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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)?

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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:

  1. I use 1/20th which is a factor of two smaller than Dave's rule of thumb;
  2. I consider that the speed is half the speed of light, which is another factor of two.

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.

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  • \$\begingroup\$ Just a note, 5MHz has a wave length of 60 meters, not 30 which is the wave length of 10MHz. I've made more than one dipole for the 30M band which is ~10.1MHz. \$\endgroup\$ – GB - AE7OO May 21 at 10:47
  • \$\begingroup\$ When I checked the dielectric constants of cables, they are roughly the same as PCBs where the speed of a signal is about half the speed of light. This was confirmed on an actual board where 15cm corresponded to a delay of 1ns on a clock that was still looking as perfect after the delay as before the delay. 5MHz has a period of 200ns. 200ns * 0.15m/ns is 30meters, not 60. This is valid for the cable. In empty space the wavelength is 60cm. (The cables dielectric is sometimes close the 3, and sqrt(3)=1.73, in which case the wavelength is about 35cm, so 30cm is pretty good). \$\endgroup\$ – le_top May 22 at 11:35
  • \$\begingroup\$ @GB-AE7OO Your dipole is an antenna in free air, where the wavelength will be closer to 60 meters. But the isolation of the cable changes the signal speed. \$\endgroup\$ – le_top May 22 at 11:41
  • \$\begingroup\$ Huh??? I think you mean the other way around. The wavelength for 10.1Mhz in vacuum/free air is give or take 30 meters. .The dielectric type and form of the coax matter for the VF, ranging from about .67 to .88 for the common coaxes. I don't have clue how you can be seeing a 10% to 40% reduction. PCB's are no where near the same. That's the reason I've been know to put down some SMA connectors(I've got trays of various ones) connectors and hook some coax up to it when the board was being squiggly on me. As far as cable size goes, thats when a VNA or TDR comes in handy. \$\endgroup\$ – GB - AE7OO May 22 at 13:30
  • \$\begingroup\$ You're right, I made a mistake. In free air, the wave is moving faster, so the length is shorter. \$\endgroup\$ – le_top May 23 at 7:33
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As a general rule of thumb, you should begin taking transmission line effects into consideration when the cable length approaches λ/10 — i.e., 1/10 of the wavelength of the highest frequency in the signal.

For example, if you have pulse rise/fall times on the order of 100 ns, you need to have good fidelity at 5 MHz, so cables longer than 6 meters should be impedance-matched.

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  • \$\begingroup\$ Would velocity factors of the coaxes dielectric factor in the calculations of wavelength ? \$\endgroup\$ – Old_Fossil May 26 '16 at 7:05
  • \$\begingroup\$ @resident_heretic: Well, yes, but we're talking about rough rules of thumb here, so the difference between 100% and 70% velocity factor gets "lost in the noise". \$\endgroup\$ – Dave Tweed May 26 '16 at 11:11

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