Industrial electronics. Question with coils and diode.

[Femme_physics]

Sorry it took me so long!


Well, I figured out when to use which formula, it all depends on the phase!

http://img560.imageshack.us/img560/5786/figuredit.jpg

And the Umax in the case of Vin

http://img812.imageshack.us/img812/1227/secondtryy.jpg [/QUOTE]

Should be

http://img641.imageshack.us/img641/8950/umaxxor.jpg

Voila! Got it. I'm on a roll!

Conventionally the turns ratio of a transformer is defined to be the ratio of secondary turns to primary turns, n = Ns/Np. Whether or not FP's k value follows this convention is something that she should check with her text or class notes.

Again, sorry it took me a while, but the answer I got was that it's the PRIMARY divided by the secondary. In fact, this is what "p" and "s" stand for (p = primary ; s = secondary)

Now, back to business

Yes, but this is only when the current in the primary circuit is flowing as you indicated.
It will still also flow the other way.

I don't see how it can flow the other way.

See?

http://img14.imageshack.us/img14/2126/bumpsc.jpg

 

[I like Serena]

Hi Fp!

Glad you got all the issues of this problem resolved.

So the max voltage is indeed 230 \sqrt 2.
It would be nice if you marked that in the graph for Vin, Vp and Vs.

And now we know that the voltage is indeed transformed down by a factor of 2.

I don't see how it can flow the other way.

See?

In the secondary circuit the current does indeed *bump* against the diode and won't flow.

However, the primary circuit does not have this problem, and the current will flow there.
And this will still induce a voltage in the secondary circuit, which will be counterbalanced by the diode, preventing the current to flow.

 

[Femme_physics]

Hi ILS! whew, refreshing to see you here again... <3 it's good to see engines are back to normal state, cap'n, even temporarily^^

It would be nice if you marked that in the graph for Vin, Vp and Vs.

Will do, as soon as I figure out how to draw them correctly

And now we know that the voltage is indeed transformed down by a factor of 2.

Yep!

In the secondary circuit the current does indeed *bump* against the diode and won't flow.

Once again I must

However, the primary circuit does not have this problem, and the current will flow there.
And this will still induce a voltage in the secondary circuit, which will be counterbalanced by the diode, preventing the current to flow.

Hold on, we're talking about Vs here. How does Vs plays a role? The way I see it, the current that flows in the first circuit, doesn't even touch the secondary coil, it just flows through the first coil. The only current that flows through Vs, is in the position where the diode allows it to flow. In reality, what I drew with the *bump* does not exist, since it's not a closed loop, so there is no current flowing and actually bumping the diode.

 

[I like Serena]

Hold on, we're talking about Vs here. How does Vs plays a role? The way I see it, the current that flows in the first circuit, doesn't even touch the secondary coil, it just flows through the first coil.

First things first, so you agree that the current flows in the primary circuit?



And no, the current in the primary circuit does not touch the secondary coil, it just flows through the primary coil.
However, the current does induce a magnetic field that does touch the secondary coil.
In turn in the secondary coil a voltage Vs is induced (across the secondary coil).

Here's a (schematical) picture:

The only current that flows through Vs, is in the position where the diode allows it to flow.

Current flowing through Vs?

Current does not flow through Vs.
Current flows through the secondary circuit.
Vs is the voltage in the secondary circuit.
A voltage difference across a resistor causes current to flow through the resistor.

In reality, what I drew with the *bump* does not exist, since it's not a closed loop, so there is no current flowing and actually bumping the diode.

Correct.

So the voltage difference across the resistors will be zero.

 

[Femme_physics]

First things first, so you agree that the current flows in the primary circuit?

Yes.

And no, the current in the primary circuit does not touch the secondary coil, it just flows through the primary coil.

I agree, that's what I thought!

However, the current does induce a magnetic field that does touch the secondary coil.
In turn in the secondary coil a voltage Vs is induced (across the secondary coil).

Duly noted, thanks for that!

Current flowing through Vs?

Current does not flow through Vs.
Current flows through the secondary circuit.
Vs is the voltage in the secondary circuit.

Oh. Ok, checked!

So, let me just take a look at the second "circuit".

http://img708.imageshack.us/img708/4882/vsss.jpg If this is correct then my graph for the wave should be correct, because when the current changes direction there is no flow. So there is indeed only half a phase. No?

So the voltage difference across the resistors will be zero.

No it won't. There is still current flowing in the other direction! So there are voltage differences across the resistors. Or "voltage drop" as we know it.

 

[I like Serena]

So, let me just take a look at the second "circuit".


If this is correct then my graph for the wave should be correct, because when the current changes direction there is no flow. So there is indeed only half a phase. No?

You seem to be mixing up voltage and current.
You're concentrating on the current and what you have drawn is indeed the graph for the current.
The graph for the voltage Vs across the coil is different.

Look at it this way.
Suppose we replace the secondary coil by a battery giving off a voltage.
If the plus pole is on the bottom, the current will *bump* against the diode and no current will flow.
The battery still gives off its voltage though.

No it won't. There is still current flowing in the other direction! So there are voltage differences across the resistors. Or "voltage drop" as we know it.

What I meant is that there are 2 states for the voltage Vs: it's positive or it's negative.
When it is positive there is indeed a voltage difference across the resistors and as you say current will flow.
When it is negative, no current will flow, so there won't be a voltage across the resistors at that time.

 

[I like Serena]

Here's an analogy with mechanics.

Suppose you have a sliding block on a slope.
When we jiggle the slope back an forth, the block will alternately slide left and right.
However, if you insert a stick in the slope on the right side of the block, it won't slide to the right, but there is still a slope.

What's the "current" in this analogy?
And what is the "voltage"?

 

[Femme_physics]

You seem to be mixing up voltage and current.
You're concentrating on the current and what you have drawn is indeed the graph for the current.
The graph for the voltage Vs across the coil is different.

Look at it this way.
Suppose we replace the secondary coil by a battery giving off a voltage.
If the plus pole is on the bottom, the current will *bump* against the diode and no current will flow.
The battery still gives off its voltage though.

Good point, so Vs is a sinus wave!

What's its Umax though...hmm... that has to be calculated.

Actually, I'd presume it's also 230V since the first circuit doesn't have any resistors. Could it be?

What I meant is that there are 2 states for the voltage Vs: it's positive or it's negative.
When it is positive there is indeed a voltage difference across the resistors and as you say current will flow.
When it is negative, no current will flow, so there won't be a voltage across the resistors at that time.

Oh, OK, agreed, makes sense

Here's an analogy with mechanics.

Suppose you have a sliding block on a slope.
When we jiggle the slope back an forth, the block will alternately slide left and right.
However, if you insert a stick in the slope on the right side of the block, it won't slide to the right, but there is still a slope.

What's the "current" in this analogy?
And what is the "voltage"?

voltage is the jiggle, block is current, I think. But I really understood what you mean with your post above it! Thanks. I'll work on the solution of the thing soon.

 

[I like Serena]

Good point, so Vs is a sinus wave!

What's its Umax though...hmm... that has to be calculated.

Actually, I'd presume it's also 230V since the first circuit doesn't have any resistors. Could it be?

Umax is not 230 V. You're talking about Urms here.

And you're forgetting that the secondary coil (where we have Vs) has less windings, causing the voltage to be lower.

voltage is the jiggle, block is current, I think. But I really understood what you mean with your post above it! Thanks. I'll work on the solution of the thing soon.

Good. :)

A little sharper:

The block is the electron.
The current is the speed of the block.

The voltage difference is the difference in height between two points on the slope (left and right of the block).

 

[Femme_physics]

Top of the morning ta ye ILS :)

Umax is not 230 V. You're talking about Urms here.

Oops, you're right *smackey the foreheady*

And you're forgetting that the secondary coil (where we have Vs) has less windings, causing the voltage to be lower.

Oops, you're also right! So since k =2

I'd do

230 / 2 = 115

So Vs(max) = 115(square root of 2)

Aye, cap'n?

Good. :)

A little sharper:

The block is the electron.
The current is the speed of the block.

The voltage difference is the difference in height between two points on the slope (left and right of the block).

Woah. Ok, mmm, as long as mechanics analogy won't be in the test, we're safe


Ok, so now I figured

Vin, Vp an Vs. Now I need Vxy

Which according to my logic

Since Vs = 115V

Vyx also equals 115V, because if we treat the diode as an ideal diode and ignore the voltage drop there, using Kirchhoff law we know that the voltage drops in the resistors (RT) must equals the voltage source. Ergo,

Vs = Vyx


*does a suspicious victory dance waiting for approval*

 

[I like Serena]

Top of the morning ta ye ILS :)

Hi Fp!

So Vs(max) = 115(square root of 2)

Aye, cap'n?

Yep!

Ok, so now I figured

Vin, Vp an Vs. Now I need Vxy

Which according to my logic

Since Vs = 115V

Vyx also equals 115V, because if we treat the diode as an ideal diode and ignore the voltage drop there, using Kirchhoff law we know that the voltage drops in the resistors (RT) must equals the voltage source. Ergo,

Vs = Vyx


*does a suspicious victory dance waiting for approval*

Not quite.
You have to distinguish the two cases, where Vs is positive, and where Vs is negative.
When Vs is positive you are right.
When Vs is negative, it is different.

 

[Femme_physics]

Hi Fp!

Yep!

Did you edit it by any chance?

Not quite.
You have to distinguish the two cases, where Vs is positive, and where Vs is negative.
When Vs is positive you are right.
When Vs is negative, it is different.

Ah, well, that's why we have graphs

http://img683.imageshack.us/img683/1575/vxywt.jpg

Did I get it?

 

[I like Serena]

Yep to all!

 

[Femme_physics]

http://img194.imageshack.us/img194/7199/indiet.jpg

Are these graphs correct? (my dashed lines really mean no line)

 

[gneill]

Hi FP, you may wish to think about the relative phase of the primary and secondary waveforms; how do the transformer 'dots' affect the phase?

While this phase difference won't make any practical difference in the operation of this simple circuit, it's worthwhile making note of it since you're drawing Voltage vs Time graphs.

 

[Femme_physics]

Hi FP, you may wish to think about the relative phase of the primary and secondary waveforms; how do the transformer 'dots' affect the phase?

Well it reverses the phase on Vs. But I don't know how to show the difference in a graph though since Vp is both directions anyway, and Vs is one direction.

I hope I understood what you meant.

 

[gneill]

Well it reverses the phase on Vs. But I don't know how to show the difference in a graph though since Vp is both directions anyway, and Vs is one direction.

I hope I understood what you meant.

The difference shows up as a reversal of polarity in the secondary voltage waveform with respect to the primary's waveform. This is equivalent to a half wavelength shift in time for the secondary voltage waveform. It boils down to shifting your graph's peaks over by a half wavelength (either direction will do!).

Note that this is only necessary if you are showing voltages in both the primary and secondary circuits on the same time axes.

All your analysis conclusions still apply, only the time-wise locations of the peaks are shifted over.

 

[Femme_physics]

The difference shows up as a reversal of polarity in the secondary voltage waveform with respect to the primary's waveform. This is equivalent to a half wavelength shift in time for the secondary voltage waveform. It boils down to shifting your graph's peaks over by a half wavelength (either direction will do!).

Note that this is only necessary if you are showing voltages in both the primary and secondary circuits on the same time axes.


All your analysis conclusions still apply, only the time-wise locations of the peaks are shifted over.

Ah, makes perfect sense I get it now.

All your analysis conclusions still apply, only the time-wise locations of the peaks are shifted over.

I'll fix it then in my next scan this evening or tomorrow. Thanks

 

[I like Serena]

Morning Fp!

Looking good!
I see you have most graphs now. Nicely labeled with the voltage.
Still Vs is along the dotted line, while Vxy is without the dotted line.

Btw, you have to consider the 2 half periods separately.

In one half your current and power are correct.

But in the other half they are zero.