Analysis of Circuits

Question 1

Define p-type and n-type Silicon

Answer 1

P-type Silicon is doped with Group 3 atoms. These atoms readily accept electrons due to their higher concentration of holes.

N-type Silicon is doped with Group 5 atoms. These atoms have one extra free electrons to be used for conduction. So N-type has free electrons for conduction at room temperature.

Question 2

Describe the features of the p-n junction

Answer 2

The p-n junction is formed when p-type and n-type silicon recombine. The free electrons from the n-type combine with the holes of the p-type creating a depletion region.

When the p-type is positive (forward bias) the electrons are repelled from the negative terminal and the depletion regions narrows.

When the p-type is negative (reverse bias) the carriers are attracted to the positive terminal causing the depletion region to widen.

Question 3

What forms the basis of the Diode

Answer 3

P-n junction

Question 4

Describe the construction of a Bi-Polar transistor

Answer 4

There are two types of BJT - pnp and npn. In each case a BJT is formed by putting two diodes back to back.

Question 5

Why is the BJT said to be bi-polar

Answer 5

The name bipolar arises as both charge carriers and holes take part in the current flow. In npn carriers dominate the flow whereas in pnp holes dominate the flow

Question 6

What are the relative levels of doping in a pnp transistor

Answer 6

The emitter is n-type and highly doped

The collected is n-type and the least doped

Question 7

What is the basic principle underlying BJT operation

Answer 7

Due to the thin base and high doping of the emitter, a small change in the base voltage (base current) leads to a large change in the collector current.

Question 8

What is the current gain

Answer 8

The current gain, hfe is the ratio of the collector to base current - Ic/Ib

Question 9

What is the minimum voltage needed for the BJT to function

Answer 9

VBE ~ 0.7V

Question 10

What are the criteria for choosing an BJT operating point

Answer 10

1. Stay wishing the safe limits
2. Ensure the max value of Ic is not exceeded
3. Choose the highest value of Vcc possible
4. Set Vce to be as close to Vcc/2 as possible. This will give the maximum voltage swing at the output.

Question 11

How can the collector resistance be found

Answer 11

By applying nodal voltage analysis we can draw a load line Ic = (Vcc - Vce)/Rc

Question 12

Define the small signal parameters

Answer 12

hfe - current gain
hre - reverse voltage transfer ratio
hie - input resistance
hoe - output admittance

Question 13

What are the key characteristics of the Basic Common Emitter

Answer 13

The basic common emitter has a base resistance to fix the operating point and no emitter resistance (gain not stabilised). The common emitter has a large negative gain and so is used as a single stage inverting amplifier

Question 14

What are the key characteristics of a stabilised common emitter circuit

Answer 14

The stabilised common emitter has a potential divider to fix the base voltage. It also has an emitter resistance to stabilise the gain. However the emitter resistance reduces the gain and so often a bypass capacitor is used (shorts the resistor in the SSM)

Question 15

What is the key assumption when analysing the stabilised common emitter

Answer 15

If the current gain is v.large/infinite then we can assume the base current is negligible. Therefore we can use potential divider to find the input voltage. If the current gain is not infinite we must draw the a Thevenin equivalent circuit.

VT = (R2/R1+R2) * Vcc
RT = R1*R2/R1+R2

Question 16

What is the benefit of adding a resistor between the base and the collector

Answer 16

The resistor acts as a negative feedback loop. This has all the benefits of negative feedback (e.g stabilised gain, increased input resistance etc.)

Question 17

What are the characteristics and role of the emitter follower

Answer 17

The emitter follower is a buffer circuit. It has a high input resistance, low output resistance and a unity gain. The operating point is stable as there is an emitter resistance.

Question 18

What are common and differential mode signals

Answer 18

Common mode signals are noise (v1=v2) whereas differential mode signals are desired (v1=-v2).

Question 19

What are differential amplifiers

Answer 19

Differential amplifiers are designed to amplify the difference between the two inputs to the circuits. Their performance is measured using the Common Mode Rejection Ratio

Question 20

How do we represent a common and differential mode signal

Answer 20

Vcm = v1+v2/2
Vdm = v1-v2/2

Question 21

Define the common mode rejection ratio

Answer 21

Differential mode gain/common mode gain

Question 22

How do you find the differential mode gain for a transistor

Answer 22

We draw the SSM of one half of the circuit neglecting the emitter resistance

Question 23

How do you find the common mode gain for a transistor

Answer 23

We draw the SSM for one half of the circuit and include twice the emitter resistance

Question 24

What is the Miller Approximation

Answer 24

The miller approximation states that if a capacitor is placed between the base and collector of a BJT we can neglect the contribution to the current at the collector.

So, when performing nodal voltage analysis at the base we include the current contribution of the miller capacitor but we ignore it at the collector.

Question 25

How do you find the 3dB frequency of a Miller Capacitor

Answer 25

To find the 3dB frequency equate the real and imaginary components of the input current expression

Question 26

What are the advantages of negative feedback

Answer 26

``` Stabilisation of the gain Increased bandwidth Increased bandwidth Increased input resistance Reduced output resistance ```

Question 27

What is the gain expression for a simple feedback system

Answer 27

G = A/1+AB | Note: A and B are dimensionless - B represents the portion of the output voltage fed back into the Op-Amp

Question 28

How are the gain, bandwidth, impedance and stability affected by the feedback system

Answer 28

1. Bandwidth is increased by (1+AB) 2. Gain is reduced by 1/(1+AB) - so additional amplifiers are required 3. Input impedance is increased by 1+AB 4. Output impedance is decreased by 1+AB 5. The fractional change in the gain is reduced by 1/(1+AB)

Question 29

What are the characteristics of an ideal Op-Amp

Answer 29

Infinite open loop gain (v+ = v-) Infinite input resistance (I+ = I- = 0 ) Zero output resistance

Question 30

To which terminal of the Op-Amp is the feedback loop

Answer 30

Always the negative terminal

Question 31

What is the gain and ip/op impedance for an inverting amplifier

Answer 31

Gain = -R2/R1 Input Impedance = R1 Output Impedance = 0

Question 32

What is the gain and ip/op impedance for a non-inverting amplifier

Answer 32

Gain = 1+R2/R1 Input impedance is infinite Output impedance is zero

Question 33

Define a Class A amplifier

Answer 33

Class A operation is when current flows in the transistor throughout the cycle. It is characterised by the drawing of constant current from the power supply. E.g. common emitter amplifier

Question 34

Define a Class B amplifier

Answer 34

Each transistor conducts for half the cycle. This results in a dead-band in the amplification. Class B operation has higher efficiency and zero power consumption.

Question 35

Define a Class AB amplifier

Answer 35

Transistors that operate for more than half a cycle but less that a whole cycle are following AB operation. This is a compromise between linearity and efficiency.

Question 36

Give an example of a Class B amplifier and outline its construction/operation

Answer 36

The complementary emitter follower is formed of two transistors - one pnp and one npn. It has the advantage of not drawing power when there is no input signal. When v is positive the pnp transistor is on when v is negative the npn transistor is on. Since both have a minimum voltage of 0.7V a dead-band forms

Question 37

What changes can be made to the Complementary Emitter Follower to improve linearity

Answer 37

The voltage drop across the diodes biases the transistors into conduction. The current source prevent cross over distortion

Question 38

How do you calculate the DC input power to a BJT

Answer 38

The product of Vcc and the steady state current (current at operating point) gives the input power

Question 39

How do we reliably generate periodic signals

Answer 39

We can use a feedback network with a frequency dependent response. This will gives us control over the amplitude and frequency of the generated signals.

Question 40

What are the conditions for oscillation

Answer 40

A is the open loop gain B is the gain of the feedback loop |AB| = 1 Im{B(jw} = 0

Question 41

What criteria are used for determining suitable values for the input and output resistance of an Op-Amp

Answer 41

For a non-ideal Op-Amp Rin should be x100 larger than the input resistance. Rout should be x100 smaller than the load resistance