Mechatronics I: Exam 1 Study Guide

The correct answer is Mechatronics. This field combines these three core disciplines to create integrated, intelligent systems.

A holistic systems approach is crucial because mechatronic systems involve tightly integrated components (mechanical, electrical, software) where the behavior of one significantly impacts others. Optimizing components in isolation often leads to suboptimal overall system performance, unforeseen interactions, or even system failure. A systems approach considers these interactions from the outset, leading to more efficient, reliable, and effective designs.

The fundamental role of a resistor in an electrical circuit is to oppose or limit the flow of electric current. It achieves this by converting some of the electrical energy into heat, thereby reducing the current that can pass through a given voltage difference.

Once a capacitor is fully charged in a DC circuit, it acts as an open circuit, meaning it blocks the flow of direct current. It maintains a constant voltage across its plates equal to the source voltage, and no more current flows through it from the DC source.

When the current through an inductor changes rapidly, the inductor generates a voltage (electromotive force, EMF) across itself. This induced voltage always opposes the change in current that caused it, a principle known as Lenz's Law. This opposition is due to the inductor's property of storing energy in a magnetic field, and it resists changes to that stored energy.

A semiconductor P-N junction diode permits significant current flow when it is "forward biased." This occurs when a positive voltage is applied to the P-type side and a negative voltage to the N-type side, which reduces the width of the depletion region. The reduced depletion region allows charge carriers (electrons and holes) to easily cross the junction, leading to a large current.

The defining characteristic of a diode's behavior is that it allows current to flow predominantly in only one direction (unidirectional flow) while blocking or severely limiting current flow in the opposite direction.

Three ideal characteristics of an Op-Amp are:
1. Infinite Input Impedance: An ideal Op-Amp draws no current from its input terminals. In practice, Op-Amps have very high, but finite, input impedance due to the internal circuitry (e.g., MOSFETs or BJTs).
2. Zero Output Impedance: An ideal Op-Amp can supply any amount of current to the load without any voltage drop across its output. Real Op-Amps have a small, but non-zero, output impedance which causes a slight voltage drop under load.
3. Infinite Open-Loop Gain: An ideal Op-Amp would have an infinitely large voltage amplification without feedback. Practical Op-Amps have a very high, but finite, open-loop gain (e.g., 10⁵ to 10⁶).
These are difficult to achieve perfectly due to physical limitations of semiconductor materials, manufacturing processes, and the need for internal components that consume power and have inherent resistances/capacitances.

The correct answer is Accuracy, precision. Accuracy relates to correctness, while precision relates to consistency.

The correct answer is c) Sensitivity. Static characteristics describe the performance of a measurement system when the input is constant or changing very slowly. Rise time, settling time, and frequency response are all dynamic characteristics.

The primary advantage is the ability to achieve intelligent, adaptive, and autonomous behavior. Computer control allows for complex decision-making, real-time adjustments, data processing, and communication, leading to systems that are more efficient, precise, and versatile than purely mechanical or electronic systems.

A common example is an Anti-lock Braking System (ABS) in a car.
Mechanical: Brake calipers, brake pads, rotors, wheels.
Electronic: Wheel speed sensors, solenoid valves, electronic control unit (ECU) circuitry.
Computer Control: The ECU's microcontroller runs algorithms that monitor wheel speed, detect wheel lock-up, and rapidly modulate brake pressure via the solenoid valves to prevent skidding and maintain steering control.

For most metallic conductors, the resistance increases as its temperature increases. This is because higher temperatures cause increased thermal vibrations of atoms, which impede the flow of electrons more effectively.

A capacitor stores energy in an electric field between its plates, while an inductor stores energy in a magnetic field created by the current flowing through its coil.

Immediately after a DC voltage source is connected to a series RL circuit, the inductor acts like an open circuit (or very high resistance) because it opposes the sudden change in current. Therefore, the current starts at zero and gradually increases exponentially towards its steady-state value (V/R) as the inductor's magnetic field builds up.

The "knee voltage" (or cut-in voltage/forward voltage drop) is the minimum forward-bias voltage required across a diode for it to begin conducting significant current. For silicon diodes, it's typically around 0.7 V, and for germanium, it's about 0.3 V. It's significant because it must be overcome by the applied voltage for the diode to function as a switch or rectifier in a circuit.

Negative feedback in Op-Amp circuits is used to stabilize the gain, reduce distortion, increase bandwidth, and make the circuit's performance less dependent on the Op-Amp's inherent characteristics (like its very high open-loop gain). It allows for precise control over the amplifier's behavior.

Accuracy refers to how close a measurement taken by the multimeter is to the true value of the quantity being measured. Resolution refers to the smallest change in the measured quantity that the multimeter can detect and display. A multimeter can have high resolution (many digits) but low accuracy if it's not calibrated correctly.

The purpose of a Wheatstone bridge circuit is to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one leg of which contains the unknown component. It is commonly used in sensor applications (e.g., strain gauges, thermistors) to convert small changes in resistance into measurable voltage changes, often for precision measurements.

The "loading effect" occurs when a measurement instrument draws current or power from the circuit or system it is measuring, thereby altering the very quantity it is trying to measure. This leads to an inaccurate reading because the act of measurement itself changes the system's behavior. For example, a voltmeter with low input impedance connected across a high-resistance circuit can significantly alter the voltage it's trying to measure.

The primary role of a power supply is to provide a stable and regulated source of electrical energy (voltage and current) to the components within an electronic circuit, ensuring they operate within their specified parameters.

In a series circuit, current is the same through all components, but the voltage drops across each component add up to the total supply voltage. In a parallel circuit, the voltage is the same across all parallel branches, but the total current from the source is divided among the branches.

The time constant ($\tau = RC$) in an RC circuit represents the time it takes for the capacitor voltage (or current) to reach approximately 63.2% of its final value during charging, or to decay to 36.8% of its initial value during discharging. It indicates how quickly the circuit responds to changes in voltage.

Inductors are used in power supply filtering (often in conjunction with capacitors to form LC filters) because they oppose changes in current. This property allows them to smooth out pulsating DC voltages (ripples) by resisting rapid current variations, thereby providing a more stable DC output.

The main function of a Zener diode is voltage regulation. When reverse-biased and the voltage across it reaches its "Zener voltage," it maintains a nearly constant voltage across its terminals, even if the current through it changes significantly. This makes it useful for creating stable reference voltages.

The "virtual short" concept states that in an ideal Op-Amp with negative feedback, the voltage difference between its inverting and non-inverting input terminals is approximately zero. This is due to the Op-Amp's extremely high open-loop gain; any tiny voltage difference at the inputs would cause a huge output, which the negative feedback then corrects by forcing the input voltages to be nearly equal.

Resolution refers to the smallest increment that an instrument can detect or display. Range refers to the minimum and maximum values that the instrument is designed to measure.

Calibration is important for measurement systems to ensure their accuracy. It involves comparing the instrument's readings against a known standard and adjusting the instrument to minimize any deviations, thereby ensuring that it provides reliable and correct measurements.

A transducer is a device that converts energy from one form to another. In mechatronic systems, they typically convert a physical quantity (e.g., temperature, pressure, light) into an electrical signal, or vice-versa. An example is a microphone (sound to electrical) or a loudspeaker (electrical to sound), or a thermistor (temperature to electrical resistance).

Bandwidth in a measurement system refers to the range of frequencies over which the system can accurately measure or respond to changes in the input signal. A wider bandwidth means the system can respond to faster changes in the measured quantity without significant distortion or attenuation.

A "pull-up" resistor connects an input pin to the positive voltage supply (e.g., 5V). It's used with push-button switches to ensure the pin has a defined HIGH state when the button is not pressed. Without it, the pin would be "floating" and susceptible to noise, leading to erratic readings. When the button is pressed, it connects the pin to ground, pulling it LOW.

An LED (Light Emitting Diode) is a diode that emits light when current flows through it in the forward direction. It requires a specific forward voltage to turn on and must have its current limited to prevent damage. A current-limiting resistor is placed in series with the LED to drop the excess voltage from the power supply and control the current flowing through the LED, protecting it from burning out.

A passive component is one that cannot generate power but can only dissipate, store, or absorb it (e.g., resistors, capacitors, inductors). An active component is one that can control current flow and can amplify or switch electrical signals, often requiring an external power source to operate (e.g., transistors, operational amplifiers, diodes, microcontrollers).

Gain in an amplifier circuit is a measure of its ability to increase the power or amplitude of a signal. It is typically defined as the ratio of the output signal power or amplitude to the input signal power or amplitude. For voltage amplifiers, it's the ratio of output voltage to input voltage (V_out/V_in).

A decoupling capacitor is placed close to the power pins of integrated circuits (ICs) in digital circuits. Its purpose is to provide a local reservoir of charge to supply instantaneous current demands from the IC during rapid switching operations and to filter out high-frequency noise on the power supply lines, thereby ensuring stable power delivery and preventing glitches.

A voltage divider circuit is a simple passive linear circuit that produces an output voltage that is a fraction of its input voltage. It consists of two series resistors (or other components with impedance) across which the input voltage is applied, and the output voltage is taken across one of the resistors. It's used to scale down a voltage to a desired level.

A short circuit is an unintended low-resistance connection across two points in a circuit, allowing a very large current to flow, often causing damage. An open circuit is a break in a circuit that prevents current from flowing, resulting in infinite resistance.

"Ground" in an electrical circuit serves as a common reference point for voltage, typically considered 0V. It provides a return path for current to the power source and helps to stabilize voltage levels throughout the circuit. It can also serve as a safety measure, providing a path for fault currents to safely dissipate.

The purpose of a fuse or circuit breaker is to protect electrical circuits from overcurrent conditions (e.g., short circuits or overloads) that could damage components or cause fires. They act as sacrificial devices (fuses) or resettable switches (circuit breakers) that automatically interrupt the current flow when it exceeds a safe level.

An ideal voltage source provides a constant voltage across its terminals regardless of the current drawn from it (it has zero internal resistance). An ideal current source provides a constant current regardless of the voltage across its terminals (it has infinite internal resistance).

First, calculate the equivalent resistance of the parallel combination (R_parallel):

1 / Rparallel = 1 / R1 + 1 / R2
1 / Rparallel = 1 / 60 Ω + 1 / 90 Ω
1 / Rparallel = 3 / 180 Ω + 2 / 180 Ω = 5 / 180 Ω
Rparallel = 180 Ω / 5 = 36 Ω

Next, calculate the equivalent resistance of R_parallel in series with R₃:

Req = Rparallel + R3
Req = 36 Ω + 30 Ω = 66 Ω

Therefore, the equivalent resistance is 66 Ω.

a) Voltage drop across R_A:

First, calculate the total resistance of the series circuit:

Rtotal = RA + RB = 3.0 kΩ + 5.0 kΩ = 8.0 kΩ

Next, calculate the total current flowing through the circuit:

I = Vsource / Rtotal = 24 V / 8.0 kΩ = 24 V / 8000 Ω = 0.003 A = 3 mA

Finally, calculate the voltage drop across R_A:

VRA = I * RA = 0.003 A * 3000 Ω = 9 V

b) Power dissipated by R_B:

Using the current calculated in part a):

PRB = I2 * RB
PRB = (0.003 A)2 * 5000 Ω
PRB = (9 * 10-6) * 5000
PRB = 0.045 W = 45 mW

Therefore:

a) Voltage across R_A: 9 V

b) Power dissipated by R_B: 45 mW

The relationship between charge (Q), current (I), and time (t) is:

Q = I * t

The relationship between charge (Q), capacitance (C), and voltage (V) is:

Q = C * V

Combining these equations to solve for time (t):

I * t = C * V
t = (C * V) / I

Given:

C = 68 μF = 68 * 10-6 F
V = 100 V
I = 0.08 A

Calculation:

t = (68 * 10-6 F * 100 V) / 0.08 A
t = 68 * 10-4 / 0.08
t = 0.0068 / 0.08
t = 0.085 s

Therefore, it will take 0.085 seconds for the voltage across the capacitor to reach 100 V.

The voltage induced across an inductor is given by the formula:

V = L * (dI/dt)

To find the rate of change of current (dI/dt), rearrange the formula:

dI/dt = V / L

Given:

V = 7.5 V
L = 30 mH = 30 * 10-3 H

Calculation:

dI/dt = 7.5 V / (30 * 10-3 H)
dI/dt = 7.5 / 0.03
dI/dt = 250 A/s

Therefore, the current through the inductor is changing at a rate of 250 A/s.

a) Current flowing through the resistor and the LED:

In a series circuit, the voltage across the resistor is the supply voltage minus the voltage drop across the LED:

VR = Vsupply - VLED
VR = 6 V - 2.2 V
VR = 3.8 V

Now, use Ohm's Law to find the current (I = V/R). Since it's a series circuit, this current flows through both the resistor and the LED:

I = VR / R
I = 3.8 V / 470 Ω
I ≈ 0.008085 A or 8.085 mA

b) Power dissipated by the LED:

The power dissipated by the LED is calculated as the product of the voltage across the LED and the current flowing through it:

PLED = VLED * I
PLED = 2.2 V * 0.008085 A
PLED ≈ 0.017787 W or 17.787 mW

Therefore:

a) Current flowing through the resistor and LED: 8.085 mA

b) Power dissipated by the LED: 17.787 mW

a) Equivalent resistance of the parallel circuit:

1 / Req = 1 / RA + 1 / RB + 1 / RC
1 / Req = 1 / 100 Ω + 1 / 200 Ω + 1 / 50 Ω
1 / Req = 2 / 200 Ω + 1 / 200 Ω + 4 / 200 Ω = 7 / 200 Ω
Req = 200 Ω / 7 ≈ 28.57 Ω

b) Total current drawn from the source:

Itotal = Vsource / Req
Itotal = 10 V / 28.57 Ω ≈ 0.35 A or 350 mA

Therefore:

a) Equivalent resistance: 28.57 Ω

b) Total current: 350 mA

a) Average rate of change of current:

dI/dt = ΔI / Δt = (Ifinal - Iinitial) / (tfinal - tinitial)
dI/dt = (1.5 A - 0.5 A) / (20 * 10-3 s) = 1 A / 0.02 s = 50 A/s

b) Magnitude of the average induced voltage:

V = L * (dI/dt)
V = (10 * 10-3 H) * (50 A/s)
V = 0.5 V

Therefore:

a) Average rate of change of current: 50 A/s

b) Average induced voltage: 0.5 V

The initial charge stored in a capacitor can be calculated using the formula:

Q = C * V

Given:

C = 100 μF = 100 * 10-6 F
V = 50 V

Calculation:

Q = (100 * 10-6 F) * (50 V)
Q = 5000 * 10-6 C = 0.005 C

Alternatively, using the average current and time:

Q = Iavg * t
Q = 0.01 A * 0.5 s = 0.005 C

Therefore, the initial charge stored in the capacitor was 0.005 C.

For an ideal inverting Op-Amp configuration, the output voltage is given by the formula:

Vout = - (Rf / Rin) * Vin

Given:

Rin = 10 kΩ
Rf = 100 kΩ
Vin = 0.2 V

Calculation:

Vout = - (100 kΩ / 10 kΩ) * 0.2 V
Vout = -10 * 0.2 V
Vout = -2 V

Therefore, the output voltage is -2 V.

When a Zener diode is operating in its breakdown region, it maintains a constant voltage across itself (the Zener voltage). In this series circuit, the voltage across the resistor will be the supply voltage minus the Zener voltage:

VR = Vsupply - VZener
VR = 9 V - 5.6 V
VR = 3.4 V

Now, use Ohm's Law to find the current flowing through the resistor:

I = VR / R
I = 3.4 V / 1.5 kΩ = 3.4 V / 1500 Ω
I ≈ 0.002267 A or 2.267 mA

Therefore, the current flowing through the resistor is approximately 2.267 mA.

Total Resistance (R_total) = R_top + R_bottom = 5 kΩ + 10 kΩ = 15 kΩ

Vbottom = Vsupply * (Rbottom / Rtotal)
Vbottom = 15 V * (10 kΩ / 15 kΩ)
Vbottom = 15 V * (10 / 15)
Vbottom = 15 V * (2 / 3)
Vbottom = 10 V

Therefore, the voltage across R_bottom is 10 V.

Charge (Q) = C * V

Q = 10 * 10-6 F * 5 V = 50 * 10-6 C = 50 μC

Average Current (I) = Q / t

I = 50 * 10-6 C / (20 * 10-3 s)
I = 50 / 20 * 10-3 A
I = 2.5 * 10-3 A = 2.5 mA

Therefore, the average charging current is 2.5 mA.

V = L * (dI/dt)
dI/dt = V / L
dI/dt = 10 V / (50 * 10-3 H)
dI/dt = 10 / 0.05 A/s
dI/dt = 200 A/s

Therefore, the rate of change of current is 200 A/s.

For an ideal non-inverting Op-Amp configuration, the output voltage is given by the formula:

Vout = Vin * (1 + Rf / R1)
Vout = 0.5 V * (1 + 18 kΩ / 2 kΩ)
Vout = 0.5 V * (1 + 9)
Vout = 0.5 V * 10
Vout = 5 V

Therefore, the output voltage is 5 V.

Number of Steps = 2¹⁰ = 1024
Resolution = V_ref / Number of Steps = 3.3 V / 1024 ≈ 0.0032226 V/step

Digital Reading = Analog Voltage / Resolution
Digital Reading = 1.65 V / (3.3 V / 1024)
Digital Reading = 1.65 * 1024 / 3.3
Digital Reading = 512

Therefore, the digital reading is 512.