Jtdcftul

Looking at the datasheet for the MC74VHC1G08, under the "features" section, it states Chip Complexity: FETs = 62.

Why does this IC need 62 transistors, while an AND gate can be made with only 6 transistors? What are the other 56 transistors being used for? My guess would be some sort of protection circuitry, but I am not sure.

There can be several reasons why more than the minimum 6 MOSFETs (4 for an NAND + 2 for an inverter) are used in this IC:

• As stated in the datasheet:

The internal circuit is composed of multiple stages, including a
buffer output which provides high noise immunity and stable output.

• The output will be made using fairly large (not minimum size) transistors. There are always "folded" meaning multiple transistors are combined into one large one where drain and source diffusion areas are shared between two transistors. This behaves as one large transistor but could be counted as many if you want a higher transistor count.




• The ESD protection at inputs and outputs of IC fabricated in modern CMOS processes often uses "grounded-gate MOSFETs" instead of the more traditional diodes.




• An "ESD clamp" circuit is needed between the supply pins, such a circuit consists of a couple of transistors.




• Digital circuits (like this AND gate) often need on-chip supply decoupling. These are called "decap cells". These are capacitors between the supply rails. These capacitors are mostly made by using the Gate-Drain/Source capacitance of Transistors.




• In CMOS processes the MOSFETs are the most "basic" components, they are also the most controlled component and most flexible ones so IC designers prefer to use a MOSFET whenever possible.

All-in-all it is "quite easy" to need 62 transistors to make a seemingly simple function like an AND gate. That's also because this IC is "a bit more" than just a simple AND gate. The AND gates in more complex circuits like CPUs, microcontrollers etc. will often only use 6 transistors. But these aren't "stand alone" AND gates like this IC.

The harder the gate of a MOSFET is driven to turn it on, the longer it will take for the MOSFET to subsequently turn off. Performance may be improved by adding circuitry to limit excess gate voltage, though doing this without increasing quiescent power dissipation is tricky.

I don't know what exact techniques are used in CMOS to prevent oversaturation, but low-power Schottky devices based on bipolar junction transistors may provide a useful analogue. Consider the two simple inverters shown below:

^{simulate this circuit – Schematic created using CircuitLab}

The inverter on the left is simpler than the one on the right, but if one runs the simulation, one will see that the adding the diode allows the circuit on the right to switch off much more quickly than the one on the left.

In the BJT-based inverters below, the Schottky diode will slightly increase the power dissipation in R3, but such increase will be tiny compared to overall power consumption. In a CMOS device, simply clamping the gate voltage would increase power dissipation, making it necessary to use other, more sophisticated, approaches.