Figure 1 shows the basic layout for a Johnson decade counter. The main outputs are shown as Q0 to Q9. Figure 2 shows part of the timing diagram for a Johnson decade counter. This timing diagram shows the output logic states against time. The counter is reset to make Q0 = 1 and then the first two pulses are applied. Complete Figure 2 to show the logic states of Q0, Q1 and Q2. [2 marks]

A-Level AQA Physics Resistance & Resistivity: Figure 1 shows the basic layout

Figure 1 shows the basic layout for a Johnson decade counter - AQA - A-Level Physics - Question 1 - 2019 - Paper 8

Question 1

Figure 1 shows the basic layout for a Johnson decade counter. The main outputs are shown as Q0 to Q9. Figure 2 shows part of the timing diagram for a Johnson decade... show full transcript

Worked Solution & Example Answer:Figure 1 shows the basic layout for a Johnson decade counter - AQA - A-Level Physics - Question 1 - 2019 - Paper 8

Step 1

Complete Figure 2 to show the logic states of Q0, Q1 and Q2

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Answer

To complete Figure 2, we need to determine the state of each output in response to the clock pulses shown.

Determine Q0 Logic State:
- The output Q0 toggles on every clock pulse. Thus:
  - For pulse 1: Q0 = 1
  - For pulse 2: Q0 = 0
Determine Q1 Logic State:
- Q1 toggles after Q0. So:
  - For pulse 1: Q1 = 0
  - For pulse 2: Q1 = 1
Determine Q2 Logic State:
- Q2 toggles after Q1. Hence:
  - For pulse 1: Q2 = 0
  - For pulse 2: Q2 = 0

This gives the following logic states:

Q0: 1, 0
Q1: 0, 1
Q2: 0, 0

Figure 1 shows the basic layout for a Johnson decade counter - AQA - A-Level Physics - Question 1 - 2019 - Paper 8

Worked Solution & Example Answer:Figure 1 shows the basic layout for a Johnson decade counter - AQA - A-Level Physics - Question 1 - 2019 - Paper 8

Complete Figure 2 to show the logic states of Q0, Q1 and Q2

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