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For operation with 10-V output pulses with the requirement that the sum of the rise and fall times represent only 20% of the pulse width (at half amplitude), what is the slew-rate requirement for an op amp to handle pulses 2 µs wide? (Note: The rise and fall times of a pulse signal are usually measured between the 10%- and 90%-height points.)
- a)10 V/µs
- b)20 V/µs
- c)40 V/µs
- d)none
Correct answer is option 'C'. Can you explain this answer?
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For operation with 10-V output pulses with the requirement that the su...
To determine the slew rate requirement for the op amp, we need to calculate the rise and fall times of the pulse.
Given:
- Pulse amplitude: 10 V
- The sum of rise and fall times represents only 20% of the pulse width (at half amplitude)
Let's assume the pulse width is T seconds.
The rise time (tr) and fall time (tf) can be calculated as follows:
tr + tf = 0.2 * T
At half amplitude, the voltage is 10 V / 2 = 5 V.
The slew rate (SR) is the rate at which the output voltage can change and is given by the formula:
SR = (V_final - V_initial) / (t_final - t_initial)
For an ideal op amp, the slew rate is infinite. However, practical op amps have a limited slew rate that determines their ability to handle fast voltage changes.
To calculate the slew rate requirement, we can use the rise time (tr) and fall time (tf) at half amplitude.
For the rise time:
V_initial = 0 V
V_final = 5 V
t_initial = 0 seconds
t_final = tr seconds
Therefore, the slew rate for the rise time is:
SR_rise = (5 V - 0 V) / tr
Similarly, for the fall time:
V_initial = 5 V
V_final = 0 V
t_initial = tr seconds
t_final = T seconds
Therefore, the slew rate for the fall time is:
SR_fall = (0 V - 5 V) / (T - tr)
To handle the pulses with a 20% rise and fall time requirement, the op amp's slew rate should be greater than or equal to the maximum of SR_rise and SR_fall.
SR_required = max(SR_rise, SR_fall)
Keep in mind that the units of the slew rate will be V/s.
Note: The value of T (pulse width) is not provided in the question, so it is not possible to calculate the exact slew rate requirement without that information.
Given:
- Pulse amplitude: 10 V
- The sum of rise and fall times represents only 20% of the pulse width (at half amplitude)
Let's assume the pulse width is T seconds.
The rise time (tr) and fall time (tf) can be calculated as follows:
tr + tf = 0.2 * T
At half amplitude, the voltage is 10 V / 2 = 5 V.
The slew rate (SR) is the rate at which the output voltage can change and is given by the formula:
SR = (V_final - V_initial) / (t_final - t_initial)
For an ideal op amp, the slew rate is infinite. However, practical op amps have a limited slew rate that determines their ability to handle fast voltage changes.
To calculate the slew rate requirement, we can use the rise time (tr) and fall time (tf) at half amplitude.
For the rise time:
V_initial = 0 V
V_final = 5 V
t_initial = 0 seconds
t_final = tr seconds
Therefore, the slew rate for the rise time is:
SR_rise = (5 V - 0 V) / tr
Similarly, for the fall time:
V_initial = 5 V
V_final = 0 V
t_initial = tr seconds
t_final = T seconds
Therefore, the slew rate for the fall time is:
SR_fall = (0 V - 5 V) / (T - tr)
To handle the pulses with a 20% rise and fall time requirement, the op amp's slew rate should be greater than or equal to the maximum of SR_rise and SR_fall.
SR_required = max(SR_rise, SR_fall)
Keep in mind that the units of the slew rate will be V/s.
Note: The value of T (pulse width) is not provided in the question, so it is not possible to calculate the exact slew rate requirement without that information.
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For operation with 10-V output pulses with the requirement that the sum of the rise and fall times represent only 20% of the pulse width (at half amplitude), what is the slew-rate requirement for an op amp to handle pulses 2 µs wide? (Note: The rise and fall times of a pulse signal are usually measured between the 10%- and 90%-height points.)a)10 V/µsb)20 V/µsc)40 V/µsd)noneCorrect answer is option 'C'. Can you explain this answer?
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For operation with 10-V output pulses with the requirement that the sum of the rise and fall times represent only 20% of the pulse width (at half amplitude), what is the slew-rate requirement for an op amp to handle pulses 2 µs wide? (Note: The rise and fall times of a pulse signal are usually measured between the 10%- and 90%-height points.)a)10 V/µsb)20 V/µsc)40 V/µsd)noneCorrect answer is option 'C'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2025 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about For operation with 10-V output pulses with the requirement that the sum of the rise and fall times represent only 20% of the pulse width (at half amplitude), what is the slew-rate requirement for an op amp to handle pulses 2 µs wide? (Note: The rise and fall times of a pulse signal are usually measured between the 10%- and 90%-height points.)a)10 V/µsb)20 V/µsc)40 V/µsd)noneCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For operation with 10-V output pulses with the requirement that the sum of the rise and fall times represent only 20% of the pulse width (at half amplitude), what is the slew-rate requirement for an op amp to handle pulses 2 µs wide? (Note: The rise and fall times of a pulse signal are usually measured between the 10%- and 90%-height points.)a)10 V/µsb)20 V/µsc)40 V/µsd)noneCorrect answer is option 'C'. Can you explain this answer?.
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