Sketching output waveforms while considering slew rate

In summary, using the frequency equation and the duty cycle equation, it can be determined that the input waveform has a time period of 1 millisecond and a duty cycle of 50%, resulting in a pulse width and amplitude of 1 V. The output voltage, with a slew rate of 0.5 volts per microsecond, would take 4 microseconds to go from -1 V to + 1V. However, this does not take into account the gain and the horizontal axis is not to scale, making it difficult to represent on a plot or observe on a digital oscilloscope. Further adjustments may be needed for a more accurate representation.
  • #1
JC2000
186
16
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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  • #2
JC2000 said:
5. Thus the output waveform for one pulse should look like :
You have not considered the gain.
 
  • #3
cnh1995 said:
You have not considered the gain.
Oof! Careless of me. Other than that bit is the rest fine?
 
  • #4
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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Likes JC2000 and cnh1995

1. What is slew rate and why is it important in sketching output waveforms?

Slew rate is the rate at which the output voltage of a circuit changes. It is important in sketching output waveforms because it determines how quickly the output voltage can change and can affect the overall performance and stability of the circuit.

2. How is slew rate calculated and what are the units of measurement?

Slew rate is calculated by dividing the change in output voltage by the change in time. The units of measurement for slew rate are volts per microsecond (V/μs).

3. How does slew rate affect the shape of an output waveform?

Slew rate affects the steepness of the slopes in an output waveform. If the slew rate is too low, the slopes will be gradual and the waveform will appear rounded. If the slew rate is high enough, the slopes will be steep and the waveform will appear more angular.

4. What happens if the slew rate is exceeded in a circuit?

If the slew rate is exceeded, the output voltage will not be able to change as quickly as desired and the waveform will appear distorted. This can lead to errors in the circuit's performance and potentially cause damage to the circuit.

5. How can slew rate be improved in a circuit?

Slew rate can be improved by using components with higher current capabilities, such as transistors with higher current ratings. Additionally, using feedback circuits or adding capacitance to the circuit can also help improve slew rate.

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