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Appendix 1

Self-study questions

The answers to all self-study questions in the book are provided here, indexed by chapter.

Chapter 1 Scales

Q1: What is the total current I_tot, as a result of adding 100mA and 3mA?

Q2: What is the total voltage V_out as a result of adding 10μV and 7mV?

Q1 answer: The total current I_tot is 103mA.

Q2 answer: The total voltage V_tot is approximately 7mV.

Equations

Q1: Add $\frac{2}{3} + \frac{1}{4}$ $\frac{2}{3} + \frac{1}{4}$ , what is this fraction (as a voltage) of an Arduino 5V power supply?

Q2: What is the reciprocal of $\frac{1}{3}$ $\frac{1}{3}$ and what is the reciprocal of $\frac{3}{4}$ $\frac{3}{4}$ ?

Q1 answer: $\frac{2}{3} + \frac{1}{4} = \frac{11}{12}$ $\frac{2}{3} + \frac{1}{4} = \frac{11}{12}$ , $\frac{11}{12}$ $\frac{11}{12}$ of a 5V supply is 4.58V.

Q2 answer: The reciprocal of $\frac{2}{3}$ $\frac{2}{3}$ is $\frac{3}{2}$ $\frac{3}{2}$ , the reciprocal of $\frac{3}{4}$ $\frac{3}{4}$ is $\frac{4}{3}$ $\frac{4}{3}$ .

Electrical charge

Q1: What is current?

Q2: What is potential difference?

Q3: What is resistivity?

Notes:

These three statements define the fundamentals of electricity and electronics, and so it is important to understand them rather than simply memorize the phrase. In the case of current, the important thing to remember is that this is the flow of charge over time – charge is the amount of electrons involved. The term voltage is more commonly used for potential difference, but the units of the quantity do not adequately describe the difference between two net charges. Resistivity is perhaps the easiest of the three to remember, as it describes the resistance of a material to the flow of current. Resistors are used throughout electronics to control current, with a useful rule of thumb being that the absence of any resistance creates an open circuit – an uncontrolled path for current to flow down (which is never desirable).

Chapter 2

Q1: Fill in the two missing labels on this open loop system.

Answer: The missing labels are sensor (left label) and actuator (right label).

Q2: (a) What type of system is this? (b) What is the missing label on the lower right of the diagram?

Answer: The missing label is feedback – this is a fundamental element in any system, and is crucial to the use of operational amplifiers in chapter 7.

Q3: Fill in the two missing labels on this open loop system.

Answer: The missing labels are source (left label) and load (right label).

Notes:

For the first question, an open-loop system is shown where sensors convert physical quantities (e.g. sound, temperature, movement) into electrical signals, whilst actuators convert electrical signals into physical quantities. The second question shows a closed-loop system, where feedback is used to control the processes that the system carries out. The third question shows an open-loop system in terms of its source and load impedances, which define how different parts of a system ‘see’ each other in electronics terms. If you have an instrument input to an amplifier, the impedance (AC form of resistance – see chapter 6) of the instrument is the source the amplifier must be able to handle as an input. Similarly, the same amplifier may have to drive an actuator like a loudspeaker as a load – this explains why loudspeakers are usually rated by their impedance in ohms (Ω).

Chapter 3 Series circuits

Use the following schematic to answer the questions below:

Q1: What is the total resistance in the circuit?

Q2: What is the total current I_tot, in the circuit?

Q3: What are the voltage drops across each of the three resistors?

Parallel circuits

Use the following schematic to answer the questions below:

Q4: What is the total resistance in the circuit?

Q5: What is the current I_tot, and the current in each branch of the circuit?

Voltage dividers

Use the following schematic to answer the questions below:

Q6: What is the voltage V_out?

Q7: What is the voltage across the first resistor, V_R1?

Chapter 4 Programming concepts

Chapter 6 LTspice simulation

Q1: Create an LTspice circuit to simulate the following schematic:

(a) What is the current flowing through the circuit?

(b) What are the voltage drops across the resistor (V_R) and the capacitor (V_C)?

Answer:

The current flowing through the circuit should be measured through R1 with a current probe in LTspice, to give the current in phase with the source (as resistors have no phase angle). If a comparison is made with the capacitor current (light grey trace below) then the difference in phase can also be seen:

This book avoids extensive discussion of phase, but it is important to note the impact of the capacitor on the phase of the current in the circuit. Adding a cursor to the LTspice current trace will give a maximum simulated value of 497.37μA through R1 at the first peak in the waveform at a time of 256.94μs (microseconds) – this is close to the value of 0.5mA calculated in worked example 6.4.1.

The simulated voltage drops can also be measured, but a simultaneous trace shows the discrepancy in the magnitude of the capacitor voltage at this frequency:

Adding a cursor to each trace gives a peak voltage of 4.99V through R1 at a time of 256.94μs, whilst the peak voltage through C1 is 31.63mV at a time of 491.32μs. This suggests a problem, as the calculated value from example 6.4.1 gives a capacitor voltage of 16mV. The problem here is the short analysis window of 0.01 seconds (used to show the frequency cycles of the input source), which does not give the simulation enough time to calculate the capacitor voltage once the circuit has stabilised at around 100ms. If the Spice Directive is changed to increase the analysis window to 100ms (.tran 200 0.1 0), the output trace shows how the capacitor voltage quickly reduces from 31mV to ~15.37mV:

Q2: Create an LTspice circuit to simulate the following schematic:

(a) What are the individual branch currents in the circuit?

(b) What is the magnitude of the total current flowing through the circuit?

Chapter 7 Amplification

Q1: Calculate the power gain Ap and decibel power gain ap (dB) of an audio system with:

(a) A DC input power of 2W and an output power of 4W.

(b) A DC input power of 1W and an output power of 5W.

Q2: Calculate the voltage gain Av and decibel voltage gain av (dB) of an audio amplifier with:

(a) An input signal of 100mV and an output signal of 1V.

(b) An input signal of 100mV and an output signal of 0.5V.

Operational amplifiers

Use the following schematic for an inverting operational amplifier to answer the questions below:

Q3: If V_in = 100mV and V_out = −0.5V, what are the values of R₁ and R₂?

Q4: If V_in = 100mV and V_out = −2V, what are the values of R₁ and R₂?

Chapter 8 First-order filters

Q1: An RC circuit is configured as a first-order low-pass filter, with a 1kΩ resistor and a 1μF capacitor. What is the cutoff frequency of the filter?

Q2: A CR circuit is configured as a first-order high-pass filter, with a 10kΩ resistor and a 10nF capacitor. What is the cutoff frequency of the filter?

Q3: Design a low-pass filter with a cutoff of 200Hz – how close can you get to this value with a resistor of 470Ω and a single capacitor?

From appendix 3, the standard values available for a capacitor would range from 1μF to 4.7μF for E3–E12 series components. Starting with the E3 series will give an idea of the range involved:

1μF capacitor:

\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π R C} = \frac{1}{2 π \times (470) \times (1 \times 10^{- 6})} \\ = \frac{1}{6.28 \times 470 \times 0.000001} = \frac{1}{0.00295} \approx 339 H z \end{array}

$\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π R C} = \frac{1}{2 π \times (470) \times (1 \times 10^{- 6})} \\ = \frac{1}{6.28 \times 470 \times 0.000001} = \frac{1}{0.00295} \approx 339 H z \end{array}$

2.2μF capacitor:

\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 p \times (470) \times (2.2 \times 1 0^{- 6})} \\ = \frac{1}{6.28 \times 470 \times 0.0000022} = \frac{1}{0.00649} \approx 154 H z \end{array}

$\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 p \times (470) \times (2.2 \times 1 0^{- 6})} \\ = \frac{1}{6.28 \times 470 \times 0.0000022} = \frac{1}{0.00649} \approx 154 H z \end{array}$

4.7μF capacitor:

\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π \times (470) \times (4.7 \times 1 0^{- 6})} \\ = \frac{1}{6.28 \times 470 \times 0.0000047} = \frac{1}{0.01387} \approx 72 H z \end{array}

$\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π \times (470) \times (4.7 \times 1 0^{- 6})} \\ = \frac{1}{6.28 \times 470 \times 0.0000047} = \frac{1}{0.01387} \approx 72 H z \end{array}$

The E3 values suggest that somewhere between 1μF and 2.2μF will be closer, so moving up to E6 the middle value is 1.5μF:

\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π \times (470) \times (1.5 \times 1 0^{- 6})} \\ = \frac{1}{6.28 \times 470 \times 0.0000015} = \frac{1}{0.00442} \approx 226 H z \end{array}

$\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π \times (470) \times (1.5 \times 1 0^{- 6})} \\ = \frac{1}{6.28 \times 470 \times 0.0000015} = \frac{1}{0.00442} \approx 226 H z \end{array}$

This would be a reasonably close cutoff frequency for most purposes, but if greater accuracy is needed then the E48 series includes a value of 1.69μF:

\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π \times (470) \times (1.69 \times 1 0^{- 6})} \\ = \frac{1}{6.28 \times 470 \times 0.00000169} = \frac{1}{0.00498} \approx 200 H z \end{array}

$\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π \times (470) \times (1.69 \times 1 0^{- 6})} \\ = \frac{1}{6.28 \times 470 \times 0.00000169} = \frac{1}{0.00498} \approx 200 H z \end{array}$

Q4: Design a high-pass filter with a cutoff of 3kHz – how close can you get to this value with a single resistor and capacitor of 0.1μF?

The previous example included a cutoff frequency value of 339Hz for a combination of a 1μF capacitor with a 470Ω resistor, so it would be prudent to begin calculations by factoring in the 0.1μF value in this question:

1μF capacitor:

\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π R C} = \frac{1}{2 π \times (470) \times (0.1 \times 1 0^{- 6})} \\ = \frac{1}{6.28 \times 470 \times 0.0000001} = \frac{1}{0.00029} \approx 3388 H z \end{array}

$\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π R C} = \frac{1}{2 π \times (470) \times (0.1 \times 1 0^{- 6})} \\ = \frac{1}{6.28 \times 470 \times 0.0000001} = \frac{1}{0.00029} \approx 3388 H z \end{array}$

Looking at the structure of the equation, the reciprocal relationship means that increasing the 2πRC value will decrease the cutoff frequency. Again beginning with the E3 series, the next value in the range will be 1kΩ:

\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π \times (1000) \times (0.1 \times 1 0^{- 6})} \\ = \frac{1}{6.28 \times 1000 \times 0.0000001} = \frac{1}{0.00063} \approx 1592 H z \end{array}

$\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π \times (1000) \times (0.1 \times 1 0^{- 6})} \\ = \frac{1}{6.28 \times 1000 \times 0.0000001} = \frac{1}{0.00063} \approx 1592 H z \end{array}$

This is too low a frequency, so go to the next series (E6) to find a resistor between 470Ω and 1kΩ – this gives a value of 680Ω:

\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π \times (680) \times (0.1 \times 1 0^{- 6})} \\ = \frac{1}{6.28 \times 680 \times 0.0000001} = \frac{1}{0.00043} \approx 2342 H z \end{array}

$\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π \times (680) \times (0.1 \times 1 0^{- 6})} \\ = \frac{1}{6.28 \times 680 \times 0.0000001} = \frac{1}{0.00043} \approx 2342 H z \end{array}$

This is still too low, so moving to E12 gives a value of 560Ω:

\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π \times (560) \times (0.1 \times 10^{- 6})} \\ = \frac{1}{6.28 \times 560 \times 0.0000001} = \frac{1}{0.00035} \approx 2843 H z \end{array}

$\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π \times (560) \times (0.1 \times 10^{- 6})} \\ = \frac{1}{6.28 \times 560 \times 0.0000001} = \frac{1}{0.00035} \approx 2843 H z \end{array}$

This value is closer than the first attempt of 3388Hz, but given that 3kHz is in an area of human hearing that is particularly sensitive (see section 8.1 for equal loudness curves) the calculation should perhaps aim for closer. This means moving to E24 and a value of 510Ω:

\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π \times (510) \times (0.1 \times 1 0^{- 6})} \\ = \frac{1}{6.28 \times 510 \times 0.0000001} = \frac{1}{0.00032} \approx 3122 H z \end{array}

$\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π \times (510) \times (0.1 \times 1 0^{- 6})} \\ = \frac{1}{6.28 \times 510 \times 0.0000001} = \frac{1}{0.00032} \approx 3122 H z \end{array}$

This is closer again, but instead of 157Hz less the value is now 122Hz more! The E48 series includes a value of 536Ω:

\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π \times (536) \times (0.1 \times 1 0^{- 6})} \\ = \frac{1}{6.28 \times 536 \times 0.0000001} = \frac{1}{0.00033} \approx 2971 H z \end{array}

$\begin{array}{l} C u t o f f F r e q u e n c y, f c = \frac{1}{2 π \times (536) \times (0.1 \times 1 0^{- 6})} \\ = \frac{1}{6.28 \times 536 \times 0.0000001} = \frac{1}{0.00033} \approx 2971 H z \end{array}$

This would seem to be a close enough value to 3kHz for most practical purposes – to get any closer, the E192 series could be used, which includes a value of 530Ω (3003Hz)!

Notes:

The values chosen in questions Q1 and Q2 are intended to show the influence of scaling factors on calculations like cutoff frequency. Chapter 1 showed how scaling factors can become confusing, and it is important to note that while the resistor in Q2 has now increased by a factor of 10 from 1kΩ to 10kΩ, the change in the capacitor from 1μF to 10nF represents a factor of 100 decrease. The reciprocal relationship in the equation means that decreasing the 2πRC value will increase the cutoff frequency, so although the resistor increases by a factor of 10, the capacitor decreases by a factor of 100 – thus the cutoff frequency increases by a factor of 10 between Q1 and Q2.

The calculations performed in Q3 and Q4 may seem tedious (as they are!) and there are online filter calculators available for many filter types (including higher-order examples) that will quickly perform these calculations for you (a spreadsheet can also be used to quickly ‘plug in’ numerous values). The questions are intended to show how a pragmatic approach to choosing component values can get close enough to the required value in many instances. Cost is always a major factor in electronics, and when you study commercial schematics for well-known audio products it can be fascinating to see where compromises have (and have not) been made. Guitar effects pedal schematics (notably by Boss) contain some very impressive solutions that are both cost-effective and musically valid – studying the filter designs used in such circuits would be a valuable extension of the basic concepts covered in this book.

A final observation on the practicalities of audio filter design when considered relative to the purely theoretical exercise of calculating first-order RC cutoff values: the simple (and cheaper) option for the filter above would seem to be adding a second series resistor (such as a 68Ω resistor from the E6 series) to the first 470Ω value to bring the overall resistance value to ~538Ω and achieve a cutoff frequency of 2958Hz. This would be mathematically valid, but every component added to a circuit will both introduce noise and attenuate the signal. In addition, the cost of the second resistor must now be added to the build and in the highly competitive audio electronics market, the price of a piece of equipment must (usually) be kept to a minimum if it is to have any chance of market success. These factors are not important when learning about audio electronics, but when moving into the practical world of equipment sales they become critical to the success of a new product.

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Table of Contents for Appendix 1 Self-study questions

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Self-study questions

Chapter 1

Scales

Equations

Electrical charge

Chapter 2

Chapter 3

Series circuits

Parallel circuits

Voltage dividers

Chapter 4

Programming concepts

Chapter 6

LTspice simulation

Chapter 7

Amplification

Operational amplifiers

Chapter 8

First-order filters

Table of Contents for
Appendix 1 Self-study questions