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I have an arbitrary waveform generator that allows me to set the load as 50 Ω or high-Z.

This generator is connected to an amplifier that has as input an unbalanced BNC connector, 10 kΩ, single-ended.

I need to measure the signal I am sending to this amplifier.

My oscilloscope has the classical internal settings for 50 Ω and 1 MΩ. How would you set the generator and the oscilloscope? At 50 Ω or high impedance?

Because I need to send the signal to two devices (oscilloscope and amplifier), I need a BNC tee. Where would be the best place to put it? Currently, I set it in the generator, then two cables route the signal to the oscilloscope and the amplifier.

I ask this because the signal is a short pulse (100 μs). For some reason, the more gain I put in the amplifier, the higher the overshoot I measure from the generator's signal, which may indicate I have undesired reflections.

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  • \$\begingroup\$ To get no reflections you need 50 Ohm on the whole way. 50 Ohm in the generator, the 50 Ohm cable and 50 Ohm at the end (as the load) were you can measure the voltage with your oscilloscope or maybe inside of the oscilloscope. A short circuit (0 Ohm) or a open connection (infinite Ohm) give reflections and everything between to but then less, except 50 Ohm. \$\endgroup\$
    – MikroPower
    Apr 6, 2023 at 22:16
  • \$\begingroup\$ 100 microseconds isn't really that short at all. That's 0.1 milliseconds. Reflections are unlikely to be a problem unless your wiring is quite long (like 10km) \$\endgroup\$
    – user253751
    Apr 6, 2023 at 23:22
  • \$\begingroup\$ @user253751 Or if they're trying to measure rise time. \$\endgroup\$
    – Hearth
    Apr 6, 2023 at 23:50
  • \$\begingroup\$ An "unbalanced BNC connector", as opposed to what, exactly? All coaxial connectors are unbalanced. \$\endgroup\$
    – Hearth
    Apr 6, 2023 at 23:52
  • \$\begingroup\$ @Hearth it's like an ATM machine \$\endgroup\$
    – user253751
    Apr 7, 2023 at 0:04

3 Answers 3

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If the amplifier has a high impedance input, then the simplest approach is to place the T right at the amplifier and ensure the shortest connection from that "stub" to the amplifier input. The source is then set to 50ohm output impedance, and the scope at the far end of the line is set to 50 ohm termination impedance and acts as the termination for what would result in a single transmission line from input to output (the amplifier as a shorted stub with high input impedance, if that is indeed the case, would be transparent to the signal propagating down the 50 ohm transmission line and would see the same signal quality as observed at the oscilloscope.). This is the following setup:

viable set up 1

Similarly, and preferred if the amplifier impedance is not a sufficiently high impedance, is put the T at the scope as close as possible to the scope input with the scope set to high impedance and then use a 50 ohm termination at the input to the amplifier at the far end of the line.

viable set up 2

In both cases we are minimizing the length of "stubs" along a drop distribution and ensuring the termination at the end of that stub is a high impedance. Between the two options, whichever of the two devices (scope or amplifier) is closer to a true "open" high impedance, should be the one that is placed at the end of the stub, while the other be at the very end of what would be essentially one straight transmission line, with a 50 ohm termination at the far end.

Note on maximum frequency which does motivate our level of concern: If working with a "square wave" clock or digital signal, it is not the data rate of the digital signal we are concerned about, but the rising and falling edges that we want to maintain that will dictate the distortion due to back and forth reflections from incorrect termination strategies. For example, if the 10% to 90% rise time of a digital signal (assuming it is that) was 1 ns, the 3 dB bandwidth of that signal is 350 MHz:

3 dB BW is related to rise time as $$f=0.35/t_r$$.

A simple Fourier Series expansion of a square wave will demonstrate the frequency content, given it is a sum of all the odd harmonics of the fundamental frequency with amplitudes going down at the inverse of the harmonic. So a 10 MHz square wave for example, if we maintain up to 350 MHz to have a 1 ns 10% to 90% rise time, will have a 35th harmonic that is \$20\log_{10}(1/35)\$ = -31 dB down. This would be the level of distortion we would introduce if we had a situation where that frequency would reflect and add to other copies of itself at different phase shifts of any significance along the transmission line - in particular where we make measurements (in this case the amplifier input and the scope input). If our application is not sensitive to such distortion, we can relax our constraints accordingly.

As far as managing reflections, the strategy is to use a transmission line of a given impedance and match the source and load on each end of that transmission line with that impedance. Any branches along the way should be minimized in length (stubs) and be a high impedance, as even if they were each terminated in 50 ohms, it would create a significant reflection at the branch. We can understand this intuitively by knowing when we terminate a transmission line with its impedance, since there is no reflection, it appears at the source as an infinitely long transmission line. Thus, equivalently, it appears at the source as a 50 ohm load. These are equivalent:

terminated transmission line

If we were, incorrectly to add a T anywhere along the line with an additional termination of 50 ohms, the T would present as a 25 ohm termination at its location (a significant mismatch in impedance as a source of reflections). Below shows a case of having a T somewhere along the line with an additional transmission line on each output of the T, each terminated in 50 ohms. As introduced above, these would be equivalent electrically to a 50 ohm termination for each transmission line right at that location. This is not a good approach but illuminates the thought process for other options.

bad case

At the location of any impedance mismatch between a transmission line or source of impedance \$Z_o\$ and load of impedance \$Z_L\$, the reflection coefficient is given as:

$$\rho = \frac{Z_L-Z_o}{Z_L+Z_o}$$

The reflection coefficient is a complex number with a signed magnitude from -1 to +1, indicating the percent of reflected signal and it's phase. For example if we had an open termination, the reflection coefficient is 1 (the entire signal is reflected back and in phase with the incident), and if we had a short termination, the reflection coefficient is -1 (the entire signal is reflected back and out of phase with the incident).

With that we can demonstrate one termination approach used in digital design called "series termination" when we have the case of a single source and single load and no branch drops along the line as depicted in the drawing below (For this I am simplifying my drawings to eliminate the ground side of the T-line).

TDR measurement

Here we have a single pulse at the source with a step change from 0 to 2V. At the first moment of the transition as measured at the source, there is no reflection and the transmission line, at that moment, is equivalent to how an infinity long transmission line would appear: a 50 ohm load to ground. Thus we have a voltage splitter and the step as it is launched onto the transmission line is a 0V to 1V step. This travels down the line until it reaches the open at the far end of the line T seconds later, which results in a complete reflection 100% and in phase. Thus we get a 0V to 1V step reflected back and propagates back to the source adding to the 1V already settled by the incident pulse. The reflection will reach the matched impedance at the source (which appears as 50 ohms to ground since a voltage source is a Thevenin short) and therefore no further reflections occur, in this case. Thus the final step will settle to 2V all along the line. How that distortion appears at various points along the line is summarized in the graphic below showing the three points A, B, and C depicted in the earlier graphic:

TDR Results

What we see from this, for the case of not having a multidrop situation, that a single series termination will result in no signal integrity issues at the input to the amplifier (which is where we care). If however we were to have a multidrop case (adding another high impedance amp at location B for example), the result could be disastrous, especially for a digital design where the "amp" is a CMOS gate that transitions when the pulse is at mid level. As an aside, the graphics above demonstrate "Time Domain Reflectometry" (TDR) where we can launch a pulse down a line and from the result measured at "A" determine the length to a short or open somewhere down the line.

If we followed through the effects of reflections with delayed copies of reflections, the worst case possible amplitude variation is given as

$$A(\phi) = 0.5|1+e^{-j\phi}| = |\cos(\phi/2)|$$

Where \$\phi\$ is the phase for a particular frequency due to the time delay between the signal and its reflected copy. So at 350 MHz for example, the wavelength in a teflon coax cable is 124 cm. If the shortest distance of the amplifier input to the transmission line when placed right at the T was 2 cm, the round trip distance would be 4 cm or a total angle for the 350 MHz component of \$(4/124) 2\pi = 0.2\$ radians (11.6 degrees). The variation in dB using the formula above would be 0.04 dB. (of a component for the case of a 10 MHz square wave that is already down 31 dB!). This is insignificant, but a 20 cm stub in comparison would be a 5 dB variation. Similarly with regards to TDR demonstration above, a 2 cm stub would correspond to a time delay T of 24 ps. If our rise time was 1 ns, there would be no measurable distortion assuming a pure open termination. (To the degree the load does not appear as an open will impact actual reflection effects that will occur). Generally I use a rule of thumb accordingly that any transmission line distance on any stubs that is less than 15° would be insignificant for most applications.

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  • \$\begingroup\$ Would it be safe to set the T exactly at the amplifier? Mostly, I am afraid of any "spikes" due to reflections in the amplifier input. To give you an idea of my application, I send single pulsed with 1V of amplitude (0.5 V of offset to guarantee it goes from 0 to 1V). The pulse width is from 10us to 100 us. Spikes in the input are not acceptable because it could damage the circuit in the output of the amplifier. What do you mean by "put the T at the scope as close as possible to the scope"? The T has BNC connectors, as the scope. So, it is just a matter of plugging the T at the scope, right? \$\endgroup\$
    – JJJJeb
    Apr 7, 2023 at 12:25
  • \$\begingroup\$ @JJJeb the point is if you terminate the very end of the transmission line in the impedance of the line, and minimize any "stubs" along that line to insignificance (if your rise time of your digital signal (assuming it is that) is 1 ns, the 3 dB bandwidth of that signal is 350 MHz. As far as reflections go, any transmission line distance less than 15° would be insignificant for most applications. In standard teflon coax transmission line cables, at 350 MHz, 15° is a length of 2.46 cm (close to 1")--- so the T should be immediately at the amplifier input in that case. I'll add that detail. \$\endgroup\$ Apr 7, 2023 at 12:32
  • \$\begingroup\$ Thanks for the detailed answer. I have a lot to unpack there. In short, the T should be placed at the amplifier, and both the scope and waveform generator at 50 ohms. Is this correct? Notice that the T position and impedance setting are the only things I can change. Adding a 50 ohms feedthrough in the BNC connector of my amplifier is also an option, but I do not see what I gain from it. \$\endgroup\$
    – JJJJeb
    Apr 7, 2023 at 14:02
  • \$\begingroup\$ @JJJJeb yes. You can place the scope or the Amp at either location. Whatever is at the end of the line, terminate in 50 ohms. Whatever is at the T, ensure it is high impedance. If one of the two is closer to being an "open", then use that at the T. Note if your Arb is like the one I use, the setting of Hi Z or 50 ohms only changes what is displayed for the voltage output and the instrument has a 50 ohm source impedance regardless. Measure the output of your arb with nothing connected and set to 2V and see what voltage you get when you change from Hi Z to 50 ohm. \$\endgroup\$ Apr 7, 2023 at 14:16
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    \$\begingroup\$ I’ve fixed a couple of LaTeX issues, but for your information when you want to write inline LaTeX on this site you should escape the dollar signs: \$…\$ (other sites luckily don’t have this requirement). \$\endgroup\$ Apr 25, 2023 at 17:08
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The output impedance setting on the waveform generator ONLY affects its internal display or setting of the output voltage it has no other affect on operation. Its output impedance will probably be 50 ohm independent of the setting.

When the generator is being used in a 50 ohm terminated setup the voltage at the output of the generator will be attenuated by a factor of two.

The generator's indication of output voltage will take account of this attenuation to show the actual output voltage after this attenuation.

When the high-impedance setting is selected the instrument will display the voltage as if there was no attenuation.

Typically low frequency use will not require a 50 ohm environment and so the high-impedance selection would be appropriate. Low frequency in this context means that the wavelength of the signal is very large compared to the size of any cables or wiring. (eg 10 times)

Since signals travel in coax at about 60-70% of the speed of light that travels about a speed of one foot per nanosecond a lab bench setup with cable lengths less than ten feet would indicate that frequencies below about 10MHz can be treated as low frequency and can use the high-impedance setting without 50 ohm terminations (assuming that the devices themselves do not have 50 ohm terminations). When dealing with signals with rapid rise times it is the bandwidth of the harmonics that is important, not just the fundamental.

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  • \$\begingroup\$ So, from your comment, I understand that both settings would be fine, although setting the scope and generator at high impedance is preferred. To give you an idea of my application, I send single pulsed with 1V of amplitude (0.5 V of offset to guarantee it goes from 0 to 1V). The pulse width is from 10us to 100 us. A 1m cable goes to the scope (so I measure the signal the generator is sending to the amplifier), and a 30cm cable goes to the amplifier input. The T is at the generator. \$\endgroup\$
    – JJJJeb
    Apr 7, 2023 at 13:22
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In general, the setup would look like this for non-critical situations:

schematic

simulate this circuit – Schematic created using CircuitLab

You should have a 50 ohm termination at the amplifier input to minimize reflections.

The oscilloscope in the middle of your line will cause an impedance bump. You'll need to test this out using the above setup sans the amplifier and using a 50 ohm terminator where the cable connects to the amplifier. This may work fine as long as the rise/fall time of the signal isn't too fast.

Are you measuring the overshoot at the amplifier input or output? If you're seeing increased overshoot at higher gains at the amplifier output, this indicates amplifier instability. If you're seeing this at the amplifier input, I would suspect Miller capacitance issues if using a transistor amplifier. To reduce Miller effect, consider using a cascode amplifier topology.

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  • \$\begingroup\$ At low frequencies - less than 10's of Megahertz, it it not usually necessary to use 50 ohm termination. Since the amplifier under test is 10kohm input this is probably a low frequency application. \$\endgroup\$ Apr 6, 2023 at 23:33
  • \$\begingroup\$ The T was right in the waveform generator. A 1m cable goes to the scope (so I measure the signal the generator is sending to the amplifier), and a 30cm cable goes to the amplifier input. As I increased the gain of the amplifier, I could see a strange effect on the signal measured with the scope. The higher the gain, the lower the was the voltage at the state region of the pulse. For a pulse of 1V, it quickly reached 1V, but then it dropped a bit during the steady-state value, so the signal looks like having an overshoot. \$\endgroup\$
    – JJJJeb
    Apr 7, 2023 at 12:35

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