Exam Deck

Question 1

Power Amplifiers: Class A Amplifier

Answer 1

Question 2

Power Amplifiers: Class A Emitter Follower

Answer 2

Question 3

When you’re looking at a BJT what does the position of the arrow indicate?

Answer 3

Indicates the position of the emitter terminal.

Question 4

What does Am represent?

Answer 4

The mid band gain

i.e. the voltage gain

and is equivalent to Vo/Vsig

Question 5

Common-emitter amplifier

Answer 5

Question 6

CS amplifier: Miller’s Theorem

Questions a) and b)

Answer 6

Question 7

Open circuit time constant method

What do you need to remember?

Answer 7

That the resistance wrt gate and drain is as follows:

𝑅_𝑔𝑑 = 𝑅’_𝑠𝑖𝑔 + 𝑅’_𝐿 + 𝑔_𝑚𝑅’_𝑠𝑖𝑔𝑅’_𝐿

Just remember this!

Bode Plot is straightforward. The magnitude until the break frequency is hit and then -20dB/decade.

Question 8

What do you need to remember about how divides itself when it faces paths with different resistance values towards ground?

Answer 8

It goes through the path of least resistance. Thus if there is a resistance R_m for one path and resistances R₁ and R₂ for the other then the current will split into

I*(R₁ + R₂)/((R₁ + R₂ + R_m) for the R_m path

and

I*(R_m)/((R₁ + R₂ + R_m) for the R₁ + R₂ path

T

Question 9

Conversion Efficiency

Answer 9

Question 10

Explain the statement that “a power amplifier can be regarded as a frequency converter”. [2 marks]

Answer 10

The statement “a power amplifier can be thought of as a frequency converter”, refers to the fact that a power amplifier converts dc power from the supply into ac signal power in the load.

Question 11

Transconductance Amplifier

Answer 11

Question 12

Power Amplifiers

Answer 12

Question 13

List three important parameters in an ideal op-amp other than gain. In a practical case, briefly explain what is the effect of having a finite open-loop gain. [8 marks]

Answer 13

Three important parameters of an ideal opamp:

1) Infinite bandwidth.
2) Infinite input resistance.
3) Zero output resistance.

Having a finite open loop gain causes the calculation of the closed loop gain to not purely depend on the ratio of external elements (resistors, capacitors). In other words, the smaller the open loop gain, the higher the deviation from the ideal closed loop gain.

Question 14

Phase-shift oscillator

Answer 14

Question 15

LC oscillator

Answer 15

Question 16

What do A and Af respectively stand for?

Answer 16

for open loop gain

and

closed loop gain

Question 17

If you get a transfer function as the following, how do you calculate the break frequency?

Answer 17

Since the denominator is already technically in the correct form, we can see that that 1/w_c = g_mL

Thus w_c = 1/g_mL

Question 18

Transconductance (series - series)

Answer 18

Question 19

Unity gain frequency of MOSFET: Do a)

Answer 19

Question 20

CE short-circuit current gain h_fe: Question b) and c)

Answer 20

Question 21

Common Collector Configuration: Resistances seen

Answer 21

Question 22

What is the common collector also known as ?

Answer 22

Emitter Follower

Voltage Follower

typically used as a voltage buffer.

Question 23

Bandwidth of Transistor circuits

Answer 23

External capacitors (coupling and bypass) limit the bandwidth at low frequencies by forming a high-pass. Smaller RC may dominate, so smaller R may dominate.

Internal capacitors limit bandwidth at high frequencies by forming a low-pass. Larger RC (lower freq.) may dominate, so larger R may dominate.

Question 24

What is the main difference when using the open- circuit time constant method on CS and CC (emitter follower) or CD (source follower) respectively?

Answer 24

CS is referring to common source. While CC and CD, are common collector and common drain, equivalent nodes on BJT and MOSFET respectively.

The main difference is that when calculating C_gd for the CS you connect a test current source I_x across the gate and drain.

On the other hand, for the source follower (CD), to find Cgs,

a test voltage Vgs is added between nodes G and S.

Question 25

Emitter Follower Question

Answer 25

Question 26

CS Amplifier

Answer 26

Question 27

List the five benefits of using negative feedback in a circuit. Which parameter is usually traded for these benefits to occur?

Answer 27

Five benefits:

i) Desensitize the gain
ii) Reduce non-linear distortion
iii) Reduce the effect of noise
iv) Control the input and output resistance
v) Extend the bandwidth

Those benefits are obtained at the expense of a reduction in gain.

Question 28

Feedback

Answer 28

Question 29

Explain how the internal capacitors Cgd and Cgs behave at mid-band frequencies (you need to justify your answer).

Answer 29

The internal capacitors Cgs, and Cgd usually have a very small capacitance and so behave like open-circuits in the midband. This is because their impedance is given by Z=1/jωC and so for C<< we have Z>>.

Question 30

State Miller’s Theorem

Answer 30

According to Miller’s theorem the bridging capacitor Cgd can be replaced by two equivalent shunt capacitances one between the Gate and the ground, denoted as C1 and another between the drain and the ground, denoted as C2.

c1 = cgd(1-k)

c2 = cgd(1 - 1/k)