Terminators, Attenuators and BNCs, Oh My!  - Page 1
Forum section: electronics / general

author: torch
author_rank: Frequent Contributor
timestamp: January 22, 2022, 01:55:20 pm
content: I am trying some measurement setups that involve co-ax cabling and a signal generator with 50ohm output impedance. Which of course results in distorted signals due to ringing and echoes and fairies and things. After some research (including videos likeI bought a couple of feed-through 50 ohm terminators and a couple of 2 watt, 50 ohm, 20dB attenuators. Then I fell down a rabbit hole.Let me say that I am still trying to wrap my head around using dB ratios in voltage measurements. So maybe I misunderstood something. But everything I read indicated that 10:1 voltage ratio = 20dB. (as opposed to 10:1 power ratio would be 10dB.) Ahah, says I, simply buy 20dB attenuators and leave the scope input set to 10x. Simple, right?Not so fast. After a lot of experimenting and head-scratching, I eventually discovered that the attenuators I bought seemed to be fairly accurate only when the scope input was set to 5:1, as below (channel 1, yellow is a 300MHz, 10x scope probe borrowed from my Tek 475 connected via BNC probe adapter to a BNC T, channel 2, purple, is the RG58 coax through the terminator and channel 3 is the Siglent probe that came with the scope connected via BNC probe tip adapter. Source is a Heathkit RF signal generator at 1MHz):Note that with channel 2 set for 5x, the voltages are all within a few millivolts.Even at 100MHz, the voltages are very similar, despite some phase-shifting due to differing cable lengths:Thinking maybe the issue has something to do with the scope, I tried it out on the 475. Of course, it only has two inputs, so channel 1 is via the Tek probe and channel 2 is the attenuated coax cable. I shifted the traces apart 1 division for clarity on this old-school monochrome analog CRT. Note that the Tek probe has the identification ring but the BNC does not, so ignore the light on channel 2.1Mhz:and 100MHz:I next thought perhaps the attenuators were mis-labelled, however the resistance measured through the centre conductors is 81.4 ohms and from centre conductor to ground is 50.4 ohms. If I did the math right, that is very close to a 20dB, 50 ohm tee configuration, with the two R1 = 40.9 ohms crossing the Tee and R2 = 10.1 ohms leg to ground.If my setup was somehow measuring power, then 20dB ssould equate to 100x, not 5x. If I put a 50 ohm terminator inline with the attenuator, then the voltage should halve so as to be 10:1, but the net termination would be 25 ohms, which kind of defeats the purpose, and the probes would also see 1/2 the voltage -- effectively 20:1.What am I missing here???

author: NaxFM
author_rank: Regular Contributor
timestamp: January 22, 2022, 02:21:35 pm
content: Yeah, dB are a bit hard to grasp at first but then they become very simple.Remember that dBs ALWAYS refer to power, ok? When dealing with voltages, to get power we have to apply that voltage across a load: the 50 ohm resistor.Let's use a 50 ohm load and dBm for my example: dBm = 10 log(Power/1mW), if we apply the definition of power:dBm = 10 log((V^2/50) /(Vref^2/50)), where Vref is the voltage you have to apply across a 50 ohm resistor to get 1mWAs you can see we can eliminate the 50 ohm and we get dBm = 10 log(V^2/Vref^2) = 10 log((V/Vref)^2). Now apply the property of the exponent inside a logarithm and we get dBm = 10 * 2 * log(V/Vref) = 20 log(V/Vref).And that's where that 20 comes from.As you can see we are still measuring power, but this time, assuming the reference resistance is exactly the same (50 ohms), we are dealing only with voltages. And that's why a 10 times increase in voltage corresponds to 20 dB increase in power, because power is proportional to voltage squared

author: NaxFM
author_rank: Regular Contributor
timestamp: January 22, 2022, 02:59:43 pm
content: "Ahah, says I, simply buy 20dB attenuators and leave the scope input set to 10x."No, you also have to terminate the attenuator, because it expects a 50 ohm impedance both at the input and the output. This is why you are reading double the voltage.Also, if the measurement is correct with your 10x probes, it means that you set your function generator for HiZ load, which is not what you want, since you are working in a 50 ohm envroiment

author: torch
author_rank: Frequent Contributor
timestamp: January 23, 2022, 04:24:29 am
content: Quote from: NaxFM on January 22, 2022, 02:59:43 pmOk, I think you found the part I missed. I never saw specific mention of this in anything I read or watched, but as I wrap my head around it, I surmise it is probably so fundamental that nobody thought it was worth mentioning. The purpose of an attenuator is to reduce the power in the system, not to terminate the system. Placing it at the end of the co-ax improves signal-to-noise ratio, but otherwise doesn't matter where it is located or why it is reducing the power.There's still a tiny part of my brain that keeps screaming "But! But! Parallel resistors!" I have to keep telling it to shut up and believe the eyes because what you said is demonstrably correct -- if I add a 50 ohm terminator after the attenuator, the voltages just work.Quote from: NaxFM on January 22, 2022, 02:59:43 pmOk, you lost me there. My function generator is a cheap Hantek. It only has a 50 ohm output. However, the effect of comparing voltages between the Hantek and the HeathKit RF signal generator used for the above photos was the same. (I switched to the HeathKit when initially wrestling with this because the Hantek's waveforms become horribly distorted at anything over 10MHz. Someday I'll buy a better function generator.) I'm not sure what the HeathKit expects as a load. There's a pair of .01uF caps between the 6AN8 pentode plate and the RF output. I have seen that tube characterized as having "relatively" high output impedance.

author: NaxFM
author_rank: Regular Contributor
timestamp: January 23, 2022, 08:36:09 am
content: Quote from: torch on January 23, 2022, 04:24:29 amI don't know about your particular function generator, but in many cases these low frequency generators have a settings where you can tell it the impedance you are connecting them to.The internal output impedance is a physical 50 ohm resistor, so it stays 50 ohms.But let's say you properly terminate everything at 50 ohmsThis way if you set the generator to output 5 volts, you will see 5 volts at the oscilloscope but the internal voltage generator before the 50 ohm output impedance will instead produce 10 volts, but you see only 5 (the correct value) because you have a resistor divider between the output impedance and your terminator, which halves the voltage.So, everything is fine and all the values are correct. You set 5 volts on the screen and see 5 volts on the scope.Now, with the same exact settings, what if you want to connect the output of the signal generator directly to the scope or to an high impedance circuit like an opamp without any termination?What voltage will you see at the scope?We now have a 10 voltage source, a 50 ohm output impedance and a massive 1Meg input impedance.The two impedances will again form a voltage divider but now the difference is so huge that all the voltage drop is across the 1Meg impedance, so you will see 10 volts at the scope and maybe think there's something wrong with the setup.To solve this problem function generators usually have a setting somewhere that allows you to set the impedance to expect at the output, this way the function generator will just calculate the correct voltage you will read at the scope and show you on the screen.It doesn't change anything internally, it just shows you what voltage you should be reading when you connect the generator to a certain impedance.In any case don't worry, you'll get used to it. Everyone finds these things a little bit confusing when starting to work with terminators and function generators.When I first got it I thought my generator was broken for a couple of days just because I didn't know how to use it properly

author: NaxFM
author_rank: Regular Contributor
timestamp: January 23, 2022, 08:58:33 am
content: If you look at the first picture, the function generator says one volt and at the multimeter i get one volt, everything is fine.But notice what it says in the middle of the signal generator screen, slightly on the right, it says : "Load Hi-Z". This means it expects to find an high impedance load, like a multimeter.Now, look at the second picutre. Why do i still get 1 volt on the multimeter but the function generator only shows 500mV? Again, look at what is says: "Load 50 ohms"It expects a 50 ohm load, so now it shows the voltage you would have if you connect it to a 50 ohm load. It now expects the voltage divider effect between two 50 ohm impedances to take place and shows you the voltage accordingly.In the last picture i don't change any setting but i connect it to a 50 ohm resistor. Now all the readings are correct because both the set impedance and the expected impedance are matched.If I had set the expexted impedance on the generator at an arbitrary value, say 123 ohms, then i would need a 123 ohm resistor connected to the output of the function generator for it to show the correct voltage.This is not to say that if you change the impedance settings you can put whatever load you want without having reflections. The output impedance is still 50 ohms and you will get reflections if you go up in frequency. That settings is just there 'cause it may be handy when you want to low frequency circuits which do now have a 50 ohm input impedance. You won't find this setting on synthetizers or high end gear, but just low frequency arbitrary wave generators, because it's something you can really use at very low frequencies where reflection is not a concern

author: torch
author_rank: Frequent Contributor
timestamp: January 23, 2022, 03:29:15 pm
content: Quote from: NaxFM on January 23, 2022, 08:36:09 amAh, yes. Mine didn't come with that feature from Hantek, but thanks to Mechatrommer (a user here on this site), I have some replacement software that will do just that.