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An ultrasound sensor produces an output that needs to be amplified to 3.0 V - AQA - A-Level Physics - Question 2 - 2018 - Paper 8

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An ultrasound sensor produces an output that needs to be amplified to 3.0 V. The amplifier used has a voltage gain of 40. Calculate the input voltage $V_{in}$ to th... show full transcript

Worked Solution & Example Answer:An ultrasound sensor produces an output that needs to be amplified to 3.0 V - AQA - A-Level Physics - Question 2 - 2018 - Paper 8

Step 1

Calculate the input voltage $V_{in}$ to the amplifier from the sensor.

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Answer

To find the input voltage, we use the formula:

Vout=VinimesextgainV_{out} = V_{in} imes ext{gain}

Given that the output voltage VoutV_{out} is 3.0 V and the gain is 40:

3.0=Vinimes403.0 = V_{in} imes 40

Rearranging gives:

Vin=3.040=0.075 VV_{in} = \frac{3.0}{40} = 0.075 \text{ V}

Step 2

Complete the circuit diagram in Figure 3 by adding and labelling two resistors, $R_{in}$ and $R_f$, so that the operational amplifier is correctly configured in its non-inverting mode.

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Answer

To configure the operational amplifier in non-inverting mode:

  • Connect RinR_{in} between the input signal and the non-inverting terminal (+).
  • Connect RfR_f between the output and the inverting terminal (-).
  • Ground the other end of RfR_f to ensure proper feedback.

Step 3

Determine, using resistors selected from the list below, how the voltage gain of 40 can be achieved by the non-inverting amplifier of Figure 3.

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Answer

The formula for the gain of a non-inverting amplifier is given by:

extgain=1+RfRin ext{gain} = 1 + \frac{R_f}{R_{in}}

Setting this equal to 40:

40=1+RfRin40 = 1 + \frac{R_f}{R_{in}}

This simplifies to:

39=RfRin39 = \frac{R_f}{R_{in}}

Choosing Rin=1kΩR_{in} = 1 kΩ gives:

Rf=39kΩR_f = 39 kΩ.

Step 4

Discuss whether this operational amplifier is suitable for amplifying the sensor’s output voltage.

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Answer

Given the ultrasound frequency of 50 kHz, we can calculate the required gain-bandwidth product:

gain×bandwidth=40×50 kHz=2.0 MHz\text{gain} \times \text{bandwidth} = 40 \times 50 \text{ kHz} = 2.0 \text{ MHz}

However, since the amplifier specification shows a maximum bandwidth of 1.0 MHz, this amplifier would not be suitable for this application, as it would not provide the necessary amplification without distorting the signal.

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