PLs. take a look at it. It is a simple inveretr at the output, the resistor in the circuit puzzled me. Would anyone can see what's the function of this resistor?
It is an iverter with fast rise time and slower fall time at the output. Since it is CMOS assuming it drives a CMOS load which presents a capacitive load, p-ch pulls up output fast but n-ch has extra resistor on top of its Ron so pulls down slower.
Another feature of this circuit is that, crowbar current reduced since during transition between Vdd and GND there is an extra resistor.
This will definitely help you as far as ESD. But it depends on the value of the resistor. For MM ESD it probably helps the NMOS if the resistor is of small value. For HBM ESD, I think this has to be like 1kOhm for it to significantly help ESD.
It is an iverter with fast rise time and slower fall time at the output. Since it is CMOS assuming it drives a CMOS load which presents a capacitive load, p-ch pulls up output fast but n-ch has extra resistor on top of its Ron so pulls down slower.
True, and if you take outputs from both sides of the resistor, you have a simple non-overlapping driver to the next stage, which limit the shot through power loss.
and if it is used as output buffer, then the resistor can improve the output impedance linearity with Vout, reduce reflection, and increase signal quality.
Short circuiting the output to VCC will kill the NMOS if this resistor is missing.
This is a push-pull stage with "open drain feature". Designed for inequal sourcing and sinking currents.