CN101562922A - High brightness LED driving power without electrolytic capacitor - Google Patents

Abstract

The invention relates to a high brightness LED driving power without an electrolytic capacitor, which comprises a primary power circuit, an output current detect filter circuit, an input voltage feed forward circuit, an output current feedback control circuit, a multiplier and a pulse-width modulation chip. The flyback converter topology is used to work in the current-interruption mode, power factor correction is automatically realized, the methods of series inductance and filter capacitor are added and used so as that output current can be close to ideal sine square wave; appropriate low-order harmonic is injected into input current to further lower peak-to-average ratio of output current, which is realized by input current feed forward method of altering the change regularity of dutyfactor in the working frequency cycle and future lowering peak-to-average ratio of output current to 1.38. The driving power with high power factor and low peak-to-average of output current can be provided for high brightness LED lamps without electrolytic capacitor. By canceling the electrolytic capacitor with large volume, short service life and high capacity, the driving power of the invention has the advantages of small volume, long service life and the like.

Description

The high brightness LED driving power of no electrolytic capacitor

Technical field

The high brightness LED driving power of no electrolytic capacitor of the present invention, the A.C.-D.C. converter of genus transformation of electrical energy device.

Background technology

(Light Emitting Diode LED) with plurality of advantages such as its energy-saving and environmental protection, efficient, long-lives, becomes the green illumination light source of a new generation to high brightness LED.Increasingly mature along with the high-brightness LED lighting technology, it will be widely used in every field, and become the first-selection of lighting source.Manufacturing high efficiency, low cost, High Power Factor, long-life driving power are the keys that guarantees LED luminescent quality and overall performance.Normal lighting occasion in the civil power input often adopts the driving power framework shown in the accompanying drawing 1, is divided into adapter and driver two parts.The function of adapter is to realize that (Power Factor Correction PFC) with AC/DC conversion (AC/DC), provides the burning voltage of 24V or 12V for the back level in the input power factor correction.Driver is made up of the special chip for driving of LED, for the steady operation of high-brightness LED provides constant current.Two-stage type LED driving power can guarantee the luminescent quality of LED preferably, but has shortcomings such as device is many, volume is big, the life-span is short.Suppose that the pfc converter input power factor is 1, then input current i _InBe and input voltage v _InSynchronous sine wave, as shown in Figure 2.This moment input power p _InBe sinusoidal quadratic form, realize constant voltage output, i.e. P _oConstant, need to adopt the bigger electrochemical capacitor of appearance value to realize the balance of input, power output.But because the life-span of electrochemical capacitor and the working life of LED differ greatly, so electrochemical capacitor becomes the principal element that influences LED driving power bulk life time.And the electrochemical capacitor volume is bigger, has influenced the further raising of driving power power density.

Summary of the invention

The objective of the invention is for overcome above-mentioned two-stage type LED driving power because of use that electrochemical capacitor can not match with the long-life of LED major defect, design a kind of high brightness LED driving power that does not use electrochemical capacitor.

Another object of the present invention is to make output current near desirable sinusoidal square wave, and promptly the output current peak-to-average force ratio is reduced to 2 the high brightness LED driving power that does not use electrochemical capacitor.

A further object of the present invention is by changing the Changing Pattern of duty ratio in power frequency period, the output current peak-to-average force ratio is reduced to the high brightness LED driving power of 1.38 the no electrolytic capacitor that volume is little, the life-span is long.

The high brightness LED driving power of common no electrolytic capacitor of the present invention, under the situation of not considering power factor and output current peak-to-average force ratio, its main power circuit (1) comprises and input voltage source v _InThe diode rectifier circuit, Flyback inductance L, switching tube Q and the diode D that connect, the positive pole of described diode rectifier circuit output is connected with the drain electrode of switching tube Q through the primary coil of Flyback inductance L, the source electrode of switching tube Q is connected with the negative pole of diode rectifier circuit output, the gate pole of switching tube Q is connected with the output of pulse width modulating chip (6), and the secondary coil of Flyback inductance L is connected with load LED through diode D.

In order to improve input power factor and to reduce the output current peak-to-average force ratio of the high brightness LED driving power of no electrolytic capacitor of the present invention, can on the basis of above-mentioned main power circuit (1), set up input filter inductance L _In, input filter capacitor C _In, output filter capacitor C _o, output inductor L _oWith the open-circuit-protection capacitor C _Prot, described input filter inductance L _InBe serially connected between the primary coil of the positive pole of diode rectifier circuit output and Flyback inductance L input filter capacitor C _InBe connected input filter inductance L _InOutput and the negative pole of diode rectifier circuit output between, output inductor L _oBe serially connected between diode D and the load LED output filter capacitor C _oWith the open-circuit-protection capacitor C _ProtBe connected to output inductor L _oTwo ends and the other end of load LED between, can make the output current peak-to-average force ratio be reduced to 2.

In order further to reduce the high brightness LED driving power output current peak-to-average force ratio of no electrolytic capacitor of the present invention, can be on the basis of foregoing circuit, set up output current detection filter circuit (2), input voltage feed forward circuit (3), output current feedback control circuit (4), multiplier (5) again, the sample resistance of described input voltage feed forward circuit (3) is connected input filter capacitor C _InTwo ends, the subtracter output E point in the input voltage feed forward circuit (3) is connected with a part input of multiplier (5), and the peak value sampling point B point in the input voltage feed forward circuit (3) links to each other with the denominator input of multiplier (5); Current transformer T primary coil in the output current detection circuit (2) is serially connected between the secondary coil and load LED of Flyback inductance L, the current transformer T secondary coil in the output current detection circuit (2) behind rectifying and wave-filtering with output current feedback control circuit (4) in error adjuster IC ₃Inverting input connect; Error adjuster IC ₃Output F point be connected with another molecule input of multiplier (5); The output of multiplier (5) is connected with the input of pulse width modulating chip (6), the drive signal of exporting with the sawtooth waveforms handing-over back in the pulse width modulating chip (6) is connected with the gate pole of switching tube Q, pulse width modulating chip (6) is the duty cycle signals driving switch pipe Q of a (1-k|sin ω t|) with Changing Pattern, wherein k is by the power factor decision of power requirement, after the k value was determined, a value was determined by power.

Described input voltage feed forward circuit (3) comprises sample resistance R ₁And R ₂, emitter follower IC ₁With subtracter IC ₂, described sample resistance R ₁And R ₂Between sampling point A point and emitter follower IC ₁In-phase input end connect emitter follower IC ₁A road of output obtains the denominator input and the subtracter IC of a certain signal and multiplier (5) at the B point behind peak value sampling ₂In-phase end link to each other, another road obtains a certain signal and subtracter IC at the C point after dividing potential drop ₂Inverting input connect subtracter IC ₂Output E point be connected with a part input of multiplier (5).

Input voltage feed forward circuit (3) is used for modulating the output duty cycle signal of pulse width modulating chip (6), make input current that suitable three times and quintuple harmonics injection be arranged, and injection rate can be regulated; Be not less than at input power factor under 0.9 the situation, the output current peak-to-average force ratio can be reduced to 1.34 from 2; Be not less than at input power factor under 0.95 the situation, the output current peak-to-average force ratio can be reduced to 1.45 from 2.

The high brightness LED driving power of no electrolytic capacitor of the present invention, employing anti exciting converter topology is operated in the discontinuous current pattern, realizes power factor correction automatically, and adopt the method that adds series inductance and filter capacitor, make output current near desirable sinusoidal square wave; Also further propose in input current, to inject an amount of low-order harmonic with further reduction output current peak-to-average force ratio, realized with the input voltage feed forward method, by changing duty ratio, the output current peak-to-average force ratio further is reduced to 1.38 at the Changing Pattern of operating frequency in the cycle.Realized under the situation of no electrolytic capacitor also owing to cancelled short big capacity electrochemical capacitor of big life-span of volume, having obvious advantages such as volume is little, the life-span is long for high-brightness LED lamp provides power factor the high and low driving power of output current peak-to-average force ratio.

Description of drawings

Accompanying drawing 1 is traditional LED driving power structured flowchart;

Accompanying drawing 2 is input voltage in the pfc converter, input current, input power and power output oscillogram;

Accompanying drawing 3 is single stage type high brightness LED driving power schematic diagrames of no electrolytic capacitor of the present invention;

Input and output power and output current wave when accompanying drawing 4 is PF=1;

Accompanying drawing 5 is a kind of electrical block diagrams of the high brightness LED driving power of no electrolytic capacitor of the present invention;

Accompanying drawing 6 is the interior primary current waveforms of a switch periods of anti exciting converter;

Accompanying drawing 7 is former secondary current waveforms of anti exciting converter;

Accompanying drawing 8 is another kind of electrical block diagrams of the high brightness LED driving power of no electrolytic capacitor of the present invention;

Accompanying drawing 9 is accompanying drawing 8 secondary currents and output current wave figure;

Accompanying drawing 10 is output currents when being direct current, input voltage and input current waveform;

Accompanying drawing 11 is output currents when being direct current, preceding 21 subharmonic distribution maps in the input current;

Accompanying drawing 12 is that input current injects triple-frequency harmonics, i _O1+3 ^*(t) with I ₃ ^*Change curve;

Accompanying drawing 13 is that input current injects triple-frequency harmonics, and output current peak-to-average force ratio and input power factor are with I ₃ ^*Change curve;

Accompanying drawing 14 is to change the schematic diagram of change in duty cycle rule to realize that the output current triple-frequency harmonics injects;

Accompanying drawing 15 is desired duty cycle functions

With match duty ratio function 1.72 (1-0.5sin ω t) curve comparison diagram;

Accompanying drawing 16 is that input current injects low-order harmonic, output current wave schematic diagram;

When accompanying drawing 17 is input current injection three times and quintuple harmonics, i _O1+3+5 ^*(t) oscillogram;

Accompanying drawing 18 is I under three kinds of situations ₃ ^*And I ₅ ^*Span;

The output current peak-to-average force ratio was with I when accompanying drawing 19 was situation I ₃ ^*And I ₅ ^*Change curved surface;

The output current peak-to-average force ratio was with I when accompanying drawing 20 was situation II ₃ ^*And I ₅ ^*Change surface chart;

The minimum peak-to-average force ratio of output current was found the solution schematic diagram when accompanying drawing 21 was situation III;

Accompanying drawing 22 is desired duty cycle functions

With match duty ratio function (2.02-1.22|sin ω t|) curve comparison diagram;

Accompanying drawing 23 is another electrical block diagram of the high brightness LED driving power of no electrolytic capacitor of the present invention;

Accompanying drawing 24 is a kind of electrical block diagrams that adopt the multiplier (5) of discrete device composition.

Main designation in the above-mentioned accompanying drawing: v _InSupply voltage.L _In, the input filter inductance.C _In, input filter capacitor.i _n, Flyback inductance primary current.Q, switching tube.D, diode.C _o, output filter capacitor.L _o, output inductor.C _Prot, open-circuit-protection electric capacity.V _o, output voltage.L, Flyback inductance.T, current transformer.

Embodiment

For easy analysis, make the following assumptions earlier:

1. all devices are ideal element, and are lossless;

2. switching frequency is much larger than the input voltage frequency.

1. the high brightness LED driving power basic thought and the circuit topology of no electrolytic capacitor

With reference to the accompanying drawings 3, the input power factor of supposing converter is 1, then input current i _InBe and input voltage v _InThe synchronous sine wave of same frequency is shown in accompanying drawing 4 (a).Input power p _InFor sinusoidal square waveform, shown in accompanying drawing 4 (b).LED has characteristic of semiconductor, and when its conducting, its both end voltage equals its conduction voltage drop, is approximately a constant pressure source, if output current i _oBe sinusoidal square wave, shown in accompanying drawing 4 (c), then power output also is sinusoidal square wave, and equates with input power is instantaneous, shown in accompanying drawing 4 (d).So just can when guaranteeing High Power Factor, remove the electrochemical capacitor in the circuit, prolong the useful life of driving power greatly.

In conjunction with above-mentioned control thought, consider that the LED driving power should have the power factor height, the power density height, characteristics such as volume is little, also need electrical isolation to guarantee user's Electrical Safety simultaneously, selection work of the present invention in the discontinuous current pattern (Discontinuous Current Mode, anti exciting converter DCM) is as the circuit topology of LED driving power, under the situation of not considering power factor (PF) and output current peak-to-average force ratio, its main power circuit (1) comprises and input voltage source v _InThe diode rectifier circuit, Flyback inductance L, switching tube Q and the diode D that connect, the positive pole of described diode rectifier circuit output is connected with the drain electrode of switching tube Q through the primary coil of Flyback inductance L, the source electrode of switching tube Q is connected with the negative pole of diode rectifier circuit output, the gate pole of switching tube Q is connected with the output of pulse width modulating chip (6), being connected with load LED through diode D of the secondary coil of Flyback inductance L.Its structure as shown in Figure 5.Accompanying drawing 6 has provided anti exciting converter former secondary current waveform in a switch periods.

Make input ac voltage be:

v _in＝V _msinωt (1)

V wherein _mBe the input ac voltage amplitude, ω is the input ac voltage angular frequency.

During switching tube Q conducting, at input voltage v _InUnder the effect, primary current i _pThe linearity of starting from scratch rises, when Q turn-offs, and i _pReach its peak value i _{P_pk}, that is:

$i_{p\_\mathrm{pk}}=\frac{V_m\left|\sin \mathrm{\ωt}\right|\·D_y}{L_p\·f_s}---\left(2\right)$

L wherein _pBe the former limit of transformer inductance value, D _yBe switching tube duty ratio, f _sBe switching frequency.

In a switch periods, primary current mean value i _{P_av}For:

$i_{p\_\mathrm{av}}=\frac{1}{2}\&CenterDot;i_{p\_\mathrm{pk}}\&CenterDot;D_y=\frac{V_m\left|\sin \mathrm{\&omega;t}\right|\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="D\; y"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">D</figure-callout>}}_y^{}}{2\&CenterDot;L_p\&CenterDot;f_s}---\left(3\right)$

In order to make the anti exciting converter that is operated in the DCM pattern realize PFC automatically, should control duty ratio D _yIn a power frequency period, remain unchanged, from formula (3) as can be seen, i _{P_av}Be proportional to the absolute value of input voltage, corresponding, input current is proportional to input voltage.

When switching tube Q turn-offs, secondary diode D conducting, the energy delivery that is stored in the former limit of transformer arrives secondary, its voltage substantially constant during owing to the LED conducting, so secondary current i _sFrom its peak value i _{S_pk}Linearity drops to zero.According to conservation of energy principle, secondary current peak value i _{S_pk}For:

$i_{s\_\mathrm{pk}}=n\&CenterDot;i_{p\_\mathrm{pk}}=\frac{n\&CenterDot;V_m\left|\sin \mathrm{\&omega;t}\right|\&CenterDot;D_y}{L_p\&CenterDot;f_s}---\left(4\right)$

N is the former secondary turn ratio of transformer in the formula.

Secondary current i _sDrop to zero time t from its peak value _rFor:

$t_r=\frac{L_s\&CenterDot;i_{s\_\mathrm{pk}}}{V_o}=\frac{L_s\&CenterDot;n\&CenterDot;V_m\&CenterDot;\left|\sin \mathrm{\&omega;t}\right|\&CenterDot;D_y}{V_o\&CenterDot;L_p\&CenterDot;f_s}=\frac{V_m\&CenterDot;\left|\sin \mathrm{\&omega;t}\right|\&CenterDot;D_y}{n\&CenterDot;V_o\&CenterDot;f_s}---\left(5\right)$

V in the formula _oBe output voltage, its value is V _o=mV _LED, i.e. the conduction voltage drop of m LED.

Then secondary current drops to zero corresponding duty ratio D _RFor:

$D_R=\frac{t_r}{T_s}=\frac{V_m\left|\sin \mathrm{\&omega;t}\right|\&CenterDot;D_y}{n\&CenterDot;V_o}---\left(6\right)$

So in a switch periods, the mean value i of secondary current _{S_av}For:

$i_{s\_\mathrm{av}}=\frac{1}{2}\&CenterDot;i_{s\_\mathrm{pk}}\&CenterDot;D_R=\frac{{\mathrm{<figure-callout\; id="2"\; label="V\; m"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">V</figure-callout>}}_m^{}{\mathrm{<figure-callout\; id="2"\; label="sin"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">sin</figure-callout>}}^2\mathrm{\&omega;t}\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="D\; y"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">D</figure-callout>}}_y^{}}{2\&CenterDot;L_p\&CenterDot;V_o\&CenterDot;f_s}---\left(7\right)$

Can release the mean value I of output current in half power frequency period by following formula _oFor:

$I_o=\frac{2}{T_{\mathrm{line}}}{\&Integral;}_0^{\frac{T_{\mathrm{line}}}{2}}{i}_{s\_\mathrm{av}}\&CenterDot;\mathrm{dt}=\frac{{V_{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">m</figure-callout>}}}^2\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="D\; y"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">D</figure-callout>}}_y^{}}{4\&CenterDot;L_p\&CenterDot;V_o\&CenterDot;f_s}---\left(8\right)$

In half power frequency period, the envelope of converter input current peak value, mean value and secondary current peak value is | sin ω t| form, and the secondary average current is sin ²ω t form is shown as accompanying drawing 7.

2. add series inductance and filter capacitor to reduce the output current peak-to-average force ratio

In the such scheme, the pulsating current peak-to-average force ratio that LED bears is very big, easily causes the damage of LED.For addressing this problem, can on the basis of the main power circuit (1) of above-mentioned Fig. 5, set up input filter inductance L _In, input filter capacitor C _In, output filter capacitor C _o, output inductor L _oWith the open-circuit-protection capacitor C _Prot, as shown in Figure 8, described input filter inductance L _InBe serially connected between the primary coil of the positive pole of diode rectifier circuit output and Flyback inductance L input filter capacitor C _InBe connected input filter inductance L _InOutput and the negative pole of diode rectifier circuit output between, output inductor L _oBe serially connected between diode D and the load LED output filter capacitor C _oWith the open-circuit-protection capacitor C _ProtBe connected to output inductor L _oTwo ends and the other end of load LED between, in circuit, seal in inductance, constitute low pass filter, with elimination electric current medium-high frequency harmonic component; Add filter capacitor simultaneously,, can make the output current peak-to-average force ratio be reduced to 2 for the high-frequency harmonic electric current provides path.

3. reduce the control thought of output current peak-to-average force ratio

In order further to reduce the output current peak-to-average force ratio, need the waveform of clear and definite output current and the relation of peak-to-average force ratio and other variable.By above analysis as can be known, the expression formula of the input current of driving power and output current is respectively:

$i_{\mathrm{in}}\left(t\right)=\frac{V_m\sin \mathrm{\&omega;t}\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="D\; y"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">D</figure-callout>}}_y^{}}{2\&CenterDot;L_p\&CenterDot;f_s}=I_1\&CenterDot;\sin \mathrm{\&omega;t}---\left(9\right)$

$i_o\left(t\right)=\frac{{{\mathrm{<figure-callout\; id="2"\; label="V\; m"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">V</figure-callout>}}_m}^2{\mathrm{<figure-callout\; id="2"\; label="sin"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">sin</figure-callout>}}^2\mathrm{\&omega;t}\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="D\; y"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">D</figure-callout>}}_y^{}}{2\&CenterDot;L_p\&CenterDot;V_o\&CenterDot;f_s}---\left(10\right)$

I ₁Be the fundamental voltage amplitude of input current, $I_1=\frac{V_m\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="D\; y"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">D</figure-callout>}}_y^{}}{2\&CenterDot;L_p\&CenterDot;f_s},$ The pass that can be released output current and input current by formula (9) and formula (10) is:

$i_o\left(t\right)=\frac{p_{\mathrm{in}}\left(t\right)}{V_o}=\frac{V_m\sin \mathrm{\&omega;t}\&CenterDot;i_{\mathrm{in}}\left(t\right)}{V_o}---\left(11\right)$

Because output voltage V _oBe direct current, by following formula as can be known, output current i _o(t) waveform and input power p _In(t) waveform unanimity, promptly an input current and a SIN function are long-pending.Optimal output current peak-to-average force ratio is 1, and this moment, output current was a direct current, and the waveform of input current should be sinusoidal inverse so, as shown in Figure 10.By this figure as can be seen, in input voltage zero passage place, input current infinity.In order to analyze input current harmonics distribution situation this moment, the maximum of input current is limited in 100 times of fundamental voltage amplitude, then it is carried out fourier decomposition, resulting preceding 21 subharmonic distribution maps as shown in Figure 11, I wherein _n ^*Be the perunit value of each harmonic current amplitude, its benchmark is the fundamental current amplitude.As can be seen, comprised a large amount of odd harmonics in the input current, by calculating, input power factor is approximately equal to zero.

By above-mentioned analysis, guaranteeing that input power factor satisfies under the condition of commercial lighting requirement, can in input current, inject an amount of odd harmonic, make output current obtain lower peak-to-average force ratio.

4. input current injects the relation of triple-frequency harmonics and output current peak-to-average force ratio and input power factor

In input current, inject triple-frequency harmonics,, can realize the reduction of output current peak-to-average force ratio by changing the waveform of output current.Below will analyze input current triple-frequency harmonics injection method, the input current behind the injection triple-frequency harmonics is:

$i_{\mathrm{in}1+3}\left(t\right)=I_1(\sin \mathrm{\&omega;t}+I_3^{*}\&CenterDot;\mathrm{<figure-callout\; id="3"\; label="sin"\; filenames="A20091002736000262.png,A20091002736000263.png"\; state="\{\{state\}\}">sin</figure-callout>}3\mathrm{\&omega;t})---\left(12\right)$

I wherein ₃ ^*The triple-frequency harmonics amplitude that is injection is about fundamental voltage amplitude I ₁Perunit value.

With formula (12) substitution formula (11), can release output current and be:

$i_{o1+3}\left(t\right)=\frac{V_m{I}_1}{V_o}\&CenterDot;\sin \mathrm{\&omega;t}\&CenterDot;(\sin \mathrm{\&omega;t}+I_3^{*}\&CenterDot;\mathrm{<figure-callout\; id="3"\; label="sin"\; filenames="A20091002736000262.png,A20091002736000263.png"\; state="\{\{state\}\}">sin</figure-callout>}3\mathrm{\&omega;t})---\left(13\right)$

Following formula is carried out integration, obtain at the mean value of half power frequency output current in the half period be:

${\mathrm{<figure-callout\; id="1"\; label="I\; o"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">I</figure-callout>}}_o=\frac{2}{T_{\mathrm{line}}}\&CenterDot;{\&Integral;}_0^{\frac{T_{\mathrm{line}}}{2}}\frac{V_m\&CenterDot;I_1}{V_o}\&CenterDot;\sin \mathrm{\&omega;t}\&CenterDot;\left(\sin \mathrm{\&omega;t}+I_3^{*}\&CenterDot;\mathrm{<figure-callout\; id="3"\; label="sin"\; filenames="A20091002736000262.png,A20091002736000263.png"\; state="\{\{state\}\}">sin</figure-callout>}3\mathrm{\&omega;t}\right)\&CenterDot;\mathrm{dt}=\frac{V_m\&CenterDot;I_1}{2\&CenterDot;V_o}---\left(14\right)$

Know the size of output current mean value and triple-frequency harmonics injection rate I by formula (14) ₃ ^*Irrelevant.In order to study output current peak-to-average force ratio and I ₃ ^*Relation, the definition output current perunit value (being peak-to-average force ratio) be:

$i_{o1+3}^{*}\left(t\right)=\frac{i_{\mathrm{<figure-callout\; id="1"\; label="o"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">o</figure-callout>}1+3}\left(t\right)}{{\mathrm{<figure-callout\; id="1"\; label="I\; o"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">I</figure-callout>}}_o}---\left(15\right)$

Accompanying drawing 12 has provided following formula in different I ₃ ^*Under the value, half interior change curve of power frequency period, obviously this sets of curves is about ω t=pi/2 symmetry.At ω t=pi/2 place, because triple-frequency harmonics is opposite with fundamental phase, along with the increase of triple-frequency harmonics injection rate, the instantaneous value at ω t=pi/2 place constantly reduces, and is corresponding, and the both sides current value increases, and makes output current be the shape of a saddle.Following surface analysis different I ₃ ^*The time output current peak value moment and the size that occur.

To ω t differentiate, make it equal 0, that is: formula (15)

$\frac{d\left({\mathrm{<figure-callout\; id="1"\; label="i\; o"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">i</figure-callout>}}_o^{1+3}\left(t\right)\right)}{d\left(\mathrm{\&omega;t}\right)}=2\mathrm{<figure-callout\; id="2"\; label="sin"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">sin</figure-callout>}2\mathrm{\&omega;t}\&CenterDot;(1+3I_3^{*}-8I_3^{*}\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="sin"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">sin</figure-callout>}}^2\mathrm{\&omega;t})=0---\left(16\right)$

Can be by following formula in the hope of output current, occur in the ω t ∈ [0, pi/2] extreme value the time be carved with 3, that is:

$\mathrm{\&omega;t}=0,\frac{\mathrm{\&pi;}}{2},\mathrm{<figure-callout\; id="1"\; label="arcsin"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">arcsin</figure-callout>}\sqrt{\frac{1+3I_3^{*}}{8I_3^{*}}}---\left(17\right)$

12 as can be known with reference to the accompanying drawings, and output current is about ω t=pi/2 symmetry, ω t ∈ (pi/2, π] in, also have corresponding with it extreme point:

$\mathrm{\&omega;t}=\mathrm{\&pi;}-\mathrm{<figure-callout\; id="1"\; label="arcsin"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">arcsin</figure-callout>}\sqrt{\frac{1+3I_3^{*}}{8I_3^{*}}},\mathrm{\&pi;}---\left(18\right)$

Obviously when ω t=0 and π, output current is 0, is minimum point, only need to discuss below therefore ω t (0, pi/2] equal in the interval

Situation with pi/2.Characteristics according to trigonometric function can know, for $\mathrm{\&omega;t}=\mathrm{<figure-callout\; id="1"\; label="arcsin"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">arcsin</figure-callout>}\sqrt{\frac{1+3I_3^{*}}{8I_3^{*}}}$ , the precondition that this formula is set up is:

$\sqrt{\frac{1+3I_3^{*}}{8I_3^{*}}}\&le;1---\left(19\right)$

Can try to achieve I by following formula ₃ ^*〉=0.2.That is to say, work as I ₃ ^*＜0.2 o'clock, extreme point $\mathrm{\&omega;t}=\mathrm{<figure-callout\; id="1"\; label="arcsin"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">arcsin</figure-callout>}\sqrt{\frac{1+3I_3^{*}}{8I_3^{*}}}$ Be non-existent.This moment i _O1+3 ^*(t) have only a maximum point ω t=pi/2,, can get the output current peak-to-average force ratio and be its substitution formula (15):

${\mathrm{<figure-callout\; id="1"\; label="I\; o"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">I</figure-callout>}}_o^{1+3}\left(t\right){|}_{\mathrm{\&omega;t}=\frac{\mathrm{\&pi;}}{2}}=2\&CenterDot;(1-{\mathrm{<figure-callout\; id="3"\; label="I"\; filenames="A20091002736000262.png,A20091002736000263.png"\; state="\{\{state\}\}">I</figure-callout>}}_3^{*})$ (I ₃ ^*＜0.2)(20(a))

Work as I ₃ ^*〉=0.2 o'clock, 12 as can be known with reference to the accompanying drawings, $\mathrm{\&omega;t}=\mathrm{<figure-callout\; id="1"\; label="arcsin"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">arcsin</figure-callout>}\sqrt{\frac{1+3I_3^{*}}{8I_3^{*}}}$ Must be maximum point, and ω t=pi/2 must be minimum point, will $\mathrm{\&omega;t}=\mathrm{<figure-callout\; id="1"\; label="arcsin"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">arcsin</figure-callout>}\sqrt{\frac{1+3I_3^{*}}{8I_3^{*}}}$ Substitution formula (15) can get the output current peak-to-average force ratio and is:

${\mathrm{<figure-callout\; id="1"\; label="I\; o"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">I</figure-callout>}}_o^{1+3}\left(t\right){|}_{\mathrm{\&omega;t}=\mathrm{<figure-callout\; id="1"\; label="arcsin"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">arcsin</figure-callout>}\sqrt{\frac{1+3I_3^{*}}{8I_3^{*}}}}=\frac{{(1+3I_3^{*})}^2}{{8I}_3^{*}}$ (I ₃ ^*≥0.2)(20(b))

Can obtain the output current peak-to-average force ratio with I by formula (20) ₃ ^*Change curve, as shown in Figure 13.With formula (20 (b)) to I ₃ ^*Carry out differentiate, can proper I ₃ ^*=0.33 o'clock, output current peak-to-average force ratio minimum was 1.51.

Injecting triple-frequency harmonics in input current can cause input power factor to reduce.According to input power factor PF computing formula:

$\mathrm{PF}=\frac{\frac{I_1}{\sqrt{2}}}{\sqrt{{\mathrm{<figure-callout\; id="2"\; label="I\; o"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">I</figure-callout>}}_o^{}+\underset{m=1}{\overset{\&infin;}{\mathrm{\&Sigma;}}}\left(\frac{I_m^2}{2}\right)}}---\left(21\right)$

Can obtain PF about I by following formula ₃ ^*Curve, as shown in Figure 13, as can be seen from the figure, work as I ₃ ^*=0.33 o'clock, PF=0.95 can satisfy the requirement of commercial illumination to PF 〉=0.9.

5. the implementation method injected of input current triple-frequency harmonics

From formula (9) as can be seen, the injection of input current triple-frequency harmonics can be by changing duty ratio D _yChanging Pattern in half power frequency period realizes, as shown in Figure 14.

Can get by formula (9) and formula (12):

$i_{\mathrm{in}1+3}\left(t\right)=\frac{V_m\&CenterDot;\sin \mathrm{\&omega;t}\&CenterDot;{\mathrm{<figure-callout\; id="1"\; label="D\; y"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">D</figure-callout>}}_y^{1+3}\left(t\right)}{2\&CenterDot;L_p\&CenterDot;f_s}=\frac{V_m\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="D\; y"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">D</figure-callout>}}_y^{}\&CenterDot;(\sin \mathrm{\&omega;t}+I_3^{*}\&CenterDot;\mathrm{<figure-callout\; id="3"\; label="sin"\; filenames="A20091002736000262.png,A20091002736000263.png"\; state="\{\{state\}\}">sin</figure-callout>}3\mathrm{\&omega;t})}{2\&CenterDot;L_p\&CenterDot;f_s}---\left(22\right)$

Can release the change in duty cycle function by following formula is:

${\mathrm{<figure-callout\; id="1"\; label="D\; y"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">D</figure-callout>}}_y\left(t\right)=D_y\&CenterDot;\sqrt{\frac{\sin \mathrm{\&omega;t}+I_3^{*}\&CenterDot;\mathrm{<figure-callout\; id="3"\; label="sin"\; filenames="A20091002736000262.png,A20091002736000263.png"\; state="\{\{state\}\}">sin</figure-callout>}3\mathrm{\&omega;t}}{\sin \mathrm{\&omega;t}}}=D_y\&CenterDot;\sqrt{1+3I_3^{*}-4I_3^{*}\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="sin"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">sin</figure-callout>}}^2\mathrm{\&omega;t}}---\left(23\right)$

With formula I ₃ ^*=0.33 substitution following formula obtains the output current smallest peaks and all than the change in duty cycle function of correspondence is:

${\mathrm{<figure-callout\; id="1"\; label="D\; y"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">D</figure-callout>}}_y\left(t\right)=D_y\&CenterDot;\sqrt{2-1.32\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="sin"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">sin</figure-callout>}}^2\mathrm{\&omega;t}}---\left(24\right)$

Change in duty cycle function shown in the formula (24) is realized relatively difficulty with analog circuit, seek below function simple relatively and that circuit is realized easily come match it.

Based on Taylor series:

$f\left(x\right)=f\left(x_0\right)+f^{\&prime;}\left(x_0\right)\&CenterDot;(x-x_0)+\frac{f^{\&prime;\&prime;}\left(x_0\right)}{2!}\&CenterDot;{(x-x_0)}^2+L+\frac{f^{\left(n\right)}\left(x_0\right)}{n!}\&CenterDot;{(x-x_0)}^n+L---\left(25\right)$

Only keep the first order derivative item, formula (24) can be written as:

${\mathrm{<figure-callout\; id="1"\; label="D\; y"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">D</figure-callout>}}_y\left(t\right)\&ap;{\mathrm{<figure-callout\; id="1"\; label="D\; y"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">D</figure-callout>}}_y\left(\sin {\mathrm{\&omega;}}_L{t}_0\right)+\frac{d\left(D_{\mathrm{<figure-callout\; id="1"\; label="y"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">y</figure-callout>}1+3}\left(t\right)\right)}{d\left(\sin {\mathrm{\&omega;}}_{L}t\right)}{|}_{\sin {\mathrm{\&omega;}}_{\mathrm{<figure-callout\; id="0"\; label="L\; t"\; filenames="A20091002736000243.png,A20091002736000261.png"\; state="\{\{state\}\}">L</figure-callout>}}{t}_{}{}_0}\&CenterDot;(\sin {\mathrm{\&omega;}}_{L}t-\sin {\mathrm{\&omega;}}_L{t}_0)$

$=[{\mathrm{<figure-callout\; id="1"\; label="D\; y"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">D</figure-callout>}}_y\left(\sin {\mathrm{\&omega;}}_L{t}_0\right)-\frac{d\left({\mathrm{<figure-callout\; id="1"\; label="D\; y"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">D</figure-callout>}}_y\left(t\right)\right)}{d\left(\sin {\mathrm{\&omega;}}_{L}t\right)}{|}_{\sin {\mathrm{\&omega;}}_{\mathrm{<figure-callout\; id="0"\; label="L\; t"\; filenames="A20091002736000243.png,A20091002736000261.png"\; state="\{\{state\}\}">L</figure-callout>}}{t}_{}{}_0}\&CenterDot;\sin {\mathrm{\&omega;}}_L{t}_0]+\frac{d\left(D_{\mathrm{<figure-callout\; id="1"\; label="y"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">y</figure-callout>}1+3}\left(t\right)\right)}{d\left(\sin {\mathrm{\&omega;}}_{L}t\right)}{|}_{\sin {\mathrm{\&omega;}}_{\mathrm{<figure-callout\; id="0"\; label="L\; t"\; filenames="A20091002736000243.png,A20091002736000261.png"\; state="\{\{state\}\}">L</figure-callout>}}{t}_{}{}_0}\&CenterDot;{\sin \mathrm{\&omega;}}_{L}t---\left(26\right)$

For simplicity, formula (26) can be written as:

D _{y_fit}(t)＝D _y·a·(1+k·|sinω _Lt|) (27)

With formula (27) substitution formula (9) and formula (10), can get:

$i_{\mathrm{in}\_\mathrm{fit}}\left(t\right)=\frac{V_m\&CenterDot;\sin {\mathrm{\&omega;}}_{L}t\&CenterDot;D_{y\_\mathrm{fit}}^2\left(t\right)}{2\&CenterDot;L_p\&CenterDot;f_s}=\frac{V_m\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="D\; y"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">D</figure-callout>}}_y^{}{\&CenterDot;a}^2}{2{\&CenterDot;L}_p\&CenterDot;f_s}\&CenterDot;\sin {\mathrm{\&omega;}}_{L}t{\&CenterDot;(1+k\&CenterDot;|\sin {\mathrm{\&omega;}}_{L}t\left|\right)}^2---\left(28\right)$

$i_{o\_\mathrm{fit}}\left(t\right)=\frac{V_{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">m</figure-callout>}}^2\&CenterDot;\mathrm{<figure-callout\; id="2"\; label="si\; n"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">si</figure-callout>}{n}^{}{}^2{\mathrm{\&omega;}}_{L}t\&CenterDot;d_{y\_\mathrm{fit}}^2\left(t\right)}{2\&CenterDot;L_p\&CenterDot;f_s\&CenterDot;V_o}=\frac{V_{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">m</figure-callout>}}^2\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="D\; y"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">D</figure-callout>}}_y^{}{\&CenterDot;a}^2}{2{\&CenterDot;L}_p\&CenterDot;V_o\&CenterDot;f_s}\&CenterDot;\mathrm{si}{n}^2{\mathrm{\&omega;}}_{L}t{\&CenterDot;(1+k\&CenterDot;|\sin {\mathrm{\&omega;}}_{L}t\left|\right)}^2.---\left(29\right)$

Consider that the PF value should be not less than 0.95, promptly

$\mathrm{PF}=\frac{\frac{I_{1\_\mathrm{fit}}}{\sqrt{2}}}{I_{\mathrm{in}\_\mathrm{fit}\_\mathrm{rms}}}=\frac{\frac{1}{\sqrt{2}}\&CenterDot;\frac{2}{\mathrm{\&pi;}}{\&Integral;}_0^{\mathrm{\&pi;}}{i}_{\mathrm{in}\_\mathrm{fit}}\left(t\right)\&CenterDot;\sin {\mathrm{\&omega;}}_{L}t\&CenterDot;d{\mathrm{\&omega;}}_{\mathrm{<figure-callout\; id="1"\; label="L\; t"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">L</figure-callout>}}t}{\sqrt{\frac{1}{\mathrm{\&pi;}}{\&Integral;}_0^{\mathrm{\&pi;}}{i}_{\mathrm{in}\_\mathrm{fit}}^2\left(t\right)\&CenterDot;d{\mathrm{\&omega;}}_{L}t}}$

$=\frac{\frac{1}{\sqrt{2}}\&CenterDot;(1+\frac{16}{3\mathrm{\&pi;}}\&CenterDot;k+\frac{3}{4}\&CenterDot;k^2)}{\sqrt{\frac{1}{2}+\frac{16}{3\mathrm{\&pi;}}\&CenterDot;k+\frac{9}{4}\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="k"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">k</figure-callout>}}^2+\frac{64}{15\mathrm{\&pi;}}\&CenterDot;{\mathrm{<figure-callout\; id="3"\; label="k"\; filenames="A20091002736000262.png,A20091002736000263.png"\; state="\{\{state\}\}">k</figure-callout>}}^3+\frac{5}{16}\&CenterDot;{\mathrm{<figure-callout\; id="4"\; label="k"\; filenames="A20091002736000272.png,A20091002736000273.png"\; state="\{\{state\}\}">k</figure-callout>}}^4}}=0.95---\left(30\right)$

That separates must k=-0.5.

According to formula (29), the mean value of half interior output current of power frequency period is

$I_{o\_\mathrm{fit}}=\frac{1}{\mathrm{\&pi;}}{\&Integral;}_0^{\mathrm{\&pi;}}{i}_{o\_\mathrm{fit}}\left(t\right)\&CenterDot;d{\mathrm{\&omega;}}_{L}t=\frac{V_{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">m</figure-callout>}}^2\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="D\; y"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">D</figure-callout>}}_y^{}\&CenterDot;a^2}{4\&CenterDot;L_p\&CenterDot;V_o\&CenterDot;f_s}(1+\frac{16}{3\mathrm{\&pi;}}\&CenterDot;k+\frac{3}{4}\&CenterDot;k^2)---\left(31\right)$

When adopting the match duty ratio, must guarantee that output current mean value is constant, can be got by formula (31) and formula (8):

$a^2\&CenterDot;(1+\frac{16}{3\mathrm{\&pi;}}\&CenterDot;k+\frac{3}{4}\&CenterDot;k^2)=1---\left(32\right)$

With k=-0.5 substitution formula (32), can get a=1.72. so formula (27) can be written as:

D _{y_fit}(t)＝D _y·1.72·(1-0.5·|sinω _Lt|) (33)

Accompanying drawing (15) is the comparison diagram of match duty ratio and desired duty cycle.

The effect comparison of the input current triple-frequency harmonics control strategy of table 1 ideal and match

6. input current injects three times and the relation of quintuple harmonics and output current peak-to-average force ratio and input power factor

Inject triple-frequency harmonics in input current after, output current reduces at ω t=pi/2 place, and this moment, its peak position was in the both sides of ω t=pi/2, shown in accompanying drawing 16 (a).If inject quintuple harmonics in input current, at ω t=pi/2 place, quintuple harmonics and fundamental voltage amplitude stack can cause the rising of peak-to-average force ratio, but on the both sides of ω t=pi/2, quintuple harmonics reduce output current, shown in accompanying drawing 16 (b).Therefore can consider to inject simultaneously in input current three times and quintuple harmonics, the former reduces the size of output current at ω t=pi/2 place, and the latter reduces the size of output current on ω t=pi/2 both sides, shown in accompanying drawing 16 (c).Two harmonic waves cooperatively interact, and output current is played the effect of peak load shifting, make it approach square wave more, thereby reduce its peak-to-average force ratio.The injection rate of discussion three times and quintuple harmonics when guaranteeing that input power factor satisfies the relevant criterion requirement, reduces the peak-to-average force ratio of output current to greatest extent below.

After injecting three times and quintuple harmonics, the expression formula of input current is:

$i_{\mathrm{in}1+3+5}\left(t\right)=I_1(\sin \mathrm{\&omega;t}+I_3^{*}\&CenterDot;\mathrm{<figure-callout\; id="3"\; label="sin"\; filenames="A20091002736000262.png,A20091002736000263.png"\; state="\{\{state\}\}">sin</figure-callout>}3\mathrm{\&omega;t}+I_5^{*}\&CenterDot;\mathrm{<figure-callout\; id="5"\; label="sin"\; filenames="A20091002736000272.png,A20091002736000273.png"\; state="\{\{state\}\}">sin</figure-callout>}5\mathrm{\&omega;t})---\left(34\right)$

I wherein ₃ ^*And I ₅ ^*Be respectively that three times of injecting and quintuple harmonics amplitude are to fundamental voltage amplitude I ₁Perunit value.

With formula (34 go into formula (11), can release output current and be:

$i_{o1+3+5}\left(t\right)=\frac{V_m{\&CenterDot;I}_1}{V_o}\&CenterDot;\sin \mathrm{\&omega;t}\&CenterDot;(\sin \mathrm{\&omega;t}+I_3^{*}\&CenterDot;\mathrm{<figure-callout\; id="3"\; label="sin"\; filenames="A20091002736000262.png,A20091002736000263.png"\; state="\{\{state\}\}">sin</figure-callout>}3\mathrm{\&omega;t}+I_5^{*}\&CenterDot;\mathrm{<figure-callout\; id="3"\; label="sin"\; filenames="A20091002736000262.png,A20091002736000263.png"\; state="\{\{state\}\}">sin</figure-callout>}3\mathrm{\&omega;t})---\left(35\right)$

By following formula can in the hope of half power frequency in the half period output current mean value be:

${\mathrm{<figure-callout\; id="1"\; label="I\; o"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">I</figure-callout>}}_o=\frac{2}{T_{\mathrm{line}}}\&CenterDot;{\&Integral;}_0^{\frac{T_{\mathrm{line}}}{2}}\frac{V_m\&CenterDot;I_1}{V_o}\&CenterDot;\sin \mathrm{\&omega;t}\&CenterDot;(\sin \mathrm{\&omega;t}+I_3^{*}\&CenterDot;\mathrm{<figure-callout\; id="3"\; label="sin"\; filenames="A20091002736000262.png,A20091002736000263.png"\; state="\{\{state\}\}">sin</figure-callout>}3\mathrm{\&omega;t}+I_5^{*}\&CenterDot;\mathrm{<figure-callout\; id="3"\; label="sin"\; filenames="A20091002736000262.png,A20091002736000263.png"\; state="\{\{state\}\}">sin</figure-callout>}3\mathrm{\&omega;t})\&CenterDot;\mathrm{dt}=\frac{V_m\&CenterDot;I_1}{2{\&CenterDot;V}_o}---\left(36\right)$

Formula (36) shows, the injection rate of the size of output current mean value and three times and quintuple harmonics is irrelevant.In order to study output current peak-to-average force ratio and I ₃ ^*And I ₅ ^*Relation, the definition output current perunit value be:

$i_{o1+3+5}^{*}\left(t\right)=\frac{i_{\mathrm{<figure-callout\; id="1"\; label="o"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">o</figure-callout>}1+3+5}\left(t\right)}{{\mathrm{<figure-callout\; id="1"\; label="I\; o"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">I</figure-callout>}}_o}---\left(37\right)$

With formula (35) and formula (36) substitution formula (37), can get:

$i_{o1+3+5}^{*}\left(t\right)=2{\mathrm{<figure-callout\; id="2"\; label="sin"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">sin</figure-callout>}}^2\mathrm{\&omega;t}\&CenterDot;[(1+3I_3^{*}+5I_5^{*})-4(I_3^{*}+5I_5^{*}){\mathrm{<figure-callout\; id="2"\; label="sin"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">sin</figure-callout>}}^2\mathrm{\&omega;t}+16I_5^{*}{\mathrm{<figure-callout\; id="4"\; label="sin"\; filenames="A20091002736000272.png,A20091002736000273.png"\; state="\{\{state\}\}">sin</figure-callout>}}^4\mathrm{\&omega;t}]---\left(38\right)$

According to formula (38) i that can draw _O1+3+5 ^*Waveform in half power frequency period, as shown in Figure 17, as can be seen from the figure, the cycle of output current is half power frequency period, and about ω t=pi/2 symmetry.

In order to obtain the peak value of output current, need to calculate the extreme point of output current, to ω t differentiate, and make it equal 0 formula (38), then have:

$\frac{d{i}_{o1+3+5}^{*}\left(t\right)}{d\left(\mathrm{\&omega;t}\right)}=2\mathrm{<figure-callout\; id="2"\; label="sin"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">sin</figure-callout>}2\mathrm{\&omega;t}[(1+3I_3^{*}+5I_5^{*})-8(I_3^{*}+5I_5^{*})\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="sin"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">sin</figure-callout>}}^2\mathrm{\&omega;t}+48I_5^{*}\&CenterDot;{\mathrm{<figure-callout\; id="4"\; label="sin"\; filenames="A20091002736000272.png,A20091002736000273.png"\; state="\{\{state\}\}">sin</figure-callout>}}^4\mathrm{\&omega;t}]=0---\left(39\right)$

The condition that following formula is set up is:

sin2ωt＝0 (40(a))

Or $(1+3I_3^{*}+4I_5^{*})-8(I_3^{*}+5I_5^{*})\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="sin"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">sin</figure-callout>}}^2\mathrm{\&omega;t}+48I_5^{*}\&CenterDot;{\mathrm{<figure-callout\; id="4"\; label="sin"\; filenames="A20091002736000272.png,A20091002736000273.png"\; state="\{\{state\}\}">sin</figure-callout>}}^4\mathrm{\&omega;t}=0---\left(40\left(b\right)\right)$

Can obtain at [0, π] interval three interior extreme point ω t=0, pi/2 and π by formula (40 (a)).With they difference substitution formulas (38), ω t=0 and π are minimum point as can be known, corresponding $i_{\mathrm{<figure-callout\; id="1"\; label="o"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">o</figure-callout>}1+3+5}^{*}=0.$

Calculate for convenient, make sin ω t=x,, can solve its substitution formula (39):

$x_1=\frac{\sqrt{3}}{6I_5^{*}}\&CenterDot;\sqrt{5I_5^{*2}+I_3^{*}\&CenterDot;I_5^{*}-I_5^{*}\&CenterDot;\sqrt{(10\&CenterDot;I_5^{*2}+I_3^{*2}+I_3^{*}\&CenterDot;I_5^{*}-3I_5^{*})}}---\left(41\left(a\right)\right)$

$x_2=\frac{\sqrt{3}}{6I_5^{*}}\&CenterDot;\sqrt{5I_5^{*2}+I_3^{*}\&CenterDot;I_5^{*}+I_5^{*}\&CenterDot;\sqrt{(10\&CenterDot;I_5^{*2}+I_3^{*2}+I_3^{*}\&CenterDot;I_5^{*}-3I_5^{*})}}---\left(41\left(b\right)\right)$

$x_3=-\frac{\sqrt{3}}{6I_5^{*}}\&CenterDot;\sqrt{5I_5^{*2}+I_3^{*}\&CenterDot;I_5^{*}+I_5^{*}\&CenterDot;\sqrt{(10\&CenterDot;I_5^{*2}+I_3^{*2}+I_3^{*}\&CenterDot;I_5^{*}-3I_5^{*})}}---\left(41\left(c\right)\right)$

$x_4=-\frac{\sqrt{3}}{6I_5^{*}}\&CenterDot;\sqrt{5I_5^{*2}+I_3^{*}\&CenterDot;I_5^{*}-I_5^{*}\&CenterDot;\sqrt{(10\&CenterDot;I_5^{*2}+I_3^{*2}+I_3^{*}\&CenterDot;I_5^{*}-3I_5^{*})}}---\left(41\left(d\right)\right)$

In [0, π] interval, sin ω t 〉=0, obviously x ₃And x ₄Be extraneous root, cast out.17 (a), x as can be known with reference to the accompanying drawings ₁Corresponding to α ₁And α ₅, x ₂Corresponding to α ₂And α ₄, α ₁And α ₂Respectively with α ₅And α ₄About ω t=pi/2 symmetry.

i _O1+3+5 ^*About ω t=pi/2 symmetry, when therefore asking extreme value, only need to consider that the extreme point in [0, pi/2] interval gets final product.In [0, pi/2] interval, i _O1+3+5 ^*4 extreme points are arranged, promptly 0, α ₁, α ₂And pi/2 (α ₃).α is discussed below ₁, α ₂Whether exist.

By formula (41) as can be seen, x ₁And x ₂The necessary condition that exists is:

$10I_5^{*2}+I_3^{*2}+I_3^{*}\&CenterDot;I_5^{*}-3I_5^{*}\&GreaterEqual;0---\left(42\right)$

When satisfying formula (42), must satisfy the following conditions formula:

$5I_5^{*2}+I_3^{*}\&CenterDot;I_5^{*}+I_5^{*}\&CenterDot;\sqrt{(10\&CenterDot;I_5^{*2}+I_3^{*2}+I_3^{*}\&CenterDot;I_5^{*}-{3I}_5^{*})}\&GreaterEqual;0---\left(43\left(a\right)\right)$

$5I_5^{*2}+I_3^{*}\&CenterDot;I_5^{*}-I_5^{*}\&CenterDot;\sqrt{(10\&CenterDot;I_5^{*2}+I_3^{*2}+I_3^{*}\&CenterDot;I_5^{*}-{3I}_5^{*})}\&GreaterEqual;0---\left(43\left(b\right)\right)$

Because sin ω t=x, so ω t should satisfy 0＜x＜1 in [0, pi/2] interval, and promptly formula (43 (a)) and formula (43 (b)) need satisfied:

$\frac{\sqrt{3}}{{6I}_5^{*}}\&CenterDot;\sqrt{5I_5^{*2}+I_3^{*}\&CenterDot;I_5^{*}-I_5^{*}\&CenterDot;\sqrt{(10\&CenterDot;I_5^{*2}+I_3^{*2}+I_3^{*}\&CenterDot;I_5^{*}-{3I}_5^{*})}}\&le;1---\left(44\left(a\right)\right)$

$\frac{\sqrt{3}}{{6I}_5^{*}}\&CenterDot;\sqrt{5I_5^{*2}+I_3^{*}\&CenterDot;I_5^{*}+I_5^{*}\&CenterDot;\sqrt{(10\&CenterDot;I_5^{*2}+I_3^{*2}+I_3^{*}\&CenterDot;I_5^{*}-{3I}_5^{*})}}\&le;1---\left(44\left(b\right)\right)$

When conditional (44 (b)) was set up, obviously formula (44 (a)) was also set up, at this moment x ₁And x ₂All exist.Satisfying formula (42) and formula (44 (b)), but when not satisfying formula (44 (a)), then x ₁Do not exist, and x ₂Exist.

The condition that formula (42) is set up is (derivation is seen appendix)

$I_3^{*}\&le;0.41,$ And $I_5^{*}<\frac{-I_3^{*}+3-\sqrt{-39I_3^{*2}-6I_3^{*}+9}}{20},$ Or $I_5^{*}>\frac{-I_3^{*}+3+\sqrt{-39I_3^{*2}-6I_3^{*}+9}}{20}---\left(45\left(a\right)\right)$

Or

$I_3^{*}\&GreaterEqual;0.41---\left(45\left(b\right)\right)$

The condition that formula (44 (a)) is set up is (derivation is seen appendix)

$I_5^{*}\&GreaterEqual;\frac{I_3^{*}}{7}---\left(46\right)$

Or $I_5^{*}<\frac{I_3^{*}}{7},$ And $I_5^{*}\&le;\frac{5}{13}{I}_3^{*}-\frac{1}{13}$

The condition that formula (44 (b)) is set up is (derivation is seen appendix)

$I_5^{*}\&GreaterEqual;\frac{I_3^{*}}{7},$ And $I_5^{*}\&GreaterEqual;\frac{5}{13}{I}_3^{*}-\frac{1}{13}---\left(47\right)$

In addition, if require input power factor PF 〉=0.9, I so ₃ ^*And I ₅ ^*Also need satisfy:

$I_3^{*2}+I_5^{*2}\&le;0.234---\left(48\right)$

In sum, according to I ₃ ^*And I ₅ ^*The value difference, 3 kinds of situations will appear in output current:

Situation I: work as I ₃ ^*And I ₅ ^*When satisfying formula (45), formula (47) and formula (48), root x ₁And x ₂All exist, this moment, output current had 7 extreme points in half power frequency period, wherein 0 and π be minimum point.Replace the rule that occurs, α as can be known according to maximum point and minimum point ₁With pi/2 be maximum point, shown in accompanying drawing 17 (a).This moment I ₃ ^*And I ₅ ^*Span shown in the dash area among Figure 18 (a), its Chinese style (45) is corresponding oval outside, corresponding two straight lines of formula (47) top, the corresponding semicircle of formula (48) inside;

Situation II: work as I ₃ ^*And I ₅ ^*Satisfy formula (45), formula (46) and formula (48), and when not satisfying formula (47), root x ₁Exist, and root x ₂Do not exist, this moment, output current had 5 extreme points in half power frequency period.This moment α ₁Be maximum point, shown in accompanying drawing 17 (b).This moment I ₃ ^*And I ₅ ^*Span shown in the dash area in the accompanying drawing 18 (b), its Chinese style (45) is corresponding oval outside, formula (47) is not satisfied owing to satisfy formula (46), then I in the corresponding semicircle of formula (48) inside ₃ ^*And I ₅ ^*Span corresponding to two straight lines belows and straight line $I_5^{*}=I_3^{*}/7$ Top and straight line $I_5^{*}=(5I_3^{*}-1)/13$ The below;

Situation III: work as I ₃ ^*And I ₅ ^*When not satisfying formula (45), root x ₁And x ₂Do not exist, this moment, output current had only 3 extreme points in half power frequency period.Because α ₁And α ₂Do not exist, pi/2 becomes maximum point so, shown in Figure 17 (c).This moment I ₃ ^*And I ₅ ^*Span oval inner, shown in the dash area among Figure 18 (c).

When output current wave was situation I, its peak value appeared at α ₁Or α ₃Moment corresponding is when peak value appears at α ₃The place is with α ₃=pi/2 substitution formula (38), this moment, output current peak-to-average force ratio function was:

${\mathrm{<figure-callout\; id="1"\; label="I\; o"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">I</figure-callout>}}_o^{1+3+5}\left({\mathrm{\&alpha;}}_3\right)=2\&CenterDot;(1-{\mathrm{<figure-callout\; id="3"\; label="I"\; filenames="A20091002736000262.png,A20091002736000263.png"\; state="\{\{state\}\}">I</figure-callout>}}_3^{*}+I_5^{*})---\left(49\right)$

When peak value appears at α ₁The place, with formula (41) substitution formula (38), this moment, the peak-to-average force ratio function of output current was:

${\mathrm{<figure-callout\; id="1"\; label="I\; o"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">I</figure-callout>}}_o^{1+3+5}\left({\mathrm{\&alpha;}}_1\right)=(\frac{I_3^{*2}}{27I_5^{*2}}+\frac{I_3^{*}}{27I_5^{*}}-\frac{1}{{9I}_5^{*}}+\frac{10}{27})\&CenterDot;\sqrt{(10\&CenterDot;I_5^{*2}+I_3^{*2}+I_3^{*}\&CenterDot;I_5^{*}-3I_5^{*})}$

$-\frac{I_3^{*2}}{18I_5^{*}}-\frac{I_3^{*2}}{27I_5^{*2}}+\frac{I_3^{*}}{{6I}_5^{*}}-\frac{25I_5^{*}}{54}+\frac{5I_3^{*}}{9}+\frac{5}{6}---\left(50\right)$

According to formula (49) and formula (50), utilize Maple software to draw out at I ₃ ^*And I ₅ ^*(accompanying drawing 18 (a)) output current peak-to-average force ratio is with I in the excursion ₃ ^*And I ₅ ^*The curved surface that changes, shown in accompanying drawing 19 (a), wherein top-surface camber and lower surface camber correspond respectively to peak point and appear at α ₃And α ₁The time the variation curved surface.The minimum point of two surfaces intersection is the minimum value of output current peak-to-average force ratio.Can read output current peak-to-average force ratio minimum by figure is 1.34, corresponding I ₃ ^*=0.465 (accompanying drawing 19 (b)), I ₅ ^*=0.135 (accompanying drawing 19 (c)).At this moment, PF=0.90 satisfies the requirement to power factor of commercial illumination.

When output current wave was situation II, its peak value appeared at α ₁Moment corresponding, with formula (41 (b)) substitution formula (38), the peak-to-average force ratio function of output current is suc as formula shown in (50) at this moment.

According to formula (38), when utilizing Maple software to draw situation II, at I ₃ ^*And I ₅ ^*(accompanying drawing 18 (b)) output current peak-to-average force ratio is with I in the excursion ₃ ^*And I ₅ ^*The curved surface that changes is shown in accompanying drawing 20 (a).Reading output current peak-to-average force ratio minimum by figure is 1.382, corresponding I ₃ ^*=0.44 (accompanying drawing 20 (b)), I ₅ ^*=0.092 (accompanying drawing 20 (c)).At this moment, PF=0.91 satisfies the requirement of commercial illumination to power factor.

When output current wave was situation III, its peak value appeared at α ₃=pi/2 place with its substitution formula (38), obtains the peak-to-average force ratio function of output current this moment, shown in (49).For asking at I ₃ ^*And I ₅ ^*The minimum peak-to-average force ratio of output current in the excursion, defined function:

$I_5^{*}=I_3^{*}+\mathrm{\&Delta;I}---\left(51\right)$

With following formula substitution formula (49), can get

${\mathrm{<figure-callout\; id="1"\; label="I\; o"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">I</figure-callout>}}_o^{1+3+5}\left({\mathrm{\&alpha;}}_3\right)=2\&CenterDot;(1+\mathrm{\&Delta;I})---\left(52\right)$

As can be seen from the above equation, when Δ I gets minimum value, corresponding peak-to-average force ratio minimum.When situation III, I ₃ ^*And I ₅ ^*Value need satisfy the scope shown in the accompanying drawing 18 (c), therefore, make straight line according to formula (51) $I_5^{*}=I_3^{*}+\mathrm{\&Delta;I},$ When this straight line and the oval dash area that constitutes were tangent, Δ I obtained minimum value, this moment output current peak-to-average force ratio minimum, as shown in Figure 21.Simultaneous formula (45) and formula (51), equation discriminant equals 0 when utilizing ellipse and straight line tangent, tries to achieve Δ I=0.301, and with its substitution formula (52), obtaining the minimum peak-to-average force ratio of output current is 1.398.Calculate the I of point of contact correspondence ₃ ^*=0.382, I ₅ ^*=0.081, at this moment, PF=0.93 satisfies the requirement of commercial illumination to power factor.

Table 2 provided under three kinds of situations smallest peaks all than and corresponding I ₃ ^*And I ₅ ^*Value, therefrom as can be seen, the output current peak-to-average force ratio minimum during situation I is 1.34, corresponding I ₃ ^*=0.465, I ₅ ^*=0.135.

Under three kinds of situations of table 2 smallest peaks all than and pairing I ₃ ^*And I ₅ ^*Value

7. the implementation method injected of input current three times and quintuple harmonics

The injection of input current three times and quintuple harmonics can be by changing duty ratio D _yChanging Pattern in half power frequency period is to realize.Can get by formula (9) and formula (12):

$\frac{V_m\&CenterDot;\sin {\mathrm{\&omega;}}_{L}t\&CenterDot;{\mathrm{<figure-callout\; id="1"\; label="D\; y"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">D</figure-callout>}}_y^{1+3+5}\left(t\right)}{2{\&CenterDot;L}_p\&CenterDot;f_s}=\frac{V_m{\&CenterDot;\mathrm{<figure-callout\; id="2"\; label="D\; y"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">D</figure-callout>}}_y^{}}{2\&CenterDot;L_p\&CenterDot;f_s}(\sin {\mathrm{\&omega;}}_{L}t+I_3^{*}\&CenterDot;\mathrm{<figure-callout\; id="3"\; label="sin"\; filenames="A20091002736000262.png,A20091002736000263.png"\; state="\{\{state\}\}">sin</figure-callout>}3{\mathrm{\&omega;}}_{L}t+I_5^{*}\&CenterDot;\mathrm{<figure-callout\; id="5"\; label="sin"\; filenames="A20091002736000272.png,A20091002736000273.png"\; state="\{\{state\}\}">sin</figure-callout>}5{\mathrm{\&omega;}}_{L}t).---\left(53\right)$

Can release the change in duty cycle function by following formula is:

${\mathrm{<figure-callout\; id="1"\; label="D\; y"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">D</figure-callout>}}_y\left(t\right)=D_y\&CenterDot;\sqrt{\frac{\sin \mathrm{\&omega;t}+I_3^{*}\&CenterDot;\mathrm{<figure-callout\; id="3"\; label="sin"\; filenames="A20091002736000262.png,A20091002736000263.png"\; state="\{\{state\}\}">sin</figure-callout>}3\mathrm{\&omega;t}+I_5^{*}\&CenterDot;\mathrm{<figure-callout\; id="5"\; label="sin"\; filenames="A20091002736000272.png,A20091002736000273.png"\; state="\{\{state\}\}">sin</figure-callout>}5\mathrm{\&omega;t}}{\sin \mathrm{\&omega;t}}}---\left(54\right)$

With I ₃ ^*=0.465, I ₅ ^*=0.135, substitution following formula and abbreviation get:

${\mathrm{<figure-callout\; id="1"\; label="D\; y"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">D</figure-callout>}}_y\left(t\right)={\mathrm{<figure-callout\; id="1"\; label="D\; o"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">D</figure-callout>}}_o\&CenterDot;\sqrt{{2.16\mathrm{<figure-callout\; id="4"\; label="sin"\; filenames="A20091002736000272.png,A20091002736000273.png"\; state="\{\{state\}\}">sin</figure-callout>}}^4\mathrm{\&omega;t}-4.56{\mathrm{<figure-callout\; id="2"\; label="sin"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">sin</figure-callout>}}^2\mathrm{\&omega;t}+3.07}---\left(55\right)$

Change in duty cycle function shown in the formula (55) can be realized with digital control.

Also can realize duty ratio shown in the formula (55) with analog circuit, but comparatively complicated, seek below function simple relatively and that circuit is realized easily come match it.

Based on formula (25) Taylor series:

Only keep the first order derivative item, formula (55) can be written as:

${\mathrm{<figure-callout\; id="1"\; label="D\; y"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">D</figure-callout>}}_y\left(t\right)\&ap;{\mathrm{<figure-callout\; id="1"\; label="D\; y"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">D</figure-callout>}}_y\left(\sin {\mathrm{\&omega;}}_L{t}_0\right)+\frac{d\left(D_{\mathrm{<figure-callout\; id="1"\; label="y"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">y</figure-callout>}1+3+5}\left(t\right)\right)}{d\left(\sin {\mathrm{\&omega;}}_{L}t\right)}{|}_{\sin {\mathrm{\&omega;}}_{\mathrm{<figure-callout\; id="0"\; label="L\; t"\; filenames="A20091002736000243.png,A20091002736000261.png"\; state="\{\{state\}\}">L</figure-callout>}}{t}_{}{}_0}\&CenterDot;(\sin {\mathrm{\&omega;}}_{L}t-\sin {\mathrm{\&omega;}}_L{t}_0)$

$=[{\mathrm{<figure-callout\; id="1"\; label="D\; y"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">D</figure-callout>}}_y\left(\sin {\mathrm{\&omega;}}_L{t}_0\right)-\frac{d\left({\mathrm{<figure-callout\; id="1"\; label="D\; y"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">D</figure-callout>}}_y\left(t\right)\right)}{d\left(\sin {\mathrm{\&omega;}}_{L}t\right)}{|}_{\sin {\mathrm{\&omega;}}_{\mathrm{<figure-callout\; id="0"\; label="L\; t"\; filenames="A20091002736000243.png,A20091002736000261.png"\; state="\{\{state\}\}">L</figure-callout>}}{t}_{}{}_0}\&CenterDot;\sin {\mathrm{\&omega;}}_L{t}_0]+\frac{d\left(D_{\mathrm{<figure-callout\; id="1"\; label="y"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">y</figure-callout>}1+3+5}\left(t\right)\right)}{d\left({\sin \mathrm{\&omega;}}_{L}t\right)}{|}_{\sin {\mathrm{\&omega;}}_{\mathrm{<figure-callout\; id="0"\; label="L\; t"\; filenames="A20091002736000243.png,A20091002736000261.png"\; state="\{\{state\}\}">L</figure-callout>}}{t}_{}{}_0}\&CenterDot;\sin {\mathrm{\&omega;}}_{L}t---\left(56\right)$

For simplicity, formula (56) can be written as:

D _{y_fit}(t)＝D _y·a·(1+k·|sinω _Lt|) (57)

With formula (57) substitution formula (9) and formula (10), can get:

$i_{\mathrm{in}\_\mathrm{fit}}\left(t\right)=\frac{V_m\&CenterDot;\sin {\mathrm{\&omega;}}_{L}t\&CenterDot;D_{y\_\mathrm{fit}}^2\left(t\right)}{2\&CenterDot;L_p\&CenterDot;f_s}=\frac{V_m\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="D\; y"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">D</figure-callout>}}_y^{}\&CenterDot;a^2}{2\&CenterDot;L_p\&CenterDot;f_s}\&CenterDot;\sin {\mathrm{\&omega;}}_{L}t\&CenterDot;{(1+k\&CenterDot;|\sin {\mathrm{\&omega;}}_{L}t\left|\right)}^2---\left(58\right)$

$i_{o\_\mathrm{fit}}\left(t\right)=\frac{V_{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">m</figure-callout>}}^2\&CenterDot;\mathrm{<figure-callout\; id="2"\; label="si\; n"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">si</figure-callout>}{n}^{}{}^2{\mathrm{\&omega;}}_{L}t\&CenterDot;d_{y\_\mathrm{fit}}^2\left(t\right)}{2\&CenterDot;L_p\&CenterDot;f_s\&CenterDot;V_o}=\frac{V_{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">m</figure-callout>}}^2\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="D\; y"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">D</figure-callout>}}_y^{}\&CenterDot;a^2}{2\&CenterDot;L_p\&CenterDot;V_o\&CenterDot;f_s}\&CenterDot;\mathrm{si}{n}^2{\mathrm{\&omega;}}_{L}t\&CenterDot;{(1+k\&CenterDot;|\sin {\mathrm{\&omega;}}_{L}t\left|\right)}^2.---\left(59\right)$

Consider that the PF value should be not less than 0.9, promptly

$\mathrm{PF}=\frac{\frac{I_{1\_\mathrm{fit}}}{\sqrt{2}}}{I_{\mathrm{in}\_\mathrm{fit}\_\mathrm{rms}}}=\frac{\frac{1}{\sqrt{2}}\&CenterDot;\frac{2}{\mathrm{\&pi;}}{\&Integral;}_0^{\mathrm{\&pi;}}{i}_{\mathrm{in}\_\mathrm{fit}}\left(t\right)\&CenterDot;\sin {\mathrm{\&omega;}}_{L}t\&CenterDot;d{\mathrm{\&omega;}}_{\mathrm{<figure-callout\; id="1"\; label="L\; t"\; filenames="A20091002736000251.png,A20091002736000271.png"\; state="\{\{state\}\}">L</figure-callout>}}t}{\sqrt{\frac{1}{\mathrm{\&pi;}}{\&Integral;}_0^{\mathrm{\&pi;}}{i}_{\mathrm{in}\_\mathrm{fit}}^2\left(t\right)\&CenterDot;d{\mathrm{\&omega;}}_{L}t}}$

$=\frac{\frac{1}{\sqrt{2}}\&CenterDot;(1+\frac{16}{3\mathrm{\&pi;}}\&CenterDot;k+\frac{3}{4}\&CenterDot;k^2)}{\sqrt{\frac{1}{2}+\frac{16}{3\mathrm{\&pi;}}\&CenterDot;k+\frac{9}{4}\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="k"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">k</figure-callout>}}^2+\frac{64}{15\mathrm{\&pi;}}\&CenterDot;{\mathrm{<figure-callout\; id="3"\; label="k"\; filenames="A20091002736000262.png,A20091002736000263.png"\; state="\{\{state\}\}">k</figure-callout>}}^3+\frac{5}{16}\&CenterDot;{\mathrm{<figure-callout\; id="4"\; label="k"\; filenames="A20091002736000272.png,A20091002736000273.png"\; state="\{\{state\}\}">k</figure-callout>}}^4}}=0.9---\left(60\right)$

That separates must k=-0.61.

According to formula (59), the mean value of half interior output current of power frequency period is

$I_{o\_\mathrm{fit}}=\frac{1}{\mathrm{\&pi;}}{\&Integral;}_0^{\mathrm{\&pi;}}{i}_{o\_\mathrm{fit}}\left(t\right)\&CenterDot;d{\mathrm{\&omega;}}_{L}t=\frac{V_{\mathrm{<figure-callout\; id="2"\; label="m"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">m</figure-callout>}}^2{\&CenterDot;\mathrm{<figure-callout\; id="2"\; label="D\; y"\; filenames="A20091002736000252.png,A20091002736000272.png"\; state="\{\{state\}\}">D</figure-callout>}}_y^{}\&CenterDot;a^2}{4\&CenterDot;L_p\&CenterDot;V_o\&CenterDot;f_s}(1+\frac{16}{3\mathrm{\&pi;}}\&CenterDot;k+\frac{3}{4}{\&CenterDot;k}^2)---\left(61\right)$

When adopting the match duty ratio, must guarantee that output current mean value is constant, can be got by formula (61) and formula (8):

$a^2\&CenterDot;(1+\frac{16}{3\mathrm{\&pi;}}\&CenterDot;k+\frac{3}{4}\&CenterDot;k^2)=1---\left(62\right)$

With k=-0.61 substitution formula (62), can get a=2.02. so formula (57) can be written as:

D _{y_fit}(t)＝D _y·2.02·(1-0.61·|sinω _Lt|) (63)

Accompanying drawing 22 is the comparison diagram of match duty ratio and desired duty cycle.

Table 3 has provided the input current three times of desirable and match and the comparative analysis of quintuple harmonics method, can find, the output current peak-to-average force ratio of match mode is slightly high, but its implementation is very simple.

Table 3 desirable and the input current three times of match and the contrast of quintuple harmonics injection method

Accompanying drawing 23 is another electrical block diagram of the present invention; This circuit mainly is on the basis of foregoing circuit, set up output current detection filter circuit (2), input voltage feed forward circuit (3), output current feedback control circuit (4), multiplier (5) again, the sample resistance of described input voltage feed forward circuit (3) is connected input filter capacitor Cin two ends, subtracter output E point in the input voltage feed forward circuit (3) is connected with a part input of multiplier (5), and the peak value sampling point B point in the input voltage feed forward circuit (3) links to each other with the denominator input of multiplier (5); Current transformer T primary coil in the output current detection circuit (2) is serially connected between the secondary coil and load LED of Flyback inductance L, the current transformer T secondary coil in the output current detection circuit (2) behind rectifying and wave-filtering with output current feedback control circuit (4) in error adjuster IG ₃Inverting input connect; Error adjuster IG ₃Output F point be connected with another molecule input of multiplier (5); The output of multiplier (5) is connected with the input of pulse width modulating chip (6), the drive signal of exporting with the sawtooth waveforms handing-over back in the pulse width modulating chip (6) is connected with the gate pole of switching tube Q, switching tube Q drain electrode and source electrode are serially connected between the negative pole of the primary coil of the Flyback inductance L in the main power circuit (1) and diode rectifier circuit output, pulse width modulating chip (6) is the duty cycle signals driving switch pipe Q of a (1-k|sin ω t|) with Changing Pattern, wherein k is by the power factor decision of power requirement, after the k value was determined, a value was determined by power.

Described input voltage feed forward circuit (3) comprises sample resistance R ₁And R ₂, emitter follower IC ₁With subtracter IC ₂, described sample resistance R ₁And R ₂Between sampling point A point and emitter follower IC ₁In-phase input end connect emitter follower IC ₁A road of output obtains the denominator input and the subtracter IC of a certain signal and multiplier (5) at the B point behind peak value sampling ₂In-phase end link to each other, another road obtains a certain signal and subtracter IC at the C point after dividing potential drop ₂Inverting input connect subtracter IC ₂Output E point be connected with a part input of multiplier (5).

As in input voltage feed forward circuit (3), injecting multiplier (5) with an amount of three times and quintuple harmonics, can adjust the duty cycle signals of pulse width modulating chip (6), can be under input power factor be not less than 0.9 situation, the output current peak-to-average force ratio can be reduced to 1.34 from 2.Be not less than at input power factor under 0.95 the situation, the output current peak-to-average force ratio can be reduced to 1.45 from 2.

The output current of the output current detection circuit of this circuit (2) is connected into the error adjuster IC in the output current feedback control circuit (4) after testing ₃Inverting input, the output F point current potential of error adjuster is v _EA, with a part input v of multiplier (5) _yConnect; In the input voltage feed forward circuit (3), the input voltage after the rectification is by sample resistance R ₁And R ₂After the dividing potential drop, be v at A point current potential _A=k _vV _m| sin ω t (k wherein _vBe the dividing potential drop coefficient), with emitter follower IC ₁In-phase input end connect emitter follower IC ₁Output signal a road through R ₃, D ₁, C ₁And R ₄Be v at the B point voltage behind the peak value sampling _B=k _vV _m, with the denominator input v of multiplier (5) _zWith subtracter IC ₂In-phase input end link to each other, another road is through R ₅, R ₆Be v at the C point voltage after the dividing potential drop _C=0.61v _A=0.61k _vV _m| sin ω t|, with subtracter IC ₂Inverting input connect subtracter IC ₂Output E point and a part input v of multiplier (5) _xConnect, B point current potential and C point current potential are through subtracter IC ₂The back is v at the current potential of output point E _x=v _B-v _c=k _vV _m(1-0.61|sin ω _LT|), the current potential of the output P of multiplier (5) is v _P=v _xV _y/ v _z=v _EA(1-0.61|sin ω _LT|), link to each other with the input of pulse width modulating chip (6), this signal is connected with the gate pole of switching tube Q with the drive signal that export the sawtooth waveforms handing-over back in the pulse width modulating chip (6), with Changing Pattern is the duty cycle signals driving switch pipe Q of a (1-k|sin ω t|), can in input current, inject an amount of three times and quintuple harmonics, be not less than at input power factor under 0.9 the situation, the output current peak-to-average force ratio can be reduced to 1.34 from 2.Be not less than at input power factor under 0.95 the situation, the output current peak-to-average force ratio can be reduced to 1.45 from 2.

Above-mentioned pulse width modulating chip can be selected models such as UC3843, UC3844, UC3525, emitter follower IC for use ₁, subtracter IC ₂With error adjuster IC ₃Select operational amplifiers such as TL074, TL072, LM358 for use, multiplier (5) can adopt integrated IC, also can adopt discrete device as shown in Figure 24 to form.

Claims (6)

1, a kind of high brightness LED driving power of no electrolytic capacitor is characterized in that main power circuit (1) comprises and input voltage source v _InThe diode rectifier circuit, Flyback inductance L, switching tube Q and the diode D that connect, the positive pole of described diode rectifier circuit output is connected with the drain electrode of switching tube Q through the primary coil of Flyback inductance L, the source electrode of switching tube Q is connected with the negative pole of diode rectifier circuit output, the gate pole of switching tube Q is connected with the output of pulse width modulating chip (6), being connected with load LED through diode D of the secondary coil of Flyback inductance L.

2, the high brightness LED driving power of no electrolytic capacitor as claimed in claim 1 is characterized in that described main power circuit (1) also comprises input filter inductance L _In, input filter capacitor C _In, output filter capacitor C _o, output inductor L _oWith the open-circuit-protection capacitor C _Prot, described input filter inductance L _InBe serially connected between the primary coil of the positive pole of diode rectifier circuit output and Flyback inductance L input filter capacitor C _InBe connected input filter inductance L _InOutput and the negative pole of diode rectifier circuit output between, output inductor L _oBe serially connected between diode D and the load LED output filter capacitor C _oWith the open-circuit-protection capacitor C _ProtBe connected to output inductor L _oTwo ends and the other end of load LED between, the output current peak-to-average force ratio is reduced to 2.

3, the high brightness LED driving power of no electrolytic capacitor as claimed in claim 2, it is characterized in that also comprising output current detection filter circuit (2), input voltage feed forward circuit (3), output current feedback control circuit (4), multiplier (5), the sample resistance of described input voltage feed forward circuit (3) is connected input filter capacitor C _InTwo ends, the subtracter output E point in the input voltage feed forward circuit (3) is connected with a part input of multiplier (5), and the peak value sampling point B point in the input voltage feed forward circuit (3) links to each other with the denominator input of multiplier (5); Current transformer T primary coil in the output current detection circuit (2) is serially connected between the secondary coil and load LED of Flyback inductance L, the current transformer T secondary coil in the output current detection circuit (2) behind rectifying and wave-filtering with output current feedback control circuit (4) in error adjuster IC ₃Inverting input connect; Error adjuster IC ₃Output F point be connected with another molecule input of multiplier (5); The output of multiplier (5) is connected with the input of pulse width modulating chip (6), pulse width modulating chip (6) is the duty cycle signals driving switch pipe Q of a (1-k|sin ω t|) with Changing Pattern, wherein k is by the power factor decision of power requirement, and after the k value was determined, a value was determined by power.

4, the high brightness LED driving power of no electrolytic capacitor as claimed in claim 3 is characterized in that described input voltage feed forward circuit (3) comprises sample resistance R ₁And R ₂, emitter follower IC ₁With subtracter IC ₂, described sample resistance R ₁And R ₂Between sampling point A point and emitter follower IC ₁In-phase input end connect emitter follower IC ₁A road of output obtains the denominator input and the subtracter IC of a certain signal and multiplier (5) at the B point behind peak value sampling ₂In-phase end link to each other, another road obtains a certain signal and subtracter IC at the C point after dividing potential drop ₂Inverting input connect subtracter IC ₂Output E point be connected with a part input of multiplier (5).

5, as the high brightness LED driving power of the described no electrolytic capacitor of claim 1-4, its essence is to use the pulsating current driving LED, and in input current, inject an amount of low-order harmonic with further reduction output current peak-to-average force ratio, available digital control realizes that also available input voltage feed forward method is realized; Such as, being not less than at input power factor under 0.9 the situation, the output current peak-to-average force ratio can be reduced to 1.34 from 2; Be not less than at input power factor under 0.95 the situation, the output current peak-to-average force ratio can be reduced to 1.45 from 2.

6,, it is characterized in that described pulse width modulating chip can select models such as UC3843, UC3844, UC3525, emitter follower IC for use as the high brightness LED driving power of described no electrolytic capacitor one of among the claim 1-5 ₁, subtracter IC ₂With error adjuster IC ₃Select operational amplifiers such as TL074, TL072, LM358 for use, multiplier (5) can adopt integrated IC, also can adopt discrete device to form.

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