# Designing a DC link Capacitor

Discussion in 'Electronics Homework Help' started by diya0908, Feb 27, 2013.

1. ### diya0908

7
0
Feb 27, 2013
hi, I would appreciate is some one could help me with this question by giving me clues about what formula i could use and how to proceed.

Ques:

The DC supply for the voltage source inverter of the VFD is derived using six pulse diode converter.
Supply voltage is 600V, 3phase, 50Hz. Xdāā=12.5 pu ,X/R ratio as 20. Model the load seen by the DC link capacitor as pure resistor. Take rated load current as 900A.
(i) Design DC link capacitor to get DC voltage with 10% ripple at rated load. (Hint : use simplified formula).
(ii) Simulate the above system. (Assume suitably any missing data)
(iii) Draw the circuit diagram.

2. ### Miguel Lopez

252
64
Jan 25, 2012
Do you mean a three phase bridge rectifier supplying a pure resistive load? 600V - 900A? Do you need to calculate the smoothing capacitor for 10% ripple?

3. ### diya0908

7
0
Feb 27, 2013
yes sir, i myself am struggling to understand the question, anyhow I feel it is a 3phase six pulse diode rectifier supplying a resistive load and having a supply voltage of 600V.

the first part of the quest asks to design a capacitor, but where will it be placed?? and how do we calculate it?

4. ### diya0908

7
0
Feb 27, 2013

Ok so now i get it, the dc link capacitor is placed after the rectifier. But the "load seen by the DC link capacitor is a pur resistor"-- could you explain me what does this mean.

5. ### Miguel Lopez

252
64
Jan 25, 2012
I guess we're talking about something like this. The value of the capacitor is not correct. I just wrote any number. Anyway in this case it means that you will supply a load of only 1,55 ohms. I think the 900A value is too high.

Althought the 3 phase bridge rectifier gives a decent DC level at the output, to have 10% of ripple it needs a huge amount of microfarads. You should check this:

http://en.wikipedia.org/wiki/Rectifier#Three-phase_bridge_rectifier
http://en.wikipedia.org/wiki/Ripple_(electrical)

It means that voltage and current will be in phase. No reactive effect (inductive or capacitive). The ripple can cause the reactances to behave different to a pure resistor.
http://en.wikipedia.org/wiki/Electrical_reactance

#### Attached Files:

• ###### 3N Rectifier.jpg
File size:
26.8 KB
Views:
265
Last edited: Feb 27, 2013
6. ### diya0908

7
0
Feb 27, 2013
yes sir, you are correct, the value of current is too high and the value of resistance is therefore low. I have gone through the links and i believe now i know how to find the value of the DC link Capacitor (C=I/Vf)

now my question is what does Xd''=12.5 and X/R ratio mean??

7. ### diya0908

7
0
Feb 27, 2013
Also i would like to know how did you calculate the value of resistance in your diagram

8. ### Miguel Lopez

252
64
Jan 25, 2012
You can use Ohm's Law, but to be honest, in the diagram I used "trial and error" in the simulation software.

R = V / I

http://en.wikipedia.org/wiki/Ohm's_law

9. ### diya0908

7
0
Feb 27, 2013
Could you tell me which program you have used to simulate the circuit.