参考设计的TIDA-00364设计问题

user1900697

Other Parts Discussed in Thread: TIDA-00364

dear TI
in Table 2. Motor Nameplate Information
Rated phase-to-phase voltage = 33V rms
为什么相电流 Rated phase current = 130 ARMS
5kw电机，线电压是33V有效值, 线电流是130A 才对啊，为什么相电流是130A?

3 年多前

0 Annie Liu 3 年多前

TI__Mastermind 33345 points

正在研究您的问题。您也可以查看下帖了解进展：https://e2e.ti.com/support/tools/sim-hw-system-design/f/simulation-hardware-system-design-tools-forum/996850/tida-00364-design-problem

0 Annie Liu 3 年多前

TI__Mastermind 33345 points

请查看回复，如有您有进一步问题，建议您在此帖中与工程师沟通

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