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Figure 6 shows an astable circuit based on a NOT logic gate - AQA - A-Level Physics - Question 4 - 2018 - Paper 8

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Figure 6 shows an astable circuit based on a NOT logic gate. The symbol in the centre of the logic gate means that the output voltage V₀ changes at two different inp... show full transcript

Worked Solution & Example Answer:Figure 6 shows an astable circuit based on a NOT logic gate - AQA - A-Level Physics - Question 4 - 2018 - Paper 8

Step 1

Calculate PRF

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Answer

To calculate the pulse repetition frequency (PRF) using the given formula:

PRF=11.4×5.1×103×10×10−9≈14 kHzPRF = \frac{1}{1.4 \times 5.1 \times 10^3 \times 10 \times 10^{-9}} \approx 14 \text{ kHz}

Step 2

Draw output voltage V₀

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Answer

On Figure 7, draw a square wave that alternates between 0 V and 5 V, corresponding to the UST and LST values.

Step 3

Calculate the new resistor for frequency modification

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Answer

To achieve a frequency 4 times the original, the new frequency (f') will be:

f′=4f=4×14 kHz=56 kHzf' = 4f = 4 \times 14 \text{ kHz} = 56 \text{ kHz}

Using the frequency formula, solve for R:

R=11.4×f′×C =11.4×56×103×10×10−9 ≈1.71 kΩR = \frac{1}{1.4 \times f' \times C}\ = \frac{1}{1.4 \times 56 \times 10^3 \times 10 \times 10^{-9}} \ \approx 1.71 \text{ kΩ}

This resistor should be added in parallel with the existing resistor to modify the PRF.

Step 4

Calculate R₁ and R₂ for 5 kHz with 75% duty cycle

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Answer

For a 5 kHz signal, the period (T) is:

T=15 kHz=0.2 msT = \frac{1}{5 \text{ kHz}} = 0.2 \text{ ms}

With a 75% duty cycle, the charging time (t_c) is:

tc=0.75×T=0.15 mst_c = 0.75 \times T = 0.15 \text{ ms}

Plugging into the charging formula:

0.15 ms=0.7×(R1+R2)×10×10−90.15 \text{ ms} = 0.7 \times (R_1 + R_2) \times 10 \times 10^{-9}

Solving for R₁ + R₂ gives:

R1+R2=21.43 kΩR_1 + R_2 = 21.43 \text{ kΩ}

For the discharging time (t_d):

td=0.25×T=0.05 mst_d = 0.25 \times T = 0.05 \text{ ms}

Using the discharging time formula:

0.05 ms=0.7×R2×10×10−90.05 \text{ ms} = 0.7 \times R_2 \times 10 \times 10^{-9}

Solving for R₂ gives:

R2=7.14 kΩR_2 = 7.14 \text{ kΩ}

Assuming R₁ and R₂ are 14.29 kΩ and 7.14 kΩ respectively are valid values for the circuit.

Step 5

Draw the wave pattern

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Answer

On Figure 9, draw a square wave that ranges from 0 V to 5 V, with a period of 0.2 ms, displaying a 75% duty cycle.

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