# Transfer function of a digital phase frequency detector

I'm confused about the transfer function of a digital phase frequency detector. Why can we say that the pfd output is proportional to the phase error?

The pfd (with charge pump) generates current pulses of fixed amplitude $I_{CP}$ like it is described for example here. For small phase deviations the length of these current pulses is proportional to the phase difference of the input signals. So the pfd output current is clearly not proportional to the phase error.

On the other hand, the transfer function of a digital PFD is said to be $K(s) = \frac{I_{CP}}{2\pi} = \frac{I_{out}(s)}{\Delta \phi(s)}$. In this sense the pfd generates a current proportional to the phase error.

Why is this not a contradiction?