I'm building a base for a PMT that outputs pulses of width <=1 µs. In Hamamatsu's PMT handbook it states on pg. 112 (emphasis mine):
When using a photomultiplier tube which is not a fast response type or using a coaxial cable with a short length, an impedance matching resistor is not necessarily required on the photomultiplier tube side.
Why does cable length effect the need for termination resistors, and at what length does the having an impedance matching resistor on the photomultiplier tube side begin to matter (for RG-174)?
Best Answer
The rule of thumb I use is that anything exceeding 1/20 of the wavelength is to be considered a transmission line. And bad terminated transmission lines have reflections that distort the signal.
To get a fast approximation of the wavelength, I consider that the speed of a signal is half the speed of light (based on experience with PCBs) and that the speed in a cable is similar. Hence, the signal travels 15 centimeters every nanosecond.
One period of 5MHz is 200ns, so the wavelength of the electrical signal is about 30 meters. One twentieth of that is 1.5 meters. The difference with Dave Tweed's calculation is that:
Therefore I find 1.5 meters in stead of 6.
Checking the dielectric constants of PVCs, I see that there is a great variance for commonly used materials. The dielectric constant of a PCB using FR4 for its material is just above 4 (with the square root being 2). I'ld say that the highest value you'll use in practice is 4 while it may be about 3 for cables.
The rule of thumb that an electrical signal travels at half the speed of light is a bit pessimistic for cables but ok - it impacts the length estimation by about 15%. Regarding the main part of the rule (1/10th or 1/20th) - it depends on how much distortion you allow. I do not remember how much it is for 1/20th but there is a theory behind it (as there is for 1/10th) and I prefer to be on the safe side.