Sketching output waveforms while considering slew rate

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Discussion Overview

The discussion revolves around sketching output waveforms while considering the slew rate in relation to a given input waveform. Participants explore the implications of the slew rate and gain on the output waveform, focusing on the technical aspects of waveform representation and measurement.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the time period of the waveform as 1 millisecond and identifies the pulse width and amplitude as 1 V based on a 50% duty cycle.
  • Another participant points out that the initial analysis does not account for the gain, suggesting that this is a critical oversight.
  • A participant acknowledges the oversight regarding gain and seeks confirmation on the rest of their analysis.
  • Further, a participant notes that the horizontal axis of the proposed waveform is not to scale, emphasizing that the rise and fall times are significantly shorter than the period of the square wave input.
  • There is a question posed about how to accurately represent the waveform on a plot and how it would appear on a digital oscilloscope.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there are competing views regarding the importance of gain and the representation of the waveform. The discussion remains unresolved with respect to the correct depiction of the output waveform.

Contextual Notes

Limitations include the lack of consideration for gain in the initial analysis and the potential misrepresentation of time scales in the proposed waveform sketch.

JC2000
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Homework Statement
Sketch the output voltage when input voltage applied to non inverting amplifier with gain 10 for a square wave input signal with peak to peak voltage ##2 V##, ##50 %## duty cycle and frequency ##1## kHz. Supply Voltage is +12 V and -12V, the op amp has slew rate 0.5 V/ microsecond and the op amp is nulled initially.
Relevant Equations
Time Period = 1 / Frequency

Duty Cycle = (Pulse Width *100)/ Time Period
1. Using the frequency equation I know the time period is 1 millisecond.

2. The duty cycle (50%) equation tells me that the pulse width and the amplitude is 1 V.

3. Thus the input waveform looks like this : (?)

IMG_20200601_170230027_2.jpg


4. As the slew rate is 0.5 volts per microsecond, the output voltage would take 4 microseconds to go from -1 V to + 1V (?)

5. Thus the output waveform for one pulse should look like :

IMG_20200601_170746425_2.jpg


(I converted milliseconds to microseconds)

Is this correct?
 
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JC2000 said:
5. Thus the output waveform for one pulse should look like :
You have not considered the gain.
 
cnh1995 said:
You have not considered the gain.
Oof! Careless of me. Other than that bit is the rest fine?
 
JC2000 said:
Oof! Careless of me. Other than that bit is the rest fine?
The horizontal axis is not to scale, even ignoring the voltage gain issue. The rise/fall times are on the order of a few us, while the period of the square wave input is on the order of a ms. Do you have any ideas for how to represent that on a plot? How would you observe this on a digital oscilloscope, for example?
 
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