Conceptualizing feed-through termination to a scope - Page 1
Forum section: electronics / testgear

author: jorgerod
author_rank: Newbie
timestamp: February 11, 2022, 01:29:39 am
content: Hello, I have a newbie question trying to conceptually understand how a 50 ohm feed-through termination works.I need to make some measurements of a50 MHz10 MHz square-wave oscillator that outputs 50 ohms. Specifically, on my SDS1104X-E I need to look at its Vp-p, overshoot, and Vrms to determine power. Typically this signal will feed into a 50-ohm matched port, and the scope only has 1Mohm input resistance mode.For this case, will I see a meaningful difference on the scope using the 10x probe vs. a direct coax connection with a 50-ohm feed-through termination? Will the voltage levels be approximately the same, or is there a manual scaling I'd need to consider with the feed-through? I know to set the instrument to 1x probe when using the direct feed.Conceptually, I am struggling to understand why the feed-through termination helps. I understand it's providing a matched load on the other end of a transmission line (coax). But that's still in series with the scope's internal 1 Mohm resistor, right? I'm not grasping how having 50 ohms in series with 1 Mohm is significant. I appreciate any help in conceptually understanding this better!

author: fourfathom
author_rank: Super Contributor
timestamp: February 11, 2022, 01:55:31 am
content: The feed-through termination isn't in series with the signal. It is a 50 Ohm resistor from signal to ground. Of course you can't use the scope probe, you connect a 50 Ohm transmission line to the feed-through termination.There may be some effect on the signal shape due to the shunt-capacitance at the scope input but at 50 MHz that's not likely to be very noticeable.For non-critical terminated measurements I just use a small BNC 50 Ohm dummy load on a BNC Tee at the scope input. The cable to the DUT goes to the other Tee port.

author: jorgerod
author_rank: Newbie
timestamp: February 11, 2022, 02:03:29 am
content: Oh, of course it's from signal to ground and not in series. That's the mental mistake I was making. Thank you!

author: alm
author_rank: Super Contributor
timestamp: February 11, 2022, 02:05:16 am
content: As fourfathom writes, the 50 Ohm termination is in parallel with the 1 MOhm // 10pF or so of the scope. So it is a transmission line terminated into 50 Ohm (its characteristic impedance), and the scope's input is connected across this 50 Ohm resistor with an extremely short connection (~1 cm).If you are talking about moderate frequencies, then the voltage levels will be very similar to terminating the source in a discrete 50 Ohm resistor, and then connecting a scope probe across that resistor. It's just that the bandwidth of the feed-through termination directly connected to the input should be better due to very short stub and lack of any other components.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 11, 2022, 05:57:57 am
content: Very little difference at merely 100MHz bandwidth -- are you expecting to see more than a badly smeared 50MHz anyway?Tim

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 11, 2022, 06:58:50 am
content: As mentioned, the terminator acts inline, leaving a stub length between it and the scope. This has the effect of introducing an impedance discontinuity, and the stub length can be represented as an equivalent LC resonant trap, thus introducing a peak or notch to the frequency response.To understand the impedance discontinuity, understand what it means to have a 50 ohm transmission line. Consider a plane cutting across the cable: if you inject a current at this point, you will find it drops V = I Zo volts (or any other combination of circumstances: a voltage source and consequent current draw, or an impedance divider). Likewise, any signals coming out of this point (whether your own signal reflected back around from a discontinuity, or an independent source at the far end), also have the same impedance (at least in the near term, so, setting aside any reflections).It's important to note how we're modifying the situation, to maintain the use of Kirchoff's laws. You might naively look at the situation and say, oh yeah that coax cable is a wire in a shield, that wire must be a node and therefore any current injected at one end squirts out instantly at the other end, and the voltages are equal. Well we know that can't be the case because the speed of light exists and the transmission line is nonzero length. That's causally impossible. But we can maintain the fiction of KVL/KCL at a cutting plane: the width of the line at this point is quite small (some ~mm), so that the delay across it is also very small, hopefully small enough to ignore (and for plain old RG-58, this works just fine up to several GHz). Instead of saying current goes into the wire at one end and immediately out the other end -- a violation of causality -- we say the current goes in the wire at one end, andout the shield at the same end.This is the definition of a port.We might use ports that are common ground -- convenient for unbalanced transmission lines (coax), bulkhead connectors, ground plane PCBs and so on. In this case, the port is a 2-terminal component, with one pin assigned ground by definition (or, perhaps with some nonzero impedance to ground in case we need to model common-mode conditions, which is to say, when our shielding and grounding isn't as ideal as we'd assumed).We can further improve the abstraction by considering a completely floating port, as if isolated by an ideal transformer. Though this isn't important in this situation, so this is just for information.Finally, we can analyze the equivalent circuit. We have a transmission line coming in, some kind of connection to the termination resistor, and more TL to the scope (and whatever its input equivalent circuit looks like).If we use the old fashioned BNC tee and terminator, then inside the tee, we have three TLs connecting.From any given direction, we start on a 50 ohm TL, and see two TLs in parallel, 50 || 50 = 25 ohms.So a tee necessarily introduces a discontinuity.Notice the terminated arm has a nonzero stub length, but is terminated at 50 ohms, so no reflection occurs, and it looks like any length of TL -- including zero and infinity. We can thus replace this arm with a regular resistor and don't have to worry about any TL effect.So we reduce the system to a signal-side TL, a shunt resistor, and a scope-side TL.The stub length between shunt resistor and scope, is not terminated, so does incur a reflection. Exactly how much, and what phase, depends on the input equivalent circuit. (It's labeled as 1M || 10pF or whatever, but it's not usually actually a capacitor, but has significant ESR, and maybe ESL or other higher-order TL effects as well. It might indeed even approach 50 ohms at very high frequencies, i.e. on the order of 1 / (2 pi (10pF) (50Ω)) = 318MHz. All depends on circuit details. Though, not that we'll ever see anything about that anyway, with a mere 100MHz scope. On the upside -- this does mean the scope input should be reasonably ideal given its bandwidth. Neat, huh?)If we know that the stub length is, say 5cm, c_0 = 0.67c, and Zo = 50Ω, then it will act as a 1/4 wave impedance transformer, converting from the 25 ohms at the tee, to Zo^2/Zin = 100 ohms at the scope, i.e. having a peak gain of 2 at the 1/4 wave frequency. Or put another way, a series L, shunt C network between termination and actual scope input. 1/4 wave in 5cm at 0.67c is 1GHz, so L = 8nH and C = 3.18pF. (Uh, give or take a geometric factor; that's close though.) And alternate dips and peaks at harmonics thereof.Whereas if we design the terminator as an inline component, the distance from TL to termination resistor doesn't really matter -- as long as its stub is matching, it can't matter -- but we can shrink the distance from terminator to scope.Fortunately, this won't really matter for scopes much slower than 1GHz. So we expect good performance regardless of method. Using a tee and terminator might be worse for more mundane reasons, like just poorly fitting BNCs. A pass-thru type can have a tiny amount less stub length, though that's still mostly dominated by the physical connector and whatever's inside the scope, so not really much we can do about it. Also, note the magnitude of components here: if the stub resonant mode has on the order of 3pF in it, then then scope's 10pF+ likely dominates, and this amounts to a tweaking of its response rather than full-out peaks and troughs at those high frequencies.It also may be worth noting that BNC itself isn't very well matched, and maybe our assumption that the tee is a three-way at 50 ohms, isn't actually true. But this will only be different by a few dB worth, so still not very significant. (It's commonly quoted that BNC is good up to about 2GHz, with SWR rising quickly >1.3 or so by then. Seems reasonable given the ~cm scale length of this hardware!)Also, if we have a faster scope (where this stub length begins to matter), they conveniently provide internal terminations, making this whole matter moot. And with good reason, eh?(Probably more than you ever wanted to know, surely more than you were expecting; I haven't actually gone through the calculation in a while so explaining this was a good refresher. Cheers!)Tim

author: NaxFM
author_rank: Regular Contributor
timestamp: February 11, 2022, 07:40:30 am
content: Be careful that with a 100MHz scope and a 50MHz square wave, even if everything is perfectly terminated all you will see is a badly shaped sine wave.It's a beginner mistake, the first time I tried to see a 25MHz square wave with my 70MHz scope i thought the function generator was broken...If you use a fourier transform on a 50MHz square wave you will see that the first harmonic is at 50MHz, but all the other harmonics which give the square wave its shape are odd multiples of 50MHz, so the third harmonic will be at 150MHz and the fifth at 250MHz.To see a good square wave you need to see at least the fifth harmonic, so in theory a scope with 250MHz bandwidth is the absolute minimum to see something relatively good.However there's a handy simple formula to calculate the bandwidth required to properly see a square wave, and it depends on the rise time.It is : BW = 0.35/Tr, where Tr is the rise time of the square wave and BW is roughly the bandwidth required to properly see it on a scopeYou can try a little experiment yourself.Set the generator to output a 10MHz square wave and measure it with the scope. It will look nice and sharp.Now enable the 20MHz bandwidth limit on the scope and measure again. You will see a mess even if the frequency is well within the bandwidth. That is because the next odd harmonics will be at 30 and 50 MHz, well outside the bandwidth of the scope, so you will see them very attenuated

author: jorgerod
author_rank: Newbie
timestamp: February 11, 2022, 11:37:33 am
content: Thanks everybody - this has been extremely helpful. And I had a typo in my question - the clock is 10 MHz, not 50, so I anticipate I'll capture enough harmonics to see some sharpness. Thank you again!