Part of the common mode amplification of the probe when connected together @8:00 could also be due to the connection point used, if the noise/source frequency is high enough the connections will act like a transmission line and there could be a phase difference over the length between the connections being side by side and not stacked at the same point, that phase difference could produce an effective shift for the differential probe.

Thank you Dave for educating us!

👍Wow, pretty advanced features that you are explaining here! Many thanks.

1kV RMS max, nice thing to measure the signal across the vacuum tube amp's output transformer primary winding. Cool way of demonstrating the common mode interference.

I wouldn't call it "phase reversal", that kind of implies change of impedance response (capacitive to inductive). It is just a phase wrap, since the screen is limited to +- 180 degree.

@EEVblog

Жыл бұрын

Yeah, poor choice of phrase.

Very nice explanation Dave, enjoyed this one very much.

I love these kind of videos. They make learn and relearn things I have dealt with for some time. 😎 Thanks and happy 2023 new year. 🥳

This is really great stuff Dave! I loved every bit of it, it's well polished.

Очень в тему! Just in time!

Merry Christmas

Dave, you should make a note that you’re measuring CMRR from input to output (the smaller, the better), while the formulas you show and what you describe display CMRR as the ratio of differential gain to common mode gain (the larger, the better). I can see an astute viewer being confused. Otherwise, great video.

@userPrehistoricman

Жыл бұрын

The former, in my opinion, shouldn't be called CMRR at all. It's common mode gain.

@AppleTheBandit

Жыл бұрын

@@userPrehistoricman I agree!

thank youuu sir it is very educative

One-of my first projects in electronics as a design tech was to build a common mode amplifier. For the life of I had no idea what there was to achieve expect build an amplifier that had a gain of 1 or less. The cathode follower ! I used an RCA low noise high frequency 6DS4.

Beginner here. I am stumbling around attempting to build me own constant current supply. So CMRR in this context is mostly about (sort of) DC voltages across an op amp or current sense IC. By "sort of" DC I mean that of course there is some variation across the shunt resistor. But here's my question: WHy, in this context is the common mode voltage defined as the average of the two inputs? The actual voltage across the inputs (shunt voltage drop) is clearly different, not the same. So why do we actually calculate a common mode voltage as the average? This seems to be different that the notion of say AC hum noise common to both inputs.

For me, a real test is to solder both inputs together on the board and then measure. Now the output can result from the difference in length and position in the twisted pair of the probe wires. However, it is not worth the effort considering the very high rejection ratio even without that.

@EEVblog

Жыл бұрын

Yes, that's a the difference between a system measurment (with probes and test fixtures etc) and a raw product circuit measurement. In this case with attached leads you can argue the leads must be included in the spec.

Too late to ask Santa this gorgeous oscilloscope

@EEVblog

Жыл бұрын

Next year.

Wanted to buy one, but country is not available off the online store........

@EEVblog

Жыл бұрын

DHL and extreme region price differences mean I can't sell into every country any more :-(

Today, Daiyve demonstrates a set up where the probe costs LESS than the oscilloscope! Just grab that R & S scope for $23,632.

@sc0or

Жыл бұрын

Why 23? Their web side says only 8. UPD Ah!.. I see now how their marketing department makes a salary. For me that looks stpd. Engineers need three separate models instead of three “upgrades”. Because certain things are better than a marketing bs.

@ales_xy

Жыл бұрын

How is the price of the scope or probe related to the subject? This video is about principles. You don't need this particular scope to measure CMRR.

@sc0or

Жыл бұрын

@@ales_xy Because this scope was few times advertised during this vid )

@userPrehistoricman

Жыл бұрын

Don't probes usually cost less than the scope?

I know this doesn't make sense, but I actually prefer looking at the scope from the cameras perspective rather than the screen capture. Wouldn't be the same for a smaller scope, but for this one...

👍 🔥

Thank you Dave! I've a question. I don't saw you set up HVP-70 with MXO4. With 1/10 attenuation on Probe, MXO4 will get 5mV/div minimum; but the screen showed 2mV/div. is this right?

Did I win the scope? I guessed 42 Send to Long Beach, CA. USA, 3rd house on the right. THANK YOU!

Can you do balanced audio with these?

Guess you need to put the whole setup in a faraday cage to reject as much noise as possible

Why Ch1 changed from 1.75V to 1.80V when changing frequencies? The attenuation in the circuit should be linear, right?

@EEVblog

Жыл бұрын

Slight non-linearity in wavegen vs freq, plus quantisation error.

Wouldn't a VNA be a better choice to test CMRR over a frequency sweep? Something like the Bode 100 or similar.

@EEVblog

Жыл бұрын

Sure, but most people don't have one. And in the case of the Bode 100 it only has a 1Vrms max output which is kinda limiting.

@SeanBZA

Жыл бұрын

@@EEVblog CMMR at 50Hz and around there would be easy, using your isolation transformer to provide a 400VAC CMMR source, so the error is at a higher level. Isolation transformer so that you get filtering of the more common noise that exists on the mains waveform these days, though a ferroresonant transformer does do a much better filtering job, though at the expense of having a lot of third harmonic noise, simply due to the non linear attenuation with output voltage as it does it's regulation.

@EEVblog

Жыл бұрын

@@SeanBZA Yes, it's simply a matter of a higher voltage source, however you want to do that source.

👍👍👍👍👍👏👏👏👏👏

I never got a satisfying answer to what "common mode" noise was. I always saw CM chokes and wondered why they were needed. My understanding is common mode is simply a signal that it applies to all input lines (without prejudice). Further, since lengths of wire are effectively antennas, they can pick up high frequency interference which a choke will filter out due to the magnetic flux, which differential signals will pass through due to the topology of the choke and its windings. This video seems to confirm my understanding. Is there any additional nuance that I'm missing?

@EEVblog

Жыл бұрын

Have you watched my common mode noise video?

@reicehamel1829

Жыл бұрын

Common mode rejection is the main purpose of a differential input its original model was a transformer not an op amp/differential input. A transformer used in signal processing like audio will have a balanced input and it will only transfer the difference in the input. since it will only transfer the difference any radiation or noise imposed on both inputs simultaneously will be rejected. This provides a form of noise cancellation and makes for a low noise clean input signal regardless of shielding or other means of RF /noise rejection.

A transformer can be used as differential probe. Did you ever measure the CMRR of e.g. an Ethernet transformer? Citation: "All the grounds of the inputs are common" That is not completely true. It may only be true if you have no ground loops.

Good but TOO LONG.

@EEVblog

Жыл бұрын

That's what she said.

I’m sorry - I’m just gonna say this subject is hocus Pocus - as well as that gadget measuring it -& that fancy microwave with a funny picture on it. Hocus Pocus.

I might be just a tad sleepy, but I didn't understand CMRR any better after watching this video?

My dog chewy at 21:15

Kamagatza 🥷

@NeverTalkToCops1

Жыл бұрын

With his sister, Sukiyaki.

@PushyPawn

Жыл бұрын

@@NeverTalkToCops1 Then the Dad, *Teriyaki*