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Resistance of an inverter from the DC link side?

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spencermun

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Hi, I'm trying to design a power inverter for solar panels. I'm out of school doing this on my own so any help would be appreciated.

It would be 450V going to a boost converter, to a DC link capacitor, and then the inverter, followed by an LCL filter.
The equation for a DC link capacitor, based on the target voltage ripple is:
C=duty/(R*Vripple*f_sw)
Where D is duty, f_sw is switching frequency, Vripple is the ripple (delta Vo/Vo) going into the inverter, and R is the resistance of the load. Well the load in this case is a switching inverter going into an LCL filter going into a 10 ohm load, so it's not intuitive as to what the resistance is.
My question is how do I calculate even a rough idea of what this resistance might be?'
If you need to know my specific design requirements, the output will be 2kVA on a 10 ohm load, target efficiency is 95%. Input is 450V (although could vary from 360V to 540V depending on sunlight) on a 10 Ohm load.
One thing I tried was ohms law: (450V)^2/(2000VA) = ~750 Ohms. I'm not sure if that's right.
 

BradtheRad

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It sounds as though this is about how to calculate the value of a smoothing capacitor?

A few uSec charge at several A, then a few uSec discharge at several A?

There is a different formula you might use, which calculates a Farad value, starting with a given timeframe, Amperes, and acceptable ripple V. You need to settle on a prospective value for all three variables, to find a Farad value.

Expect to need a gang of capacitors, in order to spread the Ampere flow going back and forth many times per second.
 

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