# Inverter question

#### PAB

Joined Jun 21, 2007
6
Can anyone help me with these questions
Problem: I have a transformer with a (pri) -star || (sec) -star/delta (producing 2 x 3 phase supplies 30 electrical degrees apart) this in turn suppling a rectifier which runs @ 100% duty cycle and converts these 6 phases to DC Voltage.

Question: i); What is the DC output voltage of the rectifier, (assuming there is no voltage drop accross the SCR's or anywhere in the system), and
ii); what is the ripple voltage and frequency of the DC output if the supply frequency is 50Hz.
iii); What capacitance (in Farads) would be required to remove this Ripple voltage.

The best way, I found out is to draw a phasor diagram, for star an delta and calculate phase angle and voltage amplitude

@ 380Vac phase to phase
(6 phases)

(sec) PhaseS STAR winding
A: 380 sin 0 = 0V
B: 380 sin 120 = 329V
C: 380 sin -120 = -329V (remmember this phase will be inverted)

(sec) Phases DELTA Winding
D: 380 sin 30 = 190V
E: 380 sin 150 = 190V
F: 380 sin -90 = -380V (remmember this phase will be inverted)

From this work out the complex number for each phase (with trig)

A: 0/_ 0 = 0

B: 329/_ 120 = (cos120 x 329) = -164.5 , (sin120 x 329) = j(284.9)

c: 329/_ -120 (remmember this phase will be inverted)
C: 329 /_ 120 = (cos120 x 329) = -164.5 , (sin120 x 329) = j(284.9)

D: 190/_ 30 = (cos30 x 190) = 164.5 , (sin30 x 190) = j(95)

E: 190/_ 150 = (cos150 x 190) = -164.5 , (sin150 x 190) = j(95)

F: 380/_ -90 (remmember this phase will be inverted)
F: 380/_ 90 = (cos90 x 380) = 0 , (sin90 x 380) = j(380)

= -329 , = j(1139.8)

from -329+j1139.8 we get (pythagoras theorem) = 1186Vdc

1186V will be the peak output voltage of the inverter across the DC bus.

remmember this is unfiltered so there will be a small ripple voltage.

To calculate this ripple

There will be an intersecting of the phases every 15degrees therefore:

(remmembering that @ 50Hz there will be 12 pulses every 0.02seconds)

per phase
(380 sin90) = 380V
(380 sin75) = 367.05V

Therefore Vripple = 12.95V or 3.4%ripple per phase

3.4% of 1186V = 40.32Vripple @ (50Hz x12)

Vripple= 40.32V @ 600Hz

Is this correct? How do I work out the Capacitance value I need to add?

#### Newton1Law

Joined Aug 22, 2007
10
This would be a twelve pulse rectifier and would have a minimum voltage of 0.966 per unit (based on the peak value of your voltage) and thus a 0.034 per unit ripple at 720 Hertzs. The capacitor you would need is approximately given by C>= 0.7 * (I load) / (Vmax ripple voltage allowed)*(Frequency). There is no amount of capacitance that would give you zero ripple.