Same Spec'ed Rigol Probes but Not The Same?

[Mechatrommer]

my probe is RP2200 with serial number on it allright. cant take a picture right now, camera's lens offline!. i'm in Batman Project, my sleep schedule has been shifted to 8am - 3pm
just now i tried to profile my FG using direct FG-BNC-PCB-BNC-DSO up to 100MHz sine signal, so it means x1 probing. is that means my reading cannot be trusted?

[A Hellene]

The rule of 1/(2πFC) indicates an additional loading of 15.9Ω!
Fortunately, probe loading is not purely capacitive; but, yes, I would not trust the reading --if any could successfully be done at those frequencies using x1 probes.

Just see the impact the x10 probes had on the 16MHz crystal oscillator of the mega8, in my previous message.


-George

[Mechatrommer]

ok george! i saw your osc measurement earlier, its sensible since x1 with more loading will attenuate the signal further. but i just retried to compare x1 and x10 reading from FG signal, shocking to me, the result are reversed!

100MHz sine signal 50ohm source impedance unterminated:
x1 direct pcb bnc = 4.36Vpp
x10 rigol probe = 3.04Vpp

100MHz sine signal 50ohm source impedance terminated:
x1 direct pcb bnc = 2.48Vpp
x10 rigol probe = 1.52Vpp

what the hell is going on?!

[A Hellene]

Ah, the magic is done by the direct connection using some serious transmission line with low attenuation --if terminated correctly!
By "x1 probing" I assumed that you would use the (extremely lossy) oscilloscope probes, set x1. 


-George

[Mechatrommer]

ok i know whats going on.... ground loop inductance.... again! ok, using spring gnd, i managed to increase the signal in x10, but still not as good as x1. maybe at this frequency, coax shielding is much more important. my intuition telling me.

100MHz sine signal 50ohm source impedance terminated:
x1 direct pcb bnc = 2.48Vpp
x10 rigol probe (spring gnd) = 2.32Vpp

maybe george, you were measuring a high impedance source, that MegOhm input impedance will affect too much on voltage divider. mine is 50ohm source only, so doesnt matter 1M or 10M ohm the attenuation will not be affected too much i think.

[Mechatrommer]

ok thanx george for clarifying

[A Hellene]

I am sorry for the delayed response; I had to refill my empty glass with poison on the rocks!


When the transmission line is not terminated, the source cannot provide any power to the load, since there is not any real load at the end of the transmission line! The measuring device, which is a parasitic load, should ideally be no load at all to the system! Thus, the extremely high input impedance of the test equipment.

Voltage is the reason of current flow; and without any current flowing the voltage reading alone is mostly meaningless --except for debugging purposes, only.

The concept of a matched source, transmission line and load impedance is to transmit as much power as possible to the load. Reading just a ghost voltage at the end of an open transmission line does not really mean anything useful!

Transferring power (or energy) is the challenge, which can only be achieved when there is as much current flow to the load as possible. And the best it can be done, in the real world, is to transfer to the load a little less than half of the power the source can deliver. Either in DC or in AC.


-George


[EDIT]: Additional information.

[Kiriakos-GR]

I love batman projects, and If you had invite me too, I would had bring my own toys too. 

[Mechatrommer]

Reading just a ghost voltage at the end of an open transmission line does not really mean anything useful!

so reading open ended (no load) battery giving a useless result? i believe what you meant by "ghost voltage" is for open ended high frequency signal where transmission line length > wavelength / 4 or according to Jim Williams > 1 inch where things like reflection and standing wave is much more prominent (power not transferred).

[vk6zgo]

Reading just a ghost voltage at the end of an open transmission line does not really mean anything useful!

so reading open ended (no load) battery giving a useless result? i believe what you meant by "ghost voltage" is for open ended high frequency signal where transmission line length > wavelength / 4 or according to Jim Williams > 1 inch where things like reflection and standing wave is much more prominent (power not transferred).

No,it is not the high tech end of things,it is more related to simple voltage divider action.

(1) With paper & a pencil,draw an ideal generator with zero internal resistance.
(This is impossible,but we can imagine it). 
For convenience,make it a 10volt  p-p source.

(2)Add a 50 ohm resistor in series with one terminal of the generator.(A real generator  with 50 ohm output impedance may be regarded as an ideal generator in series with a 50 ohm resistor).

(3) Try various resistor values in series  with the 50ohms,forming a voltage divider across the generator,
& calculate the  voltage across,& power dissipation in the extra resistor.

You will find that the power dissipation across the extra resistor is maximum when that resistor is equal to
the series resistor (50 ohm),& the voltage across it is 5 volts p-p (1/2 the open circuit voltage of the generator).

(4)The voltage across a 1Mohm resistor in the same position will read very close to the open circuit voltage.

This is extremely easy to explain face to face,& very hard to explain via a forum.

VK6ZGO 

[Mechatrommer]

i understand, no need face to face but we try to read voltage, not power, as george termed it as "ghost voltage".
if i measure terminated signal, the unterminated signal will read about twice as that, no "ghost" will flying around, or maybe some misunderstanding here

[A Hellene]

Alright, dear Mechatrommer, it was my mistake. After all, you know that I do not believe in ghosts!

The terms "voltage" as well as "ghost voltage" were not accurate but oversimplified ones. Would it be better if I used the terms "electric potential difference" and "unspecified/phenomenal voltage" instead, respectively? I was trying to start from square #1 since the measuring equipment impedance was very close to the impedance of the circuit under test, something that VK6ZGO explained very well.

The RP2200 probes are not a linear load, since they are loading the test circuit with a combination of a 10MΩ resistance in parallel to a 17±5pF capacitance. In DC their loading is straightforward: 10MΩ! But in AC, things change dramatically as the frequency rises.

For example, RP2200's load at 100MHz is equivalent to a 10MΩ ±2% resistor in parallel to a 72.3Ω .. 132.6Ω one (i.e. 17±5pF parallel input capacitance @100MHz), which equals to a load of an equivalent resistance of 72.3Ω .. 132.6Ω. Of course, this is not a load that can be ignored, since it inserts some significant measurement error to a signal source of 50Ω or 51Ω impedance.

Under these conditions, a reading of 2.32Vpp at the open end of a non terminated 50Ω transmission line using the RP2200 probes would suggest an actual presence of 3.20Vpp .. 3.92Vpp at the test points: A deviation of +38% .. +69% = +53.5%±15.5 of the actual reading --having, of course, not taken into account any possible parasitic elements (as the probe's ground current loop inductance), or standing waves or reflections due to impedance mismatch of the load, and/or the length ratio of the transmission line. Well, this is not just a deviation; this is a faulty reading.

In a similar manner, the DS1000 oscilloscope input impedance at the BNC is 1MΩ±2% with a capacitance of 18pF±3pF. Again, in DC the input capacitance is insignificant; but not in HF (my mega8 16MHz oscillator example) or, even worse, in VHF...


-George


[EDIT]: Calculation errors corrected.

[vk6zgo]

Interestingly,Mechatrommer's results, even if they show errors in absolute value,still follow closely the expected ratio between terminated & unterminated voltages.

There is a bit of error for the direct connection,but the ratio is spot on when using the X10 probe.


100MHz sine signal 50ohm source impedance unterminated:
x1 direct pcb bnc = 4.36Vpp
x10 rigol probe = 3.04Vpp

100MHz sine signal 50ohm source impedance terminated:
x1 direct pcb bnc = 2.48Vpp
x10 rigol probe = 1.52Vpp

By the way, Mechatrommer,picture (1) of reply#8 shows a lot better results for phase error.
Had you already started tweaking the probes at that stage,or was there a measurement error with the results in
your original posting?
You probably are asking a lot of your probes if they are not specified for the same phase error at 100MHz.
A lot of things quote amplitude/frequency response,but phase/frequency response is rarely quoted.



VK6ZGO

[Mechatrommer]

Interestingly,Mechatrommer's results, even if they show errors in absolute value,still follow closely the expected ratio between terminated & unterminated voltages.

i dont see any interesting point there, just headache and losing more hair.

By the way, Mechatrommer,picture (1) of reply#8 shows a lot better results for phase error.
Had you already started tweaking the probes at that stage,or was there a measurement error with the results in
your original posting?

it seems to be, but the fact that the reply #8 phase improvement at (1) and (2) is due to cold rigol startup and test. the 1st posting is the one is done after proper warm up and calibration. thats a luck, not improvement. i only tweaked the probe capacitance on (3) in picture reply #8

You probably are asking a lot of your probes if they are not specified for the same phase error at 100MHz.
A lot of things quote amplitude/frequency response,but phase/frequency response is rarely quoted.

i asked if they are different. simple yes or no. and then you asked me to switch probe which i did. then i got the answer, the answer is yes, they are different. simple as that. as all other matters revolving around us, its "money" matter. if you have complain, you shut up, take out $$$ out of your pocket and buy a brand name one.

[vk6zgo]

The "interesting" thing is that the voltage ratio is what it should be from simple voltage divider theory!
No confusion,all the other effects are important,but the "correctly terminated " source will present a voltage 1/2 that of the same source "unterminated".
Try the same test at a lower frequency,say around 1kHz,to get a good understanding on how it works.


 The Rigol is performing "above & beyond the call of duty" to be able to measure phase
difference at 100MHz,even with your direct connection,so it was definitely a good buy!

I just realised that,when you are measuring phase,you are only doing that,so you can adjust one or both probes for minimum phase error,as shown in your reply#8,ignoring the effect on the normal "Calibrator" waveform.

Switch one or both of the vertical channels to "Uncalibrated",& adjust the variable vertical amplitude control till both signals have the same amplitude on the screen.

You should now have equal amplitudes,& very close to zero phase error on your display.

You can now use this set up to measure phase difference between two separate test points.(As they are unlikely to be the same level,you will probably have to match the vertical amplitudes as before).

When you are finished,simply return the vertical channels to "Cal",& readjust the probes using the "Calibrator" waveform.

VK6ZGO