If 3-phase induction motor of 4kw, 400v, 1440rpm is driven by SVPWM technique using indirect vector control scheme, How DC link Voltage (supply voltage of inverter) can be decided
Do you mean the value of DC link voltage? If yes, the voltage should be about 560 V (400V * 20.5).
Yes sir, but how?
Is there in mathematical proof? Or any reference if u can suggest
It's a basic relation for AC voltage. 400V in this case means RMS value of voltage.
U_RMS = U_max / sqrt(2)
above formula after transformaton gives:
U_max = U_RMS * sqrt(2)
U_RMS = 400V, also U_max is about 560 V
The DC link voltage can be selected using the following relation:
Vdc>=2*Vs(max) with Vs(max) the peak value of the line-to-neutral AC-side voltage. This is expressed in d (direct) and q (quadrature) axes as: Vs(max)=sqrt (Vd^2+Vq^2), where, Vd and Vq are the dq-component of the voltage. Vd and Vq are also functions of the modulating signal m provided by the PWM.
I agree with previous answers. The value obtained could be slightly increased depending on the application to consider that AC side voltage could vary within a range (e.g. 10%), and the PWM deadband.
Sure, the AC voltage is varying. But the DC link voltage in a standard 3-phase bridge inverter should be constant. The question was about DC-link voltage. Simply 400VAC means 560V peak voltage.
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