50 Ohm Termination Error (When amplitude matters) - Page 1
Forum section: electronics / general

author: Dexter_Bunco
author_rank: Contributor
timestamp: March 31, 2018, 04:16:31 pm
content: Hello!A couple days ago I was scripting a Met/Cal procedure for a Tektronix TCPA300. For those who don't know, it's a current probe amplifier that takes a TCP series current probe and scales it to be read on an o-scope. Following the cal procedure in the service manual, it has me source 100mA from my 5700 (using a special calibration cable that goes directly from the calibrator to the amplifier without a probe), then its output gets read on a 3458 using a 50Ohm feedthru termination. Simple enough. A 100mA DC source with the Amp set to 1A/V range should result in a 5V DC readout on the DMM. Which is does. Except it doesn't. I admit I am used to taking 50Ohm terms for granted. Hook one up, assume it's good. But after my first couple times calibrating these amps failed, I looked again at the required equipment list and found that Tek was calling out a high-precision term ( ±0.1%). I checked my terminations and they are all over the place, some as much as a 2% error.This is where I expose my ignorance. I know on a surface level what’s happening when impedance matching and how the 50Ohm termination is acting like a 2:1 voltage divider but my understanding of the circuit is poor. In theory, I should be able to characterize the termination and calculate the expected voltage drop but I am not sure how to do this. I’ve tried a few different correction calculations and they get me close but the results are inconsistent.I am certain there’s a mathematical solution to this issue but admittedly I do not know where to look and google isn’t helping. Perhaps I am searching the wrong termination terminology

author: rhb
author_rank: Super Contributor
timestamp: March 31, 2018, 06:43:41 pm
content: Draw the basic equivalent circuit and solve for the all the terms algebraically. Then plug in the appropriate values.I've been fooling around comparing a Chinese 50 ohm thru terminator to a tee and terminator using one of Leo Bodnar's fast edge pulsers. The differences are noticeable on the scope as is the difference in the terminators.Very likely you need to examine the terminators with a VNA over the frequency range of interest.Part of your problem may have been what you found on the internet. I've had several instances of that lately myself.

author: ap
author_rank: Frequent Contributor
timestamp: March 31, 2018, 08:55:54 pm
content: This is not about VNA and reflections and so on. The 100mA is DC as stated, so all you need to ensure is that the load is 50 ohms accurate to 0.1%. With DC, this could be any low TC 50 ohms resistor (or e.g. two 100 ohms in parallel, for practical purposes). You can verify the accuarcy with the 3458A directly. Then you can adjust your amp accordingly.You could then, in a next step, transfer the now known DC value to the load you have (and derive a factor) and move on with AC (probably your DC resistor, if reasonably good would also work considerably high at AC).

author: Dexter_Bunco
author_rank: Contributor
timestamp: March 31, 2018, 11:25:45 pm
content: Quote from: rhb on March 31, 2018, 06:43:41 pmWell, that's part of the dilemma, I don't know what the equivalent circuit looks like, whether it's a basic op amp or something else. I was hoping that there's a standard form for these types types of amplifiers. But I have too many gaps in my knowledge to properly ask the question.Quote from: ap on March 31, 2018, 08:55:54 pmMy plan was to characterize the termination using the 3458. That’s not an issue. The struggle is that I am not sure how to calculate my expected voltage from that value. For example, one of my terminations was 50.897 Ohms. Which is fine but I am not sure what my expected voltage should be based upon this resistance value. As rhb pointed out, ideally I’d have a circuit diagram where I can just algebraically work out what my expected voltage should be. I have no such diagram.

author: ap
author_rank: Frequent Contributor
timestamp: April 01, 2018, 12:54:44 am
content: You have a voltage divider, the internal voltage (before the internal 50 ohms output resistor Ro) is 2x Uc (Uc is the calibration default voltage you want to achieve).The voltage Ut across the termination resistor Rt is as follows:Ut = 2x Uc x Rt / (Ro+Rt)

author: Dexter_Bunco
author_rank: Contributor
timestamp: April 01, 2018, 01:14:44 am
content: Quote from: ap on April 01, 2018, 12:54:44 amFantastic! So in my particular case Ut = 2*5*50.897/(50+50.897)UT = 508.97 / 100.897UT = 5.0445 VI'll check my numbers on Monday to make sure this is consistent with what I was seeing. Thanks!