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2P5 Electronics > Analysis of Circuits > Flashcards

Analysis of Circuits Flashcards

1
Q

Define p-type and n-type Silicon

A

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.

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2
Q

Describe the features of the p-n junction

A

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.

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3
Q

What forms the basis of the Diode

A

P-n junction

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4
Q

Describe the construction of a Bi-Polar transistor

A

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

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5
Q

Why is the BJT said to be bi-polar

A

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

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6
Q

What are the relative levels of doping in a pnp transistor

A

The emitter is n-type and highly doped

The collected is n-type and the least doped

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7
Q

What is the basic principle underlying BJT operation

A

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.

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8
Q

What is the current gain

A

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

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9
Q

What is the minimum voltage needed for the BJT to function

A

VBE ~ 0.7V

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10
Q

What are the criteria for choosing an BJT operating point

A
  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.
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11
Q

How can the collector resistance be found

A

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

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12
Q

Define the small signal parameters

A

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

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13
Q

What are the key characteristics of the Basic Common Emitter

A

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

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14
Q

What are the key characteristics of a stabilised common emitter circuit

A

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)

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15
Q

What is the key assumption when analysing the stabilised common emitter

A

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
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16
Q

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

A

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

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17
Q

What are the characteristics and role of the emitter follower

A

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.

18
Q

What are common and differential mode signals

A

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

19
Q

What are differential amplifiers

A

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

20
Q

How do we represent a common and differential mode signal

A 
Vcm = v1+v2/2
Vdm = v1-v2/2
21
Q

Define the common mode rejection ratio

A

Differential mode gain/common mode gain

22
Q

How do you find the differential mode gain for a transistor

A

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

23
Q

How do you find the common mode gain for a transistor

A

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

24
Q

What is the Miller Approximation

A

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.

25
Q

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

A

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

26
Q

What are the advantages of negative feedback

A 
Stabilisation of the gain
Increased bandwidth 
Increased bandwidth
Increased input resistance
Reduced output resistance
27
Q

What is the gain expression for a simple feedback system

A

G = A/1+AB

Note: A and B are dimensionless - B represents the portion of the output voltage fed back into the Op-Amp

28
Q

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

A
  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)
29
Q

What are the characteristics of an ideal Op-Amp

A

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

30
Q

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

A

Always the negative terminal

31
Q

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

A

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

32
Q

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

A

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

33
Q

Define a Class A amplifier

A

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

34
Q

Define a Class B amplifier

A

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.

35
Q

Define a Class AB amplifier

A

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.

36
Q

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

A

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

37
Q

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

A

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

38
Q

How do you calculate the DC input power to a BJT

A

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

39
Q

How do we reliably generate periodic signals

A

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.

40
Q

What are the conditions for oscillation

A

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

41
Q

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

A

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

2P5 Electronics (1 decks)

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