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The figures show diagrams of line signals: the first in the open-drain + pull up mode, the second push-pull + pull-down.

enter image description here enter image description here

There are several questions:

  1. Why does a push pull output have a faster slope than an open drain output?
  2. Why would a sawtooth (open drain) waveform be converted to a square wave when delays are added?
  3. Well, in general, what are the differences between these modes?
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  • \$\begingroup\$ What delays you mean? \$\endgroup\$
    – Justme
    May 18 at 15:10
  • \$\begingroup\$ @Justme a delay of a certain number of clock cycles that occurs between switches. \$\endgroup\$
    – 1i10
    May 18 at 15:22
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    \$\begingroup\$ Delays don't convert it to a square wave. The rising edge will be just as slow as before. It may just appear to you like that if you zoom out with a scope, because the slow rise is a smaller percentage of the whole period. \$\endgroup\$
    – Justme
    May 18 at 16:09

1 Answer 1

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Open-drain depends on a pullup resistor to provide the high state, while the low state is actively created by a transistor. So the pullup network has relatively high impedance to VDD and very low impedance to GND because when the transistor fully turns on, it is effectively a short to GND.

The pullup resistor (e.g. 10k) will be much slower to charge up the output node since the pullup resistor limits the rate at which the output node can be charged from low to high and that's why you see an RC type of reponse on the rising edge while the high to low transition happens much faster due to having an output transitor that pulls the line low, with pretty small impedance.

Push-pull configuration doesn't rely on a pullup resistor to create the high state, it actually uses another transistor in addition to the one used for creating the low state. So both paths to VDD and GND are low impedance when they need to be. This allows to much faster transition from low to high.

enter image description here

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