CN103236798A - Three-level inverter dead time compensation control method - Google Patents

Abstract

A three-level inverter dead time compensation control method is used for compensating the decrease of system performance caused by dead time. According to the topological structure of a three-phase three-level PWM (pulse width modulation) inverter, the method first sets corresponding dead time and determines the on-off delays of power elements in the inverter; the method then considers the tube voltage drops of the power transistors of the inverter and the tube voltage drops of clamping diodes of the inverter and calculates the common voltage expression of the output neutral point of the inverter; afterwards, according to the sector with three-phase current value and a reference voltage vector, the method determines the relational coefficient between voltage error caused by the tube voltage drops of the power transistors, the tube voltage drops of the clamping diodes and the on-off delays of the power transistors and current, and thereby the total dead time compensation time of each phase of the three-level PWM inverter is worked out. The method takes the dead time, the on-off delays of the power elements and the tube voltage drops into full consideration, solves the problem of voltage and current distortion caused by the dead time effect in the three-level inverter, and enhances system performance.

Description

A kind of three-level inverter dead area compensation control method

Technical field

The present invention relates to a kind of three-level inverter dead area compensation control method, be used for compensation because the decline of the systematic function that the dead band causes is specially adapted to high-power ac motor speed control by variable frequency field.

Background technology

Three level topological structures have advantages such as output capacity is big, output voltage is high, current harmonic content is little, make that three level structures have obtained in high-power ac motor speed control by variable frequency field using widely.Straight-through for preventing the inverter brachium pontis, must in the triggering signal of same brachium pontis complementation, add the dead band, to guarantee that with behind the switching tube reliable turn-off on the brachium pontis complementary with it switching tube could conducting.But because the introducing of Dead Time causes the system control performance variation, output voltage and electric current distort, and particularly may cause motor generation mechanical resonant during low speed, in order to overcome above-mentioned shortcoming, need compensate the dead band.Existing dead-zone compensation method has usually opening with turn-off time and tube voltage drop of the Dead Time of two level, switching device is compensated, or to the Dead Time of three level, switching device open with turn-off time and tube voltage drop in certain or certain two kinds of dead time effect that cause compensate.(Jong-Woo Choi wherein, Seung-Ki Sul.Inverter Output Voltage Synthesis Using Novel Dead Time Compensation.IEEE Transactions on Power Electronics, 1996,11 (2): 221-227.) provided a kind of method of dead area compensation preferably that is applied in the two-level inverter device, this method has been considered Dead Time, IGBT turns on and off the time, the factors such as tube voltage drop of IGBT and fly-wheel diode, but this method is applied in the three-level inverter, because there are evident difference in three-level inverter and two level operation principles, thereby this method is only applicable to the dead area compensation of two-level inverter, and (Wei Xuesen, Yan Changhui, Ma Xiaoliang, Deng. based on the three-level inverter dead-zone compensation method research of FPGA. power electronic technology, 2005,39 (5): 24-17.) provided a kind of three-level inverter dead-zone compensation method based on FPGA, but owing to not considering the influence to dead area compensation of IGBT tube voltage drop and switching time, therefore after adopting this method to compensate, the output voltage loss that still can exist the dead time effect that caused by switching delay and tube voltage drop to cause.In addition, (Jin Shun, Zhong Yanru. consider in the time of a kind of novelty that midpoint potential balance and burst pulse are eliminated and three level space voltage vector pulse duration modulation methods of dead area compensation. Proceedings of the CSEE, 2005,25 (6): 60-66.) based on space vector modulation, propose a kind of three level dead area compensation strategies, but do not consider the conduction voltage drop of switching device, thereby caused the loss of voltage of output voltage not to be compensated.And (Song Wenxiang, open the rising sun. based on three-level inverter dead band and the tube voltage drop compensation policy of current phasor. electrician's electric energy new technology, 2012,31 (2): three 47-51.) given level dead area compensation strategies are not considered the switch time-delay of power device, so still can have the dead time effect that is caused by switching delay after the compensation.

Summary of the invention

Technology of the present invention is dealt with problems: overcome and do not considered in the existing inverter dead-zone compensation method that power device opens the defective of turn off delay time and tube voltage drop, propose a kind of novelty, simple three-level inverter dead area compensation control method, improved systematic function.

Technical solution of the present invention is: a kind of three-level inverter dead area compensation control method, and concrete steps are:

1. according to the topological structure of three-phase tri-level PWM inverter, consider that switching device turn-offs required time of delay fully, corresponding Dead Time T is set _d

2. according to the power device of three-phase tri-level PWM inverter, determine that it opens time-delay T _OnWith turn off delay time T _Off

3. consider the power tube tube voltage drop of three-phase tri-level PWM inverter and the tube voltage drop of clamping diode, calculate the output voltage to neutral universal expression formula of three-phase tri-level PWM inverter;

4. the tube voltage drop of considering power tube tube voltage drop and clamping diode becomes the characteristics that increase greatly with electric current, power device during with conducting and clamping diode equivalence are slope resistance, calculate the three-phase tri-level power device of PWM invertor operation when the service area and the tube voltage drop of clamping diode, afterwards the output voltage to neutral universal expression formula of the three-phase tri-level PWM inverter that 3. obtains of step of updating;

5. according to the sector at three-phase electricity flow valuve and place, reference voltage vector angle, determine to open voltage error that turn off delay time causes and the coefficient of relationship k between the electric current by the tube voltage drop of power tube tube voltage drop and clamping diode and power tube;

6. the Dead Time T that 1. arranges according to step _d, the power device 2. determined of step open time-delay T _OnWith turn off delay time T _OffAnd the coefficient of relationship k that 5. obtains of step, calculate each mutually total dead area compensation time T of three-phase tri-level PWM inverter _{Com_a}, T _{Com_b}, T _{Com_c}, by the compensation of system feedback control realization to the dead band.

Described step is opened voltage error that turn off delay time causes and the definite method of the coefficient of relationship k between the electric current is by the tube voltage drop of power tube tube voltage drop and clamping diode and power tube in 5.:

(i) when the reference voltage vector angle is in 0 °～60 °,

$k=\frac{({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})}{2i_{\mathrm{as}}+i_{\mathrm{bs}}-i_{\mathrm{cs}}}{M}_i$

(ii) when the reference voltage vector angle is in 60 °～120 °,

$k=\frac{({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})}{i_{\mathrm{as}}}{M}_i$

(iii) when the reference voltage vector angle is in 120 °～180 °,

$k=\frac{({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})}{2i_{\mathrm{as}}-i_{\mathrm{bs}}+i_{\mathrm{cs}}}{M}_i$

(iv) when the reference voltage vector angle is in 180 °～240 °,

$k=\frac{({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})}{-2i_{\mathrm{as}}-i_{\mathrm{bs}}+i_{\mathrm{cs}}}{M}_i$

(v) when the reference voltage vector angle is in 240 °～300 °,

$k=\frac{(V_{\mathrm{<figure-callout\; id="0"\; label="ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">ce</figure-callout>}0}+V_{d0})}{-i_{\mathrm{as}}}{M}_i$

(vi) when the reference voltage vector angle is in 300 °～360 °,

$k=\frac{({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})}{-2i_{\mathrm{as}}+i_{\mathrm{bs}}-i_{\mathrm{cs}}}{M}_i$

V in the formula _Ce0, V _D0Be respectively the tube voltage drop threshold values of power tube and the tube voltage drop threshold values of clamping diode; i _As, i _Bs, i _CsBe respectively the electric current of threephase load, M _i=[2sign (i _As)-sign (i _Bs)-sign (i _Cs)], wherein $\mathrm{sign}\left(i_{\mathrm{as}}\right)=\left\{\begin{array}{cc}1& (i_{\mathrm{as}}>0)\\ -1& (i_{\mathrm{as}}<0)\end{array}\right.,$ $\mathrm{sign}\left(i_{\mathrm{bs}}\right)=\left\{\begin{array}{cc}1& (i_{\mathrm{bs}}>0)\\ -1& (i_{\mathrm{bs}}<0)\end{array}\right.,$ $\mathrm{sign}\left(i_{\mathrm{cs}}\right)=\left\{\begin{array}{cc}1& (i_{\mathrm{cs}}>0)\\ -1& (i_{\mathrm{cs}}<0)\end{array}.\right.$

Principle of the present invention is: at neutral point clamp type three-phase tri-level PWM inverter band induction machine topological structure (as shown in Figure 2), neutral point clamp type three-phase tri-level inverter comprises A, B, C three-phase brachium pontis, every phase brachium pontis is made up of 4 switching tubes, 4 fly-wheel diodes and two clamping diodes, P represents positive bus-bar among the figure, N represents negative busbar, and O represents mid point; A, b, c represent the output of three-phase brachium pontis; S represents the central point of three phase electric machine; S _A1, S _A2, S _A3, S _A44 switching tubes of expression A phase, D _A1And D _A22 clamping diodes of expression A phase.Be that example is analyzed dead area compensation controlling party ratio juris (direction of rated current flow direction motor is positive direction) mutually with A, suppose the voltage U between A phase output terminal a and the motor three phase winding central point s _AsFor just, S _A2Conducting always, S _A4Turn-off always.Fig. 3 (a) is by A phase voltage reference value U _AsCompare (symmetric regular-sampled) A phase S that obtains with carrier wave _A1Pipe and S _A3Pipe is at the ideal pulse that does not arrange under the situation of dead band.Fig. 3 (b) is provided with Dead Time T for electric current is timing _dCompensation policy under the situation.S ideally _A1Pipe should be at T ₁Constantly open-minded, T ₂Turn-off constantly; S _A3Pipe should be at T ₁Turn-off T constantly ₂Constantly open-minded.Compensation T _{Com_a}After time, S _A1Pipe is at T ₁+ T _d-T _{Com_a}The moment postpones open-minded, at T ₂+ T _{Com_a}Postpone constantly to turn-off; S _A3Pipe is at T ₁-T _{Com_a}Turn-off in advance constantly, at T ₂+ T _d+ T _{Com_a}Constantly postpone open-minded.Fig. 3 (c) is for considering the T that delays time that opens of power device _OnWith turn off delay time T _Off, the voltage U of the A phase alignment O of actual output _AoFig. 3 (d) is the voltage U of equivalence output A phase alignment O _Ao, opening time-delay and being defined as switch controlling signal and opening to the time interval that the actual power device is opened, it has comprised the delay of opening of signal delay time of drive circuit and power device, the definition of turn off delay time is with to open time-delay similar.Need to prove the make-up time T among Fig. 3 _Com, for the A phase, be T _{Com_a}, for the B phase, be T _{Com_b}, for the C phase, be T _{Com_c}

S among Fig. 3 (a) _A1The desirable service time of pipe is T ₂-T ₁, and S among Fig. 3 (d) _A1The equivalent actual service time of pipe is T ₂-T ₁-T _d+ 2T _{Com_a}-T _On+ T _Off, the dead area compensation strategy shown in Fig. 3 (e), can get A mutually the expression formula of the error time between actual service time and desirable service time be:

T _{err_a}＝sign(i _as)T _ma (1)

Wherein

T _ma＝-T _d+2T _{com_a}-T _on+T _off

$\mathrm{sign}\left(i_{\mathrm{as}}\right)=\left\{\begin{array}{cc}1& (i_{\mathrm{as}}>0)\\ -1& (i_{\mathrm{as}}<0)\end{array}\right.$

If T _aWith

For at U _AsFor timing is illustrated in S in the switch periods _A1Actual effectively ON time and the desirable ON time of pipe, wherein

By given modulating wave U _AsDo to determine T than the high level time that produces pulse with modulating wave _aFor

The basis on superpose Dead Time, open the high level time of the actual pulse after the turn-off time; At U _AsBe illustrated in S in the switch periods when negative _A4Actual effectively ON time and the desirable ON time of pipe.Then have

$T_a=T_a^{*}+\mathrm{sign}\left(i_{\mathrm{as}}\right)T_{\mathrm{ma}}---\left(2\right)$

In like manner can get other two-phase time expression formulas is

$T_b=T_b^{*}+\mathrm{sign}\left(i_{\mathrm{bs}}\right)T_{\mathrm{mb}}---\left(3\right)$

$T_c=T_c^{*}+\mathrm{sign}\left(i_{\mathrm{cs}}\right)T_{\mathrm{mc}}---\left(4\right)$

When the reference phase voltage when negative, S _A3Conducting always, S _A1Turn-off, it is consistent with formula (1)～(4) in like manner to analyze the gained conclusion always.By formula (1)～(4) as can be seen, because Dead Time, IGBT turn on and off the influence of time-delay, actual effectively ON time is bigger with desirable ON time gap, can be by changing make-up time T _Com(be T mutually for A _{Com_a}, be T mutually for B _{Com_b}, be T mutually for C _{Com_c}) come the departure time.

There is potential difference in the reality because power device two ends, and deviation can appear in output voltage to neutral and ideal value.The tube voltage drop of supposing each IGBT pipe is identical, and the tube voltage drop of fly-wheel diode and clamping diode is identical, and neutral point voltage balance is V _Dc1=V _Dc2=V _Dc/ 2.Be example mutually with A, work as i _As＞0 o'clock, the voltage of actual output point a alignment O was

$V_{\mathrm{ao}}=\frac{V_{\mathrm{dc}}}{2}-{2V}_{\mathrm{ce}}$ (work as S _a=1 o'clock) (5)

V _Ao=-V _d-V _Ce(work as S _a=0 o'clock) (6)

$V_{\mathrm{ao}}=-\frac{V_{\mathrm{dc}}}{2}-{2V}_d$ (work as S _a=-1 o'clock) (7)

In the formula, V _CeBe the IGBT tube voltage drop; V _dBe diode tube pressuring drop.S _aFor the switch function of A phase, work as S _A1Pipe and S _A2Pipe conducting, S _A3Pipe and S _A4Pipe turn-offs, when output a is connected to positive bus-bar P, and S _a=1; Work as S _A2Pipe and S _A3Pipe conducting, S _A1Pipe and S _A4Pipe turn-offs, when output a is connected to mid point O, and S _a=0; Work as S _A3Pipe and S _A4Pipe conducting, S _A1Pipe and S _A2Pipe turn-offs, when output a is connected to negative busbar N, and S _a=-1.

Work as i _As＜0 o'clock, the voltage of actual output point a alignment O was

$V_{\mathrm{ao}}=\frac{V_{\mathrm{dc}}}{2}+{2V}_d$ (work as S _a=1 o'clock) (8)

V _Ao=V _d+ V _Ce(work as S _a=0 o'clock) (9)

$V_{\mathrm{ao}}=-\frac{V_{\mathrm{dc}}}{2}+{2V}_{\mathrm{ce}}$ (work as S _a=-1 o'clock) (10)

Suppose that the sense of current is constant in the sampling period, considering under tube voltage drop, on off state and the sense of current situation, consideration formula (5)～(10), actual output voltage to neutral is expressed general formula and is

$V_{\mathrm{ao}}=S_a(\frac{1}{2}{V}_{\mathrm{dc}}+V_d-V_{\mathrm{ce}})-\mathrm{sign}\left(i_{\mathrm{as}}\right)(V_{\mathrm{ce}}+V_d)---\left(11\right)$

Because tube voltage drop can be big and increase along with the change of electric current, the power device equivalence in the time of can be with conducting becomes a slope resistance, and the tube voltage drop that then operates in the power device in operate as normal district is respectively

V _ce＝V _ce0+r _ce|i _as| (12)

V _d＝V _d0+r _d|i _as| (13)

Wherein, V _Ce0, V _D0Be the tube voltage drop threshold values; r _Ce, r _dSlope resistance during for conducting.

Bring formula (12), (13) into formula (11), obtain

$V_{\mathrm{ao}}=S_a(\frac{1}{2}{V}_{\mathrm{dc}}+V_d-V_{\mathrm{ce}})-({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})\mathrm{sign}\left(i_{\mathrm{as}}\right)-(r_{\mathrm{ce}}+r_d)i_{\mathrm{as}}$

Voltage reference value U between A phase output terminal a and motor three phase winding central point s _{As_ref}Be timing, S _a1 the time of getting in one-period is

-1 the time of getting is 0, so S in the one-period _aMean value be As reference voltage U _{As_ref}When negative, S _a-1 the time of getting in one-period is

1 the time of getting is 0, so S in the one-period _aMean value be $-\frac{T_a^{*}+T_{\mathrm{ma}}\mathrm{sign}\left(i_{\mathrm{as}}\right)}{T_s}.$ So, S in the one-period _aMean value be $\left[\frac{T_a^{*}+T_{\mathrm{ma}}\mathrm{sign}\left(i_{\mathrm{as}}\right)}{T_s}\right]\mathrm{sign}\left(U_{\mathrm{as}\_\mathrm{ref}}\right).$

Convolution (2)～(4) get the equivalent voltage that outputs to mid point in the one-period:

$V_{\mathrm{ao}}=\left[\frac{T_a^{*}+T_{\mathrm{ma}}\mathrm{sign}\left(i_{\mathrm{as}}\right)}{T_s}\right](\frac{1}{2}{V}_{\mathrm{dc}}+V_d-V_{\mathrm{ce}})\mathrm{sign}\left(U_{\mathrm{as}\_\mathrm{ref}}\right)-({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})\mathrm{sign}\left(i_{\mathrm{as}}\right)-(r_{\mathrm{ce}}+r_d)i_{\mathrm{as}}$

In like manner get the equivalent voltage V that other two-phases output to mid point _BoAnd V _Co

Have for three symmetrical loads

V _as+V _bs+V _cs＝0

i _as+i _bs+i _cs＝0

The three-phase voltage expression formula is

$\left\{\begin{array}{c}{V}_{\mathrm{as}}=V_{\mathrm{ao}}+V_{\mathrm{os}}\\ V_{\mathrm{bs}}=V_{\mathrm{bo}}+V_{\mathrm{os}}\\ V_{\mathrm{cs}}=V_{\mathrm{co}}+V_{\mathrm{os}}\end{array}\right.$

Be example mutually with A, can get actual phase voltage and be

$\begin{array}{c}{V}_{\mathrm{as}}=V_{\mathrm{ao}}-\frac{1}{3}(V_{\mathrm{ao}}+V_{\mathrm{bo}}+V_{\mathrm{co}})=\frac{1}{3T_s}(\frac{1}{2}{V}_{\mathrm{dc}}+V_d-V_{\mathrm{ce}})M_{\mathrm{as}}+\\ \frac{1}{3T_s}(\frac{1}{2}{V}_{\mathrm{dc}}+V_d-V_{\mathrm{ce}})M_u-\frac{1}{3}({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})M_i-(r_{\mathrm{ce}}+r_d)i_{\mathrm{as}}\end{array}---\left(14\right)$

Wherein

$M_{\mathrm{as}}=[2\mathrm{sign}\left(U_{\mathrm{as}\_\mathrm{ref}}\right)T_a^{*}-\mathrm{sign}\left(U_{\mathrm{bs}\_\mathrm{ref}}\right)T_b^{*}-\mathrm{sign}\left(U_{\mathrm{cs}\_\mathrm{ref}}\right)T_c^{*}]$

M _u＝[2sign(U _{as_ref})sign(i _as)T _ma-sign(U _{bs_ref})·sign(i _bs)T _mb-sign(U _{cs_ref})sign(i _cs)T _mb]

M _i＝[2sign(i _as)-sign(i _bs)-sign(i _cs)]

Because V _d-V _CeValue much smaller than V _Dc/ 2, so with V _d-V _CeIgnore, will

$T_a^{*}/T_s=\left|U_{\mathrm{as}\_\mathrm{ref}}\right|/(V_{\mathrm{dc}}/2)$

$T_b^{*}/T_s=\left|U_{\mathrm{bs}\_\mathrm{ref}}\right|/(V_{\mathrm{dc}}/2)$

$T_c^{*}/T_s=\left|U_{\mathrm{cs}\_\mathrm{ref}}\right|/(V_{\mathrm{dc}}/2)$

Bringing formula (14) into gets

$V_{\mathrm{as}}=\frac{1}{6}{V}_{\mathrm{dc}}[2\mathrm{sign}\left(U_{\mathrm{as}\_\mathrm{ref}}\right)\frac{\left|U_{\mathrm{as}\_\mathrm{ref}}\right|}{V_{\mathrm{dc}}/2}-\mathrm{sign}\left(U_{\mathrm{bs}\_\mathrm{ref}}\right)\frac{\left|U_{\mathrm{bs}\_\mathrm{ref}}\right|}{V_{\mathrm{dc}}/2}-\mathrm{sign}\left(U_{\mathrm{cs}\_\mathrm{ref}}\right)\frac{\left|U_{\mathrm{cs}\_\mathrm{ref}}\right|}{V_{\mathrm{dc}}/2}+$

$\frac{1}{6}{V}_{\mathrm{dc}}[2\mathrm{sign}\left(U_{\mathrm{as}\_\mathrm{ref}}\right)\frac{\left|U_{\mathrm{as}\_\mathrm{err}}\right|}{V_{\mathrm{dc}}/2}\mathrm{sign}\left(i_{\mathrm{as}}\right)-\mathrm{sign}\left(U_{\mathrm{bs}\_\mathrm{ref}}\right)\frac{\left|U_{\mathrm{bs}\_\mathrm{err}}\right|}{V_{\mathrm{dc}}/2}\mathrm{sign}\left(i_{\mathrm{bs}}\right)-\mathrm{sign}\left(U_{\mathrm{cs}\_\mathrm{ref}}\right)\frac{\left|U_{\mathrm{cs}\_\mathrm{err}}\right|}{V_{\mathrm{dc}}/2}\mathrm{sign}\left(i_{\mathrm{cs}}\right)]$

$-\frac{1}{3}({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})[2\mathrm{sign}\left(i_{\mathrm{as}}\right)-\mathrm{sign}\left(i_{\mathrm{bs}}\right)-\mathrm{sign}\left(i_{\mathrm{cs}}\right)]-(r_{\mathrm{ce}}+r_d)i_{\mathrm{as}}$

Because of tube voltage drop and to open turn off delay time all relevant with the size of electric current, so by tube voltage drop with open the output voltage error U that turn off delay time causes _{As_err}Also with current related, establish

|U _{as_err}|＝k·|i _as|

Then

$\begin{array}{c}{V}_{\mathrm{as}}=U_{\mathrm{as}\_\mathrm{ref}}+k\&CenterDot;\frac{1}{3}\left[2\mathrm{sign}\left(U_{\mathrm{as}\_\mathrm{ref}}\right)\right|i_{\mathrm{as}}|\mathrm{sign}\left(i_{\mathrm{as}}\right)-\mathrm{sign}\left(U_{\mathrm{bs}\_\mathrm{ref}}\right)|i_{\mathrm{bs}}|\mathrm{sign}\left(i_{\mathrm{bs}}\right)\\ -\mathrm{sign}\left(U_{\mathrm{cs}\_\mathrm{ref}}\right)\left|i_{\mathrm{cs}}\right|\mathrm{sign}\left(i_{\mathrm{cs}}\right)]-\frac{1}{3}({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})M_i-(r_{\mathrm{ce}}+r_d)i_{\mathrm{as}}\end{array}---\left(15\right)$

Wherein,

$U_{\mathrm{as}\_\mathrm{ref}}=\frac{1}{6}{V}_{\mathrm{dc}}[2\mathrm{sign}\left(U_{\mathrm{as}\_\mathrm{ref}}\right)\frac{\left|U_{\mathrm{as}\_\mathrm{ref}}\right|}{V_{\mathrm{dc}}/2}-\mathrm{sign}\left(U_{\mathrm{bs}\_\mathrm{ref}}\right)\frac{\left|U_{\mathrm{bs}\_\mathrm{ref}}\right|}{V_{\mathrm{dc}}/2}-\mathrm{sign}\left(U_{\mathrm{cs}\_\mathrm{ref}}\right)\frac{\left|U_{\mathrm{cs}\_\mathrm{ref}}\right|}{V_{\mathrm{dc}}/2}]$

By formula (15) as can be seen, (r _Ce+ r _d) item is similar to stator resistance, so in induction machine, stator resistance equals real electrical machinery stator resistance and (r _Ce+ r _d) and.Slope resistance when the equivalence stator resistance is defined as actual stator resistance and adds the power device conducting.

$r_s^{\&prime;}=r_s+(r_{\mathrm{ce}}+r_d)$

In order to make actual phase voltage with identical with reference to phase voltage, need to satisfy

$k\&CenterDot;\frac{1}{3}\left[2\mathrm{sign}\left(U_{\mathrm{as}\_\mathrm{ref}}\right)\right|i_{\mathrm{as}}|\mathrm{sign}\left(i_{\mathrm{as}}\right)-\mathrm{sign}\left(U_{\mathrm{bs}\_\mathrm{ref}}\right)|i_{\mathrm{bs}}|\mathrm{sign}\left(i_{\mathrm{bs}}\right)-$

$\mathrm{sign}\left(U_{\mathrm{cs}\_\mathrm{ref}}\right)\left|i_{\mathrm{cs}}\right|\mathrm{sign}\left(i_{\mathrm{cs}}\right)]-\frac{1}{3}({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})M_i=0$

U in the formula _{As_ref}, U _{Bs_ref}, U _{Cs_ref}Symbol the unknown, can corresponding obtain the different expression formula of k according to the difference of sector that reference voltage vector drops on, the value of k is with the three-phase current value is relevant with the reference voltage vector angle at this moment.

When the reference voltage vector angle is in 0 °～60 °,

$k=\frac{({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})}{{2i}_{\mathrm{as}}+i_{\mathrm{bs}}-i_{\mathrm{cs}}}{M}_i$

When the reference voltage vector angle is in 60 °～120 °,

$k=\frac{({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})}{i_{\mathrm{as}}}{M}_i$

When the reference voltage vector angle is in 120 °～180 °,

$k=\frac{({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})}{{2i}_{\mathrm{as}}-i_{\mathrm{bs}}+i_{\mathrm{cs}}}{M}_i$

When the reference voltage vector angle is in 180 °～240 °,

$k=\frac{({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})}{-{2i}_{\mathrm{as}}-i_{\mathrm{bs}}+i_{\mathrm{cs}}}{M}_i$

When the reference voltage vector angle is in 240 °～300 °,

$k=\frac{({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})}{-i_{\mathrm{as}}}{M}_i$

When the reference voltage vector angle is in 300 °～360 °,

$k=\frac{({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})}{-2i_{\mathrm{as}}+i_{\mathrm{bs}}-i_{\mathrm{cs}}}{M}_i$

Cause $T_{\mathrm{ma}}=\frac{\left|U_{\mathrm{as}\_\mathrm{err}}\right|}{V_{\mathrm{dc}}/2}{T}_s=\frac{k\left|i_{\mathrm{as}}\right|}{V_{\mathrm{dc}}/2}{T}_s=-T_d+2T_{\mathrm{com}\_a}-T_{\mathrm{on}}+T_{\mathrm{off}}$

So obtain mutually total make-up time of A be

$T_{\mathrm{com}\_a}=\frac{T_{\mathrm{on}}-T_{\mathrm{off}}+T_d}{2}+k\frac{\left|i_{\mathrm{as}}\right|}{V_{\mathrm{dc}}}{T}_s---\left(16\right)$

In like manner can get mutually total make-up time of B, C is

$T_{\mathrm{com}\_b}=\frac{T_{\mathrm{on}}-T_{\mathrm{off}}+T_d}{2}+k\frac{\left|i_{\mathrm{bs}}\right|}{V_{\mathrm{dc}}}{T}_s---\left(17\right)$

$T_{\mathrm{com}\_c}=\frac{T_{\mathrm{on}}-T_{\mathrm{off}}+T_d}{2}+k\frac{\left|i_{\mathrm{cs}}\right|}{V_{\mathrm{dc}}}{T}_s---\left(18\right)$

The present invention's advantage compared with prior art is: the present invention is directed to the voltage and current distortion problem that dead time effect produces in the three-level inverter, fully take into account the reason that turn off delay time and tube voltage drop etc. cause dead time effect of opening of Dead Time, power device, dead area compensation is analyzed, described method can make the three-level inverter output current wave obviously improve, actual output line voltage also raises to some extent, realized the full remuneration of three level dead time effects, it is minimum that dead time effect is dropped to.

Description of drawings

Fig. 1 three-level inverter dead area compensation of the present invention control method flow chart.

Fig. 2 three-level inverter dead area compensation of the present invention control method schematic diagram,

Each point waveform in Fig. 3 three-level inverter dead area compensation of the present invention control method schematic diagram, wherein Fig. 3 (a) is the desirable driving pulse the when dead band is not set, Fig. 3 (b) is for arranging Dead Time T _dCompensation policy under the situation, Fig. 3 (c) opens the pulse of time-delay and turn off delay time for considering power device, and Fig. 3 (d) is the A phase voltage U of actual output _Ao, Fig. 3 (e) is i _As＜0 o'clock dead area compensation strategy.

Fig. 4 three-level inverter control system of the present invention block diagram.

Fig. 5 the method for the invention is used for the I of system _α, I _βCurrent waveform and Li Sa thereof are as figure, wherein Fig. 5 (a) is for when the invertor operation frequency is f=2Hz, add the preceding oscillogram of dead area compensation algorithm, when Fig. 5 (b) is f=2Hz, oscillogram behind the adding dead area compensation algorithm when Fig. 5 (c) is f=5Hz, adds the preceding oscillogram of dead area compensation algorithm, when Fig. 5 (d) is f=5Hz, the oscillogram behind the adding dead area compensation algorithm.

Embodiment

A kind of three-level inverter dead area compensation control method, as shown in Figure 1, concrete steps are:

1. according to the topological structure of three-phase tri-level PWM inverter, consider the time of delay that switching device turn-offs fully, present embodiment arranges corresponding Dead Time T _d=10 μ s;

2. according to the power device of three-phase tri-level PWM inverter, determine that it opens time-delay T _On=7 μ s and turn off delay time T _Off=7 μ s;

3. consider the power tube tube voltage drop of three-phase tri-level PWM inverter and the tube voltage drop of clamping diode, calculate the output voltage to neutral universal expression formula of three-phase tri-level PWM inverter;

As shown in Figure 2, be example mutually with A, work as i _As＞0 o'clock, the voltage of actual output point a alignment O was

$V_{\mathrm{ao}}=\frac{V_{\mathrm{dc}}}{2}-{2V}_{\mathrm{ce}}$ (when Sa=1)

V _Ao=-V _d-V _Ce(work as S _a=0 o'clock)

$V_{\mathrm{ao}}=-\frac{V_{\mathrm{dc}}}{2}-{2V}_d$ (work as S _a=-1 o'clock)

In the formula, V _CeBe the IGBT tube voltage drop; V _dBe diode tube pressuring drop.

Work as i _As＜0 o'clock, the voltage of actual output point a alignment O was

$V_{\mathrm{ao}}=\frac{V_{\mathrm{dc}}}{2}+{2V}_d$ (work as S _a=1 o'clock)

V _Ao=V _d+ V _Ce(work as S _a=0 o'clock)

$V_{\mathrm{ao}}=-\frac{V_{\mathrm{dc}}}{2}+{2V}_{\mathrm{ce}}$ (work as S _a=-1 o'clock)

By shown in Figure 3, S among Fig. 3 (a) _A1The desirable service time of pipe is T ₂-T ₁, and S among Fig. 3 (d) _A1The equivalent actual service time of pipe is T ₂-T ₁-T _d+ 2T _{Com_a}-T _On+ T _Off, the dead area compensation strategy is shown in Fig. 3 (e), and the expression formula that can get A phase error time is:

T _{err_a}＝sign(i _as)T _ma

Wherein

T _ma＝-T _d+2T _{com_a}-T _on+T _off

$\mathrm{sign}\left(i_{\mathrm{as}}\right)=\left\{\begin{array}{cc}1& (i_{\mathrm{as}}>0)\\ -1& (i_{\mathrm{as}}<0)\end{array}\right.$

If T _aWith At U _AsFor timing is illustrated in S in the switch periods _A1Actual effectively ON time and the desirable ON time of pipe, wherein By given modulating wave U _AsDo to determine than the high level time that produces pulse with modulating wave.T _aFor

The basis on superpose Dead Time, open the high level time of the actual pulse after the turn-off time; At U _AsRepresent S in another switch periods when negative _A4Actual effectively ON time and the desirable ON time of pipe then have

$T_a=T_a^{*}+\mathrm{sign}\left(i_{\mathrm{as}}\right)T_{\mathrm{ma}}$

In like manner can get other two-phase time expression formulas is

$T_b=T_b^{*}+\mathrm{sign}\left(i_{\mathrm{bs}}\right)T_{\mathrm{mb}}$

$T_c=T_c^{*}+\mathrm{sign}\left(i_{\mathrm{cs}}\right)T_{\mathrm{mc}}$

4. the tube voltage drop of considering power tube tube voltage drop and clamping diode becomes the characteristics that increase greatly with electric current, power device during with conducting and clamping diode equivalence are slope resistance, calculate the three-phase tri-level power device of PWM invertor operation when the service area and the tube voltage drop of clamping diode, afterwards the output voltage to neutral universal expression formula of the three-phase tri-level PWM inverter that 3. obtains of step of updating;

Suppose that the sense of current is constant in the sampling period, considering under tube voltage drop, on off state and the sense of current situation, can get actual A and export voltage to neutral mutually and express general formula and be

$V_{\mathrm{ao}}=S_a(\frac{1}{2}{V}_{\mathrm{dc}}+V_d-V_{\mathrm{ce}})-\mathrm{sign}\left(i_{\mathrm{as}}\right)(V_{\mathrm{ce}}+V_d)$

Because tube voltage drop can be big and increase along with the change of electric current, the power device equivalence in the time of can be with conducting becomes a slope resistance, and the tube voltage drop that then operates in the power device in operate as normal district is respectively

V _ce＝V _ce0+r _ce|i _as|

V _d＝V _d0+r _d|i _as|

Wherein, V _Ce0, V _D0Be the tube voltage drop threshold values; r _Ce, r _dSlope resistance during for conducting.According to top various, can get:

$V_{\mathrm{ao}}=S_a(\frac{1}{2}{V}_{\mathrm{dc}}+V_d-V_{\mathrm{ce}})-({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})\mathrm{sign}\left(i_{\mathrm{as}}\right)-(r_{\mathrm{ce}}+r_d)i_{\mathrm{as}}$

As reference voltage U _{As_ref}Be timing, S _a1 the time of getting in one-period is

-1 the time of getting is 0, so S in the one-period _aMean value be

As reference voltage U _{As_ref}When negative, S _a-1 the time of getting in one-period is

1 the time of getting is 0, so S in the one-period _aMean value be

So, S in the one-period _aMean value be $\left[\frac{T_a^{*}+T_{\mathrm{ma}}\mathrm{sign}\left(i_{\mathrm{as}}\right)}{T_s}\right]\mathrm{sign}\left(U_{\mathrm{as}\_\mathrm{ref}}\right).$

Therefore, can get the equivalent voltage that A in the one-period outputs to mid point mutually is:

$V_{\mathrm{ao}}=\left[\frac{T_a^{*}+T_{\mathrm{ma}}\mathrm{sign}\left(i_{\mathrm{as}}\right)}{T_s}\right](\frac{1}{2}{V}_{\mathrm{dc}}+V_d-V_{\mathrm{ce}})\mathrm{sign}\left(U_{\mathrm{as}\_\mathrm{ref}}\right)-({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})\mathrm{sign}\left(i_{\mathrm{as}}\right)-(r_{\mathrm{ce}}+r_d)i_{\mathrm{as}}$

In like manner get the equivalent voltage V that other two-phases output to mid point _BoAnd V _Co

Have for three symmetrical loads

V _as+V _bs+V _cs＝0

i _as+i _bs+i _cs＝0

The three-phase voltage expression formula is

$\left\{\begin{array}{c}{V}_{\mathrm{as}}=V_{\mathrm{ao}}+V_{\mathrm{os}}\\ V_{\mathrm{bs}}=V_{\mathrm{bo}}+V_{\mathrm{os}}\\ V_{\mathrm{cs}}=V_{\mathrm{co}}+V_{\mathrm{os}}\end{array}\right.$

Be example mutually with A, can get actual phase voltage and be

$V_{\mathrm{as}}=V_{\mathrm{ao}}-\frac{1}{3}(V_{\mathrm{ao}}+V_{\mathrm{bo}}+V_{\mathrm{co}})=\frac{1}{3T_s}(\frac{1}{2}{V}_{\mathrm{dc}}+V_d-V_{\mathrm{ce}})M_{\mathrm{as}}+$

$\frac{1}{3T_s}(\frac{1}{2}{V}_{\mathrm{dc}}+V_d-V_{\mathrm{ce}})M_u-\frac{1}{3}({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})M_i-(r_{\mathrm{ce}}+r_d)i_{\mathrm{as}}$

Wherein

$M_{\mathrm{as}}=[2\mathrm{sign}\left(U_{\mathrm{as}\_\mathrm{ref}}\right)T_a^{*}-\mathrm{sign}\left(U_{\mathrm{bs}\_\mathrm{ref}}\right)T_b^{*}-\mathrm{sign}\left(U_{\mathrm{cs}\_\mathrm{ref}}\right)T_c^{*}]$

M _u＝[2sign(U _{as_ref})sign(i _as)T _ma-sign(U _{bs_ref})·sign(i _bs)T _mb-sign(U _{cs_ref})sign(i _cs)T _mb]

M _i＝[2sign(i _as)-sign(i _bs)-sign(i _cs)]

Because V _d-V _CeValue much smaller than V _Dc/ 2, so with V _d-V _CeIgnore, will

$T_a^{*}/T_s=\left|U_{\mathrm{as}\_\mathrm{ref}}\right|/(V_{\mathrm{dc}}/2)$

$T_b^{*}/T_s=\left|U_{\mathrm{bs}\_\mathrm{ref}}\right|/(V_{\mathrm{dc}}/2)$

$T_c^{*}/T_s=\left|U_{\mathrm{cs}\_\mathrm{ref}}\right|/(V_{\mathrm{dc}}/2)$

So can get:

$V_{\mathrm{as}}=\frac{1}{6}{V}_{\mathrm{dc}}[2\mathrm{sign}\left(U_{\mathrm{as}\_\mathrm{ref}}\right)\frac{\left|U_{\mathrm{as}\_\mathrm{ref}}\right|}{V_{\mathrm{dc}}/2}-\mathrm{sign}\left(U_{\mathrm{bs}\_\mathrm{ref}}\right)\frac{\left|U_{\mathrm{bs}\_\mathrm{ref}}\right|}{V_{\mathrm{dc}}/2}-\mathrm{sign}\left(U_{\mathrm{cs}\_\mathrm{ref}}\right)\frac{\left|U_{\mathrm{cs}\_\mathrm{ref}}\right|}{V_{\mathrm{dc}}/2}]+$

$\frac{1}{6}{V}_{\mathrm{dc}}[2\mathrm{sign}\left(U_{\mathrm{as}\_\mathrm{ref}}\right)\frac{\left|U_{\mathrm{as}\_\mathrm{err}}\right|}{V_{\mathrm{dc}}/2}\mathrm{sign}\left(i_{\mathrm{as}}\right)-\mathrm{sign}\left(U_{\mathrm{bs}\_\mathrm{ref}}\right)\frac{\left|U_{\mathrm{bs}\_\mathrm{err}}\right|}{V_{\mathrm{dc}}/2}\mathrm{sign}\left(i_{\mathrm{bs}}\right)-\mathrm{sign}\left(U_{\mathrm{cs}\_\mathrm{ref}}\right)\frac{\left|U_{\mathrm{cs}\_\mathrm{err}}\right|}{V_{\mathrm{dc}}/2}\mathrm{sign}\left(i_{\mathrm{cs}}\right)]$

$-\frac{1}{3}({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})[2\mathrm{sign}\left(i_{\mathrm{as}}\right)-\mathrm{sign}\left(i_{\mathrm{bs}}\right)-\mathrm{sign}\left(i_{\mathrm{cs}}\right)]-(r_{\mathrm{ce}}+r_d)i_{\mathrm{as}}$

5. according to the sector at three-phase electricity flow valuve and place, reference voltage vector angle, determine to open voltage error that turn off delay time causes and the coefficient of relationship k between the electric current by the tube voltage drop of power tube tube voltage drop and clamping diode and power tube;

6. the Dead Time T that 1. arranges according to step _d, the power device 2. determined of step open time-delay T _OnWith turn off delay time T _OffAnd the coefficient of relationship k that 5. obtains of step, calculate each mutually total dead area compensation time T of three-phase tri-level PWM inverter _{Com_a}, T _{Com_b}, T _{Com_c}, realize compensation to the dead band by system feedback control, it embodies formula and is:

So obtain mutually total make-up time of A be

$T_{\mathrm{com}\_a}=\frac{T_{\mathrm{on}}-T_{\mathrm{off}}+T_d}{2}+k\frac{\left|i_{\mathrm{as}}\right|}{V_{\mathrm{dc}}}{T}_s$

In like manner can get mutually total make-up time of B, C is

$T_{\mathrm{com}\_b}=\frac{T_{\mathrm{on}}-T_{\mathrm{off}}+T_d}{2}+k\frac{\left|i_{\mathrm{bs}}\right|}{V_{\mathrm{dc}}}{T}_s$

$T_{\mathrm{com}\_c}=\frac{T_{\mathrm{on}}-T_{\mathrm{off}}+T_d}{2}+k\frac{\left|i_{\mathrm{cs}}\right|}{V_{\mathrm{dc}}}{T}_s$

Described step in 5. power tube tube voltage drop and the tube voltage drop of clamping diode and by definite method that power tube is opened the coefficient of relationship k of the voltage error that causes between turn off delay time and the electric current be:

(i) when the reference voltage vector angle is in 0 °～60 °,

$k=\frac{({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})}{{2i}_{\mathrm{as}}+i_{\mathrm{bs}}-i_{\mathrm{cs}}}{M}_i$

(ii) when the reference voltage vector angle is in 60 °～120 °,

$k=\frac{({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})}{i_{\mathrm{as}}}{M}_i$

(iii) when the reference voltage vector angle is in 120 °～180 °,

$k=\frac{({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})}{2i_{\mathrm{as}}-i_{\mathrm{bs}}+i_{\mathrm{cs}}}{M}_i$

(iv) when the reference voltage vector angle is in 180 °～240 °,

$k=\frac{({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})}{-{2i}_{\mathrm{as}}-i_{\mathrm{bs}}+i_{\mathrm{cs}}}{M}_i$

(v) when the reference voltage vector angle is in 240 °～300 °,

$k=\frac{({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})}{-i_{\mathrm{as}}}{M}_i$

(vi) when the reference voltage vector angle is in 300 °～360 °,

$k=\frac{({\mathrm{<figure-callout\; id="0"\; label="V\; ce"\; filenames="HSA00000884836300041.png"\; state="\{\{state\}\}">V</figure-callout>}}_{\mathrm{ce}}+V_{d0})}{-{2i}_{\mathrm{as}}+i_{\mathrm{bs}}-i_{\mathrm{cs}}}{M}_i$

V in the formula _Ce0, V _D0Be respectively the tube voltage drop threshold values of power tube and the tube voltage drop threshold values of clamping diode; i _As, i _Bs, i _CsBe respectively the electric current of threephase load, M _i=[2sign (i _As)-sign (i _Bs)-sign (i _Cs)], wherein $\mathrm{sign}\left(i_{\mathrm{as}}\right)=\left\{\begin{array}{cc}1& (i_{\mathrm{as}}>0)\\ -1& (i_{\mathrm{as}}<0)\end{array}\right.,$ $\mathrm{sign}\left(i_{\mathrm{bs}}\right)=\left\{\begin{array}{cc}1& (i_{\mathrm{bs}}>0)\\ -1& (i_{\mathrm{bs}}<0)\end{array}\right.,$ $\mathrm{sign}\left(i_{\mathrm{cs}}\right)=\left\{\begin{array}{cc}1& (i_{\mathrm{cs}}>0)\\ -1& (i_{\mathrm{cs}}<0)\end{array}\right..$

In order to verify beneficial effect of the present invention, built the three-level inverter control system hardware platform based on DSP and CPLD, as shown in Figure 4, mainly comprise major loop and control loop two large divisions among the figure.Major loop is by three-phase alternating-current supply, three-phase commutation bridge, pre-charge circuit, neutral point clamp type three-phase tri-level inverter, motor load, voltage sensor and current sensor.Control loop comprises that DSP and CPLD control board, 12 road PWM optical fiber transmissions and receiving loop, 12 road PWM faults transmit and receive loop, power module.Three phase mains links to each other with rectifier bridge in the major loop, rectifier bridge converts three-phase alternating-current supply to DC power supply and uses for three-level inverter, rectifier bridge links to each other with three-level inverter, inverter links to each other with motor, and the DC power supply that three-level inverter is exported rectifier bridge converts the three-phase alternating current of expected frequency f to and supplies with to motor load; The transmission of 12 road PWM optical fiber receives with fault and links to each other with three-level inverter with the DSP control board.Control loop sends by optical fiber and receiving loop control three-level inverter, and PWM optical fiber sends and receiving loop is controlled conducting and the shutoff of three-level inverter switching device with DSP control board (the present invention's employing the be TMS320LF28335) control signal that calculates; The optical fiber receiving loop sends fault-signal to DSP, and voltage, current sensor will be gathered three-level inverter busbar voltage and output current, and signal is delivered to the DSP control board; The DSP control board is handled signal and the transmission that receives and is realized the midpoint potential balanced algorithm, and power module provides power supply for the The whole control loop.

Carried out the experiment that open loop V/F controls at hardware platform shown in Figure 4, the rectification side adopts not control rectifying circuit, and input line voltage is 380V, and inverter adopts the SVPWM modulation system, and load is the 30kW asynchronous machine.Switching frequency is 1kHz, and the sampling time is 1ms, and the Dead Time of setting is 10 μ s.Fig. 5 is for flowing into A, B, the C three-phase current of motor, the I that obtains after the 2s/3s conversion _α, I _βWaveform and their Li Saru (lissajous) figure.Fig. 5 (a) and (b) add waveform before and after the dead area compensation algorithm when being respectively 2Hz wherein.Add dead area compensation algorithm after-current waveform and be clearly better I _α, I _βLissajous figures also round.Fig. 5 (c) and Fig. 5 (d) add the waveform before and after the dead area compensation algorithm when being respectively 5Hz.As seen from the figure, behind the adding dead area compensation, current waveform improves.Adding the dead area compensation algorithm front and back of considering switch time-delay, tube voltage drop, the output line voltage of actual measurement is as shown in table 1.

Output line voltage before and after table 1 dead area compensation

As can be seen from Table 1, after adding the dead area compensation algorithm, actual measurement line voltage also has the lifting of about 4V, has compensated the loss of voltage that is caused by dead time effect.

The content of describing that do not elaborate in the specification of the present invention belongs to this area professional and technical personnel's known prior art.

Claims (2)

1. three-level inverter dead area compensation control method, it is characterized in that: concrete steps are:

1. according to the topological structure of three-phase tri-level PWM inverter, consider that switching device turn-offs required time of delay fully, corresponding Dead Time T is set _d

2. according to the power device of three-phase tri-level PWM inverter, determine that it opens time-delay T _OnWith turn off delay time T _Off

3. consider the power tube tube voltage drop of three-phase tri-level PWM inverter and the tube voltage drop of clamping diode, calculate the output voltage to neutral universal expression formula of three-phase tri-level PWM inverter;

4. the tube voltage drop of considering power tube tube voltage drop and clamping diode becomes the characteristics that increase greatly with electric current, power device during with conducting and clamping diode equivalence are slope resistance, calculate the three-phase tri-level power device of PWM invertor operation when the service area and the tube voltage drop of clamping diode, afterwards the output voltage to neutral universal expression formula of the three-phase tri-level PWM inverter that 3. obtains of step of updating;

5. according to the sector at three-phase electricity flow valuve and place, reference voltage vector angle, determine to open voltage error that turn off delay time causes and the coefficient of relationship k between the electric current by the tube voltage drop of power tube tube voltage drop and clamping diode and power tube.

6. the Dead Time T that 1. arranges according to step _d, the power device 2. determined of step open time-delay T _OnWith turn off delay time T _OffAnd the coefficient of relationship k that 5. obtains of step, calculate each mutually total dead area compensation time T of three-phase tri-level PWM inverter _{Com_a}, T _{Com_b}, T _{Com_c}, by the compensation of system feedback control realization to the dead band.

2. three-level inverter dead area compensation control method according to claim 1 is characterized in that: described step is opened voltage error that turn off delay time causes and the definite method of the coefficient of relationship k between the electric current is by the tube voltage drop of power tube tube voltage drop and clamping diode and power tube in 5.:

(i) when the reference voltage vector angle is in 0 °～60 °,

$k=\frac{(V_{\mathrm{ce}0}+V_{d0})}{{2i}_{\mathrm{as}}+i_{\mathrm{bs}}-i_{\mathrm{cs}}}{M}_i$

(ii) when the reference voltage vector angle is in 60 °～120 °,

$k=\frac{(V_{\mathrm{ce}0}+V_{d0})}{i_{\mathrm{as}}}{M}_i$

(iii) when the reference voltage vector angle is in 120 °～180 °,

$k=\frac{(V_{\mathrm{ce}0}+V_{d0})}{{2i}_{\mathrm{as}}-i_{\mathrm{bs}}+i_{\mathrm{cs}}}{M}_i$

(iv) when the reference voltage vector angle is in 180 °～240 °,

$k=\frac{(V_{\mathrm{ce}0}+V_{d0})}{-{2i}_{\mathrm{as}}-i_{\mathrm{bs}}+i_{\mathrm{cs}}}{M}_i$

(v) when the reference voltage vector angle is in 240 °～300 °,

$k=\frac{(V_{\mathrm{ce}0}+V_{d0})}{-i_{\mathrm{as}}}{M}_i$

(vi) when the reference voltage vector angle is in 300 °～360 °,

$k=\frac{(V_{\mathrm{ce}0}+V_{d0})}{-{2i}_{\mathrm{as}}+i_{\mathrm{bs}}-i_{\mathrm{cs}}}{M}_i$

V in the formula _Ce0, V _D0Be respectively the tube voltage drop threshold values of power tube and the tube voltage drop threshold values of clamping diode; i _As, i _Bs, i _CsBe respectively the electric current of threephase load, M _i=[2sign (i _As)-sign (i _Bs)-sign (i _Cs)], wherein $\mathrm{sign}\left(i_{\mathrm{as}}\right)=\left\{\begin{array}{cc}1& (i_{\mathrm{as}}>0)\\ -1& (i_{\mathrm{as}}<0)\end{array}\right.,$ $\mathrm{sign}\left(i_{\mathrm{bs}}\right)=\left\{\begin{array}{cc}1& (i_{\mathrm{bs}}>0)\\ -1& (i_{\mathrm{bs}}<0)\end{array}\right.,$ $\mathrm{sign}\left(i_{\mathrm{cs}}\right)=\left\{\begin{array}{cc}1& (i_{\mathrm{cs}}>0)\\ -1& (i_{\mathrm{cs}}<0)\end{array}\right..$

Images