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Mock Paper

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3. In the circuit of figure 3, the transistors are identical and have the parameters:

b = 150, CBE = 10 pF, CBC = 4 pF.

- What name is given to the amplifier configuration shown in figure 3?
- Cascode amplifier
- What is the key advantage of this circuit compared with the common-emitter amplifier?
- Higher upper cut-off frequency

(c)Calculate the quiescent base, emitter and collector voltages for the two transistors and the collector current of Q1.

Quiescent assumptions:

IB = 0, VBE = 0.5 V

24V dropped across three 100kW resistors, 8V across each:

- = 150, CBE = 10 pF, CBC = 4 pF.

(d)Calculate the mid-band gain, the input impedance and the output impedance of the circuit.

- = 150, CBE = 10 pF, CBC = 4 pF.

(e)Calculate the lower and upper cut-off frequencies of the amplifier for a source impedance of 1 kW.

Lower cut-off

- = 150, CBE = 10 pF, CBC = 4 pF.

(e)Calculate the lower and upper cut-off frequencies of the amplifier for a source impedance of 1 kW.

Upper cut-off

- = 150, CBE = 10 pF, CBC = 4 pF.

In the circuit shown in figure 4, assume that the forward biased diode voltage is 0.5 V.

(f)Calculate the thermal noise voltage that appears across the resistor in terms of V/ÖHz

T = 300 K

k = 1.38×10-23 J/K

q = 1.6×10-19 C

(g)Calculate the additional noise voltage that appears across the resistor due to shot noise in the diode current (again in V/ÖHz). Hence calculate the total noise voltage across the resistor.

T = 300 K

k = 1.38×10-23 J/K

q = 1.6×10-19 C

4

The total maximum power dissipation of a class-B power amplifier is calculated to be 15 W. Given output transistors with a specification of TJmax = 150 °C, qJA = 40 °C/W, qJC = 1.5 °C/W:

i.Calculate the power dissipated by each output transistor.

ii.Calculate the minimum specifications for heatsinks that could be used for each transistor.

iii.Calculate the minimum specification for a single heatsink that could be used by both transistors.

Single heatsink must be twice the specification:

- (b)In a common-emitter amplifier:
- Explain why the base-collector capacitance of the transistor usually has the most influence over the upper cut-off frequency of the amplifier.
- In a common-emitter amplifier, the base-collector capacitance is multiplied by the amplifier gain (plus one) due to the Miller effect. Consequently, it would usually appear to be an order of magnitude larger than the base-emitter capacitance
- Suggest three ways in which the upper cut-off frequency of an amplifier can be increased.
- Decrease the gain (reduces Miller effect)
- Use a transistor with lower junction capacitances
- Adopt a cascode configuration

(c)Explain, using supporting diagrams, how generalised impedance converter(s) can be used to simulate:

i.A grounded inductance

i.e. the circuit has the same input impedance as an inductance of CR2 Henries

(c)Explain, using supporting diagrams, how generalised impedance converter(s) can be used to simulate:

ii.A floating inductance

Again, the circuit has the same impedance as an inductance of CR2 Henries

(c)Explain, using supporting diagrams, how generalised impedance converter(s) can be used to simulate:

iii.A frequency dependent negative resistance

i.e. a frequency dependant negative resistance