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Andy aka
Andy aka
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Does anyone have an explanation?

Yes, this is what will happen when a squarewave input is processed by an RC differentiator: -

enter image description here

The output vertical edges are constrained to an amplitude of 3.3 volts but, once an edge has occurred, \$V_{OUT}\$ fairly rapidly discharges to 0 volts due to the charging of the capacitor via the resistor. The next edge (of the opposite polarity) drives \$V_{OUT}\$ in the opposite direction.

The above picture is 10 kHz but, as frequency rises the p-p will tend to eventually become the same as the input waveform. At 100 kHz \$V_{OUT}\$ is about 5 volts p-p: -

enter image description here

Does anyone have an explanation?

Yes, this is what will happen when a squarewave input is processed by an RC differentiator: -

enter image description here

The output vertical edges are constrained to an amplitude of 3.3 volts but, once an edge has occurred, \$V_{OUT}\$ fairly rapidly discharges to 0 volts due to the charging of the capacitor via the resistor. The next edge (of the opposite polarity) drives \$V_{OUT}\$ in the opposite direction.

The above picture is 10 kHz but, as frequency rises the p-p will tend to eventually become the same as the input waveform.

Does anyone have an explanation?

Yes, this is what will happen when a squarewave input is processed by an RC differentiator: -

enter image description here

The output vertical edges are constrained to an amplitude of 3.3 volts but, once an edge has occurred, \$V_{OUT}\$ fairly rapidly discharges to 0 volts due to the charging of the capacitor via the resistor. The next edge (of the opposite polarity) drives \$V_{OUT}\$ in the opposite direction.

The above picture is 10 kHz but, as frequency rises the p-p will tend to eventually become the same as the input waveform. At 100 kHz \$V_{OUT}\$ is about 5 volts p-p: -

enter image description here

added 132 characters in body
Source Link
Andy aka
Andy aka
  • 505.2k
  • 35
  • 402
  • 892

Does anyone have an explanation?

Yes, this is what will happen when a squarewave input is processed by an RC differentiator: -

enter image description here

The output vertical edges remain limitedare constrained to an amplitude of 3.3 volts but, because once an edge has happened the outputoccurred, \$V_{OUT}\$ fairly rapidly discharges to 0 volts, due to the charging of the capacitor via the resistor. The next edge (oppositeof the opposite polarity) drives an edge\$V_{OUT}\$ in the opposite direction.

The above picture is 10 kHz but, as frequency rises the p-p will tend to eventually become the same as the input waveform.

Yes, this is what will happen when a squarewave input is processed by an RC differentiator: -

enter image description here

The edges remain limited to 3.3 volts but, because once an edge has happened the output fairly rapidly discharges to 0 volts, the next edge (opposite polarity) drives an edge in the opposite direction.

The above picture is 10 kHz but, as frequency rises the p-p will tend to eventually become the same as the input waveform.

Does anyone have an explanation?

Yes, this is what will happen when a squarewave input is processed by an RC differentiator: -

enter image description here

The output vertical edges are constrained to an amplitude of 3.3 volts but, once an edge has occurred, \$V_{OUT}\$ fairly rapidly discharges to 0 volts due to the charging of the capacitor via the resistor. The next edge (of the opposite polarity) drives \$V_{OUT}\$ in the opposite direction.

The above picture is 10 kHz but, as frequency rises the p-p will tend to eventually become the same as the input waveform.

Source Link
Andy aka
Andy aka
  • 505.2k
  • 35
  • 402
  • 892

Yes, this is what will happen when a squarewave input is processed by an RC differentiator: -

enter image description here

The edges remain limited to 3.3 volts but, because once an edge has happened the output fairly rapidly discharges to 0 volts, the next edge (opposite polarity) drives an edge in the opposite direction.

The above picture is 10 kHz but, as frequency rises the p-p will tend to eventually become the same as the input waveform.