Consider a quantum system with two levels separated by an energy $E = \hbar \omega$.
In a sense, this is the smallest possible physical system from the point of view of information content.
When a two level system can be accurately controlled, measured, and interacted with other two level systems, it's reasonable to call it a "quantum bit" or "qubit".
To operate a qubit, we need a device that can write information to it.
In other words, we need a device that can control the qubit's quantum state.
This controller device will inevitably be classical in nature, because at some level it has to be something that we can control with our hands/eyes/etc.
Such a classical system has, by nature, some degree of noise.
So we can ask a simple question: given a classical controller with a certain maximum coherent power output and a certain amount of noise, how does the write speed compare with the rate by which our control system's noise corrupts the qubit quantum state?
Let the power output of the controller be $P$ and its noise spectral density be $S(\omega)$.
We also need to know how strongly the controller is coupled to the quantum system.
In the case of superconducting qubits, the controller always has some resistance, and so coupling the controller to the qubit causes some energy decay in the qubit even if the controller had no noise.
The dimensionless quantity
$$Q \equiv \omega \times \left( \frac{\text{energy stored in qubit}}{\text{energy loss rate to controller}} \right)$$
captures the coupling strength between the controller and the qubit.
The stuff in parentheses is actually just the energy decay lifetime $T_1$ that the controller imparts on the qubit, so
$$Q = \omega \times T_1 \, .$$
If the coupling is strong, the qubit feels the controller's resistance more, energy leaks out faster, and $Q$ goes down.
If the coupling is weak, the qubit feels the controller's resistance less, the qubit keeps it energy longer, and $Q$ goes up.
The same argument goes for noise: if the qubit-controller coupling is weak and $Q$ is high, the qubit feels the controller's noise less strongly.
But there's a tension here: weak coupling and high $Q$ means that the signal we're trying to send in from the controller hits the qubit less strongly, so our write time will go up.
Working through the control and noise theory gives two equations:
$$T_\text{write} = \frac{\pi}{2} \sqrt{\frac{Q \hbar}{P}}
\qquad \text{and} \qquad
T_\text{noise} = \frac{Q \hbar}{S(\omega)}$$
where $T_\text{write}$ is the time it takes to flip the qubit from $\lvert 0 \rangle$ to $\lvert 1 \rangle$ (i.e. the write time) and $T_\text{noise}$ is the time it takes for the controller's noise causes the qubit's state to become random.
Generally speaking, we want $T_\text{write}$ to be small (fast write speed) and $T_\text{noise}$ to be large (long time before noise kills the qubit), so we want $T_\text{noise}/T_\text{write}$ to be big.
Well, we have
$$\frac{T_\text{noise}}{T_\text{write}}
= \frac{2}{\pi} \frac{1}{S(\omega)} \sqrt{\frac{P}{Q \hbar}} \, . \tag{$\star$}
$$
Therefore, to maximize $T_\text{noise}/T_\text{write}$, we want
What about $Q$?
It looks like we want low $Q$, but that's not really true because lowering $Q$ means we're lowering the time the qubit survives before leaking its energy into the controller (remember $Q = \omega T_1$).
In real applications, we pick the coupling strength between the qubit and controller by choosing a $Q$ large enough such that $T_1$ is several orders of magnitude larger than the time we need to do quantum computations, e.g. $T_1 \approx 10^4 \times T_\text{write}$.
Once you've picked $Q$, the power and noise of your controller determine the ratio of $T_\text{noise} / T_\text{write}$.
Fortunately, commercial equipment gets us $P$ and $S$ values that are good enough.
Note that the value of $\hbar$ comes into Equation ($\star$), and we're kind of lucky that the value is such that it doesn't prevent quantum computing from working.
