What happens if PID loop is too fast?
When you execute a PID with higher sampling rate you can encounter two things:

The integrator may not work properly if the new delta is very tiny. This is due to the final resolution of floating point number. Using a double precision float numbers significantly improves this problem until certain point (sampling frequency).

The differentiator increases its output due to the input noise (also quantization noise). The output follows \$\dfrac{d\varepsilon}{dt}\$ so shorter sample time (higher freq.) means higher output for the same step change (quantization). A way to solve this phenomena is to use 1st order low pass filter on the D-part. However large filter times would also make the D-part unresponsive at high dynamics - right the opposite for what the D-Part is used for. Another countermeasure is to use a dead band on D-part input - a small change in the input signal is therefore neglected ( a good way to eliminate quantization noise)


Let we define a PID function as:
$$y=K_p \cdot(\varepsilon+\dfrac{1}{T_i}\int\varepsilon dt+ T_d \dfrac{d\varepsilon}{dt})$$
Then discrete form:
$$y=K_p\cdot(\varepsilon + \dfrac{\Delta T}{T_i}\Sigma\varepsilon + T_d\dfrac{\Delta\varepsilon}{\Delta T} )$$
Often is written also in this form:
$$y=k_p\varepsilon + k_i\Sigma\varepsilon + k_d\Delta\varepsilon$$
If your PID routine uses last form, you should re-calcuate ki,kd at every change of kp or sample time.