Difference between open drain and push-pull modes

Question

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

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?

Answer 1

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.