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AQA A-Level Physics: Electronics

6 exam-style questions with full mark schemes and model answers. Write your own answer and the AI examiner marks it against the mark scheme.

Learn this properly: Semiconductor Devices and Junction Transistors
Question 16 marksExplain

An operational amplifier (op-amp) is connected as an inverting amplifier. The input signal is applied through an input resistor RinR_{in}Rin to the inverting (-) input, the non-inverting (+++) input is held at 0 V, and a feedback resistor RfR_fRf is connected from the output back to the inverting input.

Explain how this circuit works as an inverting amplifier. Your answer should refer to:

  • the assumptions that the op-amp draws negligible input current and has a very large open-loop gain;
  • why the inverting input behaves as a virtual earth sitting at approximately 0 V;
  • how the negative feedback through RfR_fRf leads to a voltage gain of G=RfRinG = -\frac{R_f}{R_{in}}G=RinRf.

(6 marks)

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Question 26 marksCalculate

An op-amp is wired as an inverting amplifier and powered from supply rails of +15+15+15 V and 15-1515 V. The resistor values are listed below.

QuantityValue
Input resistor RinR_{in}Rin22 kΩ
Feedback resistor RfR_fRf220 kΩ
Supply rails±15\pm 15±15 V

(a) Calculate the voltage gain of the amplifier. (1 mark)

(b) A steady input of +0.80+0.80+0.80 V is applied. Calculate the output voltage, and state whether the output is within the range of the supply rails. (2 marks)

(c) The input is now increased to +2.0+2.0+2.0 V. Calculate the output voltage the gain formula predicts, explain why the actual output cannot reach this value, and state the actual output voltage. (3 marks)

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Question 35 marksCalculate and interpret

A temperature-warning circuit uses a potential divider made from a negative-temperature-coefficient (NTC) thermistor in series with a fixed 10 kΩ resistor, across a 5.0 V supply. The voltage measured across the fixed resistor is fed to the non-inverting (+++) input of an op-amp comparator. The inverting (-) input is held at a fixed reference of 2.5 V. The comparator is powered from ±15\pm 15±15 V rails.

The thermistor's resistance at two temperatures is given below.

ConditionThermistor resistance
Cold (5 °C)30 kΩ
Hot (60 °C)2.5 kΩ

(a) Calculate the voltage at the non-inverting input (across the 10 kΩ resistor) when the thermistor is cold, and state the comparator output voltage (+Vs+V_s+Vs or Vs-V_sVs). (2 marks)

(b) Repeat the calculation for the hot condition and state the comparator output. (2 marks)

(c) State, with a reason, whether this circuit switches its output high when the surroundings get hot or when they get cold. (1 mark)

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Question 45 marksCalculate and explain

A designer is building an automatic security light that must switch on in darkness. A light-dependent resistor (LDR) is connected in series with a fixed 10 kΩ resistor across a 6.0 V supply, and the voltage across the LDR is fed to the non-inverting (+++) input of an op-amp comparator. The comparator output drives a transistor that switches a relay, and the relay turns the lamp on when the comparator output is high (+Vs+V_s+Vs). The LDR resistance varies as shown.

ConditionLDR resistance
Bright daylight1.0 kΩ
Darkness100 kΩ

(a) Calculate the voltage at the non-inverting input in bright daylight and in darkness. (2 marks)

(b) The reference voltage on the inverting input is set to 3.0 V. Explain how the system responds as it gets dark, and explain why a reference of 3.0 V is a sensible choice. (3 marks)

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Question 54 marksCalculate

A microphone pre-amplifier uses an op-amp connected as a non-inverting amplifier, powered from ±15\pm 15±15 V rails. The feedback resistor is Rf=90 kΩR_f = 90 \ \text{k}\OmegaRf=90 kΩ and the resistor from the inverting input to earth is Rin=10 kΩR_{in} = 10 \ \text{k}\OmegaRin=10 kΩ.

(a) Calculate the voltage gain of the amplifier. (1 mark)

(b) A signal of amplitude 0.500.500.50 V is applied at the input. Calculate the amplitude of the output signal. (1 mark)

(c) Express the voltage gain of the amplifier in decibels, using GdB=20log(VoutVin)G_{dB} = 20\log\left(\frac{V_{out}}{V_{in}}\right)GdB=20log(VinVout). (2 marks)

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Question 63 marksState

Real operational amplifiers are often analysed by treating them as ideal op-amps.

State three properties of an ideal operational amplifier. (3 marks)

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