# How this DIY High-Speed Probe doesn't reflect signals?

I am reading The Art of Electronics, page 809 and 810, where they show this probe:

Which is a DIY probe, made out of a RG-178 coaxial cable, in series with a 953 ohms resistor.

I wonder why the 953 ohms resistor is needed, won't it reflect any signals entering the 50 ohm, instantaneous impedance, coaxial cable? Is it for limiting the current out of low frequency signals? Or is it just for creating a 20x probe (1k/50R = 20)? Then what's the benefit of a 20x probe and not 10x? For me it looks like a big inductor for high-frequency signals, and the book doesn't specify any special, low inductance, resistors to be used.

• 953 is close to the ideal 950. some resistors are bulk film, some are laser trimmed in a spiral. What is the time constant of 10 nanoHenry and 1,000 ohms? 10 picoSeconds, right? – analogsystemsrf Nov 11 '19 at 3:23
• I know why it's 953 ohms and not 950 ohms, that's not my question, I am questioning why there's a resistor at all. – mFeinstein Nov 11 '19 at 3:25
• So the resistor is to limit the current on the capacitance of the coax? Won't it cause reflections for impedance mismatch? Isn't the loading a part of the 50ohms impedance of the coax? – mFeinstein Nov 11 '19 at 3:37
• @mFeinstein Have you considered the fact that the HC signal already has proper source and termination impedances in the existing board and that this is merely a "tap" used to observe? If you look over HC application notes, you'll see that they use $1\:\text{k}\Omega$, by the way. But you can look at Neil's answer here on EESE for some thoughts. – jonk Nov 11 '19 at 4:03

The resistor will indeed reflect the incoming signals back, but those signals exist in a few mm of IC pin + resistor pin, which have an infinitesimally small inductance. Thus those reflections will not be enough to produce any measurable overshoot.

On the other side of the resistor, a source with ~1kOhm resistance feeding a 50 Ohm cable will create a voltage divider.

On the scope side there is a potential that the cable inductance could create an overshoot because of a reflection, but there will be none as the 50 Ohm cable will match the 50 Ohm impedance of the input.

• Although many gave me good insights, I will accept yours as the answer since you were the only one that were concise about the actual content of the core of my question, thanks! – mFeinstein Nov 11 '19 at 17:56
• If I am not mistaken, the small size of the back signal-path will not distort much also because the path won't be as much as a transmission line, as the size of the path won't be close to the size of the rising edge of the signal (unless they are very high-speed signals). – mFeinstein Nov 11 '19 at 17:59

The point of this probe is that the scope has its 50Ω termination enabled. Since the scope input impedance is matched to the cable, there is no reflection from the scope. Once the load side is matched, we don't need to worry about any source-side impedance mismatch; reflections have already been suppressed.

The point of the 950Ω resistor on the input side is simply so that the scope probe presents a 1kΩ load to the circuit, which is a pretty reasonable load, versus a 50Ω load, which is decidedly unreasonable for a measurement device we're adding onto a signal.

Apparently "Art of Electronics" doesn't explain this well (I haven't read that book so I'm just taking your word for it). This type of probe is explained pretty well in the book "High Speed Digital Design: A Handbook of Black Magic", which I have read and I highly recommend. There is also this article by Dr. Howard Johnson, who is one of the authors of the book I just mentioned, which talks about this type of probe and why it's actually superior to any scope probe you can buy.

• Yeah, they don't explain at all, just show some options. The "High Speed Digital Design: A Handbook of Black Magic" is on my waiting list. Wouldn't the probe present a 1050Ω instead? As is the 1kΩ impedance in series with the 50Ω impedance from the cable at high frequencies? – mFeinstein Nov 11 '19 at 5:09
• I prefer to think of the load as "1kΩ ±5%" ;-) Actually, if we're just concerned about the loading, then the difference between a 1kΩ load and a 1.05kΩ load is really inconsequential -- they should both qualify as "has very little effect on the circuit they're measuring". Really the only difference is "super-convenient value that everyone has laying around" (1kΩ) versus "value that provides a nice round number integer-ratio divider" (950Ω). – Mr. Snrub Nov 11 '19 at 8:18
• Yes sure, I was just checking if I wasn't missing something out, calculating the value wrongly – mFeinstein Nov 11 '19 at 15:41