What is the purpose of the phase shift test in a black box?

[Turion]

Homework Statement

So in the future, I will have to find out what is inside a black box:

A Black Box is actually a grey metal box with two terminals sticking out of it. Inside are two passive components, one is a resistor and the other may be a capacitor or an inductor, and they way they are connected may be either in series or in parallel. The point of the test is to use a multimeter, a function generator, and an oscilloscope to determine what components are inside, how they are connected, and their values, to within about 10%. There are four possible cases; RC-series, RC-parallel, RL-series and RL-parallel.

Homework Equations

The Attempt at a Solution

Here is the solution to it: http://drunkenengineer.tripod.com/Year1/ele202/blackbox/blackbox.html

The issue is that I'm having trouble understanding why the guide uses those techniques/tests.

I understand the ohmmeter test. The reason why the ohmmeter would give a reading of infinity if it is a rc-series box is because capacitors act as open circuits to DC.

Questions:

1. Phase Shift Test: we have to make a circuit. It consists of a signal generator, an oscilloscope, a black box, and the attached symbol. The attached symbol is just a signal generator, right? And functions generators are the same as signal generators, right? The shape of the circuit looks weird. What is the motive behind the circuit designed like this?
2. Inductor test: I don't understand the circuit diagram. A and B are two different channels, right? So where would the other ends of the channels be connected? What is the motive behind the circuit designed like this?

 

[rude man]

Their hookup diagram is somewhat strange all right.

They seemingly show Ya and Yb both hooked to ground and to each other! But probably what is meant is the cables from the signal generator and the oscilloscope are all coax with the outer conductors (the braids) going to a common tie-point we can call "ground". You don't need to actually connect the braids to a water pipe or other "earth ground".

Actually, the 'scope inputs should be via 1 Meg or 10 Meg 'scope probes with the signal generator coax inner conductor = Ya and the braid connected to "ground". The clip wires from the 'scope probes will also go to ground as will the low side of the 20 ohm resistor.

There is a mistake in one of the four phasor diagrams. Under "RL parallel" it shows V_L lagging V_R by 90 deg whears in reality the two are obviously in phase (phase angle = 0). And it shows V_L getting smaller for larger w which is also wrong. But all four bottom waveforms are correct so concentrate on those.

EDIT: the more I look at the phasors, the more mistakes I see. I would ignore the phasor diagrams completely.

 

[Turion]

Thank you so much for responding. Sorry if I seem like a beginner. This is my first course in Electric Circuit Analysis and the rate that they're teaching concepts makes it hard to understand what is exactly going on.

Their hookup diagram is somewhat strange all right.

They seemingly show Ya and Yb both hooked to ground and to each other! But probably what is meant is the cables from the signal generator and the oscilloscope are all coax with the outer conductors (the braids) going to a common tie-point we can call "ground". You don't need to actually connect the braids to a water pipe or other "earth ground".

I'm not sure if I understand the diagram. I've redrawn it using what I think I understand. It is attached as lab-test.png. So channel A essentially measures the voltage across the black box, and channel B measures the voltage across the the 20 ohm resistor, right? I still don't understand what that symbol means with the sine curve inside it (see attachment in first post). I've inserted a filler symbol for the lab-test.png diagram I drew.

What exactly do you mean by "cables from the signal generator and the oscilloscope are all coax"? What does it mean for cables to be coax? Does it mean that they are connected together? The braids are just ground, right?

In the instructions, it reads:

Display both channels on the scope and trigger off Channel A, set both channels to AC coupling. Adjust the time divisions to display the largest full wave of channel A. Play with the frequency until you get a good separation between the channel A and channel B waveforms.

Why exactly are we setting it to AC coupling? Don't you use DC coupling to measure both DC and AC? AC coupling is only when there is an AC voltage superimposed on a DC voltage, right? There is no DC source in the circuit.

Then it reads:

VA = channel A = voltage across test circuit
VB = channel B = current through test circuit

As I said earlier, I thought "channel B measures the voltage across the the 20 ohm resistor." I don't understand how they converted it to current.

There is a mistake in one of the four phasor diagrams. Under "RL parallel" it shows V_L lagging V_R by 90 deg whears in reality the two are obviously in phase (phase angle = 0). And it shows V_L getting smaller for larger w which is also wrong. But all four bottom waveforms are correct so concentrate on those.

EDIT: the more I look at the phasors, the more mistakes I see. I would ignore the phasor diagrams completely.

So just to make sure I understand you correctly, you're saying to ignore the stuff outlined in red in the attachment (Untitled.png), right?

 

[rude man]

Thank you so much for responding. Sorry if I seem like a beginner. This is my first course in Electric Circuit Analysis and the rate that they're teaching concepts makes it hard to understand what is exactly going on.



I'm not sure if I understand the diagram. I've redrawn it using what I think I understand. It is attached as lab-test.png. So channel A essentially measures the voltage across the black box, and channel B measures the voltage across the the 20 ohm resistor, right? I still don't understand what that symbol means with the sine curve inside it (see attachment in first post). I've inserted a filler symbol for the lab-test.png diagram I drew.

Your diagram confuses me. The original is better. I would say redraw the original but leave out the "ground" wires. You show two generators when there is only one. Yes, channel 1 to the 'scope is the voltage across the box, but actually it includes the 20 ohm resistor but you are to think of this as a wire as far as the voltage is concerned. Ch. 2 to the 'scope is the voltage across the 20 ohm which represents the current thru the box (=V_ch.2/20).

The symbol you speak of is the signal generator. A function generator is the same thing. It's called 'function' because it can generate signals other than sine waves (e.g. triangular, square, etc.) but you will be using strictly sine waves.

What exactly do you mean by "cables from the signal generator and the oscilloscope are all coax"? What does it mean for cables to be coax? Does it mean that they are connected together? The braids are just ground, right?

Sorry. "Coax" is short for "coaxial". The cable consists of an inner conductor and an outside shield or braid. Have you ever looked at 75 ohm TV or cable tv coax? Same thing. The 'scope probes are also coax. The braid connection is usually a short bit of wire with an alligator clip. The "hot" side is connected by clips built into the probes, usually.

In the instructions, it reads:

?

Why exactly are we setting it to AC coupling? Don't you use DC coupling to measure both DC and AC? AC coupling is only when there is an AC voltage superimposed on a DC voltage, right? There is no DC source in the circuit.

You are not looking for dc response. You are using a function generator that can produce low-frequency sine waves but not dc (make sure the "bias" is set to zero offset which means zero Vdc.

Then it reads:

?

As I said earlier, I thought "channel B measures the voltage across the the 20 ohm resistor." I don't understand how they converted it to current.

Current = voltage/resistance. You are measuring the current by measuring the voltage across the 20 ohm and converting it to current: i = V/R.

So just to make sure I understand you correctly, you're saying to ignore the stuff outlined in red in the attachment (Untitled.png), right?[/QUOTE]

Yes, absolutely. Whoever came up with that section must have been in an awful hurry.

 

[Turion]

Hmm... it seems that I have a major problem understanding this diagram.

You show two generators when there is only one.

So you are saying that the original diagram only has one function generator? I've attached the original diagram with the two function generators that I see.

 

[rude man]

Hmm... it seems that I have a major problem understanding this diagram.

So you are saying that the original diagram only has one function generator? I've attached the original diagram with the two function generators that I see.

There is only one generator. Why they show it twice is another mystery. Maybe they wanted to emphasize the schematic nature of the generator.

Remove the wiring where it says : ... added for simplicity ...". Those wires look like they're shorting everything out! But be aware that you need to run the low side of the generator, the 20 ohm resistor and the two scope probe braids to a common tie-point called "ground".

 

[Turion]

Alright, here is my next attempt to redraw it. Is this any good? I've removed ground.

 

[rude man]

Still not right.

Do you have PowerPoint? I could maybe whip up something.

 

[rude man]

Never mind. Look at this.

 

[Turion]

Never mind. Look at this.

Really? That's it? They made it look much more complicated to be honest. They did such a poor job drawing something so simple. So it looks like the resistor is in series with the black box?

I think I get that part of the guide now. Thanks!

 

[Turion]

One quick question: you wrote "B channel measures i = V/20 ohms."

You mean channel B measures V and we vertically compress the sine wave by a factor of 20 to get current, right?

 

[rude man]

One quick question: you wrote "B channel measures i = V/20 ohms."

You mean channel B measures V and we vertically compress the sine wave by a factor of 20 to get current, right?

You can't talk of " ... compressing a sine wave by a factor of 20 ...". You are measuring a voltage across the 20 ohm and by Ohm's law, i = V/R.You read V on the scope B channel and you know 20 ohms so you know i.

 

[rude man]

Really? That's it? They made it look much more complicated to be honest. They did such a poor job drawing something so simple. So it looks like the resistor is in series with the black box?

Yes. If you think about it, it's going to modify the impedance inside the black box but the idea is it's so small that it's almost just a wire compared to the resistance inside your blk box.