 Ω + 1𝑗𝑗𝑘

Current = 𝑉|𝑍| = 230|𝑍| = 10 𝜇𝑚

|𝑍| =23010−5

= 2.3 × 107Ω

As both resistances are very small, |𝑍| ≈ � 1𝑗𝑗𝑘

� = 1𝑗𝑘

= 2.3 × 107

Then, 𝐶 = 12𝜋𝜋×2.3×107

= 138 𝑝𝑝

This shows that the capacitance shouldn’t be more than 138 pF for safety.

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Q19. A receptacle tester has the circuit shown below. Indicate which LEDs will be ON or OFF for fault conditions given in Table.

Conditions 1 2 3 Hot open Neutral open H and N reversed, N open Ground open Hot and ground reversed

Solution:

Conditions 1 2 3 Hot open OFF OFF OFF Neutral open ON OFF OFF H and N reversed, N open OFF ON OFF Ground open OFF OFF ON Hot and ground reversed ON ON OFF

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Lecture 6 (Lung, Spirometer): Q20. What is internal and external respiration? Solution: Internal respiration is the exchange of gases between the bloodstream and nearby cells. External respiration is the exchange of gases between the lungs and the bloodstream.

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Q21. Below are three different volumes related to human lung. What are i) the definitions and ii) typical values for an adult lung? Tidal volume, Functional Residual Capacity (FRC), Dead Space. Solution: Tidal Volume: volume of gas inspired or expired during each normal respiratory cycle. 500 ml. FRC: Amount of gas remaining in the lungs at the resting expiration. 2400 ml. Dead Space: Volume of air that is not available for gas exchange. 150 ml.

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Q22. How does spirometer work? Solution: A spirometer is used to measure lung volume. A simple spirometer can only measure the volume inspired or expired, i.e., change in volume. A spirometer has a mouthpiece. One can use the mouthpiece to breathe. The volume of the gases inside the spirometer will change as the patient breathes. This volume change is proportional to the lung volume change. There is a mechanical linkage between the volume change of the spirometer and the measurement reading output (display). A simple spirometer cannot be used to measure gas volume which remains inside the lung such as, FRC, RV etc.

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Q23. a) Figure below shows a spirometer. In its simplest form, the bellow is mechanically articulated to a biased potentiometer such that the wiper arm voltage is proportional to volume. Derive the relationship of the voltage and volume. b) If the spirometer has a radius of 10 cm, a maximum height of 10 cm, and 𝑉𝐵𝐵 = 10 𝑣𝑐𝑠𝑐𝑖, plot the air volume versus the output voltage.

Solution: (a) 𝒉 is the maximal distance change that the bellows can change. So the maximum volume change is 𝑽𝒎𝒎𝒙 = 𝝅𝒓𝟐𝒉. Since the spirometer volume (𝑽𝒔) is proportional to the output voltage (𝑽𝒐𝒐𝒐), 𝑽𝒔 = 𝒌𝑽𝒐𝒐𝒐 If we adjust linkage such that at 𝑽𝒎𝒎𝒙 the potentiometer output has 𝑽𝒐𝒐𝒐 = 𝑽𝑩𝑩.

𝑽𝒎𝒎𝒙 = 𝒌𝑽𝑩𝑩 Then the constant, 𝒌 = 𝑽𝒎𝒎𝒙

𝑽𝑩𝑩

As the changes in distance of bellows and output of potentiometer are proportional. We have

𝑽𝒔 =𝑽𝒎𝒎𝒙𝑽𝑩𝑩

𝑽𝒐𝒐𝒐

(b) As 𝒓 = 𝟏𝟏 𝒄𝒎,𝒉 = 𝟏𝟏 𝒄𝒎,𝒎𝒏𝒂 𝑽𝑩𝑩 = 𝟏𝟏 𝒗𝒐𝒗𝒐𝒔, we have

𝒌 =𝝅𝒓𝟐𝒉𝑽𝑩𝑩

=𝝅 × (𝟏𝟏 𝒄𝒎)𝟐 × (𝟏𝟏 𝒄𝒎)

(𝟏𝟏 𝒗𝒐𝒗𝒐𝒔) = 𝟏.𝟑𝟏𝟑 𝒗𝒍𝒐𝒏𝒓/𝒗𝒐𝒗𝒐

So, 𝑽𝒔 = 𝟏.𝟑𝟏𝟑𝑽𝒐𝒐𝒐

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Q24. Briefly explain nitrogen washout method for estimation of lung residual volume (RV)? Solution Nitrogen washout method for estimating lung residual volume is using a modified spirometer setup and a nitrogen analyzer. Here, the patient starts inhaling pure oxygen through a one way tube. Then he exhales air into the spirometer through another one way tube. When exhaling he expels a mixture of oxygen, nitrogen and carbon-dioxide into the spirometer. So after many breathing cycle the nitrogen content in the lung and in the spirometer will come to equilibrium. By doing a mass balance of nitrogen before and after the experiment, one can calculate the RV. Provided the person started the breathing cycle at the moment when the lung volume was at RV. Lets say at the beginning (time 𝑐1, and the lung volume is 𝑉𝑅𝑉), the total number of 𝑁2 moles in the lung is given by,

𝑝𝐿𝐿2(𝑐1)𝑉𝑅𝑉𝑇𝐿

𝑃𝑅

At the beginning the total number of 𝑁2 moles in the spirometer is NIL (0). At the end (time 𝑐2, and the lung volume is 𝑉𝑅𝑉), the total number of 𝑁2 moles in the lung is,

𝑝𝐿𝐿2(𝑐2)𝑉𝑅𝑉𝑇𝐿

𝑃𝑅

At the end (time 𝑐2, and the spirometer volume is 𝑉𝑠(𝑐2), the total number of 𝑁2 moles in the spirometer is,

𝑝𝑠𝐿2(𝑐2)𝑉𝑠(𝑐2)𝑇𝑠

𝑃𝑅

By doing a mass balance (since no 𝑁2 has gone out of the lungs other than spirometer),

𝑝𝐿𝐿2(𝑐1)𝑉𝑅𝑉𝑇𝐿

𝑃𝑅− 𝑝𝐿𝐿2(𝑐2)

𝑉𝑅𝑉𝑇𝐿

𝑃𝑅

= 𝑝𝑠𝐿2(𝑐2)𝑉𝑠(𝑐2)𝑇𝑠

𝑃𝑅

𝑉𝑅𝑉 =𝑇𝐿𝑇𝑠�

𝑝𝑠𝐿2(𝑐2)𝑉𝑠(𝑐2)𝑝𝐿𝐿2(𝑐1) − 𝑝𝐿𝐿2(𝑐2)�

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Page 25 of 27

Lecture 7 (Heart, pacemaker, oximeter, Defibrillator): Q25. When light passes through an ink solution in water the light intensity decreases by 6.02 dB. You are asked to take this solution and modify it only using water, so that when light passes through the new solution its intensity gets attenuated only by half. What would you do? Solution

𝑚(𝜆) = −10𝑠𝑐𝑙10 �𝐼𝐼0� 𝑖𝑐 𝑑𝐵

6.02 = −10𝑠𝑐𝑙10 �𝐼𝐼0�

𝑠𝑐𝑙10 �𝐼𝐼0� = −6.02/10

𝐼0 = 3.99𝐼 ≈ 4𝐼 Therefore, at present the ink solution is attenuating the light by 4 times. To make a new solution which will attenuate only half, we need to dilute the ink solution by two times. So take the ink solution, mix with equal volume of water and the new solution will have half the concentration, therefore, it will attenuate light only by half.

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Page 26 of 27

Q26. When a pacemaker is needed? Solution A pacemaker is needed when 1) the SA node ceases to function, or become unreliable, 2) The pulse from the SA node cannot reach the heart muscle due to blockage by damaged tissue.

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Page 27 of 27

Q27. What is Defibrillator used for? Solution A defibrillator is used to restore a normal cardiac rhythm from a ventricle fibrillation or atrial fibrillation.