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?
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?
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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
dear Annie
Equation 6
为什么计算每个开关管导通损耗的时候,是相电流的有效值除以根号2,再除以n,而不是直接用相电流的有效值平方乘以电阻