Super Basic Question and perhaps something more involved - Page 1
Forum section: electronics / beginners

author: Moose42
author_rank: Newbie
timestamp: September 13, 2024, 05:44:22 pm
content: Hi all!I am new here and new to having my own scope (finally in my mid 50s). I studied some physics decades ago but have not used it for EM so much, so all that stuff from Jackson has fallen out of my brain. My main goal here is to teach myself to explain anything that I see on the scope. Here are the questions:1) (NOTE: I ordered a signal generator) While I wait for that ... I connected the center wire of a short length (~30 cm) of coax to the 1 kHz, 300 mV calibrator lug on my new-to-me Tek 465B. I connected the coax to a tee and into the scope (nothing on the other end of the tee). I see a 300 mV pp square wave. Then (just for fun) I put a 50-ohm terminator on the tee and the pp voltage drops to ~ 250 mV. Now the scope has a 1 MOhm input impedance, right? But I can't work out the simple math that knocks that 300 mV down to 250 mV. What's going on? (Pics attached)2) Every video I have watched on terminations seems to pedagogically show a single pulse traveling down a transmission line to an impedance mismatch and a reflection occurring at the end. But I must have not been listening to the part regarding why frequency matters. Why don't we worry about reflections at low frequency? (Is that true?) and what is low? Below 1 Mhz? The 465B calibrator is at 1 kHz, is that low?3) Different topic: The Volts/div control on Ch 2 of the 465B doesn't work (I opened it up and it looks like the grey barrel is cracked) Is this something a novice like me can fix? I mean in terms of taking it apart, fixing the control and putting it back together again? What is the best resource / webpage / book / person ?Thanks in advance!Moose

author: rstofer
author_rank: Super Contributor
timestamp: September 13, 2024, 06:27:02 pm
content: I run into reflections when working with fast rise-time signals. FPGAs have very fast rise times and one solution is to insert a 330 Ohm resistor in series with the signal. If you look at the schematics for some of the FPGA development boards from Digilent, you will see that they use this method.You probably won't see these reflections on a pure sine wave.Fourier Analysis is the subject if you want to search for the math.

author: Manul
author_rank: Super Contributor
timestamp: September 13, 2024, 06:37:30 pm
content: Hi, and welcome to the forum.1. Scope input has very high impedance at low frequencies, so we can just ignore that and say it's infinite. So you measured 300mV with input Z = infinite and 250mV with Z = 50 Ohms. That tells that the output impedance of calibrator is somewhere around 10 Ohms. Basically you can solve it like you solve a resistor divider. If signal source would be 50 Ohms, level would drop exactly half.2. Frequency matters in a sense that there is some ratio between wavelength and transmission line length. Generally if transmission line length is shorter than 1/10 of wavelength, then we don't worry as much about reflections. When calculating wavelength, keep in mind that signal speed in cables and traces is significantly less than the speed of light. Anyway, it's all relative, if you take a 1m cable, 100kHz can be unterminated, but 100Mhz should be terminated. If you take 1000m of cable, you should terminate even at 100kHz.So should you terminate 1kHz square wave? 1kHz seems like a very low frequency, but the bandwidth required to accurately represent it is quite much higher, because it needs to pass all higher harmonics (many times the fundamental frequency of 1kHz). As you know, square wave is a sum of fundamental frequency and many odd multiples of it. Bandwith can be estimated by rise and fall time of square edges. If rise/fall time is 10ns, bandwidth is maybe 30-60 Mhz. So it's not anymore a very low frequency. If there is impedance mismatch and the cable is not super short you will start to see reflections near edges (which have high frequency content).I don't know about a scope repair, I'm not familiar with this model.

author: Moose42
author_rank: Newbie
timestamp: September 13, 2024, 07:35:26 pm
content: Thanks to both for the replies!

author: TimFox
author_rank: Super Contributor
timestamp: September 13, 2024, 08:33:54 pm
content: If you have narrow pulses at a relatively low pulse-repetition frequency, then you can see the reflections on a normal CRO from a mis-terminated transmission line, if the "electrical length" (in seconds) of the line is longer than the width of the pulses themselves.For other ratios of pulse width and period, you can still see reflections if the rise time of the pulse is short, so the reflection happens after the incoming rise is complete.

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: September 13, 2024, 11:11:12 pm
content: The square wave provided for calibration is meant to be used with a 10X probe as far as I remember.For Parts and fixing, I would join the Tek Scopes group on Groups.io, you probably need a new switch.....maybe someone will have oneI have a Tek 475 the series of scopes is a kinda classicYou need to get proper 10X probes that match the capacitance of the scopes input. The cheap ones from China are pretty good, at least the ones I have ordered are really nice. Almost ALWAYS use a 10X probe.

author: Moose42
author_rank: Newbie
timestamp: September 14, 2024, 03:16:47 pm
content: Thanks folks for the tips. I will check out the Tek forum. Also I started playing around with LTspice and was able to reproduce the voltage divider behavior of adding the 50-ohm BNC T (see pic). Also, with the scope powered off, I measured a DC resistance of 10 ohms between the calibrator connecting and scope ground.

author: Manul
author_rank: Super Contributor
timestamp: September 14, 2024, 04:06:56 pm
content: You are doing great.

author: Moose42
author_rank: Newbie
timestamp: September 14, 2024, 05:38:07 pm
content: Thanks Manul! Looks like you are in Lithuania, no? I’m an American working in Austria but my dad was born in Kaunas.

author: Manul
author_rank: Super Contributor
timestamp: September 14, 2024, 06:43:12 pm
content: Nice. Yes, I'm from Lithuania. Feel free to PM me if you wish to talk.

author: gcewing
author_rank: Regular Contributor
timestamp: September 15, 2024, 03:20:19 am
content: Quote from: Moose42 on September 14, 2024, 03:16:47 pmThat's probably a fluke, though -- in general you can't measure the output impedance of a circuit that way. You have to measure the output with different loads (as you effectively did here) and calculate it.