Oscilloscope probe question...

[cvriv]

This is probably a very noob question. I made a special BNC wire connector thing so I can connect my function generator to my bread board. Well, when I used the oscilloscope probe to check the connector to see if I made it properly, I got a diminished waveform. The frequency was spot on but the Vpp is a lot less. The function generator is pumping out a 1Vpp sine wave @ 1kHz. The oscilloscope was only reading 120mVpp. The probe says 10:1 on it. I think that has something to do with it, but what i'm thinking isn't right either as 120mVpp x 10 = 1.2Vpp. What's going on here?

The oscilloscope and probes I used it class gave me nearly exact values. Is my oscilloscope not setup properly or something? If I connect the function generator directly to the oscilloscope using a BNC to BNC wire, the oscilloscope reads exactly what the function generator is putting out.     

[tautech]

Tee your FG output then connect you scope and breadboard to a leg each of the tee then you should be able to identify if the load from your breadboard is pulling the FG output down.

[MosherIV]

Hi

"I made a special BNC wire connector thing so I can connect my function generator to my bread board. Well, when I used the oscilloscope probe to check the connector to see if I made it properly, I got a diminished waveform. The frequency was spot on but the Vpp is a lot less. The function generator is pumping out a 1Vpp sine wave @ 1kHz. The oscilloscope was only reading 120mVpp."

It is to do with impedance matching. When you deal with high(ish) freq yours inputs and outputs must be impedance matched otherwise you get strange effects like waveform cancels out or drop in amplitude. Impedance matching reduces line reflections. Dave did do a youtube video on it but I cannot remember which one it is.

Simple answer is to use the T-piece both on the FG and on the oscilloscope AND put a terminator on the oscilloscope end. This matches the impedance for you.
Your circuit will have to be impedance matched as well. Again this means that you need a 50 or 75ohm terminating resistor as part of your input circuit.

[cvriv]

Thanks:) Of course I don't have any damn Tee's because they haven't frigging arrived from china yet! I bought most everything off of Amazon. They don't really tell you where everything is coming from. I am still waiting for a few tid bits of which include my BNC tees:(

[cvriv]

I checked the input / output settings of the OSC and FG. I set the output of the FG to 50ohms and the input to the OSC to 50ohms. They were both set to high before, the 1Mohm range. When I connect the OSC and FG together, using the BNC to BNC, the OSC reads correctly. If I check with the probe now... nothing happens. The input / output impedance are the same. How will using a BNC tee fix this?

I'm not understanding whats going on here.

[MosherIV]

"I'm not understanding whats going on here. "

Do you remember with analogue TV, if you used a small TV ariel for portable TVs, sometimes you got a ghost image on the TV?
The ghost image looked like the same image but faded and moved to one side of the main image.
This was due to the TV image being reflected off things like building.

A similar thing happens with electrical signals going through cables. The same signal can be reflected back (bounces) through the cable.
The reflected signal interacts with the original signal, in your case reducing the signal amplitude.

The terminators stop the reflected signals. To understand how and why involves some deep maths (which I never really understood anyway).

Now when you directly connect the FG to the osc, they are terminated correctly - hence you see the correct signal.

However, now that both FG and Osc are expecting correct termination, the Osc probe is not correctly terminated - hence you do not see the correct signal.

The T adapters and terminators are meant to be used in conjunction with passing the signal to your circuit and allow you to view the signal at the same time.

Just found Dave's video

https://www.eevblog.com/forum/blog/eevblog-652-oscilloscope-function-generator-measurement-trap/

[alanb]

I checked the input / output settings of the OSC and FG. I set the output of the FG to 50ohms and the input to the OSC to 50ohms. They were both set to high before, the 1Mohm range. When I connect the OSC and FG together, using the BNC to BNC, the OSC reads correctly. If I check with the probe now... nothing happens. The input / output impedance are the same. How will using a BNC tee fix this?

I'm not understanding whats going on here.

Are you using the  10M ohm probe (or the lead from that probe) with the 50 ohm setting on the scope and FG? If so it wont work. When using 50ohms you need to have a plain coax lead as well as having terminators to prevent reflections.  The lead for the  10M probe is resistive as well as having resistance in the probe body and this will cause attenuation of your signal.

 

[cvriv]

I'll watch that video you posted. I remember my professor teaching us about maximum output where the load should be as close to the internal resistance as possible to maximize power to the load. I know that the internal resistance and the load resistance is a voltage divider. If matched perfectly, the voltage across the load should be just about half of the source voltage. The higher the load resistance, the more voltage is dropped across the load.

One thing I learned is that I need to specify the correct input ratio. It was set to x1 the whole time. One reason why the probe wasn't working correctly.

I noticed that when I set the ratio to x1 and use the BNC to BNC cable, the input/ out impedance of the the FG and OSC doesnt matter as long as they are the same. The cable ideally has zero resistance.

When I set the FG to high and the OSC to 50ohm and use the probe with the input ratio @ x10... the OSC read nothing. This is because most of the voltage is being dropped across the FG internal resistance. If i set the output of the FG to 50ohm and the OSC to 1Mohm, the OSC reads more than twice what the FG output vpp is... which doesn't make sense to me. How can the OSC read more than what the FG is putting out? No matter how high the OSC impedance is, it should technically never be able to to reach the source voltage. Can someone explain this?

So what I did was specify HighZ output from the FG, which I think is 10kohm, and 1Mohm input impedance for the OSC. I then added 3.7Mohm to the signal via the bread board. I then checked the OSC and it now reads 1.04Vpp, which is just about what the OSC read with the BNC to BNC cable (1.032Vpp).

So....... I need to sleep. My brain hurts. I have to think about this real good. It makes sense but doesn't.

[MosherIV]

"How can the OSC read more than what the FG is putting out? No matter how high the OSC impedance is, it should technically never be able to to reach the source voltage. Can someone explain this?"
It is very difficult to make a simple explanation.
It goes something like this : remember I said that the signals bounce/reflect - well when the reflected signal meets the source signal, the wave can cancel or add. You can get a signal that looks bigger that the originating source because the waves add together to make what looks like a larger signal.

Like I said, there is serious maths involved with the understanding. I remember it is to do with transmission line theory and tau/pi networks. Transmission lines can be simulated by hypothetical LCR networks in a Tau or Pi network, that is the transmission line is a combination of a resistor and inductor and has losses represented by resistor to ground. The line also has capacitance, represented by capacitors from the line to ground (which are going to act like filters).

After reading all the theory - the short answer is :
use co-ax (50 or 75ohm)
terminate the co-ax with the correct terminator (50ohm for 50 ohm co-ax)

[cvriv]

I think I understand why the voltage displayed by the oscilloscope can read higher than the source.

An amazing old video explaining wave reflection:


So what you are saying is that I should ditch my x10 probe? Just use a coax cable and a 50 ohm terminator? I just checked my order for the BNC tee and it says nothing about 50 ohm impedance. Are all BNC tee 50 ohms or no? Im thinking no.

[MosherIV]

"I think I understand why the voltage displayed by the oscilloscope can read higher than the source.

An amazing old video explaining wave reflection:"
Yep, you got it. Great, I remember a demo like that but could never find a video like that, so I will use that in future - thanks.

"So what you are saying is that I should ditch my x10 probe? Just use a coax cable and a 50 ohm terminator?"
Erm, it's complicated :
1. When you connect the scope, Fg and your circuit up and to see what the FG is injecting into the circuit - you need to use the co-ax cables+T+Terminators (the 50 or 75ohm)
2. When you connect just the FG to your circuit (FG need to have low output impedance and the circuit needs to be high input impedance) then you do not need the co-ax+T+terminators. The scope can now be in high input impedance and your should be able to probe around your circuit to see what is going on.
I hope that makes sense.

"I just checked my order for the BNC tee and it says nothing about 50 ohm impedance. Are all BNC tee 50 ohms or no? Im thinking no."
It depends, some T pieces have built in terminators and some do not. There is no easy visual indication as far as I know.
If the T piece has built in terminators, they get disconnected when you connect something to them, so there is no harm in connecting a terminator to it.

[rstofer]

I think I understand why the voltage displayed by the oscilloscope can read higher than the source.

So what you are saying is that I should ditch my x10 probe? Just use a coax cable and a 50 ohm terminator? I just checked my order for the BNC tee and it says nothing about 50 ohm impedance. Are all BNC tee 50 ohms or no? Im thinking no.

Mostly, Tees are just Tees.  The usual practice is to put a 50 Ohm terminator in one branch of the Tee with a terminator that looks like a cap.  Leg of the Tee toward the scope, one branch has the incoming cable and the other branch has either a terminator or a coax to the breadboard and another Tee with the terminator.  Or you can have the terminating resistor in your project.  You only want to terminate the cable once.  I wouldn't use the scope internal termination if I was terminating the cable at the project.

http://www.showmecables.com/product/BNC-Male-Terminator-5-Watt-50-Ohm.aspx

[cvriv]

How do you not use the internal scope termination? I can select either 50 or 1M ohms. I think I am understanding this. I just need to organize my thoughts. I was up all night messing around with this.

So, can you say that I did right by adding 3.7M ohms to the line via resistors to get the probe to probe correctly? What if I want to probe something that has a really high resistance? Something higher than the 3.7M I had to add. Or what if I and probing around within a circuit, does it matter where I am probing within the circuit? There are different total resistances at different points in a circuit. Shit... this is blowing my mind.

[MosherIV]

"How do you not use the internal scope termination? I can select either 50 or 1M ohms."
If you are using the scope probe - use the 1MOhm.
If you are using co-ax with Ts - use the 50 Ohm.

"So, can you say that I did right by adding 3.7M ohms to the line via resistors to get the probe to probe correctly?"
Sorry but the answer is : depends on your circuit.
Generally, outputs need to be low impedance and inputs need to be high impedance.

"What if I want to probe something that has a really high resistance?"
Bingo. Ding Ding Ding - you are getting it.
If you try to probe something that has the same or high impedance as the probe - the probe is going to affect the measurement.

"Or what if I and probing around within a circuit, does it matter where I am probing within the circuit?"
See answer to above.
In general the answer is no. If the output of 1 device is low impedance AND the input of the next device is high impedance AND the 1st device output is capable of driving the new impedance with the scope probe then the answer is not much. The higher the freq you go, the harder it is to satisfy this, so with really high freq you need to go to FET probes or active probes which both have really high impedances.

[tggzzz]

Or what if I and probing around within a circuit, does it matter where I am probing within the circuit? There are different total resistances at different points in a circuit. Shit... this is blowing my mind.

At your level of understanding, it all reduces to two resistors acting as a potential divider. One resistor is the circuit's "output" resistance, the other is the probe's resistance.

To ensure you don't inadvertently make dangerous mistakes, I suggest you look at the safety references in https://entertaininghacks.wordpress.com/library-2/scope-probe-reference-material/ That will also give you pointers to aspects you will need to understand once you have got over your current hurdle.

BTW, most electronic engineers will assume "osc" = oscillator. The contraction of "oscilloscope" is "scope".

[StillTrying]

So what I did was specify HighZ output from the FG, which I think is 10kohm,

You seem to keep thinking the FG's High Z and LowZ refers to it's own output impedance, it doesn't it refers to the impedance of the connected load.

LowZ is a special case for driving a 50R load (perhaps though a 50R cable). If the load is not exactly 50R the voltages at the load will be off.

If you're using your scopes internal 50R termination and a direct cable keep the voltages very low, it's very easy to burn the scopes internal 50R resistor, and very hard to fix!

[cvriv]

Ok. Im back at it.

I ran the 1kHz signal @ 1Vpp through three 2M ohm pots in series on the bread board. I dialed in the 3.7M ohm I was using last night and got the 1.04Vpp with my x10 probe, but when I dialed in more or less resistance, nothing really happened with the Vpp on the scope. So I began to remove pots. I ended up with one pot dialed in at 270k ohms. That gave me exactly 1Vpp and when I dialed in more or less resistance, there was significant change with the Vpp. So what is this telling me? Does this mean I'm impedance matching?

When I use a 100 ohm pot I get the same result as I did when I used to much resistance. I get 1.04Vpp and when I dial up and down the resistance, there's not much change at all.

I ordered some 50 ohm BNC terminators just because.

Also, the florescent bulb I'm using on my desk is creating some crazy feedback. Is there a bulb that does create this feedback? Led bulbs???   

[rstofer]

Well, you're getting close to what's going on.  Here's a suggestion:  Draw the circuit.  You think of the signal generator as a pure voltage source in series with some resistor.  That resistor might be 50 Ohms if the generator is set to drive 50 Ohms or it might be some high number - maybe it is in the datasheet.  You can find out what it is by driving an adjustable resistor and putting in the amount of resistance that divides the output voltage in half.  The potentiometer value is the same as the source impedance.  Thevenin Equivalent Circuit works here:

https://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem

If you set the generator for 50 Ohms and measure the output voltage without a terminator, it will be twice what it will be when you terminate the cable.  Again, this is easily explained by Thevenin.

Continue on drawing the sketch, noting where various values of resistance are located, including the scope probe.  From Ohm's Law you will be able to work through the voltage divider calculations.

From your written description, I wasn't able to follow along with enough assurance that I would offer up a solution.  A sketch is a handy thing to talk over.  Engineers are always drawing a picture.  I can't even have a conversation about a technical subject without having a pen in my hand.

[cvriv]

Well, you're getting close to what's going on.  Here's a suggestion:  Draw the circuit.  You think of the signal generator as a pure voltage source in series with some resistor.  That resistor might be 50 Ohms if the generator is set to drive 50 Ohms or it might be some high number - maybe it is in the datasheet.  You can find out what it is by driving an adjustable resistor and putting in the amount of resistance that divides the output voltage in half.  The potentiometer value is the same as the source impedance.  Thevenin Equivalent Circuit works here:

https://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem

If you set the generator for 50 Ohms and measure the output voltage without a terminator, it will be twice what it will be when you terminate the cable.  Again, this is easily explained by Thevenin.

Continue on drawing the sketch, noting where various values of resistance are located, including the scope probe.  From Ohm's Law you will be able to work through the voltage divider calculations.

From your written description, I wasn't able to follow along with enough assurance that I would offer up a solution.  A sketch is a handy thing to talk over.  Engineers are always drawing a picture.  I can't even have a conversation about a technical subject without having a pen in my hand.

I'm familiar with Thevenin's Theorem, Max Power, Voltage Dividing... this is all so familiar to me. I think Im losing understanding with the equipment resistances. I'm going to eat and then comb through your post and do what you said. I'll do some math and drawing and then will report back.   

[David Hess]

...

Also, the florescent bulb I'm using on my desk is creating some crazy feedback. Is there a bulb that does create this feedback? Led bulbs???

Some electronic ballasts for florescent bulbs are worse than others and LED bulbs are often just as bad.  I like linear florescent bulbs with passive ballasts because they are quieter and yield fewer hard shadows.  In extreme cases, incandescent halogen bulbs are the quiet option.

[MosherIV]

Hi


"I ran the 1kHz signal @ 1Vpp through three 2M ohm pots in series on the bread board. I dialed in the 3.7M ohm I was using last night and got the 1.04Vpp with my x10 probe, but when I dialed in more or less resistance, nothing really happened with the Vpp on the scope. So I began to remove pots. I ended up with one pot dialed in at 270k ohms. That gave me exactly 1Vpp and when I dialed in more or less resistance, there was significant change with the Vpp. So what is this telling me? Does this mean I'm impedance matching?

When I use a 100 ohm pot I get the same result as I did when I used to much resistance. I get 1.04Vpp and when I dial up and down the resistance, there's not much change at all."

If the picture is truely representative of what you are measuring, then the FG is driving into an infinite load (the pot is only connected to one side of the FG, if you were to connect 100ohm across the FG, it probably will not like it very much when you turn the pot towards a short circuit).
It should show whatever the FG outputs plus any noise picked up by the leads because FG is drining into infinite ohms.

I guess the other guy is telling to draw the circuit so you can figure this out for yourself.

[cvriv]

I'm trying to figure this out. I started drawing a diagram but I don't know what is what.

Assuming the function generator is the source, 1Vpp @ 1kHz. Lets say I specify the FG out to be 10k ohms. I measured the resistance of the probe and got 9M ohms which is what it's supposed to be. Then I specify the input of the scope to be 1M ohms.

Is the load a combination of the probe, scope, and total resistance of the circuit?

I had drawn a diagram but it's doesn't make any sense. I feel like I'm close but so far away.

Question: The probe resistance is included with the load?

Question: The scope input resistance... included with the load as well? No?

The FG output resistance, this is consider the internal source resistance right?

[tggzzz]

Lets say I specify the FG out to be 10k ohms.

It is an unusual generator that allows you to change its output impedance. Normally it is fixed, typically 50ohms or 600ohms.

I had drawn a diagram but it's doesn't make any sense. I feel like I'm close but so far away.

The schematic you have drawn looks strange. From left to right and top to bottom I would expect to see the generator's ideal voltage source (which you can't directly measure), the generator's output output resistance (50 or 600 ohms), then the circuit under test, then the scope+probe with one side earthed. Traditionally the scopes and probes resistance are combined, and I would expect that to be

• 1Mohm for a *1 probe
• 10Mohm for a *10 probe
• 50ohm if you have a 50ohm terminator (more accurately 50ohm in parallel with 1Mohm)

Don't forget to include the difference between peak-to-peak and rms voltages in your observations.

[StillTrying]

Here's what page 35 of the FG's manual says:

To Set the Output Load
For either of CH1 Output and CH2 Output on the Front panel, the Generator has a built-in 50R series impendence.
If the actual load does not match the set one, the displayed amplitude and offset are incorrect.
This function is used to match the displayed voltage with the expected one.

Steps for setting the Load of each channel:
(1) Press Utility and choose Output Setup. Press F1 to select CH1Load, or press
F2 to select CH2Load; press it again to select HighZ or *R ("*" represents a value).
(2) To change the load value, after selecting R, turn the knob to change the value, press / direction key to move the cursor left and right; or press the number keys to input the desired value. Press F3 or F4 to select the unit. The load range is 1 R - 10 K.

Note:
For either of CH1 Output and CH2 Output on the Front panel, the waveform generator has a fixed 50 R Series impendence. No matter what Value the set parameter is, if the real load is different from the set one, the displayed voltage will not equal the real voltage.

So that's cleared that up then. 

From 5 minutes with the manual I couldn't tell what setting "Load: HighZ" does to the output amplitude.
It would be nice and simple if it just halved the output amplitude.
What does it try to do if the load is higher than 10 K?

http://www.saelig.com/supplier/owon/AG1012F-AG1022F-AG2052F-AG2062F-Waveform-Generator-USER_MANUAL_V2.6.pdf

[tggzzz]

In other words, the output impedance is fixed and unchangeable.

Fiddling with that control just inaccurately changes the displayed voltage.

Why "inaccurately"? Because loads are not purely resistive, they have reactance. That generator can output 50MHz sine waves. At that frequency a typical "high impedance *10" scope probe will have an impedance of around 200ohms. A better choice of (cheap robust) probe would be a *10 "low impedance Z0" probe, which would have an impedance of around 500ohms.

Summary: too much complexity and too many options confuses beginners.

[ProBang2]

BTW: The probe (at breadboard and in schematics) is in series connected.
IMO not a very good idea...

[MosherIV]

Hi

Sorry, been away most of the day at a prestigeous openning event
(http://www.tnmoc.org/news/news-releases/fujitsu-innovation-hub-opens-block-h)

"I had drawn a diagram but it's doesn't make any sense. I feel like I'm close but so far away. "
Your diagram does not look like what the circuit you showed in the photo.

It looks to me like this

The FG output simply floats hence you just get the same signal no matter where you probe.

[Keysight DanielBogdanoff]

If i set the output of the FG to 50ohm and the OSC to 1Mohm, the OSC reads more than twice what the FG output vpp is... which doesn't make sense to me. How can the OSC read more than what the FG is putting out? No matter how high the OSC impedance is, it should technically never be able to to reach the source voltage. Can someone explain this?

This is a pretty common brain teaser people struggle with. Essentially, it comes down to a resistive divider circuit. Basically, the internal resistance of a function generator is 50ohms. If the termination setting in the function gen is set to High Z mode, the majority of the voltage drop will come from the High Z system impedance (High impedance in series with 50 ohms means that the 50 ohm impedance is basically negligible).

In 50 ohm mode, the function gen expects you to put 50ohms externally. So the divider is then 50 ohms (internal) in parallel with 50 ohms (external). That means 1/2 of the voltage drop will be on your 50 ohm termination and the other half will be on it's 50 ohm (internal). So, to get the correct voltage at the board, the function gen puts out double the voltage.

To demonstrate this, connect the function generator to the scope directly. Put the function generator at 1V, and both the scope and the function gen in High Z mode. You'll see 1V on the scope. Then, put the function generator in 50ohm mode and the scope still in HighZ mode. You'll see 2V on the scope. Then, put the scope in 50 ohm mode, and you'll go back to seeing 1V on the scope.

[cvriv]

Sorry for the late reply. I've been tired lately. Been stressing myself out trying to figure this out.

@ StillTrying - When you change the load impedance of this FG, it does change the amplitude. I read the manual, regarding all this, for the scope and the FG. Doesn't mean I understand what I am reading though Also, it won't let you specify an impedance greater or less than the range of 1 - 10K ohms.

@ ProBang2 - You are saying that the way I have the probe attached to everything on the breadboard is series connected? Why isn't it a good idea? How else would it be connected and why?

@ MosherIV - I have no idea what you had drawn. I recognize the source but have no idea what the other thing is.

@ Keysight_DanielBogdanoff - Ok, I understand what you are saying. I had no idea that's what the function generator was doing, doubling the output despite not saying so. I thought it was something the scope was doing.

I understand the whole voltage divider thing you are explaining and I did what you said using a BNC to BNC cable and saw it. I've been seeing this the whole time but just couldn't understand what was happening.

Now... when I use the x10 probe, what happens then? The probe itself says 10M ohm for 1M ohm inputs. I measured the resistance of the probe with a DMM and got 9M ohms. The front of the scope says 1M ohms inputs. I'm thinking the 10M ohms is a combination of the probe and the input.

When I use the probe to probe the function generator, which is set to 50 ohm impedance with 1Vpp output, the scope shows me double. The scopes probe ratio setting is set to x10 and the input set to 1M ohm. This is basically what I was getting when I used the BNC to BNC cable instead of the probe. The function generator is putting out double, despite not saying so, and the scope is reading just that.

Sooooo... all the voltage is being dropped across the.... probe and input impedance of the scope? That kind of makes sense, but when I set the input impedance of the scope to 50 ohms, I get nothing on the scope. Why is that? Initially, I thought that all the voltage was being dropped across the probe, leaving nothing for the input impedance (50 ohm) of the scope, but that doesn't make sense, because if that was true, I wouldn't read double the voltage when the input was set to 1M ohm.

I noticed if I change the scopes probe ratio setting to x5, the scope reads what the function generator is telling me, 1Vpp. This is interesting because it's cutting the output to the function generator in half. I don't know if this is the correct thing to do.

I have more questions about circuit total resistance and how it effects what I see on the scope, but I'll wait to understand whats going on with the generator, scope, and probe. 

[MosherIV]

Hi

"@ MosherIV - I have no idea what you had drawn. I recognize the source but have no idea what the other thing is. "
You were asked to draw the schematic after the photo you posted.

I looked at what you posted and it had no resemblance to what you had in the photo.

So I drew out what you had in the photo, just the FG connected to a variable resistor (or potentiometer).
What I was trying to draw (in paintbrush) was a variable resistor (or potentiometer).
BUT the pot is only connected at one point to the FG, so it does not matter where you probe, you will see the same thing.

I have re-drawn it in TinyCad.

It looks like you are trying to measure and prove things that are very hard to measure in real life.
You are trying to measure what effect measuring has on a circuit, the equipment taking the measurement is affecting the circuit.
How can you measure the effect of the effect of the measuring equipment ?
(It is possible, the people who make the test equipment like Tektronics and Keysight do it but it takes lots of equipment)

I think what is not clear is that there are 2 modes of use for the oscilloscope :
1. Direct probe measurement, where (hopefully) the impedance of the probe is negligible and you can just probe around the circuit and see what is going on). Here you do not need to worry about impedance matching (just leave the probe in x10).
2. Where line reflections are interfering with the measurement, you use a transmission line type arrangement to nullify the line reflections. This is where the impedance matching come into play (no probe, co-ax+T+terminators).

(Also FYI if it is not clear, 'impedance' is another way of saying 'AC resistance', DC resistance is constant where as AC resistance changes with frequency).

[cvriv]

Ok, I get what you mean by drawing a circuit diagram. I was doing something completely different. I fully understand what you had drawn.

So why is my scope reading double the Vpp then? I mean I know why it's reading that, but how do I get it to read exactly what the generator is putting out when probing with the probe? I mentioned setting the ratio to x5 works, but said I dont thinks thats the right way and you said to leave it set to x10. Also, why do I not get anything on the scope when I set the scope to 50 ohms? This is really the only thing I'm trying to figure out right now.

I know now that the gen puts out double when it's set to 50 ohms. I know that for sure now. With the x10 probe and the scope set to 1M, I am reading double the Vpp. Thats because all of the voltage is being dropped across the scope and probe. How do I get the scope to read what the generator is telling me it's giving, AKA not double the output.

[cvriv]

LOL. Ok. I don't think I was probing correctly before. LOL. I was probing shit. When I was probing the pot before, the probe was in series and someone here mentioned that. I didn't pick up on it until now. The crazy thing is that I'm suppose to know this! LOL. I am now probing parallel to the circuit

Ok, so I had drawn another diagram of my little circuit and it's what I'm experiencing. I'm experiencing a bit more than double on the scope. I took a measurement with my DMM and it read EXACTLY what was on the function generator screen. I set the FG to output 3Vrms and thats what the DMM read. The scope read a bit more than double that. How to get the scope to read the exact amount, not double.

[edp]

I think you have cracked it..... You mentioned both RMS and peak to peak - they are not the same.  Look up the relationship between RMS and p-p for a sine wave and I think it will start to make more sense.

[MosherIV]

Hi

Lets answer the easy one :

"why do I not get anything on the scope when I set the scope to 50 ohms?"
The 50 Ohms mode is not intended to be used for the scope probe, it is intended to be used when looking at a 'transmission line network' - the co-ax-T-Terminator is effectively a 'transmission line network'.
The 50 Ohm input selector put a 50 ohm resistor across the input to the oscilloscope and 50 ohms is a very low resistance and will pretty much short out any input when used incorrectly.

[StillTrying]

Well you can't use LowZ 50 ohm mode with a 1Meg or 10Meg probe can you.

Edit I always ignore 'A new post has been posted', - it turned out was MosherIV posting the same thing. 

[tggzzz]

LOL. Ok. I don't think I was probing correctly before. LOL. I was probing shit. When I was probing the pot before, the probe was in series and someone here mentioned that. I didn't pick up on it until now. The crazy thing is that I'm suppose to know this! LOL. I am now probing parallel to the circuit

There's an old English proverb: "more haste, less speed".

When flummoxed, stop.
Come back later and start again without preconceptions. 
Draw a picture of what you think you want, check the picture matches what you've actually done.
Create a lab book entry with columns for every setting and measurement.

All that will help you understand, and will help other people understand what you have done and are trying to do.

Which do you think is easier for someone else to understand:
"the meter reading is double", or
"the meter reading is 1.23, I change X and the meter reading is now 2.46 (i.e. doubled)"

In explicitly thinking about and writing down your procedures, you will often find that you have explained it to yourself. If that doesn't work then you will have a good basis for avoiding the traps mentioned here: https://entertaininghacks.wordpress.com/library-2/good-questions-pique-our-interest-and-dont-waste-our-time-2/

[w2aew]

Ok, I get what you mean by drawing a circuit diagram. I was doing something completely different. I fully understand what you had drawn.

So why is my scope reading double the Vpp then? I mean I know why it's reading that, but how do I get it to read exactly what the generator is putting out when probing with the probe? I mentioned setting the ratio to x5 works, but said I dont thinks thats the right way and you said to leave it set to x10. Also, why do I not get anything on the scope when I set the scope to 50 ohms? This is really the only thing I'm trying to figure out right now.

I know now that the gen puts out double when it's set to 50 ohms. I know that for sure now. With the x10 probe and the scope set to 1M, I am reading double the Vpp. Thats because all of the voltage is being dropped across the scope and probe. How do I get the scope to read what the generator is telling me it's giving, AKA not double the output.

Maybe my video on the function generator output (and load impedance setting) might help you:

[MosherIV]

Hi

Going back to your question (the hard one to answer now that I have time to think about it) :
"how do I get it to read exactly what the generator is putting out when probing with the probe?"

Looking at your new picture and diagram.....
Your 'input impedance' of 50 Ohms is too low - remember I said that outputs need to be low impedance and INPUTS NEED TO BE HIGH IMPEDANCE. So, if you make the FG drive into 10K (upto 100K) then it should look something like the FG output. Do not make the input MegOhms because then the impedance of the scope probe will start to do funny things.

I think the reason is that the 50 Ohm is creating a impedance miss-match and causing line reflections. It is counter intuitive, that is why I never truly understood high freq stuff in analogue electronics.

[rstofer]

Ok, I get what you mean by drawing a circuit diagram. I was doing something completely different. I fully understand what you had drawn.

So why is my scope reading double the Vpp then? I mean I know why it's reading that, but how do I get it to read exactly what the generator is putting out when probing with the probe? I mentioned setting the ratio to x5 works, but said I dont thinks thats the right way and you said to leave it set to x10. Also, why do I not get anything on the scope when I set the scope to 50 ohms? This is really the only thing I'm trying to figure out right now.

I know now that the gen puts out double when it's set to 50 ohms. I know that for sure now. With the x10 probe and the scope set to 1M, I am reading double the Vpp. Thats because all of the voltage is being dropped across the scope and probe. How do I get the scope to read what the generator is telling me it's giving, AKA not double the output.

Maybe my video on the function generator output (and load impedance setting) might help you:

That video is right to the point.  The FG 'assumes' a load impedance and calculates the output voltage knowing that its internal impedance is 50 Ohms.

So, if you want to get the right answer in the trivial case of the FG connected only to a scope, you can either install a terminator on the coax at the scope or set the scope for 50 Ohm input impedance (vs 1 Mohm) or set the FG for Hi-Z.  Some FGs limit the setting to something like 10k because, beyond this point, the value doesn't matter relative to the 50 Ohm internal impedance.  None of the voltage will be dropped internally (or very little) so they don't have to account for it.

All of this has only to do with the trivial case of connecting the FG to the scope.  Not an interesting scenario.  If the signal is ultimately going to go to a project, you still have two choices:  First, your project can provide the 50 Ohm load.  Usually this is a 50 Ohm resistor with the project connected in parallel (across the resistor).  This would be the case if the signal was headed toward an op amp, for example.  If there is a need to watch the signal with a scope, put a T in the middle of the coax run, connect to the scope and leave the scope on Hi-Z.  It won't load the FG but the project will.  The termination is at the project, not the scope.

The second approach would be to take the signal off the coax into a high impedance circuit with a T and terminate the T.  This might be the case with a test lead adapter (mini-hooks).  This stub should be kept short to avoid reflectance.  But the coax itself is still terminated and the termination still matches what the FG expects.   Again, you can tap the scope into the coax along the way as long as it is Hi-Z.

I suppose you could run the coax to your project as a Hi-Z tap and then continue on to the scope and have it terminate the coax with its internal 50 Ohm load.

Any way you cut it, you should terminate the coax at the far end of the cable and you should terminate it only once.

As long as the project or the coax provides the termination, the scope can probe anywhere along the run and get the right answer.

All subject to change at high frequencies...

[cvriv]

I swear to god! I went back to it and tried again and now the scope isn't reading double! It's dead on now. I double checked and triple checked to make sure everything is right. Just to verify:

The function generator is set to 50 ohm impedance and 1Vpp.

The scope probe ratio is set to x10 and the scope input impedance is set to 1M ohm.

The pot is dialed in at 50.2 ohms.

I measured the Vrms across the pot with the DMM and got 350mVrms, which is correct. This means there's 350mVrms across the function generators 50 ohm impedance as well.

I put the probe back on the pot exactly the way it's shown in the last pic I uploaded and now..... it's showing 1Vpp and 350mVrms! What the hell! LOL. I swear to god it was showing double of everything! Before when it was showing double, I was to to up the FG's imperdance to "HighZ" to prevent the FG from doubling the output, but now I'm not.

I'm going to try again in a bit just make sure I'm not losing my mind here. I know the difference between Vpp and rms. RMS = Vpk/ root 2. I'm goign to watch that video posted just for the hell of it. I think I have been pretty close to understanding the whole time, I just think that I was confusing myself but not hooking up the probe correctly and because I didnt know what was happening while messing around with the settings.

[cvriv]

Here... see for yourself. The pot is set to 50 ohms. When I remove the pot and probe just the function generator, I then get double the output on the scope. If I change the FG's Z to HighZ I then get 1:1 output on the scope. Problem friggin solved! Thank god. I get this now.