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calculation of start capacitor


martinez

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Hello Martinez

 

Aim to keep the capacitor current the same. That means you need to reduce the value of the capacitance as the frequency is increased, and you need to reduce the capacitance as the voltage is increased.

 

The current = v x 2 x pi x f x C

 

At the same voltage but 50Hz rather than 60Hz, I would select 100 x 5 / 6 = 83 uF

 

Best regards,

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Hello Martinez

 

The current through a capacitor always leads the voltage by 90 degrees.

There will be a variation in the effective phase shift due to the inductance/resistance of the start winding, but I do not believe that this will have a significant effect.

 

Best regards,

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