I am assuming you are trying to get the DC bus voltage for a grid-tied inverter. The answer to this depends on the type of converter used. The most common one will be voltage source converter, which I will based on answer on.
For voltage source converter, the DC-bus voltage will decide the maximum AC voltage that can be generated. A general guideline will be that the minimum DC voltage should be equal to the maximum line-to-line voltage of the grid.
For three-phase grid, this will be the peak value of the line-to-line voltage. For eg, 415V system will require minimum Vdc of 415*sqrt(2) = 587 V. (the multiplication of sqrt(2) is to convert from rms to peak value).
For single-phase system, the maximum line-to-line voltage needs to be generated by the inverter is simply the grid's line-to neutral voltage. So, for a 240V sistem, the minimum Vdc will be 240*sqrt(2) = 340 V.
I am assuming you are trying to get the DC bus voltage for a grid-tied inverter. The answer to this depends on the type of converter used. The most common one will be voltage source converter, which I will based on answer on.
For voltage source converter, the DC-bus voltage will decide the maximum AC voltage that can be generated. A general guideline will be that the minimum DC voltage should be equal to the maximum line-to-line voltage of the grid.
For three-phase grid, this will be the peak value of the line-to-line voltage. For eg, 415V system will require minimum Vdc of 415*sqrt(2) = 587 V. (the multiplication of sqrt(2) is to convert from rms to peak value).
For single-phase system, the maximum line-to-line voltage needs to be generated by the inverter is simply the grid's line-to neutral voltage. So, for a 240V sistem, the minimum Vdc will be 240*sqrt(2) = 340 V.
Normally we select the dc bus voltage value based on the type of converter used as already explained by Hangseng Che.
Because We have one of PV Plant having capacity 215 KW. currently one of it's inverter values are AC voltage 397 L-L, and DC voltage 570Vdc aprox
if i take peak value of 397 it becomes 569 which is DC. So according to theory and practical as well it's true to calculate dc bus bar voltage with the help of formula shared above.
i agree with above answers. but DC bus voltage level shall be atleast +30V on V Line to Line * 1.414 Example for 415V AC, we need DC bus = 415 *1.414+30 = 616V.
I need someone to comment if my approach is wrong in any sense.
I assumed the inverter topology as a 3-phase 3 leg full bridge inverter.
The peak of the average inverter pole voltage is m*Vdc/2 and the average common mode voltage (It is measured between load neutral and dc link mid point) is zero. Thus, to have a control over the current through interfacing inductor, the pole voltage should not be less than the peak of grid phase voltage i.e., 325 V. For worst case, Vdc/2 = 325 V and it gives the required minimum dc link voltage as 650 V.
Thank you.
Dear Lokesh,
Yes we have developed 5kw 3phase solar grid tie inverter with SVPWM + 3-phase 3 leg full bridge + two level . There Neural is mid point of DC link. We need to apply around 600V DC link voltage to produce 400V AC RMS. i think this will help you to clear your doubt. Please send mail me [email protected] for any clarifications.
Practically, it is 1.5 times of the actual line-line voltage.
Example 415*1.5=622.5
For three phase three wire (3P3W) distribution network the reference DC link voltage of any voltage source converter can be calculated with the help of below mentioned formula.
2*1.414*PCC Voltage (line to line)/1.737*m.
Where, m is the modulation index and, for 3P3W distribution network, it is selected 1.
The DC link voltage is choosen greater than 1.63 times the Line to Line voltage according to following formula.
Vdc >(1.15 * VL * √2)÷m
m=1-modulation index
So it should be greater than 646V.
You can refer---
EMADI A., “Energy Efficient Electric Motors”, CRC Press, page no. 393, 2004.
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