3-phase rectifier questions

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,446
Hello...

I have some questions about 3-phase rectifiers.

First of all:

1- At the load, the voltage we measure there is a Line Voltage or a Phase Voltage?

2 - Then we have derived the equations for the Load Vavg for Half and Full wave 3-phase rectifiers and I ended up writing this on my notes books:

Half wave
- Not controlled:
\[\displaystyle{VLoad_{avg} = \frac{3\cdot\sqrt{3}}{2\cdot \pi}\cdot Vmax }\]

- Controlled:
\[\displaystyle{VLoad_{avg} = \frac{3\cdot\sqrt{3}}{2\cdot \pi}\cdot Vmax \cdot \cos\left ( \alpha \right )}\]


Full wave
- Not controlled
\[\displaystyle{VLoad_{avg} = \frac{3\cdot\sqrt{3}}{\pi}\cdot Vmax }\]

- Controlled:
\[\displaystyle{VLoad_{avg} = \frac{3\cdot\sqrt{3}}{\pi}\cdot Vmax \cdot \cos\left ( \alpha \right )}\]

But when I read Mohan's Book
upload_2018-2-18_18-34-34.png

and I can't see the relationship between equation 6-36 and the ones I have from my notes book! I mean, why I have one expression in my notes book, and Mohan's book equations are different?
 

MrAl

Joined Jun 17, 2014
6,410
Hi,

I like to call them either the Line to Line voltages or the Line to Neutral voltages.
You must know how to calculate the two from each other im sure.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,446
Hi,

I like to call them either the Line to Line voltages or the Line to Neutral voltages.
You must know how to calculate the two from each other im sure.
It's from an integral
\[\displaystyle{V_{avg} = 3\cdot \frac{1}{T}\cdot \int_{0}^{T}\, V\left ( \theta \right )\, d\theta}\]

I think that it is from that expression!
 

MrAl

Joined Jun 17, 2014
6,410
Hello...

I have some questions about 3-phase rectifiers.

First of all:

1- At the load, the voltage we measure there is a Line Voltage or a Phase Voltage?

2 - Then we have derived the equations for the Load Vavg for Half and Full wave 3-phase rectifiers and I ended up writing this on my notes books:

Half wave
- Not controlled:
\[\displaystyle{VLoad_{avg} = \frac{3\cdot\sqrt{3}}{2\cdot \pi}\cdot Vmax }\]

- Controlled:
\[\displaystyle{VLoad_{avg} = \frac{3\cdot\sqrt{3}}{2\cdot \pi}\cdot Vmax \cdot \cos\left ( \alpha \right )}\]


Full wave
- Not controlled
\[\displaystyle{VLoad_{avg} = \frac{3\cdot\sqrt{3}}{\pi}\cdot Vmax }\]

- Controlled:
\[\displaystyle{VLoad_{avg} = \frac{3\cdot\sqrt{3}}{\pi}\cdot Vmax \cdot \cos\left ( \alpha \right )}\]

But when I read Mohan's Book
View attachment 146346

and I can't see the relationship between equation 6-36 and the ones I have from my notes book! I mean, why I have one expression in my notes book, and Mohan's book equations are different?

Hello again,


I think i see what is happening here. I had to go through one calculation to see it.

If we call the line to neutral voltage VLN and the line to line voltage VLL then
VLL=sqrt(3)*VLN

With VLN=100v peak, that makes VLL=173.205 volts peak to six significant digits. That is the highest voltage that can be found in the system.
Now if we multiply that by 1.35, we get 233.827v peak, but the RMS value then is 233.827/sqrt(2)=165.341 volts RMS. He does not say this though and that's a fault i think. In fact, why multiply by sqrt(2) at all.

That (corrected) way looks ok, however i do not care for that way of doing things. The more straight forward method is to simply solve for the period of one 'hump' and integrate over that period., The crossover point of any two phases is 0.5 times the peak, so it's easy to solve for the two times t1 and t2, and for a 50Hz system these come out to:
t1=1/12 seconds,
t2=5/12 seconds.

We can then calculate the average voltage for sin(wt) over that period and the result is 1.65341 times the peak which matches the above calculation.

Just to be clear, with 100v peak line to neutral i get 173v line to line, and that leads to an average DC of 165v with three phase rectification. These numbers are rounded down to just 3 significant digits for simplicity.

If you can find some examples in that same book then you can follow that work too.
 
Last edited:

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,446
My problem and my confusion of all times since I started this graduation is the number of types of voltages found in 3-phase systems. And adding to that, the different voltage names each book decides to use.

What I mean is:
Vpeak, Vpeak-to-peak, Vavg, Vrms, Vamp. Then we have relations between Line-to-Line Voltages and Line to Neutral voltages (Line voltages and phase voltages as I am used to listen to my teachers) and on top of that we also have delta, Y and Z setups that also relate themselves with each other. And last but not the least we have formulas ready to use for triangular, sawtooth, rectangular, sinusoidal, pulse waveforms. That's a lot of formulas for so many voltages that I end up confusing myself with all them!

So now you know my struggle. And there is also another thing that helps my confusion which are some formulas we use in some classes and that are a bit different to other classes and as we are given those formulas, we end up not knowing why the small differences.

For instance, in the formula I posted above, I'm not sure why we have that '3' and the 1/T before the integral. I just know that it's something like that and I'm not even sure if the formula I posted is correct!

And in this case I also struggle to understand when we need to use a \[\sqrt{3}\] or when we don't need to use it!
 

MrAl

Joined Jun 17, 2014
6,410
My problem and my confusion of all times since I started this graduation is the number of types of voltages found in 3-phase systems. And adding to that, the different voltage names each book decides to use.

What I mean is:
Vpeak, Vpeak-to-peak, Vavg, Vrms, Vamp. Then we have relations between Line-to-Line Voltages and Line to Neutral voltages (Line voltages and phase voltages as I am used to listen to my teachers) and on top of that we also have delta, Y and Z setups that also relate themselves with each other. And last but not the least we have formulas ready to use for triangular, sawtooth, rectangular, sinusoidal, pulse waveforms. That's a lot of formulas for so many voltages that I end up confusing myself with all them!

So now you know my struggle. And there is also another thing that helps my confusion which are some formulas we use in some classes and that are a bit different to other classes and as we are given those formulas, we end up not knowing why the small differences.

For instance, in the formula I posted above, I'm not sure why we have that '3' and the 1/T before the integral. I just know that it's something like that and I'm not even sure if the formula I posted is correct!

And in this case I also struggle to understand when we need to use a \[\sqrt{3}\] or when we don't need to use it!

Hello,

Well sqrt(3) relates the line to neutral voltages VLN to the line to line voltages VLL.
VLL=sqrt(3)*VLN

I understand what you mean though, that happens with choice of variables too in different writings by different authors. Somehow you have to figure out a way to relate their stuff to what you know already.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,446
Hello,

Well sqrt(3) relates the line to neutral voltages VLN to the line to line voltages VLL.
VLL=sqrt(3)*VLN

I understand what you mean though, that happens with choice of variables too in different writings by different authors. Somehow you have to figure out a way to relate their stuff to what you know already.
So, about the formula I posted in post #3, is correct? If so, where does the '3' comes from?
 

MrAl

Joined Jun 17, 2014
6,410
So, about the formula I posted in post #3, is correct? If so, where does the '3' comes from?
Hello,

That '3' could come from assuming that T is one complete period.

But i think what you need to do is think about what average is, and plot the two three phase responses, one with no rectification and one with.
Take a look at the two plots here and see what you can make of it. Note the places where the phases become equal at the same times.
In one plot i had shown this as a horizontal light gray line.
The average will be somewhere between the level where they are the same and the peak.
The AC response is shown in pic 1 and the full wave rectified in pic 2. The full wave rectified response is all above zero of course.
 

Attachments

Last edited:

MrAl

Joined Jun 17, 2014
6,410
Hello again,

Here is a more direct approach.

We start with the three phases, A,B,C, then subtract A-B, B-C, A-C and then take the absolute value of those three.
We then solve for the two times t1 and t2 as shown in the attachment, and then integrate. The times there are shown for abs(A-B).
We then divide by t2-t1.
This results in the previous factor 1.65399.
The integration result is sqrt(3)/(2*pi), then divided by t2-t1 we get 3*sqrt(3)/pi.
 
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