IA: 1P3: Analysis of Circuits and Devices Flashcards

Question

Plot a graph of AC voltage and current through a capacitor

Answer 1

P = vi Where v and i are the RMS values

Answer 2

Xc = 1 / ωC

Answer 3

* If the imaginary component of Zₜ is positive, it is an inductor * If the imaginary component of Zₜ is negative, it is a capacitor

Answer 4

Through a series resistance which represents the resistance of the wire

Answer 5

Through a parallel resistance which represents the leakage current through the dielectric

Answer 6

amplitude = 2√(2) Phase = -53.1 degrees

Answer 7

It points to the highest potential / more positive side of the source

Answer 8

It refers to the reference point for phase measurements relevant to that source

Answer 9

An ideal voltage source has a constant voltage V₀ (or amplitude v) across its terminals and no resistance. A real voltage source would have a source resistance/impedance in series with the ideal voltage source

Answer 10

An ideal current source always produces the required current. An ideal current source will have an infinite internal resistance (or impedance).

Answer 11

An ideal current source will have infinite internal resistance, a non-ideal source will have an internal resistance Rₛ **in parallel** with the source. ## Footnote internal resistance is in parallel!

Answer 12

Any **linear** two-terminal network may be replaced by a voltage source Vₜₕ **in series** with a resistance Rₜₕ ## Footnote **linear** means that this theorem does not apply to **power**

Answer 13

Any linear two-terminal network may be replaced by a current source Iₙ **in parallel** with a resistance Rₙ ## Footnote **linear** means that this theorem does not apply to **power**

Answer 14

1. You would convert the parallel circuits into their norton equivalents 2. Combine the parallel current sources into a single combined current source 3. Combine the parallel resistances into a single combined resistance 4. Take the thevenin equivalent of the circuit and determine the resistance across the load

Answer 15

The sum of the voltages around a closed loop must be zero

Answer 16

V₀ = -V₁ - V₂ - V₃ - V₄ V₀ + V₁ + V₂ + V₃ + V₄ = 0

Answer 17

The current flowing into any point of a circuit must be equal to the current which flows out of that point. In other words, the sum of current at every node must be zero

Answer 18

* Mesh analysis * Nodal analysis ## Footnote They can be applied to both AC and DC circuits

Answer 19

1. Identify mesh currents which will cause every component to have at least one mesh current passing through it 2. Consider each loop and form simultaneous equations using kirchoffs voltage law 3. Solve to find the mesh currents 4. You then have all the required values to solve the question

Answer 20

1. Label each of the nodes 2. Assign one node as zero volts (if the circuit is grounded then there is already a defined earth point) 3. Create an expression for each of the currents in terms of voltage using Ohm's law 4. Use Kirchoff's current law to determine the unknown voltage

Answer 21

Circuits used to measure unknown resistances, impedances, capacitances, or inductances by balancing 2 legs of a symmetric circuit. A bridge circuit consists of four branches that form a quadrilateral, with a source of voltage or current applied across one pair of opposite corners and a detection device connected across the other pair. **At the balance point there will be no current through G and the 2 sides are at the same voltage**

Answer 22

tends to zero

Answer 23

A resistor and a capacitor in series with each other, with Vout across the capacitor

Answer 24

Above the point where frequency starts to decrease the impedance of the capacitor. This point is defined as where the magnitude of the **voltage gain is reduced by a factor of √(2)**, this is often known as the **0.7 or 70% voltage point**. Similarly it can be defined in terms of power as the point at which the** ratio of input to output power is reduced by a factor of 2**, or in the decibel scale the log ratio is reduced by roughly 3 decibels.

Answer 25

a log₁₀ plot of the magnitude power/voltage gain against log₁₀ frequency

Answer 26

* Gain: Gain > 1 → positive Decibels * Attenuation: Gain < 1 ← Negative Decibels

Answer 27

The same as that on a low pass filter

Answer 28

**dB/decade** Decibels are the units for the gain, a decade represents a tenfold increase or decrease in frequency.

Answer 29

The slope of the stop band is limited to 20dB/decade, many applications require a much higher cut off slope and therefore a higher order filter may be required

Answer 30

There is a 90 degree phase difference between voltage (v) and current (i). The **voltage** is **leading** by 90 degrees

Answer 31

180 degrees

Answer 32

The quality factor: it is ratio of the reactance to the resistance

Answer 33

* Q = 1/ω₀CR * Q = ω₀L/R * Q = |v₂|/|v₁| **(only at resonance)**

Answer 34

* Imperfections in L or C increase the resistance and can reduce Q

Answer 35

* Q > 10 is good * Q < 1 is very bad

Answer 36

the range of frequencies over which the circuit can resonate effectively

Answer 37

Take the norton equivalent of the of the voltage source v1 and R1. You can then make all the components be in parallel by replacing R with an equivalent parallel impedance, RL.

Answer 38

## Footnote Where RL is the parallel equivalent impedance of the series impedance R

Answer 39

## Footnote Where Rᴄ is the parallel equivalent impedance of the series impedance R

Answer 40

A circuit designed to amplify the voltage of an input signal while preserving its waveform characteristics

Answer 41

A = voltage gain ## Footnote the whole circuit can be represented by just 3 parameters in this model

Answer 42

**Amplifiers which have a linear gain**, some amplifiers may have a gain which is a function of frequency and therefore cannot be represented by this model

Answer 43

You often want a high input impedance (Rin) and low output impedance (Rout) as:

Answer 44

Cascading amplifiers is when you connect two or more amplifier stages in series to achieve greater overall gain whilst avoiding other problems (such as the miller effect)

Answer 45

A circuit designed to increase the current of an input signal

Answer 46

## Footnote Note: unlike a voltage amplifier, the output impedance is now connected in parallel with the current source

Answer 47

For a power amplifier driving a load RL with an output impedance Rout, the **maximum power will occur when RL = Rout**

Answer 48

Match the load to the output impedance

Answer 49

**The amplifier power supply**, a DC source supploed the extra power delivered to the load as well as any power that might be dissipated in the internal circuit of the amplifier

Answer 50

Since the additional power results from **the amplifiers power supply, these power supplies (V+ and V-) set the maximum positive and negative voltage swing at output**. If the gain is too high and the voltage tries to exceed the power supplies, it will be "clipped", leading to distortion in the signal shape

Answer 51

The capacitors will form basic RC filters, with C1 forming a high pass filter and C2 forming a low pass filter. This will restrict the bandwidth of the circuit and therefore the filter pair is often called a bandpass filter. ## Footnote The capacitors could also have been placed at the output of the amplifier rather than the input

Answer 52

It is assumed that the capacitors C1 and C2 have sufficiently different values and their effect on the frequency response is far apart enough so that they can be considered separately: * Therefore when finding the low 0.7 frequency, you can completely ignore C2, treating it as an open circuit * When finding the high 0.7 frequency, you can completely ignore C1, treating it as a short circuit * When finding the midband gain you can perform the calculation using either scenario (usually much easier to use C1 as it is in series)

Answer 53

A semiconductor is a material that has an electrical conductivity between that of a conductor and an insulator. Its conductivity can be controlled by various factors, such as temperature and doping. For example: Silicon is an insulator at low temperatures, but as temperature is increased it becomes more conducting. Its properties can also be dramatically altered by the incoroporation of dopant atoms in the crystal.

Answer 54

Semiconductors, such as silicon, have stable covalently bonded structures. The electrons do not have enough energy to break free and move throughout the crystal at room temperature and so the semiconductor behaves like an insulator. As temperature increases, electrons can become free and so can move through the crystal, they also leave behind a hole on the silicon atom from where it came and this hole can also move throughout the crystal structure and so contributes to the current. Therefore, as temperature increases, resistance decreases, and conductivity increases

Answer 55

Dopants are impurities added to the structure, this can add extra free electrons or holes to the structure. For silicon (which has 4 valence electrons) to add extra free electrons we dope with a pentavalent metal (P, As, Sb), and to add extra holes we dope with a trivalent metal (B, Ga, In). These extra free electrons/holes increase the conductivity of the semiconductor

Answer 56

An n-type semiconductor is a type of extrinsic semiconductor which has been doped in order to increase the number of free electrons (negative charge carriers) and so increase the electrical conductivity. In the case of silicon, it is doped with a pentavalent metal.

Answer 57

A p-type semiconductor is a type fo extrinsic semiconductor which has been doped in order to increase the number of holes (positive charge carriers) and so increase the electrical conductivity. In the case of silicon, it is doped with a trivalent metal.

Answer 58

* An intrinsic semiconductor is pure semiconductor material with equal numbers of electrons and holes as charge carriers. Example: Pure silicon * An extrinsic semiconductor is a doped semiconductor with impurities to increase conductivity. It has an imbalance of charge carriers (either more electrons in n-type or more holes in p-type)

Answer 59

A structure where the doping of a semiconductor abruptly changes from p-type to n-type. When the junction is formed, electrons and holes will diffuse from n to p-type and from p to n-type, respectively. This results in a region where there are no free electrons or holes, known as the depletion region. This produces an electric field which opposes difusion, the potential associated with this electric field is known at the contact potential.

Answer 60

There is no net flow of current across the junction

Answer 61

Reverse biased is when you apply the external voltage such that the n-type will become more positive, causing more electrons to be removed from it and so increasing the size of the depletion region (hence holes are also removed from the p-type). The net current is known as the reverse saturation current, I₀.

Answer 62

A photodiode uses a reverse biased p-n junction as the reverse saturation current can be increased by exposure to light. Therefore photodiodes are engineered to allow light to access the junction.

Answer 63

Forward biased is when you apply the external voltage such that the n-type will become more negative, causing fewer electrons to be removed from it and so decreasing the size of the depletion region (hence fewer holes are also removed from the p-type). The net current is then predominantly due to the diffusion current and is usually far larger than the reverse saturation current

Answer 64

The arrow indicates the direction of conventional current flow under forward biasing

Answer 65

When a diode is forward biased, the electrons moving across the P-N junction can fall into empty holes from the P-type layer. This involves a drop from the conduction band to a lower orbital and so the electrons release energy in the form of photons. This happens in any diode, but only certain materials will produce photons of visible light.

Answer 66

You have to use graphical techniques to determine the operating point: Plot the line for the current through the **diode** and plot this on the diode V-I graph, you can often use components in series with the diode and mesh analysis to determine the equation of this current. This plotted line is known as the **load line** ## Footnote Note: the equation of this line is unique for each circuit!!

Answer 67

As rectifiers

Answer 68

* Field-effect transistor (FET) * Bipolar junction transistor (BJT)

Answer 69

A high-speed, low power transistor which is **controlled by the input voltage** as it uses an electric field to control the flow of current

Answer 70

A metal oxide semiconductor FET

Answer 71

* n-channel enhancement * n-channel depletion * p-channel enhancement * p-channel depletion

Answer 72

It can be positive or negative

Answer 73

it controls the flow of current from the drain to the source

Answer 74

* n-type: Arrow points towards the gate * Enhancement mode: gaps in symbol

Answer 75

If a voltage is applied between the drain (D) and source (S) when there is no gate voltage, then no drain current will flow as one of the p-n junctions will always be reverse biased and there are no charge carriers in the space between the drain and source. If a positive gate voltage (VGS) is then applied, there will be an electric field between the gate and bottom of the device, attracting electrons towards the region between the gate oxide and p-type substrate (most of these electrons come from the n-type regions). If VGS is high enough (above the threshold value, VT) enough electrons will accumulate to cause an inversion layer to form between the drain and source, forming an n-type conducting channel within the p-type. If you then a voltage between the drain and source (VDS) a current ID will flow from drain to source.

Answer 76

* **ID** (current between drain and source) against **VDS** (voltage between drain and source) * **ID** (current between drain and source) against **VGS** (voltage between the gate and ground) **in saturation**

Answer 77

* n-type: Arrow points towards the gate * Depletion mode: NO gaps in symbol

Answer 78

A depletion mode MOSFET has a built-in n-type channel. The application of a positive gate voltage has the same effect as previously, however a negative gate voltage repels charge away from the cchanne, decreasing the conductivity.

Answer 79

## Footnote Note: VGS is now negative to turn off the FET

Answer 80

A type of amplifier configuration using a field-effect transistor (FET), where the source terminal is grounded and so is common to both the input and the output circuits.

Answer 81

Ideally, it has infinite input impedance, however this is not very practical. Therefore normally the input impedance is usually very large (~MΩ)

Answer 82

## Footnote Note: The operating point has been set such that VDS = VDD/2

Answer 83

**VDS = VDD/2** This is so there is the maximum voltage swing at the output. The value of RD then creates a load line and operating point

Answer 84

Different limiting factors which sets the safe operating region: * A: non-linear, low resistance region * B: Maximum limit of VGS * C: Maximum power dissipation limit, (VDS * ID)max, hyperbolic curve * D: Maximum values of VDS * E: Safe operating region ## Footnote Ideally the load line should cross diagonally through the safe operating region and the operating point should be as central as possible

Answer 85

A self-biased MOSFET amplifier is a type of amplifier circuit where the biasing of the MOSFET is achieved through feedback provided by a resistor connected to the source terminal.

Answer 86

A self-biasing MOSFET amplifier uses a resistor (RS) in series with the source terminal to create a voltage drop, which sets the gate-source voltage (VGS). This establishes the operating point of the amplifier without requiring an external bias voltage source. The MOSFET is biased through feedback provided by RS, where the voltage drop across the resistor, caused by the source/drain current (ID), determines VGS.

Answer 87

Practical depletion mode MOSFET amplifier circuits use the self-biased amplifier where VGS is or needs to be negative

Answer 88

VGS must be positive if an enhancement mode MOSFET is being used. Therefore, the gate voltage is set by 2 resistors, R1 and R2 which form a potential divider, and then VGS is set by the difference between teh gate voltage and the source voltage (set by VS = RS * ID) ## Footnote Note: The input impedance is now set by the parallel combination of R1 and R2

Answer 89

From DC biasing we have set the operating point. We can then see how the circuit operates with a small signal (v₁) applied to the input. When we apply an input voltage to the amplifier, we are adding a small AC voltage to the existing DC voltage set up by the biasisng process. This means we are effectively shifitng the operating point slightly (but only by a small amount as it is a small signal) and changing Vɢs, Iᴅ, and Vᴅs accordingly.

Answer 90

With lower case variables. For example: * DC: Vɢs, Vᴅs * AC: Vgs, Vds

Answer 91

As the AC input voltage is applied (Vgs) it causes a change in Vɢs (Vgs = ΔVɢs). As Vɢs changes we move along the load line set by the DC operating point. This causes a change in both Iᴅ and Vᴅs, giving us a change in the output voltage, Vds (Vds = ΔVᴅs)

Answer 92

Using the small signal model/equivalent circuit

Answer 93

1 / rd = partial derivative of Iᴅ with respect to Vᴅs (constant Vɢs)

Answer 94

gₘ = partial derivative of Iᴅ with respect to Vɢs (constant Vᴅs)

Answer 95

It replaces the FET with a controlled current source and resistances.

Answer 96

It is the rate at which the drain current (Iᴅ) changes with respect to the gate-to-source voltage (Vɢs)

Answer 97

It is the rate at which the drain-to-source voltage (Vᴅs) change with drain current (Iᴅ)

Answer 98

It allows us to replace the FET in the circuit with its equivalent model and then calculate small signal properties such as gain, input resistance (Rᵢₙ) and output resistance (Rₒᵤₜ).

Answer 99

Vᴅᴅ is a DC component, therefore (if we assume it is ideal) it has zero series resistance. Hence, for small AC signals it will act as a direct path to ground. Therefore it can be thought of as a very large capacitor and will have zero impedance (so just a wire connecting directly to ground).

Answer 100

As a short circuit (zero impedance)

Answer 101

Vᴅᴅ is a DC component and therefore it behaves like a short circuit directly to ground in the small signal model. Therefore we can connect Rᴅ from the drain directly to ground.

Answer 102

It is a DC component, therefore it behaves as a short circuit in the AC small signal model

Answer 103

It allows you to express the amplifier with just three parameters:

Answer 104

* For a depletion mode MOSFET, Rɢ holds the gate at 0V. Therefore Vɢs will be negative. * For an ehancement mode MOSFET, the gate is biased by a potential divider of two circuits. Therefore Vɢs can be positive.

Answer 105

It is a common condition with this circuit to simplify the analysis by assuming that rd is very large.

Answer 106

Using the assumption that rd →∞ (and is therefore ignored): * Gain = - gₘRᴅ / (1 + gₘRₛ) * Rᵢₙ = Rɢ = R₁||R₂ * Rₒᵤₜ ≈ Rᴅ More precise expressions for gain and output resistance are in the bottom right - this is when rd is not ignored and is included in the expressions.

Answer 107

* Use the original circuit to determine v₁, Iᴅ, and Rs * Use the small signal model to determine Rᴅ (using the SSM expression for gain)

Answer 108

It acts as a buffer circuit - a commonly used means of separating different circuits that may interact with each other. It isolates different stages. This is because: * The gain ≈ 1 * It has a very high input resistance * It has a very low input resistance

Answer 109

Introduce a **bypass capacitor** in parallel with Rs. This allows the AC current to bypass the resistor. To analyse this we replace Rs with the impedance Zs (Zs = Rs||Cs). In order for the bypass capacacitor to be effective, Zs should be very small (and can be neglected in the SSM). This will be the case if ωRsCs >> 1. The gain then becomes -gₘRᴅ

Answer 110

* A positive gain means that the signal has been amplified/attenuated * A negative gain means that the signal has been amplified/attenuated **and inverted by 180°**

Answer 111

When connecting 2 amplifier stages, there is a problem with connecting the output of the first circuit with the input of the second circuit as they were designed with different operating points and will have different DC voltages. A **decoupling capacitor** can be inserted between the 2 circuits to prevent DC current flowing from one circuit to another. If you know the values of Rₒᵤₜ and Rᵢₙ for each circuit then you can calculate a suitable decoupling capacitor using the general amplifier model. The cutoff frequency is given by: f = 1 / 2πC(Rₒᵤₜ₁ + Rᵢₙ₁). This allows you to find a suitable capacitor for the AC frequency. Decoupling capacitors are also commonly used at the inputs and outputs of each amplifier circuit to protect either loads or input sources.

Answer 112

**The Miller effect is when a capacitance that connects the inputs of an amplifier to its output is effectively magnified due to the amplifiers gain (This is the miller capacitance).** This can be harmful as it limits the high-frequency performance of the amplifiers and slows down the response time.

Answer 113

When a real FET is made there are unwanted or parasitic capacitances which occur due to the small distances between the electrodes connected to the silicon that makes the transistor. The capacitance between the gate and the drain (Cgd) can be between the input and output and therefore cause the miller effect.

Answer 114

**Cₘ = (1 - A)C** (for an inverting amplifier) Cₘ = Miller capacitance A = gain C = original capacitance ## Footnote NOTE: This assumes an **inverting amplifier**. The equation can also be written as "Cₘ = (1 + |A|)C". ONLY for an inverting amplifier (where the gain is negative) it can be written as Cₘ = (1 - A)C

Answer 115

1. Choose a suitable FET (gₘ, rd, price etc) 2. Choose a DC operating point - biasing (Iᴅ, Vᴅs, Vɢs) 3. Calculate resistors (Rᴅ, Rɢ (or R₁ and R₂), Rs 4. Draw the SSM for the circuit 5. Calculate gain, Rᵢₙ, Rₒᵤₜ 6. Adjust accordingly (but check biasing is still okay)

Answer 116

An Operational Amplifier. They are complex analogue integrated circuits that are used to amplify voltage. They can work with both DC and AC input voltages. They usually have 2 inputs (+) and (-), an output, and are usually supplied by a dual power supply (5 terminals).

Answer 117

* Infinite open-loop gain, A = ∞ * Infinite input resistance, Rᵢ = ∞ * Zero output resistance, Rₒ = 0

Answer 118

The difference between the input voltages is **always** zero: V(+) - V(-) = 0

Answer 119

There is no current into either input **always**: i(+) = i(-) = 0

Answer 120

* Gain = - R₂/R₁ * Rᵢₙ = R₁

Answer 121

* Gain =1 + R₂/R₁ * Rᵢₙ = ∞

Answer 122

It is the OpAmp version of the source follower. It has unity gain, high input resistance, and low output resistance (a buffer).

Answer 123

* Gain =1 * Rᵢₙ = ∞ * Rₒᵤₜ = 0

Answer 124

It is the same circuit but the resistances have been replaced by impedances.

Answer 125

It is a type of amplifier that type of amplifier that converts current into voltage. If the input to an amplifier is a current and the output is a voltage, then the gain is effectively an impedance (v/i).

Answer 126

In optical communications to convert a light signal into a voltage signal, i.e. photodiodes.

Answer 127

They are sensitive to the miller effect

Answer 128

A voltage comparator is a circuit that compares two input voltages and outputs a digital signal indicating which one is higher.

Answer 129

The positive and negative power supply rails (gain is infinite but limited by the available voltage)

Answer 130

It is an OpAmp used without feedback

Answer 131

From the OpAmp model we can see that the output voltage v₀ is A(v(+) - v(-)). Hence if v+ > v- then the output will be +∞ but if v+ < v- then the output will be -∞. Clearly this is not realistic as the maximum voltage available will be set by the positive and negative supply rails (+V and -V), therefore the outputs are limited to +V and -V and so the output will switch polarity depending on which input is larger than the other. v(ref) (v-) is set by the potential divider.

Answer 132

The circuit acts like it has negative resistance: increasing voltage causes decreasing current