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Consider an inverting amplifier with a nominal gain of 1000 constructed from an op amp with an input offset voltage of 3 mV and with output saturation levels of ±10 V. What is (approximately) the peak sine-wave input signal that can be applied without output clipping?
- a)7 mV
- b)10 mV
- c)13 mV
- d)9mV
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Consider an inverting amplifier with a nominal gain of 1000 constructe...
The maximum that can be sent without clipping is 10V – 1000 X 3mV or 7V.
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Consider an inverting amplifier with a nominal gain of 1000 constructe...
±12V. The input signal to the amplifier is a sinusoidal waveform with a peak-to-peak amplitude of 10 mV. Determine the maximum and minimum output voltage swings of the amplifier.
To determine the maximum and minimum output voltage swings, we need to consider the input offset voltage and the output saturation levels.
The maximum output voltage swing can be determined by considering the maximum input voltage that will cause the output to saturate at +12V. Given that the nominal gain of the amplifier is 1000, the maximum input voltage can be calculated as follows:
Maximum input voltage = Maximum output voltage / Gain
= 12V / 1000
= 0.012V or 12 mV
However, since the input signal has a peak-to-peak amplitude of 10 mV, the maximum input voltage will be limited to half of the peak-to-peak amplitude, which is 5 mV.
Therefore, the maximum output voltage swing will be 12V for an input voltage of 5 mV.
Similarly, the minimum output voltage swing can be determined by considering the minimum input voltage that will cause the output to saturate at -12V. Using the same calculations as above, the minimum output voltage swing will also be 12V for an input voltage of -5 mV.
In summary, the maximum and minimum output voltage swings of the amplifier are ±12V for an input voltage swing of ±5 mV.
To determine the maximum and minimum output voltage swings, we need to consider the input offset voltage and the output saturation levels.
The maximum output voltage swing can be determined by considering the maximum input voltage that will cause the output to saturate at +12V. Given that the nominal gain of the amplifier is 1000, the maximum input voltage can be calculated as follows:
Maximum input voltage = Maximum output voltage / Gain
= 12V / 1000
= 0.012V or 12 mV
However, since the input signal has a peak-to-peak amplitude of 10 mV, the maximum input voltage will be limited to half of the peak-to-peak amplitude, which is 5 mV.
Therefore, the maximum output voltage swing will be 12V for an input voltage of 5 mV.
Similarly, the minimum output voltage swing can be determined by considering the minimum input voltage that will cause the output to saturate at -12V. Using the same calculations as above, the minimum output voltage swing will also be 12V for an input voltage of -5 mV.
In summary, the maximum and minimum output voltage swings of the amplifier are ±12V for an input voltage swing of ±5 mV.
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Consider an inverting amplifier with a nominal gain of 1000 constructed from an op amp with an input offset voltage of 3 mV and with output saturation levels of ±10 V. What is (approximately) the peak sine-wave input signal that can be applied without output clipping?a)7 mVb)10 mVc)13 mVd)9mVCorrect answer is option 'A'. Can you explain this answer?
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