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TIDA-00364: Design problem

Annie Liu
Genius 10295 points
Part Number: TIDA-00364

Hi Team,

in Table 2. Motor Nameplate Information
Rated phase-to-phase voltage = 33V rms

Customer wanted to know why the phase current Rated phase current = 130 ARMS?

The motor is 5kw, the customer's understanding is that the line voltage is 33V effective value, The line current should be 130A.

Thanks,

Annie

  • 0
    TI__Expert 3390 points

    Hello Annie,

    thank you for your interest in TIDA-00364.  

    That specific AC induction motor mentioned is only used to test the TIDA-00364 power stage.  It’s an example motor which has a nominal phase voltage of 33Vrms and  a nominal phase current of 130Vrms with a rated power of 5kW.

    I assume your question is how it gets 5kW?

    The rated power of an AC induction motor is calculated as: Vphase-to-phase x sqrt(3) x Iphase x cos(phi), where cos(phi) is the power factor and phi the phase shift between motor phase voltage and motor phase current.

    Applying that to this specific AC induction motor, we get a cos(phi) of ~0.68. Input power = 33Vrms x 130Arms x 1.73 x 0.68 = 5kW

    The specification of the TIDA-00364 power stage with regards to input voltage and current is as follows, see design guide Table 1.

    - Maximum phase current: 130Arms

    - DC input voltage: 48V nominal, 42V min, 56VDC max.

    For sinusoidal PWM modulation of a 3-phase inverter we get the theoretical maximum phase to phase voltage [Vrms]: Vbus/2 x sqrt(3) / sqrt(2).
    For 48V the maximum phase to phase voltage is 29.3Vrms.

    For space vector modulation with third harmonics we multiply with a factor of 1.154, so we get a higher phase-to-phase output voltage, hence 33.92Vrms at 48VDC

    Please let me know if this answer your question. 

    Thanks.

    Regards,

    Martin Staebler