SCR 3-phase full-bridge Rectifier Circuit

Thread Starter

claytonagould

Joined Mar 14, 2017
2

Circuit diagram attached, my question is that if the line voltages are given as such:
Vab = 200sin(2*pi*50t)
Vbc = 200sin(2*pi*50t - 120)
Vca = 200sin(2*pi*50t + 120)
How would i go about solving for the source Voltages Van, Vbn, and Vcn?
Edit: Picture feature not working, circuit attached.
 

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wayneh

Joined Sep 9, 2010
16,292
Look up star and delta configurations. In a star, the center point is "zero". I believe this is a delta but it's late.
 

MrAl

Joined Jun 17, 2014
7,196

Circuit diagram attached, my question is that if the line voltages are given as such:
Vab = 200sin(2*pi*50t)
Vbc = 200sin(2*pi*50t - 120)
Vca = 200sin(2*pi*50t + 120)
How would i go about solving for the source Voltages Van, Vbn, and Vcn?
Edit: Picture feature not working, circuit attached.
Hi,

Do you actually want to 'solve' this or just calculate it?

To calculate, there is the constant sqrt(3) which relates the two.

To solve, you can use the fact that:
Vab=Van-Vbn

and work that equation in terms of how the sinusoidals subtract.
 

Thread Starter

claytonagould

Joined Mar 14, 2017
2
Hi,

Do you actually want to 'solve' this or just calculate it?

To calculate, there is the constant sqrt(3) which relates the two.

To solve, you can use the fact that:
Vab=Van-Vbn

and work that equation in terms of how the sinusoidals subtract.
Thanks for the reply,
my end goal would be to solve for Van, Vbn, and Vcn in terms of sinusoidals but I keep getting stuck in the math when I try to substitute using the three equations that I came up with.
 

MrAl

Joined Jun 17, 2014
7,196
Thanks for the reply,
my end goal would be to solve for Van, Vbn, and Vcn in terms of sinusoidals but I keep getting stuck in the math when I try to substitute using the three equations that I came up with.
Hi,

Well you'll have to show some work so i can see what you are doing and where you are getting stuck.

The basic procedure, well one way to proceed, is to use exactly what is known:
Vab=Van-Vbn

and using the sinusoidal representations for Van and Vbn you'll get a result which will probably not be in the form you want, so you have to apply some trig identities for adding or subtracting sines and cosines to get it into the form you want which would be A*sin(wt+Ph) where you will have solved for A and for Ph, and the value of A is most important because that tells you the relationship between amplitudes of the line to neutral voltage and the line to line voltage.

In your case you have this:
Vab=A*sin(wt)-A*sin(wt-120)

or possibly this in the more general case:
Vab=A*sin(wt)-B*sin(wt-120)

If you solve those you'll have all the solutions you want, once you get it back into the form of C*sin(wt+Ph).
 
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