Electronics 2 Exam Instructions

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Questions and Answers

If an amplifier's AC load line operation extends 15µA above and below its Q-point base current of 40µA, how does this affect the signal handling capability, assuming symmetrical clipping?

  • Reduces maximum possible signal swing. (correct)
  • The AC load line does not affect signal handling.
  • Increases maximum possible signal swing.
  • Maintains the signal swing, but shifts the Q-point.

What does the stability of voltage gain in a BJT amplifier primarily ensure?

  • The amplifier operates at the highest possible gain.
  • The amplifier can amplify signals of any frequency without distortion.
  • The voltage gain remains constant regardless of temperature or transistor variations. (correct)
  • The amplifier consumes minimal power.

In a voltage-divider bias configuration, what is the main advantage of designing it such that it is 'stiff'?

  • The power consumption of the circuit is significantly reduced.
  • The base voltage is relatively independent of the transistor's beta. (correct)
  • The collector current becomes more sensitive to temperature variations.
  • The base voltage is highly dependent on the transistor's beta.

In a BJT amplifier circuit, if the DC beta ($\beta_{DC}$) increases due to a temperature change, and the bias is not particularly stiff, how will this primarily affect the collector current ($I_C$)?

<p>$I_C$ will increase, possibly leading to saturation or clipping. (D)</p> Signup and view all the answers

What is the primary effect of increasing the value of a bypass capacitor connected in parallel with the emitter resistor in a common-emitter amplifier?

<p>It increases the amplifier's voltage gain at AC. (D)</p> Signup and view all the answers

What is the most likely consequence of removing the emitter bypass capacitor in a common-emitter amplifier circuit?

<p>Reduced voltage gain and increased stability. (A)</p> Signup and view all the answers

Why is it important to consider the 'r'e' (AC emitter resistance) of all transistors in a multi-stage amplifier circuit when calculating the overall gain?

<p>Because <code>r'e</code> affects the input impedance and voltage gain of each stage, thus influencing the overall gain. (A)</p> Signup and view all the answers

In a class AB amplifier, what is the primary purpose of the biasing network that sets a small quiescent current?

<p>To reduce crossover distortion by ensuring that both transistors are slightly conducting even without an input signal. (A)</p> Signup and view all the answers

What is crossover distortion in a power amplifier, and which class of amplifier is most susceptible to it?

<p>A distortion occurring during the transition between the conduction of transistors in the output stage; Class B. (C)</p> Signup and view all the answers

Which characteristic distinguishes a Class A power amplifier from Class B and Class AB amplifiers?

<p>Lower distortion than Class B. (B)</p> Signup and view all the answers

In a negative feedback amplifier, if the feedback network returns a fraction of the output signal to the input, what effect does this generally have on the amplifier’s bandwidth?

<p>Bandwidth increases. (C)</p> Signup and view all the answers

Which of the following is a primary advantage of using negative feedback in an amplifier circuit?

<p>Reduced harmonic distortion. (C)</p> Signup and view all the answers

In an amplifier circuit, what is the effect of the Miller capacitance, and how does it typically impact the high-frequency response?

<p>It increases the effective input capacitance, worsening the high-frequency response. (A)</p> Signup and view all the answers

What is the primary reason for using cascading amplifier stages in electronic circuit design?

<p>To achieve a higher overall voltage gain than can be achieved with a single stage. (B)</p> Signup and view all the answers

How does the addition of a source/emitter follower stage typically affect the overall performance of a multi-stage amplifier?

<p>It decreases the output impedance and increases the current gain. (A)</p> Signup and view all the answers

Flashcards

Stability of Voltage Gain

The extent to which the voltage gain of a BJT amplifier remains constant despite variations in temperature, transistor parameters, or operating conditions.

Stiff Voltage-Divider Bias

A bias circuit where the base voltage is largely independent of the transistor's beta, providing stable operation.

Cross-Over Distortion

A distortion that occurs in class B amplifiers when the input signal is near zero, causing the transistors to be cut off and the output signal to be missing a portion around the zero-crossing point.

Cut-off Region

Region where both transistors are off, creating a 'dead zone' in signal reproduction.

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Active Region

Region where the transistor acts as a variable resistor, amplifying the input signal.

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Q-Point

The point on a transistor's characteristic curve that represents the DC bias conditions (IC and VCE) when no input signal is applied.

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Characteristic Curves

A graphical representation of the relationship between collector current (IC) and collector-emitter voltage (VCE) for different base currents (IB).

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DC (BDC) and AC (Bac) Beta

Ratio of collector current (IC) to base current (IB) in a BJT under DC conditions (BDC) or varying signal conditions (Bac).

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Study Notes

  • This is a fill-in paper test for Electronics 2, module code EEELE2A.
  • The test is set for March 15, 2024, and the duration is 90 minutes.
  • Only non-programmable calculators are allowed.
  • The marks total 37, and the question paper consists of 10 typed pages.

Instructions Summary

  • A valid student card is required.
  • Student number needs to be written clearly.
  • Answers have to be numbered correctly.
  • SMS/texting language and TIP-EX are not allowed.
  • Answers must be in pen, sketches in pencil.
  • All work must be submitted before leaving.
  • Cell phones and other mobile devices must be switched off and kept away.

Disciplinary Actions

  • Possession of undocumented material can lead to disciplinary hearing/disqualification.
  • Communication with others (excluding the invigilator) is prohibited.
  • Offensive matter is prohibited.
  • Committing irregularities is prohibited.
  • Unsatisfactory conduct is prohibited.

Prohibitions

  • Leaving the assessment venue without permission.
  • Asking for explanations of questions.
  • Borrowing anything from other candidates.

Question 1

  • This pertains to the characteristic curves of a certain amplifier.
  • The AC load line operation extends 10 µA above and below the Q-point base current of 35 µA.

Question 1.1

  • Determine the resulting peak-to-peak values of the collector current and collector-to-emitter voltage from the graph.

Question 1.2

  • Determine dc beta (BDC) and ac beta (Bac) from the graph.
  • Explain what is meant by the stability of the voltage gain for a BJT amplifier.

Question 2

  • The amplifier circuit is provided in Figure 2.

Question 2.1

  • Determine whether the voltage-divider bias circuit is stiff and justify the answer.
  • Calculate VB, IE, and VCE.

Question 2.2

  • Amplifier circuit is shown in Figure 3, with Bac = BDC = 150 for all transistors.
  • All transistor ré considerations are a must.
  • IE1 is stated as 6.468mA in Figure 3.
  • Calculate Vout.

Question 3

  • Given the load line in Figure 4, label the Q-points in the figure with their corresponding power amplifier class
  • Given the power amplifier circuit in Figure 5, draw a fully labelled and improved version of this circuit to eliminate cross-over distortion.

Equations

  • IE = IC + IB
  • Ic/IB = BDC
  • VBE = 0.7 V
  • IB = (VBB - VBE) / RB
  • VCE = VCC - ICRC
  • VCB = VCE - VBE
  • A = RC / re
  • VCE(cutoff) VCC
  • IC(sat) (VCC - VCE(sat)) / RC
  • IB(min) IC(sat) / BDC
  • IC = BDCIB
  • VB = VCC * (R2 / (R1 + R2))
  • VE = VB - VBE
  • IC = IE = VE / RE
  • VC = VCC - ICRC
  • RIN(BASE) = BDCRE
  • IE = (VTH - VBE) / (RE + RTH/BDC)
  • IE = (-VEE - VBE) / (RE + RB/BDC)
  • VCE = VCC - ICRC
  • IC = BDC(VCC - VBE) / (RE + RB/BDC)
  • IE = (VCC - VBE) / (RB/BDC)
  • VCE = VCC - ICRC
  • re = 25mV / IE
  • Ap = A'Ai
  • R = R ∥ R ∥ R in(base) 1 2 in(base)
  • Rout = Rc
  • Av = RC / re
  • Rin(source) = 1/gm
  • A0 = RC /RE1
  • For FET: R = RG (VGS) / IGSS in

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