FIELD OF THE INVENTION

The present invention relates to devices for controlling electrical drives, in particular to frequency converters for alternating-current motors.

TECHNICAL PROBLEM

There has been considerable effort by researchers and engineers to develop modern electromotor drives controlled by frequency converters. The aim is to minimize the number of semiconductor components while achieving the best possible performances and suppression of higher current harmonics. There are two contradictory requirements - first aimed at minimizing the number of components and reducing the cost of producing the device _/ and the second aimed at achieving the best possible performances related to current harmonics.

Also there is a wish to minimize the number of electrolytic capacitors in standard frequency converters in the input DC circuit.

The first disadvantage of the proposed frequency converters which comprise a "booster" capacitor is that a voltage that generates on the "booster" capacitors is higher than that of the grid voltage. This voltage is connected to the semiconductor switches, wherein in case of switching, the dynamic losses of semiconductor switches are higher than the losses of the ordinary frequency converters.

Another disadvantage is that through the motor windings doesn't only flow the current that drives the machine, but also:

- With respect to the converters shown in Figures 1 to 4, the current Ii / 3 also flows through the motor winding ¹

- With respect to converters shown in Figures 8 to 10, the grid current also flows through the motor winding. Due to these additional currents, it is necessary to load the machine, depending on the frequency converter configuration with up to 95% power.

PRIOR ART

According to the research carried out by the inventor, such inventions have not been found. The state of the art consists of the standard frequency converters or inverters. The US patent 8174853 B2 deals with variable frequency converters with pulse modulation, which solution, with respect to the required control electronics, is more expensive than the one revealed in the present invention.

BRIEF SUMMARY OF THE PRESENT INVENTION:

Therefore, to address these problems, the present patent introduces a new type of frequency converter having a booster capacitor which is at higher voltage than that of the supply grid.

The present invention reveals two types of frequency converters:

a) Converters with input DC link

b) Converters which don't have an input DC link, but are powered from the grid via the motor winding (without rectifying diodes).

The present invention proposes a solution of a frequency converter with at least one booster capacitor with higher voltage than that of the supply grid. The frequency converters shown in Rgures 1 to 4 draw approximately constant current (with constant load) from the source. This current has a small proportion of higher harmonics, so it is quite suitable in respect with the load of the source. Also, the electrolytic capacitor at input is not needed. A "booster" capacitor is required (for lkW, the capacitance of such capacitor is about 4-5pF).

The frequency converters shown in Figures 8 to 10 are quite suitable for the grid, as from it they draw from sinusoidal current with small presence of higher harmonics. It might be necessary to place an RFI filter at the input. For the present converter, the two "booster" capacitors are required (each about 4-5 F for lkW motor power).

The frequency converters shown in Figures 5 to 7 have pretty large booster capacitors, about 150-30 Q iF (2 pieces) per 1 kW of motor power. These converters are suitable for use as soft starters. When the motor rotates at 50Hz, then the "booster" capacitors can be used to improve the power factor of the grid. This is to be done in such a way that the command for such sinusoidal grid current is given (controlled by the power electronics of the converter) hat sinusoidal grid current precedes the individual phase voltage of the grid (current precedes the voltage = capacitance load).

Owing to the use of "booster" capacitor at the output of the frequency converter, additional currents flow through the electric motor, so it is possible to load an electric motor depending on the performance of the frequency converter, as shown in Figures 1 to 4, with a maximum power of 95%, respectively according to configurations shown in pictures of 8 to 10 with a maximum power of 90%.

FREQUENCY CONVERTER WITH DC LINK. This type of frequency converter is shown in Figure 1. Power is supplied via a diode three-phase bridge.

Here the input voltage +Vi has a small voltage ripple so it could almost be said that it is constant. The current Ii is divided into three equal parts of Ii / 3 that go to the windings of the motors U, V, W.

The motor windings are:

Iw= l ^{/3 +} /r ^sin( e ' - ^l20 ° ⁾ Where is the frequency at which the electric motor rotates.

The current in each branch Ii / 3 is regulated in such a way that the voltage on the capacitor CI (voltage + Vc) is (at least) greater for the value of the highest among the voltages Uu, Uv or Uw (i.e. for the value of the voltage of the motor winding).

Voltages induced in the motor are:

C/ _if = C/ _I - sin( _fi)l ./ + _? i + l20 ^e )

The power required for motor operation:

i ^> = 3-[/ _| /V2 -/ _] /V2 - cosp

Ρ = 3·υ ·Ι·οο$φ

That said power is supplied by a diode bridge, so that

Therefore, from these last two equations is obtained:

Eg. it can be said that at the 50Hz control frequency (the frequency at which the motor rotates)

Ii/3=230/530-I-G.85=G.37 I.

The flow control circuit for the phase U is given in Figure 4. Voltage + Vc is divided by a resistance divider and compared with the reference voltage Vcref. The comparison difference is directed further to the amplifier and then the signal is added to the phase current reference iU (t).

As the feedback, signal Iu is compared with said sum of signals, which is then amplified and sent to the hysteresis switch which further switches on the high or low level IGBT.

Control loops (regulation circuits) for the windings V and W are the same as those in the winding U, with adequate current references which have shift + 120 ⁰ and -120°. The current of the grid in the simulation for the 0.55kW motor is given in Figure 11. It is visible that the current Ii is approximately constant and has a small proportion of higher harmonics.

The frequency converter shown in Figure 1 can only be run in motor mode because there is the diode bridge at the input which can only draw power from the grid. The frequency converter shown in Figure 2 is equipped with a battery to be used as the voltage source, and if needed, the braking energy can be fed back into the battery. It is also possible to have a generator operation mode when it charges the battery. While braking and in generator operation mode, the current Ii changes the direction. The frequency converter shown in Figure 3 has a fully controlled reversible rectifier with IGBTs at the input point. While braking or during generator operation mode, the IGBTs - due to the constancy of the current Ii (see Figure 11) - can work at a grid's frequency and thus feed energy back to the grid. Therefore, because the current Ii is constant, it is not necessary that the IGBTs at the network input work at high frequency, but it is necessary to switch at a frequency of the grid, in a way that each IGBT conducts 120°.

FREQUENCY CONVERTER WITH MOTOR WITH ONE SET OF WINDING

This system is shown in Figure 5 and comprises an AC motor (with windings U, V,

W), power supply grid (LI, L2, L3), IGBT bridge (Q1..Q6) and "booster" capacitors CI and C2. The frequency converter itself comprises an IGBT bridge (Q1..Q6), "booster capacitors" (CI and C2), and control electronics that controls the IGBT bridge in accordance with the requirements for obtaining a certain speed of the motor rotation. Generally, one of observed phase can be shown as in Figure 6.

Here we have:

-uo = current voltage of the observed grid phase

- io = grid current component of the observed grid phase

-ul = current induced EMF (electromagnetic force) at the winding of the observed phase

-il = current corresponding to component ul in the observed phase

-L = motor leakage inductance of the observed grid phase.

The components uo/io work at the grid frequency (ie, 50 or 60 Hz), while the frequency of the component ul/il is the frequency at which the motor is to drive (eg 30 Hz).

It should also be noted that the total current through the observed phase is I = io-il. If fs is the defined as the chopping frequency of the bridge, then T = 1 / fs is associated time of the IGBTs chopping Q1-Q6.

D is the of the high level IGBT duty ratio (namely, the one that connects the observed phase to + Vc voltage).

The value for D ranges from 0 to 1.

Therefore, an equation can be set according to Figure 6:

D = l/(Vc+ VeHVe+ _u + _Ur L^)

For the current Ic it can be written that:

Where D2, D4 and D6 represent the duty ratio of the impulse of the switches Q2, Q4, and Q6

As:

Using the equation we get

Ic=l/(Vc+Ve)-[ a + b + c- d- e- f], where

^=VAiou^io^io,- u- ~ -VA ^Q - ^Q =^

b ⁼ Uou ^' iou ⁺ Uov-iov ⁺ Uow ^' iow = ^y2 'Uo'Io- ^COS( P ^{c =} Uw- iou ⁺ u _w ^■ iov ⁺ u _w ^' iow ^=3/2 'U _l 'I ₀ - ^C0S K<B - ω) ¹ - φ, ~ «] d=Uou ^' u ⁺ Uoy-iw ⁺ uow ^' iw= ^{l '} Uo- ' ^co ^o ^~ ω ^ί+ φ -a]

e = Uw · u ⁺ Wi ^■ v + U«r ^' w = ^{312 '} Ui ^' /i ' ^cos ψ _χ

f = L-(I _u -AI _u +I _v .AI _y +I _Jy -AI _jr )/T = 3/2ia) -W L^^I^md -^t-a) Here a is the shift between the voltage system LI, L2, L3 and the voltage at the windings U, V, W (namely of the generated frequency voltage

The previous equation showed the case where the directions of the rotating fields Ll-

L2-L3 and U-V-W are the same.

When they are opposite, from equation is obtained:

a = the same expression as above

b = the same expression as above ^c = Uw · iou ⁺ Mir · lav + U w · low = ^{~312 "} U i Io ' ^os ^ + β¾)' + ψ _λ ^{+ α} 1

^{d +} ω)* ⁺ φ e= the same expression as above . Ι

_{σ |} .8Ϊη(( _{) + ωι} )/ ₊ α)

When it concerns the motor mode and if the conditions are balanced, the power which is delivered to the grid must be equal to the power of the motor ie, b = e, wherein these terms in the equation are neutralized and remain the time- varying terms c, d and f - whose average value at theirs periodes is zero.

Therefore we get that Ic = 0 (for periods ω-ωΐ or ω + ωΐ).

The same can be written Namely le=0 as well . UvUo Ii'Io ^{are tne} values of the sinusoidal voltages and currents.

This type of converter has been tested by the programme LT Spice for generator and motor mode for performances with same direction of the rotating fields L1-L2-L3 and U-V-W, as well as for the opposite direction of rotating fields. Simulation has shown that such values of harmonized currents and voltages are possible when the voltage on the capacitor CI does not increase but is equal to the maximum grid voltage, plus the voltage of the motor winding (that applies to the voltage + Vc; for the voltage - Ve and the capacitor C2 this applies equally, but here the voltage is negative).

The weak point of this converter is that it feeds a current harmonic (subharmonics) into the grid, which corresponds to the voltage frequency of the motor ul. Another weak point is that CI and C2 should be relatively large (at a 0.55kW motor, it should be CI = C2 = 150..330pF) because the capacitors should "swallow" the terms c, d and f in the formula for Ic (respectively for Ie).

The advantage of this embodiment of the frequency converter is a small number of components and the possibility to operate in both generator and motor mode.

Figure 6 shows the motor operating mode, while Figure 7 shows the generator operating mode.

If this frequency converter is used for soft starting of the asynchronous motor, when the frequency of 50Hz (ul) is reached, the capacitors CI and C2 can be used to improve the "cos φ" of the grid.

Then the IGBT bridge needs to regulate the current and the individual phases of the io-il (which now both have the grid frequency) that it precedes the individual phase voltage of the grid (as if it were about the capacitance load).

The microprocessor and its accompanying electronics must control the IGBT bridge in such a way that the power of the motor and the power of the grid are equal, for which the voltage + Vc and -Ve serve as indicator, ie when it does not increase, then the basic condition for the stable operation of this frequency converter is fulfilled.

FREQUENCY CONVERTER COMPRISING MOTOR WITH TWO SETS OF WINDINGS. It is obvious that a converter with one set of windings has the current Ic, respectively Ie, which is time-varying and different from zero, but its average value is zero.

The converter also feeds the sub harmonics into the grid, which is quite inconvenient. These two problems can be solved by applying one more stator winding of the motor. In that case, this is a machine with two sets of winding (U, V and W and U ', V and W ').

Generally, one phase is then supplied with current, as shown Figure 8. Here it can be seen that double number of the semiconductor switches is required, i.e. 12. Now the control electronics must control the IGBTs in such a way that the currents il and il ' are equal in value but opposite in terms of direction, so that currents il and il' will not be fed into the grid, but would be closed within the circuit winding-winding'- IGBT-C-C-IGBT.

Figure 9 a) b) c) shows the connection example on the pair of windings W and W. Figure 9a) shows the currents and winding directions (pay attention to the dot on the winding, ie sense / direction).

In Figure 9b can be seen that the effect of the currents iow and iow' neutralize within the machine (their total ampere-turns are equal to zero).

In Figure 9c can be seen that the effect of the currents ilw and ilw ' shall be summed within the machine (their total ampere-turns are summed).

If we apply the formula

Ic = 1 / (Vc + Ve) ^■ [a + b + c - d - e - f] Then terms c, d and f have sinusoidal and cosine terms in which the shift a occur between LI, L2, L3 in one case of winding U, V, W and in other case of winding U ', V, W.

This shift is 0 ° for U, V, W, wherein for U ', V, W 'is equal to 180°. Since the sinus and cosine terms are now equal in sum but with opposite prefix (sin (x + 180°) = - sinx; cos (x + 180°) = - cosx), their effect is neutralized.

Due to this neutralization, the capacitors CI and C2 can be very small in amount. The simulation performed in LTspice programme showed that CI = C2 = 4.4 F is sufficient for the l.OkW motor with 30Hz control and 50Hz power supply.

The complete energy section is shown in Figure 10.

The feature of this system is:

- uses a double three phases winding

- uses 12 semiconductor switches

- feeds / draws sine current from the grid

- It can work as a motor and a generator

- The motor / generator can rotate in both directions

- No dissipative element is required for braking; braking energy returns to the grid

- no need for any DC link

- No need for electrolytic capacitors, but for very small 'booster' capacitors (about 2x4 F per kilowatt of installed power, ie the power of the machine).

MATHEMATICAL EXPRESSIONS FOR CONVERTER SHOWN IN FIGURE 1.

T = i +—

2=~(Vru _w -L-^)

uyir -3 0 I ur.tr J

L ₌ U & L

phase U→ phaseV→m=+\ phaseW→ m = -1 cos0 + /-sin©

sin0 =— '(β*-β-* = cos0— . -sin© 2j

2 L J

-^--T

Λ ; . A/ _ L . f _E J-""i2°° . to . _e a> _+e - iio° , -j-a ^{t ll 1} 2j

MATHEMATTCAL EXPRESSIONS FOR CONVERTER SHOWN IN FIGURE 5.

^ ^c Y ₊ y ^' ty £ ^' {i°u ⁺ iov ⁺ iow iiu~iw / ^' in')} ⁺

~V +1/ ^' (uov-hu + Uo ^y - + Uo ^' iwh jr jr ^' («it/ ^' hu ⁺ Uiv ^m / ⁺ U w ^' iiw) ~ ∑ ₀ - ^M V c V E V c V E ¹ u jy

2, = /, - ^sin (ft), ^* ' ^{+ m} · ¹²⁰ ° + ^α ) = J7- ' ^eJ " ^■ ^ ^'ω '' - e ^" -^"" · e ^";a■ e ^" *^}

_Ml = JJ _y ■ sin (ft ₎ ,■/ + ψ _] + m -120° + a) = " . _e ^Ja ■ _e ∞ _ _e --w . _e - ^j _a . _e -¾^> 0 = / ₀ -sin (^, ^■ / + n - 120°) = Ifi. ( _e ™ . e ^j ® ' - e^"™ ^" ·ε ^~ίω *} W ₀ = iJ ^(ft) ^■ ' + 0> +»-120*) = ^.{^". _e ^<a, ^ ⁾ - _e - ^w .e ^"i<a ''^ ^>

rotating fields of the same direction :

U→n = 0,m = 0

V→n = +Y,m-+\

(F→n = -l,m = -l rotating fields of opposite directions :

U→« = 0,m = 0

V→n = +l,m = -\

Rotating fields have the same direction :

eJim+n>nW _ _e ; 0+0)120 ^« _+β /(1+1).110' _+e H-\-\)-l2V _ _e jV _+e JM0' _+e -/M0° _. Q u ,w

e- _m + _n >120' _ _eJ / 0+0) liO' _+£ - 1+1) 120° _{+ g} -H- 1-1)110° _ _+e ~J ^2K ' + _β ' ² °° = 0

e - _m+n ).i20- ₌ ^o- _+e (y _{+ £} ,,ο° ₌₃

uy,w

Rotating fields have opposite directions

£ _e H→W _{= e} J0> _{+ e} ,0 ^« _+e 0° ₌ 3

u,v,w

£ _e - < _m+n) .i20° ₌ ^ _+e ;o° ₊ ^ ₌ 3 e (m-n) l20* _ _C O+OH 0° _+e /<-l-l)-l20' _{+ β} ,.(1+1).120° _Q ^ _e R-m+>>-\7V> _ _e 0+O)120° _{+ £} L+1>L20" _{+ £} -1-1).120° _ Q

U.VW

Rotating fields have the same direction :

∑ _u 4 -^ι^^ο+0-3. _β ^. _β ^'^_ ₃ .^. _β -{(-¾>^1

' J

Rotating fields have opposite directions:

'.Yjr " 4j I J ¾ ^U 1 "'o ^{= "} ~ 2 ^{' '} ^° ^{' C0S} [( ^{ω +} <*>> ) ^{' / +} Ψι ⁺ "]

Rotating fields have the same direction :

∑ M _o 'i ₁ ^sa -C/ _o -/i- ^00S [( ^® -fi>,)-'+9'-«]

Rotating fields have opposite directions:

u ∑,v,w ^r ο'Ι -∞*[{ ^ω+ ω)·*+φ+<*\ ^{1 = ~} h j ₀ ^_J °>

, = /,- sin ( _& -t+ m · 120° + a) = · {e ^, """° ^° - ' - c ^{"jm ηο°} · e- ^ar - e-'^"}

4j

If the rotating fields have the opposite directions , then :

u ∑yjr / \=fwA^ ^sin [( ^il ' ⁺ 6>i) ^, ' ^+a ]

If the rotating fields have the same direction , then :

If the rotating fields have the opposite directions , then :

∑ / , · ' ₀ = - ω·Ι ₀ ·Ι _} -sin [(fl> + J ·/ +a] u.v,w

If the rotating fields have the same direction , then :

If the rotating fields have the opposite directions , then :

3 3

°-2-«rA^o ^,sin [( ^w+ «,)- ^/+a ]-2-^ ^* o^, ^'sin [(^ ^{+ *i+ir} ] ^{+ 0}

Σ ^(ί _ο -ι ·( _ο - ^ϊ \ ^β ·( ^ω+ ωι)· ο· ^/ ₁ · ^8ίη [( ^β+ ω0· ^ί+β ]

FREQUENCY CONVERTER WITH ONE BOOSTER CAPACITOR FOR ONE- PHASE POWER SUPPLY

In the embodiment of the frequency converter with one booster capacitor for one- phase inlet voltage, the energy for the motor is supplied from the booster capacitor CI. The alternating voltage of the motor windings has the amplitude U _a . The voltage +Vi cannot be lower than U _0i therefore, when the voltage level in the is lower than the said voltage, than the diode bridge does not conduct. When the diode bridge does conduct, then the CI is charged with the inlet current Ii, which is evenly distributed as a DC component through the motor windings. As this is the same current through all of the three windings, the effect of this component is neutralized within the motor.

Figure 16 shows the frequency converter with one booster capacitor (Ri and Ci are added at the input; Ci is here for the purpose of closing the component ΔΙϊ; this component is generated when the IGBT bridge switches between OV and + Vc). In the simulation of the work, a motor model of 0.55kW was used, which runs at 30Hz with input voltage ~ 230V and with Ua = 150V. It was simulated by Ci = l F and CI = 7pF with a 227W network load.

Figure 17 shows the voltage + Vi / Vmax where Vmax is the amplitude of the input single-phase voltage. The conduction angle (time) of the diode bridge is a, which ranges from the angle βΐ to the angle β2.

Figure 18 shows an idealized view of the current Ii, i.e. the current of the diode bridge,

The power that comes from the grid is:

P = 1/π · x)dx,

This after integration results as:

P = 2/π ^■ / ₍ · Umax · cos(jt/2 - a/2), i.e. this is the power that comes from the grid. The power consumed by the motor is:

P = 3 · i/ _a /V2 · lac ^■ cos?) = 3 ^■ Umax · sin frr/2 - a/2)/V2 · I _ac · cosy

( lac is effective value of the alternating component of the motor winding current).

By equating the above-stated formulas is obtained

(/3 = lac · π/(2 · V2) · tg ( τ/2 - a/2) · cos<p, i.e. it is a pulse DC component through one motor winding.

Total effective current that heats the winding consists of lac and Ii / 3 and it amounts:

I ² = + α/π · (/ ₍ /3) ² = & · [1 + α · π/Β · tg\n/2 - a/2) · cos ² <p].

Because of the heating of the stator, the value of this current must be lower than or equal to the nominal current of the motor, therefore, the value of the alternating current with which we can load the motor can be deduced from:

l _a = l/V ¹ + α · π/Β · tg ² (n/2 - a/2) · cos ² <p .

Based on this formula, we may draw a diagram shown in Rgure 19, which shows how much the motor may be loaded depending on the frequency at which it is driven. It is assumed that the motor drives under the law Ua / f = const.

Figure 20 shows the waveforms presented by the above-mentioned simulation. It is shown the voltage + Vc (it can be seen how the voltage changes with respect to how the CI is charged and discharged), the diode bridge current Ii, and the voltage of the grid (if it were rectified).

REDUCING THE CURRENT RIPPLE. In order to carry out the periodical ripple reduction, it is necessary to notice the medium-frequency component of the current within the current with which the proposed frequency converters load the grid. The frequency of this current is equal to the modulation frequency. This current (whose frequency is equal to the modulation frequency) can be reduced / eliminated by using the known active or passive methods. The use of active method is shown in Figure 21. At the potential + Vi, toward the neutral point, the current Ii + i (t) flows, where i(t) is an undesirable component that needs to be reduced. Therefore, the capacitor Ci is placed, which is by its other side connected to the amplifier / generator. This current generator detects the current (t) over a sensor (eg resistor) and sends the current i'(t) into the capacitor, which is near equal or equal to i(t). So the current i(t) is closed in the circuit GND-amplifier-Ci+Vi and is further directed to the motor and igbt bridge.

Earlier in Figure lib the current iLl (t) was shown, ie the grid current. By using the two bipolar transistors (npn and pnp) and sources +20V and -20V toward the electronic zero, this method is simulated by the programme LTSpice. As a result, the current iLl(t) is obtained, shown in the figure below (Figure 22). The source of +20V was at average load of 0.842W, and the source of -20V was at load of 0.838W. In simulation there was Ci = lpF.

It is apparent in Figure 22 that good results are obtained even by application of the low power.

MOTOR WINDINGS PREFERABLE FOR THE PROPOSED FREQUENCY CONVERTERS. Figure 23 shows the windings of the AC motor for the converter with one booster capacitor. It is preferable that the motor in this application has a two-layer winding. Possible combinations are shown in figure. The windings are shown with component Ii / 3, which is the component that needs to be annulled with respect to the magnetic conditions in the machine. Oh the windings shown on the upper and left side, as well as on those shown on the right side, it is apparent that the influence of current Ii is annulled with respect to the magnetic conditions in the machine, because at each point on the rim of the machine there are current components Ii / 3 which are opposite, so their influence is neutralized.

Figure 24 shows the winding of the AC motor for the converter with two-booster capacitors. In the upper left hand section there are components visible, effect of which is summed in respect to the magnetic conditions in the machine, while in the upper right hand section the components are shown whose effect is annulled with respect to magnetic conditions in the machine. The two-layer winding of the machine is required. On the bottom side, the components are shown in phases U, V and W.

BRIEF DESCRIPTION OF THE DRAWINGS:

The figures describe the embodiments of the frequency converter and the system parts in accordance with the essence of the invention and serve as an example, but not as a limitation. The reference signs on the figures denote the parts described, or the similar parts. For the parts or sets of the parts that are not denoted by the reference signs on certain figures, they are deemed to be marked by a reference sign corresponding to the one by which the identical part is marked.

Figure 1 shows a proposed frequency converter with diode rectifier at input

Figure 2 shows a proposed frequency converter powered by a battery / direct current source

Figure 3 shows a proposed frequency converter with reversible rectifier at input Figure 4 shows the scheme of the voltage control on the booster capacitor and the phase current U

Figure 5 is a frequency converter without an input DC link

Figure 6 is a view of the currents and voltages in motor mode for general converter phase from Figure 5.

Rgure 7 is a view of the currents and voltages in generator mode for general converter phase from Figure 5.

Rgure 8 is a view of the currents in general converter phase without the input DC link; this applies to machine with two sets of windings Figure 9 a) b) c) is an overview of the currents and voltages in the windings W and W on double winding machine

Figure 10 is an overview of the complete energy circuit of the converter, as observed jn Figures 8 and 9.

Rgure 11 is a diagram of the grid current in the simulation for 0.55 kW motor, for converter from Fig.1.

Figure ll.b is a diagram of the grid current in the simulation for 0.55 kW motor, for the converter from Rg.l., based on the second simulation with more realistic model for semiconductor switches and more direct application of the mathematic formulas. Figure 12 is an overview of the grid current and voltages LI, L2, L3 at 30Hz drive of the converter motor from fig. 10, based on the first simulation.

Figure 12. b is an overview of the grid current and voltages LI, l_2, 13 at 30Hz drive of the converter motor from fig. 10, based on the second simulation with more realistic model for semiconductor switches and more direct application of the mathematic formulas

Figure 13 is a FFT view of a single-phase current according to Figure 12, based on the first simulation

Rgure 13.b is a FFT view of a single-phase current according to Figure 12.b, based on the second simulation with more realistic model for semiconductor switches and more direct application of the mathematic formulas

Rgure 14 is an overview of the rotor voltages of asynchronous motor (at R2'/ s) for Fig.10 (and Rg.12), based on the first simulation

Figure 14. b is an overview of the rotor voltages of asynchronous motor (at R2'/ s) for Fig.10. b (and Fig.12. b), based on the second simulation with more realistic model for semiconductor switches and more direct application of the mathematic formulas Rgure 15 is an overview of the capacitor voltages (+Vc, -Ve), based on the first simulation Figure 15.b is an overview of the capacitor voltages (+Vc, -Ve), based on the second simulation with more realistic model for semiconductor switches and more direct application of the mathematic formulas

Rgure 16 is an overview of the present frequency converter (Ri and Ci are added at the input; Ci is here for the purpose of closing the component ΔΙϊ; this component is generated when the IGBT bridge switches between OV and + Vc). In the simulation of the work, a motor model of 0.55kW was used, which runs at 30Hz with input voltage ~ 230V and with Ua = 150V. It was simulated by Ci = lpF and CI = 7pF with a 227W network load.

Figure 17 is an overview of the voltage + Vi / Vmax where Vmax is the amplitude of the input single-phase voltage.

Figure 18 shows an idealized view of the current Ii, i.e. the current of the diode bridge.

Figure 19. Diagram which shows the permissible load the motor depending of the frequency at which the motor is to drive

Figure 20. is an overview of the waveforms of the +Vc voltage change, wherein the voltage changes with respect to how the CI is charged and discharged); it is also overview of the diode bridge current Ii and the voltage of the grid.

Figure 21. shows the active method for the reduction of ripple.

Figure 22 shows the good results in the ripple reduction obtained by application of the low power.

Rgure 23 shows the windings of the AC motor for the converter with one booster capacitor.

Rgure 24 shows the winding of the AC motor for the converter with two-booster capacitors. A LIST OF REFERENCE SIGNS USED IN THE DRAWINGS

10- Phase LI

20-phase L2

30-phase L3

100 current il_l

101 current il_ 2

102 current il_ 3

103- current Im

104- current Ii

110- voltage + Vm

111- voltage + Vi

112- connector on the U2 voltage

113- connector on the V2voltage

114- connector on the W2 voltage

115- voltage Uu

116 -voltage Uv

117- voltage Uw

118- connector on the Ul voltage

119- connector on the Vlvoltage

120- connector on the Wl voltage

130- Motor winding resistance

131 - semiconductor switch or IGBT

132- leakage inductance of electric motor

133- ground

134- diode

135- voltage + Vc

136- current Iu

137- current Iv -current Iw

-booster capacitor CI

- DC source

- amplifier

- voltage Vcref

- hysteresis switch

- GATE LOW

- GATE HIGH

- node before the amplifier input

-current Iu in the loop before the amplifier input current iU (t)

- current Iu

- signal comparison

- capacitor C2

- current Ie

- voltage -Ve

- ground

- circuit of two IGBTs

- voltage ul

- voItage uO

- current I

- current il

currents iO

- winding connector U

- winding connector V

- winding connector W

- motor winding

- current Iu ^* 185- current Iv '

186- current Iw '

187- winding connector U'

188- winding connector V

189- winding connector W ¹

190- IGEnr winding '

200-voltage uO

202- current 2i0

203- voltage ul

204- current iO

205- current il

206- voItage ul '

207 current il '

208 current iO

210- Ieakage inductance of the winding L

211- ieakage inductance of the winding L '

212- winding W

213- winding W ¹

214- voltage uOw

215- current 2i0w

216 current iOw

217 current ilw

218- current ilw ¹

219- current iOw ¹

300- resistance divider DETAILED DESCRIPTION OF AT LEAST ONE WAY OF CARRYING OUT THE INVENTION

The given description of at least one possible embodiment does not in any way limit the invention.

This invention is realized in such a way that the components are mounted on the Printed Circuit Board as shown in the scheme in Figure 1 (for one possible

embodiment). The source of supply with the phases LI 10, l_2 20 and L3 30 supplies the current circuits, whereby the following currents flow: current il_l 100, current il_ 2 101, current iL 3 102, current Im 103 and current II 104. At output, after the diode 134, the voltage + Vm 110 and afterwards the voltage + Vi 111 occur. For each winding of the electric motor 130, 132, the connectors U2 112, V2 113 and W2 114 are carried out at one end, so that the respective voltages Uu 115, Uv 116 and Uw 117 and connectors are on the other end Ul 118, VI 119, and Wl 120. Each winding has an equivalent motor winding resistance 130 and motor leakage inductance 132. The motor windings are followed by a circuit of two IGBTs 131. After the IGBTs 131 at voltage + Vc 135 follows a booster capacitor CI 140 which is at the other end connected to the ground 133. Through the windings of the electric motors, which are shown as the motor leakage inductance 132, the currents Iu 136, Iv 137 and Iw 138 flow.

In another configuration, a DC source 141 may be used.

The regulation circuit is carried out with amplifiers, wherein the reference voltage Vcref 150 is present, which compares 158 with voltage + Vc of the resistance divider 300, and the comparison difference goes further to amplifier 142. To the output signal from the amplifier is added the reference for the phase current iU (t) 156. node loop 154 before the input of the amplifier; and then the amplified signal goes to the hysteresis switch 151, which further turns on the high or low level IGBT 131. The hysteresis switch 151 has two outputs GATE HIGH 153 and GATE LOW 152, which outputs are connected to the IGBTs 131 into which the current 157 is divided. In another embodiment, along with the booster capacitor CI 140 there is also a capacitor C2 160, wherein the current Ic 164 flows through capacitor CI, while the current Ie 161 flows through the capacitor C2. The capacitor C2 is at the potential - Ve 162.

Figures 6 and 7 show a simplified scheme for one general phase with current 1 168, sources ul 166 and uO 167, as well with the corresponding currents il 169 and i0 170 with ground N 163 and circuit of two IGBTs 165.

In a further embodiment, a 3-phase double-winding machine 183 with 12 semiconductor switches 131 is used. Here are also the winding connector U 180, the winding connector V 181, the winding connector W 182, the winding connector U Ί87, the winding connector V 188, and the winding connector W Ί89. In addition, here also flow the currents Iu ¹ 184, Iv' 185 and the current Iw' 186. The second winding has the semiconductor switches 190, and through these windings flow currents Iu ¹ 184, current Iv Ί85 and current Iw' 186,

In a simplified view for one general phase, the voltage uO 200 is a source of power supply, the voltages present on two pairs of winding are ul 203 and ul' 206, and the current that flow through are 2i0 202, iO 204, il 205, iO 208 and il '207. Further elements are leakage inductance of the winding L 210 and L' 211, and the

semiconductor switches 131.

Furthermore, one pair of windings may be W 212 and W '213, with source uOw 214 and the currents flowing 2i0w 215, iOw 216, ilw 217, iOw'219 and ilw' 218.