• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

How to see phase shift due to inductor in scope?

Status
Not open for further replies.

suhasm

New Member
I recently purchased an analog oscilloscope. I wish to see the phase shift between V and I due to an inductor on it. How do i do that?

I will connect the lead of one channel across the inductor. That should give me the V graph.

But how do i get the current graph?
Can i connect a resistor in series with the inductor as a ItoV converter and then take the voltage across it?
 

OutToLunch

New Member
if you have a differential probe, know the inductance and the DCR of the inductor in question, then you can set up an RC across the inductor with:

RC = L/DCR

then differentially measure the voltage across the capacitor and that will represent the voltage across the distributed DCR of the inductor. The current would then be Vmeasured/DCR.

Measuring across a resistor in series with the inductor would work as well, but keep in mind that you are introducing an element to the circuit that will not be there except for measurement purposes.
 

Willbe

New Member
Yes, you need a two channel scope. You might need to isolate the power grounds.
 

crutschow

Well-Known Member
Most Helpful Member
Yes. If you connect a small resistor to ground in series with the inductor, then you can read the voltage across the resistor to give you current.

Size the resistor to give you a reasonable display amplitude at the most sensitive setting on your scope input. That will make the resistor as small as possible so it will have minimum effect on the circuit operation.
 

MikeMl

Well-Known Member
Most Helpful Member
Here is how I would do it: Say you have a 1mH inductor and you want to see I and V at 1000Hz. I assume you have a function generator with one side of its output grounded (like a BNC jack). You have a two-channel scope where the ground side of the two inputs is also grounded. Connect a 0.1 or 1 Ohm resistor in series with the inductor.

The voltage across the resistor V(current) is proportional to the current through the resistor. This is also the current through the inductor. It will be a small voltage (on purpose, see below). RANT: This is an example of OHM's LAW as I use it!!!!

Because of the common grounds, the voltage at Chan A is the sum of V(current) + V(voltage). Since V(current)<<V(Voltage), then Chan A is a good approximation of the voltage across the inductor.

Note the phase shift between the two traces....
 

Attachments

MrAl

Well-Known Member
Most Helpful Member
Hi,

What i have done in the past is connect a small value resistor (like 0.1 ohm) in
series with the inductor, then clamp the ground lead onto the center tap formed
with this resistor and the inductor, then measure voltage across the inductor
with one channel and inverted voltage across the resistor with the other channel.
If you have an 'invert' button on your scope you can invert the resistor channel
to see the actual wave, or else you have to invert it in your mind or just add
180 degrees in your mind.

The resistor does insert some inaccuracy into the measurement because if the
esr of the inductor is small then what you measure is actually the current through
an inductor in series with a resistor rather than just an inductor, but it usually
works to some degree so you can get some idea how the phase is shifted.
 

MikeMl

Well-Known Member
Most Helpful Member
Hi,

What i have done in the past is connect a small value resistor (like 0.1 ohm) in
series with the inductor, then clamp the ground lead onto the center tap formed
with this resistor and the inductor, then measure voltage across the inductor
with one channel and inverted voltage across the resistor with the other channel.
...
But what do you do if both your scope and your function generator are grounded to each other through their third prong on their respective AC line cords? Your generator would have to have floating outputs to do what you suggest.
 
Last edited:

Mikebits

Well-Known Member
Here is how I would do it: Say you have a 1mH inductor and you want to see I and V at 1000Hz. I assume you have a function generator with one side of its output grounded (like a BNC jack). You have a two-channel scope where the ground side of the two inputs is also grounded. Connect a 0.1 or 1 Ohm resistor in series with the inductor.

The voltage across the resistor V(current) is proportional to the current through the resistor. This is also the current through the inductor. It will be a small voltage (on purpose, see below). RANT: This is an example of OHM's LAW as I use it!!!!

Because of the common grounds, the voltage at Chan A is the sum of V(current) + V(voltage). Since V(current)<<V(Voltage), then Chan A is a good approximation of the voltage across the inductor.

Note the phase shift between the two traces....
Brilliant... :)
 

MrAl

Well-Known Member
Most Helpful Member
But what do you do if both your scope and your function generator are grounded to each other through their third prong on their respective AC line cords? Your generator would have to have floating outputs to do what you suggest.

Yeah, so?

My scope is not grounded to the ac line, and my function generator
has an internal power supply that uses a transformer and rectifier
so it is isolated also.
 

MikeMl

Well-Known Member
Most Helpful Member
My scope is not grounded to the ac line, and my function generator
has an internal power supply that uses a transformer and rectifier
so it is isolated also.
My HP function generator has a BNC output jack, so is referenced to it's own chassis. Even if you cut the third prong off the AC line cord, that does not isolate the ground side of the generator's output.

There is significant capacitance between the primary and secondary winding's in its power transformer. Since Neutral (white wire) of the AC line is grounded, while Line (black wire) has 120V AC on it, the winding capacitance acts as a capacitive voltage divider, putting about 60V of 60Hz as a common-mode signal between the generator output and earth ground.

Now the scope has exactly the same problem. Attempts at "floating" a scope chassis by cutting off the third prong usually also introduce a lot of 60Hz noise into measurements. You may get away with this at low audio frequencies, but the capacitive coupling through the 'scope's power transformer becomes significant at frequencies above a few tens of kHz.
 

MrAl

Well-Known Member
Most Helpful Member
Hi again,

Very interesting, and sorry your equipment gives you a problem like that.
My generator body is made of plastic, and apparently i dont get too much
hum into the measurements because i have used this technique for years.
If i do get hum then it must be drowned out by the actual signal.

Maybe my generator capacitance generates 60v and the scope also generates
60v and so it cancels out :)

BTW, a certain class power supply has double isolation so that helps too.

You might want to try an isolation transformer with your equipment.
 

Hayato

Member
The phase shift deppends on the resistor.

The phase of an impedance, Z, is given by Arctan(Im(Z)/Re(Z)).

So, lets say, if you have a 1H inductor and a 1 Ohm resistor, your phase shift will be Arctan(1/1) = 45°

If you don't have a resistor:
Arctan(1/0) = 90°

If your resistor is 1000 ohms:
Arctan(1/1000) ~= 0°
 

MikeMl

Well-Known Member
Most Helpful Member
Yes, but XL=2×pi×f×L, which is what you should be comparing to the resistance. For a 1mH inductor at 1kHz, XL=6.28Ω. The arctan of 6.28/0.1 is 89 deg, meaning that the resistor shifts the phase by less than 1 deg.
 

Hayato

Member
Indeed, I considered 1 rad/s.

What I'm trying to say is that "phase" is relative to the circuit. If a inductor shifts the phase by 50° in "Circuit A" it does not mean that it will shift the phase by 50° in "Circuit B".

And, ideally speaking, resistors do not shift phase. Only reactive components do.
 

chris54

New Member
On a slightly different note If the resistor is of a known value and we intorduce an inductor of an unknown value could one fixure out the value of the inductor based on the phase shift.
 

The Electrician

Active Member
There is significant capacitance between the primary and secondary winding's in its power transformer. Since Neutral (white wire) of the AC line is grounded, while Line (black wire) has 120V AC on it, the winding capacitance acts as a capacitive voltage divider, putting about 60V of 60Hz as a common-mode signal between the generator output and earth ground.

Now the scope has exactly the same problem. Attempts at "floating" a scope chassis by cutting off the third prong usually also introduce a lot of 60Hz noise into measurements. You may get away with this at low audio frequencies, but the capacitive coupling through the 'scope's power transformer becomes significant at frequencies above a few tens of kHz.
It's not the transformer inter-winding capacitance that's the problem.

A typical 100 to 200 VA power transformer will have about 200 pF inter-winding capacitance.

But, when I measure the capacitance between chassis and either prong on the line cord, I get 15,000 pF for an old Tek 465, and 3500 pF for a newer Agilent digital scope. It's those Y capacitors that do you in.

But, even with 120 VAC across 15,000 pF, that's only about 700 uA @ 60Hz injected into the .1 ohm sense resistor; you'd hardly notice it.
 

MrAl

Well-Known Member
Most Helpful Member
Indeed, I considered 1 rad/s.

What I'm trying to say is that "phase" is relative to the circuit. If a inductor shifts the phase by 50° in "Circuit A" it does not mean that it will shift the phase by 50° in "Circuit B".

And, ideally speaking, resistors do not shift phase. Only reactive components do.
Oi Tudo bem :)

Im pretty sure that what Mike meant there was that the 'addition' of the
resistor shifts the phase, not the resistor itself, meaning that you
will get a different overall phase shift depending on the value of the
resistor.
 

Hayato

Member
Oi Tudo bem :)

Im pretty sure that what Mike meant there was that the 'addition' of the
resistor shifts the phase, not the resistor itself, meaning that you
will get a different overall phase shift depending on the value of the
resistor.
:)

Yes, we were talking about the same thing.
 

Mikebits

Well-Known Member

MrAl

Well-Known Member
Most Helpful Member
Hello,

That's Brazilian for "hello, how are you, or what's up?".
 
Status
Not open for further replies.

Latest threads

EE World Online Articles

Loading
Top