Do disconnected input pins act as “antennas” capacitors that get gradually charged, building up higher and higher voltage?

Question

Disconnected input pin acts as an antenna affected by surrounding electromagnetic noise, and can "float" between on and off. Does the input pin also act as a capacitor that gets gradually charged by this "noise," increasing voltage?

Answer 1

Am assuming a CMOS or HCMOS logic gate below...

Many spec sheets for complementary-symmetry logic gates give input capacitance and also leakage current. Both these are significant in your scenario.

At 25 degrees C, leakage current could be in either direction, resulting in a logic low or a logic high. For HCMOS, a typical spec might be +/- 100 nA. Higher temperature invariably results in higher leakage.
Leakage currents for microcontrollers may seem high at \$\pm 1 uA\$, but are measured at maximum temperature (perhaps 85C). I've measured room-temperature leakage well below 1 nA on a microcontroller with \$ \pm 1 uA \$ spec. That suggests input resistance is very high indeed, making a floating input very sensitive to any lurking electric field.

Typical HCMOS input capacitance is 3 pf. This would be with no trace attached.
A charged object passing close to an open pin can easily switch logic states. For example hand-waving near a chip may easily do this because of triboelectric charges on your clothing.

Such static charges meet the definition of noise. Nearby traces might inject noise too, especially if their logic state changes. In this case, capacitance of the floating input pin is significant.
Is this dangerous to the gates inside? Likely not, since no-contact electric fields cause only tiny currents to flow - easily clamped by built-in protection networks.
An unwanted result of floating input occurs when the input gate floats near the threshold voltage (roughly half-Vdd). Shoot-through current then flows from Vdd, through partially-ON MOSFETs to ground. That's wasted power, causing internal heating.

Answer 2

What follows is regarding integrated circuit inputs:

Like a capacitor? No, not exactly. The floating pin does have some parasitic capacitance (between the PCB, components, even the air), but this is an electrical connection to an input; thus any accumulated charge quickly goes to the input and doesn't have time to build excessively. Every device/pin/input/output also has some parasitic resistance, intended or not, which helps to mitigate reception of such noise somewhat. Generally parasitic resistances (leakage) is minimized as much as possible as this ultimately just wastes power.

Sometimes a high-value resistor is intentionally placed between the input and \$V_{DD}\$ or \$V_{SS}\$, a so-called "pull-up" or "pull-down" resistor. This input, if left floating, will then always revert to high or low, depending on which type is used. Sometimes these are added externally by the electronics engineer, or they can be pre-fabricated internally to an integrated circuit (integrated pull-ups, etc.)

This trace/wire/component/input, still "picks up" a small portion of the RF energy, proportional to the square of the distance. In other words, every three units of distance = 9x less energy induced. So the effect is generally rather small, but can still influence how sensitive circuitry works, especially with very high-value (or no) pull resistors.

So unless really close to an intentional radiator (i.e. transmitting equipment) the induced voltage levels will be low and alternate back-and-forth. The net effect is zero DC voltage change, but there will be some small AC voltage however. This is because all RF is periodic and bipolar - meaning that an RF event (wave, pulse, "packet") must start and stop, and the start is one direction of energy flow, while the stop is always the opposite direction. It is not possible to change this start/stop behavior; it is a fundamental aspect of electromagnetic waves. Thus all RF transmissions natively do zero DC work; they operate strictly in the AC domain. Even an EMP pulse, which is simply a massively-strong RF wave, does zero DC work. (It being so strong however, implies that it may grossly overload any type of input protection and render nearby devices useless.)

An input pin, as pgibbons explains, generally has protection diodes to prevent any voltage entering there from exceeding \$V_{DD}\$ and \$V_{SS}\$. If that pin attempts to go higher or lower than these (such as by bringing the device near to the transmitter), the coupled energy is essentially diverted to the power rails, making that energy go to the entire circuit (and thus saving this chip from damage.) The low level of energy is unlikely to be enough to power the circuit any, let alone power it excessively, so this is the standard way to protect input pins.

As for other devices, such as a MOSFET, this is a similar (but different) situation. A MOSFET has a "gate" input, which is essentially one leg of a capacitor. There usually are no protection diodes on the gate of a MOSFET, and it has extremely low parasitic leakage resistance by design, so it could be charged to whatever voltage nearby it can pickup. Most MOSFETs have a gate voltage limit, and when this is exceeded, the device fails. The 2N7000 series are notorious for failing when not handled with proper ESD procedures, because their gate input impedance is so high (and leakage/loss/parasitic resistance so high) that just touching it could couple more voltage (from your skin) than it is rated for. However, this MOSFET is perfectly safe sitting in the open air (with many radio stations, TV, Wi-Fi transmitters etc.) all bombarding the gate input - because zero DC work can be done to it.

Answer 3

Depending on the frequency and length of the traces, it is unlikely that they will act like antennas but they can. You are right that there is still noise/fields that will affect them. The voltage can build up but not by much for 2 reasons. #1 there is always some leakage, the higher the voltage the higher the leakage. #2 there are protection diodes (Zener/TVS) and/or clamping diodes, that even if the voltage built up to beyond VCC (which is extremely unlikely because of leakage, but if it did) the voltage would be clamped to ground / VCC.

Update, I added a picture. What is the voltage for point A? That will be hard to tell. Even if nothing connects to A, there is some reverse leakage from the diodes. Because the pin is floating, it doesn't take much to change the voltage on it. That little bit of reverse leakage can easily bring this pin to 2V. Probably starting at about 1V it will see it as a logic high (depending on processor etc.)