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FOC SVPWM function pwmc_ Setphasevoltage parsing

2022-04-23 06:47:00 【tilblackout】

One 、 Sector judgment

wUAlpha = Valfa_beta.qV_Component1 * ( int32_t )pHandle->hT_Sqrt3;//hT_Sqrt3 by 2/√3
wUBeta = -( Valfa_beta.qV_Component2 * ( int32_t )( pHandle->hPWMperiod ) ) * 2;
wX = wUBeta;
wY = ( wUBeta + wUAlpha ) / 2;
wZ = ( wUBeta - wUAlpha ) / 2;

ST The official sector is as follows ：

By way of anti Park Transform to get vector $V_{\alpha }$ and $V_{\beta }$ , First of all, we need to know which sector the combined vector is in . Take the vector to be synthesized in the first sector as an example , When $V_{\alpha }>0$ And $V_{\beta }<0$ when , The vector may be in I or II A sector , Demand only tan The angle can determine which sector the synthesized vector is in .

Finally, the rules are summarized as follows ：

It can be known from the above table , Judge $V_{\beta }$ 、 $\sqrt{3}V_{\alpha }-V_{\beta}$ 、 $-\sqrt{3}V_{\alpha }-V_{\beta}$ And 0 The size relationship of , Then calculate the three-phase duty cycle according to this variable , So this is multiplied by the period T.
           $wX=-V_{\beta }T$
           $wY= \frac{\sqrt{3}V_{\alpha }T-V_{\beta }T}{2}$
   $wZ= \frac{-V_{\beta }T-\sqrt{3}V_{\alpha }T}{2}$
   1、wY<0： namely $\sqrt{3}V_{\alpha }<V_{\beta}$ , The target vector is 3、4、5 A sector
               (1)wZ<0, Then in sector 5
               (2)wZ≥0：①wX≤0 when , In sector 4 ②wX>0 when , Then in sector 3
       2、wY>0： namely $\sqrt{3}V_{\alpha }>V_{\beta}$ , The target vector is 1、2、6 A sector
               (1)wZ≥0, Then in sector 2
               (2)wZ<0：①wX≤0 when , In sector 6 ②wX>0 when , Then in sector 1

Two 、 Vector time calculation

Here, take the first sector as an example to explain .

（1） First, it is necessary to ensure that the vector synthesized by adjacent vectors is within the boundary of regular hexagon , Otherwise, the voltage output waveform of the inverter will be distorted .

The amplitude of the basic space vector , That is, the value of phase voltage , The size is $\frac{2V_{dc}}{3}$ , When the action time of non-zero vector is 0 when , The synthetic vector size of adjacent basic space vectors is $\frac{2V_{dc}}{3}\times sin60^{\circ}$ . Divide the two , That is, the amplitude of the basic space vector after normalization : $\frac{1}{sin60^{\circ}}= \frac{2}{\sqrt{3}}$ .

（2） The next step is to calculate the vector action time

$V_{out}\times \cos\theta= |V_{4}|\times\frac{T_{4}}{T}+ |V_{6}|\times \frac{T_{6}}{T} \times\cos60^{\circ} = V_{\alpha }$

$V_{out}\times \sin\theta= |V_{6}|\times \frac{T_{6}}{T} \times\sin60^{\circ} = -V_{\beta }$ （ST The official benchmark is the fourth quadrant , So there's a minus sign ）

$|V_{4}|=|V_{6}|= \frac{2}{\sqrt{3}}$

Finally, we can find ：

$T_{4}= \frac{\sqrt{3}V_{\alpha }T+V_{\beta }T}{2}$ , $T_{6}= -V_{\beta }T$

Find out $T_{4}= -wZ$ , $T_{6}= wX$ , And zero vector action time $T_{0}=T-T_{4}-T_{6}$ .

Next, analyze the duty cycle in the code ：

The seven segment formula of the first sector SVPWM The wave generation order is ：0-4-6-7-6-4-0

Now we ask for A、B、C Three phase CCR value wTimePhA、 wTimePhB and wTimePhC, and TIM Set to center alignment mode , In the end, we can get ：

$wTimephA= (T - \frac{T+Z-X}{2})\div2 = \frac{T-Z+X}{4}$

$wTimephB= wTimephA - T_{4}\div2 = wTimephA+\frac{Z}{2}$

$wTimephC= wTimephB-T_{6}\div2= wTimephB-\frac{X}{2}$

Similarly, the seven segment formula of other sectors can be obtained SVPWM Duty cycle and wave waveform of ：
- In the table tA、tB、tC Indicates the entire high level time , and CCR Values are different .

Why different sectors A、B、C The waveforms of the phases are different ？

First, let's look at the definition of six sectors , This is not by 0~6 In sequence , Instead, it ensures that there is only one bit difference between two adjacent sectors , This can reduce MOSFET The number of switches .

Finally, it can be concluded that different sectors have the following switching order , Drawn in binary, it corresponds to different sectors in front of SVPWM 7 Segment diagram

A sector	Seven paragraphs SVPWM The order
I	0-4-6-7-7-6-4-0
II	0-2-6-7-7-6-2-0
III	0-2-3-7-7-3-2-0
IV	0-1-3-7-7-3-1-0
V	0-1-5-7-7-5-1-0
VI	0-4-5-7-7-5-4-0

Finally, take the first sector as an example , Analysis related configuration codes ：

static void MX_TIM1_Init(void)
    TIM_InitStruct.CounterMode = LL_TIM_COUNTERMODE_CENTER_UP;
    TIM_InitStruct.Autoreload = ((PWM_PERIOD_CYCLES) / 2);

uint16_t PWMC_SetPhaseVoltage( PWMC_Handle_t * pHandle, Volt_Components Valfa_beta )
    pHandle->hSector = SECTOR_1;
    wTimePhA = ( int32_t )( pHandle->PWMperiod ) / 4 + ( ( wX - wZ ) / ( int32_t )262144 );
    wTimePhB = wTimePhA + wZ / 131072;
    wTimePhC = wTimePhB - wX / 131072;
    pHandle->hCntPhA = ( uint16_t )wTimePhA;
    pHandle->hCntPhB = ( uint16_t )wTimePhB;
    pHandle->hCntPhC = ( uint16_t )wTimePhC;
    pSetADCSamplingPoint = pHandle->pFctSetADCSampPointSect1;
    return ( pSetADCSamplingPoint( pHandle ) );

static uint16_t R3_1_F30X_WriteTIMRegisters( PWMC_Handle_t * pHdl )
    TIMx->CCR1 = pHandle->_Super.hCntPhA;
    TIMx->CCR2 = pHandle->_Super.hCntPhB;
    TIMx->CCR3 = pHandle->_Super.hCntPhC;

（1）CCR Directly equal to hCntPhA、hCntPhB、hCntPhC

From the top seven paragraphs SVPWM As can be seen from the picture of ,a、b、c Three phase PWM The waveform is centrally aligned ：

In count up or count down mode ,PWM The period is equal to ARR; In the center alignment mode ,PWM The period is equal to 2ARR.

Here, the counting order of the central alignment mode is from ARR~0~ARR, Instead of 0~ARR~0：

① from ARR Reduced to CRR： Low level

② from CCR Reduced to 0： High level

③ from 0 Add to CCR： High level

④ from CCR Add to ARR： Low level

（2）hPWMperiod and T

hPWMperiod = T. In center alignment mode PWM The period is 2 times ARR, therefore TIM_ARR Set up in order to hPWMperiod/2. And we calculated earlier T4 and T6 When , Used in the formula T It represents a whole cycle of T.

（3）131072 and 262144

We calculated earlier that wTimephA、wTimephB、wTimephC, as follows ：

$wTimephA= (T - \frac{T+Z-X}{2})\div2 = \frac{T-Z+X}{4}$

$wTimephB= wTimephA - T_{4}\div2 = wTimephA+\frac{Z}{2}$

$wTimephC= wTimephB-T_{6}\div2= wTimephB-\frac{X}{2}$

because ADC The collected current is left aligned , So it is $Q_{15}$ Format , Calculation PWM When comparing values, change to $Q_{0}$ Format , So the calculation here needs to move to the right first 15 position , namely $2^{^{15}}=32768$ . Plus wUAlpha and wUBeta The definition is multiplied by one more 2, therefore X、Z And divide by one more 2.

in addition $131072 = 2^{17}$ 、 $262144 = 2^{18}$ , So I'll change the division in the code to shift right 17 Bit and 18 position , So as to speed up the operation .