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.