Output of Clarke Transform Stationary Frame is Double the Input

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Output of Clarke Transform Stationary Frame is Double the Input

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derekcook
Senior Contributor I

Hello, 

It looks like the amplitude of the Alpha frame of the Clarke Transform is double the instantaneous phase current A. Any idea why this would be?

 

Below I have plotted the Isa from the Feedback Output, and the Alpha from the Clarke output scaled from -1 to 1. The update rate of my trender is 20ms, so the sinusoidal wave is a bit choppy since this is occurring at 250us.

 

The green wave is the Alpha stationary frame, the blue wave is the Isa feedback output. The amplitude of the green wave (Alpha) is -0.4 to 0.4 normalized, and the blue(Isa)  is -0.2 to 0.2 normalized.

 

 pastedImage_1.png

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linestream-adam
Senior Contributor I

Derek,

The form of the Clarke transform used by KMS is in the image below:

2018-08-03 21_12_30-Alpha–beta transformation - Wikipedia.png

The only time when the A phase will exactly equal the alpha phase is when the angle is 0, because in that case all of the current is flowing through the A phase and the value in the B and C phases is half and the opposite sign of the current in the A phase. This is a special case condition and it not the common condition.

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derekcook
Senior Contributor I

Adam, 

So why would the alpha frame consistently be double the A, B and C phase Current? I am reading the A, B and C [phase current going into the Clarke block, and the alpha frame is consistently double coming out.

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linestream-adam
Senior Contributor I

The ABC phase currents are stored in a Q14 container, while the alpha beta currents are stored in Q24 container. I suspect there is a q-factor issue.

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derekcook
Senior Contributor I

Hey Adam, 

It looks like the Clarke output, Feedback Output Isa, Isb, and Isc , and park output are all _sq variables? I am treating them as such by diving by 16384 to convert them to float. Is this not correct?

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linestream-adam
Senior Contributor I

The alpha beta currents should be divided by 32768 to convert to float. I was incorrect in my previous post, they are Q15 not Q24.

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derekcook
Senior Contributor I

Great! That explains the problem. The D and Q currents must also be a Q15? This is why they are off by a factor of 2 as well. Also do the D and Q current represent peak current? So they would be a sqrt(2) greater than the RMS phase current?

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linestream-adam
Senior Contributor I

D and Q currents are also stored as Q15. The Q current should be sqrt(2) greater than RMS phase current. The D current should be approximately 0. 

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