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The answer lies in Carrier Mobility of Silicon. A CMOS stage has a P channel device from Vdd and an N channel device to Vss. Note the much higher mobility of electrons vs. holes. The rise time at the output depends primarily on how fast the P channel device can turn on, and the fall time is determined primarily by how fast the N channel device can turn on. ...

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For some reason it is easier to get $R_{on}$ small for N-MOS than for P-MOS transistors. The N-MOS transistor is in charge of setting the ouput low, i.e. discharging the line capacitance and gate capacitances of the connected inputs which is faster when $R_{on}$ is smaller.

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Only if an input could be left floating. For example, if you select between several outputs to a single input, all analog inputs are tied to something (or you believe they can never be selected) and the mux is always enabled then you don't need a resistor. Chances are that if you are using it as a digital de-multiplexer this condition won't be true and you ...

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There is no way to "deactivate" an output of the CD4017B; outputs are always either very close to VDD, or they are very close to VSS. So yes, connecting a high output to a low output will create a near-short-circuit between VDD and VSS. Use an OR gate (or equivalent thereof) if you need to activate an input if either or both of two outputs are high.

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CMOS inverters can also be used as linear amplifiers. In this case, the device is biased at the input with a dc voltage of app. at 50% of the supply voltage. Then, the upper transistor acts as a high-resistive load resistance. This can be easily verified using the output characteristics of both units in a common diagram, see here: ...

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Take a closer look at the PMOSFET. Notice something different from the NMOSFET? ... That's right, the source is tied high. Since VGS of the PMOSFET needs to go more negative in order to turn it on, it will be on when the NMOSFET is off and vice versa.

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No, in CMOS both, the N-channel and the P-channel MOSFET are enhancement types. The N-channel is enhanced by positive voltage at the gate with respect to its source. The P-channel, however, is enhanced by negative voltage at the gate with respect to its source. Thus a H at the input turns on the N-channel MOSFET (positive voltage between its gate and ...

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This is the problem with developing a real-world circuit from a "simplified" block diagram. There are no component values, the circuit connections are representative only and may not show much more than the direction a sign is going. The assumption is you will select the specific devices to use and determine the connections to be made. In general, a MOS ...

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In addition to Nicolas' answer, you will normally find these on the input stage as well. What type of diode is used can be assessed from the 'absolute maximum ratings' table in the datasheet. If Vin is: V+ + 0.3V then the devices are schottky types V+ + 0.5V then the devices are PN junction diodes These diodes can give rise to unusual effects: I have had ...

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1-2) They are here to protect the device from electrostatic discharge (ESD), by channeling the charge to power rails. ESD can be understood as a very high voltage, very high impedance source. By connecting that source to VCC/Gnd with a diode (very low impedance) the ESD source is depleted from its charge without harming the device. 3) AFAIK, just normal ...

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