Electrical Engineering (EE) Exam > Electrical Engineering (EE) Questions > (Q.8-Q.10) Consider a Miller integrator with ...
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(Q.8-Q.10) Consider a Miller integrator with a time constant of 1ms and an input resistance of 10 kΩ. Let the op amp have VOS (offset voltage) = 2 mV and output saturation voltages of ±12 V.
Q. Assuming that when the power supply is turned on the capacitor voltage is zero, how long does it take for the amplifier to saturate?
- a)3s
- b)6s
- c)9s
- d)none
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
(Q.8-Q.10) Consider a Miller integrator with a time constant of 1ms an...
Use vO = VOS
(VOS/CR)t.
(VOS/CR)t.
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(Q.8-Q.10) Consider a Miller integrator with a time constant of 1ms an...
8. If the input voltage is a square wave with a frequency of 1 kHz and an amplitude of 1 V, what is the output voltage?
Solution:
The Miller integrator circuit integrates the input voltage over time, which means that the output voltage will be the integral of the input voltage waveform. For a square wave input, the output voltage will be a triangular waveform.
The time constant of the circuit is 1 ms, which means that the output voltage will take 5 time constants or 5 ms to reach 63.2% of its final value.
The peak-to-peak value of the triangular waveform can be calculated using the formula Vpp = 2 * Vm / f, where Vm is the amplitude of the square wave and f is the frequency.
Vpp = 2 * 1 V / 1 kHz = 2 mV
The peak value of the triangular waveform can be calculated as Vp = Vpp / 2 = 1 mV.
The time period of the square wave is 1/f = 1 ms, which is equal to the time constant of the circuit. Therefore, the output voltage will reach its final value after 5 cycles of the square wave.
The area of each cycle of the square wave is (1/2) * Vm * T, where T is the time period. Therefore, the area of 5 cycles is 5 * (1/2) * 1 V * 1 ms = 2.5 Vms.
Since the time constant of the circuit is 1 ms, the output voltage will be proportional to the area under the input waveform. Therefore, the output voltage will be:
Vo = -(1/RC) * area = -(1/10kΩ * 1ms) * 2.5 Vms
Vo = -0.25 V
The negative sign indicates that the output waveform is inverted with respect to the input waveform. Therefore, the output voltage of the Miller integrator with a square wave input of 1 kHz and 1 V amplitude is -0.25 V.
9. If the input voltage is a sine wave with a frequency of 10 kHz and an amplitude of 100 mV, what is the output voltage?
Solution:
For a sine wave input, the output voltage of the Miller integrator will be a cosine waveform with a phase shift of -90 degrees.
The time constant of the circuit is 1 ms, which means that the output voltage will take 5 time constants or 5 ms to reach 63.2% of its final value.
The peak-to-peak value of the output waveform can be calculated using the formula Vpp = 2 * Vm / f, where Vm is the amplitude of the input sine wave and f is the frequency.
Vpp = 2 * 100 mV / 10 kHz = 20 mV
The peak value of the output waveform can be calculated as Vp = Vpp / 2 = 10 mV.
The phase shift of the output waveform can be calculated using the formula Φ = -arctan(1/2πfRC), where R is the input resistance and C is the capacitance.
Φ = -arctan(1/2π * 10 kHz * 10 kΩ * 1 nF) = -63.4°
Therefore, the output voltage
Solution:
The Miller integrator circuit integrates the input voltage over time, which means that the output voltage will be the integral of the input voltage waveform. For a square wave input, the output voltage will be a triangular waveform.
The time constant of the circuit is 1 ms, which means that the output voltage will take 5 time constants or 5 ms to reach 63.2% of its final value.
The peak-to-peak value of the triangular waveform can be calculated using the formula Vpp = 2 * Vm / f, where Vm is the amplitude of the square wave and f is the frequency.
Vpp = 2 * 1 V / 1 kHz = 2 mV
The peak value of the triangular waveform can be calculated as Vp = Vpp / 2 = 1 mV.
The time period of the square wave is 1/f = 1 ms, which is equal to the time constant of the circuit. Therefore, the output voltage will reach its final value after 5 cycles of the square wave.
The area of each cycle of the square wave is (1/2) * Vm * T, where T is the time period. Therefore, the area of 5 cycles is 5 * (1/2) * 1 V * 1 ms = 2.5 Vms.
Since the time constant of the circuit is 1 ms, the output voltage will be proportional to the area under the input waveform. Therefore, the output voltage will be:
Vo = -(1/RC) * area = -(1/10kΩ * 1ms) * 2.5 Vms
Vo = -0.25 V
The negative sign indicates that the output waveform is inverted with respect to the input waveform. Therefore, the output voltage of the Miller integrator with a square wave input of 1 kHz and 1 V amplitude is -0.25 V.
9. If the input voltage is a sine wave with a frequency of 10 kHz and an amplitude of 100 mV, what is the output voltage?
Solution:
For a sine wave input, the output voltage of the Miller integrator will be a cosine waveform with a phase shift of -90 degrees.
The time constant of the circuit is 1 ms, which means that the output voltage will take 5 time constants or 5 ms to reach 63.2% of its final value.
The peak-to-peak value of the output waveform can be calculated using the formula Vpp = 2 * Vm / f, where Vm is the amplitude of the input sine wave and f is the frequency.
Vpp = 2 * 100 mV / 10 kHz = 20 mV
The peak value of the output waveform can be calculated as Vp = Vpp / 2 = 10 mV.
The phase shift of the output waveform can be calculated using the formula Φ = -arctan(1/2πfRC), where R is the input resistance and C is the capacitance.
Φ = -arctan(1/2π * 10 kHz * 10 kΩ * 1 nF) = -63.4°
Therefore, the output voltage
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(Q.8-Q.10) Consider a Miller integrator with a time constant of 1ms and an input resistance of 10 kΩ. Let the op amp have VOS(offset voltage) = 2 mV and output saturation voltages of ±12 V.Q. Assuming that when the power supply is turned on the capacitor voltage is zero, how long does it take for the amplifier to saturate?a)3sb)6sc)9sd)noneCorrect answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about (Q.8-Q.10) Consider a Miller integrator with a time constant of 1ms and an input resistance of 10 kΩ. Let the op amp have VOS(offset voltage) = 2 mV and output saturation voltages of ±12 V.Q. Assuming that when the power supply is turned on the capacitor voltage is zero, how long does it take for the amplifier to saturate?a)3sb)6sc)9sd)noneCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for (Q.8-Q.10) Consider a Miller integrator with a time constant of 1ms and an input resistance of 10 kΩ. Let the op amp have VOS(offset voltage) = 2 mV and output saturation voltages of ±12 V.Q. Assuming that when the power supply is turned on the capacitor voltage is zero, how long does it take for the amplifier to saturate?a)3sb)6sc)9sd)noneCorrect answer is option 'B'. Can you explain this answer?.
(Q.8-Q.10) Consider a Miller integrator with a time constant of 1ms and an input resistance of 10 kΩ. Let the op amp have VOS(offset voltage) = 2 mV and output saturation voltages of ±12 V.Q. Assuming that when the power supply is turned on the capacitor voltage is zero, how long does it take for the amplifier to saturate?a)3sb)6sc)9sd)noneCorrect answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about (Q.8-Q.10) Consider a Miller integrator with a time constant of 1ms and an input resistance of 10 kΩ. Let the op amp have VOS(offset voltage) = 2 mV and output saturation voltages of ±12 V.Q. Assuming that when the power supply is turned on the capacitor voltage is zero, how long does it take for the amplifier to saturate?a)3sb)6sc)9sd)noneCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for (Q.8-Q.10) Consider a Miller integrator with a time constant of 1ms and an input resistance of 10 kΩ. Let the op amp have VOS(offset voltage) = 2 mV and output saturation voltages of ±12 V.Q. Assuming that when the power supply is turned on the capacitor voltage is zero, how long does it take for the amplifier to saturate?a)3sb)6sc)9sd)noneCorrect answer is option 'B'. Can you explain this answer?.
Solutions for (Q.8-Q.10) Consider a Miller integrator with a time constant of 1ms and an input resistance of 10 kΩ. Let the op amp have VOS(offset voltage) = 2 mV and output saturation voltages of ±12 V.Q. Assuming that when the power supply is turned on the capacitor voltage is zero, how long does it take for the amplifier to saturate?a)3sb)6sc)9sd)noneCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
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