CN103813591A - CRM Flyback LED (Light-Emitting Diode) driver with low output current peak-to-average ratio - Google Patents

Abstract

Translated fromChinese

本发明公开了一种低输出电流峰均比的CRM Flyback LED驱动器，包括主功率电路和锯齿波比较及开关管驱动电路、输入电压前馈电路、第一乘法器、第二乘法器和输出电流反馈电路，所述输入电压前馈电路如下：第一分压跟随电路的输出端分别与加法电路和减法电路的一个输入端连接；第二分压峰值取样电路的输出端与减法电路的另一个输入端连接；第三分压放大电路的输出端分别与加法电路的另一输入端和第一乘法器的第三输入端连接；加法电路的输出端与第一乘法器的第二输入端连接，减法电路的输出端与第一乘法器的第一输入端连接，第一乘法器的输出端与第二乘法器的第一输入端连接，第二乘法器输出端接入锯齿波比较及开关管驱动电路的输入端。本发明降低了输出电流峰均比。

The invention discloses a CRM Flyback LED driver with a low output current peak-to-average ratio, comprising a main power circuit, a sawtooth wave comparison and a switch tube drive circuit, an input voltage feedforward circuit, a first multiplier, a second multiplier and an output current Feedback circuit, the input voltage feed-forward circuit is as follows: the output terminal of the first voltage division follower circuit is connected to one input terminal of the addition circuit and the subtraction circuit respectively; the output terminal of the second voltage division peak sampling circuit is connected to the other input terminal of the subtraction circuit The input terminal is connected; the output terminal of the third voltage dividing amplifier circuit is connected with the other input terminal of the adding circuit and the third input terminal of the first multiplier respectively; the output terminal of the adding circuit is connected with the second input terminal of the first multiplier , the output of the subtraction circuit is connected to the first input of the first multiplier, the output of the first multiplier is connected to the first input of the second multiplier, and the output of the second multiplier is connected to the sawtooth comparison and switch The input terminal of the tube drive circuit. The invention reduces the peak-to-average ratio of the output current.

Description

The CRM Flyback LED driver of low output current peak-to-average force ratio

Technical field

The present invention relates to the CRM Flyback LED driver in the A.C.-D.C. converter field of electrical energy changer, particularly a kind of low output current peak-to-average force ratio.

Background technology

High brightness light emitting diode LED (light-emitting diode), because of advantages such as its high efficiency, environmental protection and long-lives, has been described as next generation's " green " light source.LED can adopt unidirectional pulsating current to drive, by controlling the mean value control LED output light flux of pulsating current, if but drive current peak-to-average force ratio is too high, and LED will damage.The advantages such as CRM Flyback LED driver has input and output isolation, it is simple to control and loss is little, but when within half input cycle, switching tube determine ON time, output current peak-to-average force ratio is larger, LED is normally worked unfavorable.

Summary of the invention

The object of the present invention is to provide a kind of CRM Flyback LED driver of low output current peak-to-average force ratio, guaranteeing that CRM Flyback LED driver PF value is greater than under the prerequisite of application requirements, reduces output current peak-to-average force ratio.

The technical solution that realizes the object of the invention is: a kind of CRM Flyback LED driver of low output current peak-to-average force ratio, comprise main power circuit and control circuit, and described main power circuit comprises input voltage source v_in, electromagnetic interface filter, diode rectifier circuit RB, transformer T₁, switching tube Q_b, diode D_b, filter capacitor C_o, filter inductance L_owith load LED, wherein input voltage source v_inbe connected with the input port of electromagnetic interface filter, the output port of electromagnetic interface filter is connected with the input port of diode rectifier circuit RB, the output negative pole of diode rectifier circuit RB is reference potential zero point, the first winding N of the output cathode of diode rectifier circuit RB and transformer T1_pdifferent name end connect, the first winding N of transformer T1_pthe drain electrode of Same Name of Ends access switching tube Qb, the source electrode of switching tube Qb is connected with reference potential zero point, transformer T₁the second winding N_zdifferent name end be connected zero point with reference potential, transformer T₁tertiary winding N_ssame Name of Ends and diode D_banodic bonding, diode D_bnegative electrode access respectively filter capacitor C_oone end and filter inductance L_oone end, filter capacitor C_oother end access reference potential zero point, filter inductance L_othe other end be connected with the anode tap of load LED, the both end voltage of load LED is output voltage V_o;

Described control circuit comprises sawtooth waveforms comparison and switch tube driving circuit, the first dividing potential drop follow circuit, the second dividing potential drop peak sample circuit, the 3rd dividing potential drop amplifying circuit, add circuit, subtraction circuit, the first multiplier, the second multiplier, output current feedback circuit; The wherein output of sawtooth waveforms comparison and switch tube driving circuit and switching tube Q_bgate pole connect; The input of the first dividing potential drop follow circuit and input voltage sampled point V_gthe output cathode that is diode rectifier circuit RB connects, and the output terminals A of the first dividing potential drop follow circuit is connected with input of add circuit and an input of subtraction circuit respectively; The input of the second dividing potential drop peak sample circuit and input voltage sampled point V_gthe output cathode that is diode rectifier circuit RB connects, and the second output C of dividing potential drop peak sample circuit and another input of subtraction circuit are connected; The 3rd input of dividing potential drop amplifying circuit and the output voltage V of main power circuit_oanodal connection, the output D of the 3rd dividing potential drop amplifying circuit respectively with another input of add circuit and the 3rd input v of the first multiplier_zconnect; The second input v of the output E of add circuit and the first multiplier_yconnect the output F of subtraction circuit and the first input end v of the first multiplier_xconnect the first input end v of the output of the first multiplier and the second multiplier_aconnect the output access sawtooth waveforms comparison of the second multiplier and the input of switch tube driving circuit, the second input v of the output of output current feedback circuit and the second multiplier_bconnect.

Compared with prior art, its remarkable advantage is in the present invention: (1), keeping whole input voltage range internal power factor PF value to be greater than under the prerequisite of application requirements, has effectively reduced output current peak-to-average force ratio; (2) reduce the spoilage of LED, extended the life-span of institute's driving LED; (3) there is safe and reliable advantage, have a extensive future.

Accompanying drawing explanation:

Fig. 1 is Flyback pfc converter main circuit schematic diagram.

Fig. 2 is the inductive current oscillogram of CRM Flyback pfc converter.

Fig. 3 is power factor PF value and V_m/ nV_ograph of relation.

Fig. 4 is output current peak-to-average force ratio and V_m/ nV_ograph of relation.

Fig. 5 is g (ω ideal curve and matched curve figure t) in half power frequency period.

Fig. 6 is wide input range lower critical inductance value curve chart.

Fig. 7 is switching frequency change curve in half power frequency period under different input voltages, and wherein (a) input voltage is 85V, and (b) input voltage is 175V, and (c) input voltage is 265V.

Fig. 8 is primary current peak I_pkwith effective value I_rmscurve chart under wide region input.

Fig. 9 is the electrical block diagram of the CRM Flyback LED driver of the low output current peak-to-average force ratio of the present invention.

Embodiment

The operation principle of 1CRM Flyback pfc converter

Fig. 1 is Flyback pfc converter main circuit.

Without loss of generality, definition input ac voltage v_inexpression formula be

v_in(t)＝V_msinωt (1)

Wherein V_mfor input voltage peak value, ω=2 π f_linefor input voltage angular frequency, f_linefor input voltage frequency

Voltage after input rectifying is so

v_g＝V_m|sinωt| (2)

Fig. 2 is the inductive current waveform of a switch periods inner conversion device.In the time of switching tube Q conducting, diode D cut-off, former limit inductance L_pthe voltage at two ends is v_g, current i L_pby zero beginning with v_g/ L_plinear rising of slope, i so_lppeak value iL_{p_pk}for

$i_{\mathrm{Lp}\_\mathrm{pk}}=\frac{V_m\left|\sin \mathrm{\ωt}\right|}{L_p}{t}_{\mathrm{on}}---\left(3\right)$

Wherein t_onfor the ON time of Q

When Q turn-offs, diode D conducting, by secondary inductance L_scurrent i_lsafterflow, now secondary inductance L_sthe voltage at two ends is V_o, secondary inductance current i_lswith V_o/ L_sslope from secondary current peak value i_{ls_pk}decline, i_lsdrop to zero time t_offfor

$t_{\mathrm{off}}=\frac{i_{\mathrm{Ls}\_\mathrm{pk}}}{V_o/L_s}=\frac{{\mathrm{ni}}_{\mathrm{Lp}\_\mathrm{pk}}}{n^2{V}_o/L_p}=\frac{V_m\left|\sin \mathrm{\ωt}\right|}{{\mathrm{nV}}_o}{t}_{\mathrm{on}}---\left(4\right)$

Wherein n is the transformer primary secondary turn ratio, L_sfor transformer secondary inductance, i_{ls_pk}for secondary inductance current peak.

Because Flyback converter is operated in CRM pattern, therefore, in the time that the electric current of diode D drops to zero, switching tube Q is open-minded, starts new switch periods.

Can obtain duty ratio d by formula (4) is

d(t)＝t_on/(t_on+t_off)＝nV_o/(nV_o+V_m|sinωt|) (5)

By formula (3) and (5), in a switch periods, the mean value i of former limit inductive current_{lp_av}for

$i_{\mathrm{Lp}\_\mathrm{av}}=\frac{1}{2}{i}_{\mathrm{Lp}\_\mathrm{pk}}d\left(t\right)=\frac{n{V}_o{V}_m\left|\sin \mathrm{\ωt}\right|}{{2L}_p({\mathrm{nV}}_o+V_m|\sin \mathrm{\ωt}\left|\right)}{t}_{\mathrm{on}}---\left(6\right)$

So, input current i_infor

$i_{\mathrm{in}}\left(t\right)=\frac{n{V}_o{V}_m\sin \mathrm{\ωt}}{{2L}_p({\mathrm{nV}}_o+V_m|\sin \mathrm{\ωt}\left|\right)}{t}_{\mathrm{on}}---\left(7\right)$

In a switch periods, the mean value i of secondary current_{ls_av}be output current i_ofor:

$i_o\left(t\right){=i}_{\mathrm{Lp}\_\mathrm{av}}=\frac{1}{2}{i}_{\mathrm{Lp}\_\mathrm{pk}}(1-d\left(t\right))=\frac{n{\left(V_m\right|\sin \mathrm{\ωt}\left|\right)}^2}{{2L}_p({\mathrm{nV}}_o+V_m|\sin \mathrm{\ωt}\left|\right)}{t}_{\mathrm{on}}---\left(8\right)$

So, in half power frequency period, output current mean value I_ofor

$I_o=\frac{1}{T_{\mathrm{line}}/2}{\∫}_0^{T_{\mathrm{line}}/2}{i}_o\left(t\right)\mathrm{dt}---\left(9\right)$

Wherein T_linefor the input voltage cycle.

By formula (1) and formula (7), can obtain the average value P of input power in half power frequency period_in

$P_{\mathrm{in}}=\frac{1}{T_{\mathrm{line}}/2}{\∫}_0^{T_{\mathrm{line}}/2}{v}_{\mathrm{in}}\left(t\right)i_{\mathrm{in}}\left(t\right)\mathrm{dt}=\frac{1}{\mathrm{\π}}{\∫}_0^{\mathrm{\π}}\frac{1}{2}\frac{{\left(V_m\sin \mathrm{\ωt}\right)}^2}{L_p}{t}_{\mathrm{on}}\frac{{\mathrm{nV}}_o}{{\mathrm{nV}}_o+V_m\left|\sin \mathrm{\ωt}\right|}d\mathrm{\ωt}--\left(9\right)$

Suppose that transducer effciency is 100%, input power equals power output so, i.e. P_in=P_o.Can obtain switching tube ON time t by formula (10)_on

$t_{\mathrm{on}}=\frac{2\mathrm{\&pi;}{L}_p{P}_o}{{\mathrm{nV}}_o{\mathrm{<figure-callout\; label="V\; m"\; filenames="HDA0000467476040000021.png,HDA0000467476040000032.png"\; state="\{\{state\}\}">V</figure-callout>}}_m^{}}\&CenterDot;\frac{1}{{\&Integral;}_0^{\mathrm{\&pi;}}\frac{{\left(\sin \mathrm{\&omega;t}\right)}^2}{{\mathrm{nV}}_o+V_m\left|\sin \mathrm{\&omega;t}\right|}\mathrm{d\&omega;t}}---\left(11\right)$

By formula (7), formula (10) and formula (11) can be in the hope of the expression formula of power factor PF value

$\mathrm{PF}=\frac{P_{\mathrm{in}}}{\frac{1}{\sqrt{2}}{V}_m{I}_{\mathrm{in}\_\mathrm{rms}}}=\frac{P_o}{\frac{1}{\sqrt{2}}{\mathrm{<figure-callout\; label="V\; m"\; filenames="HDA0000467476040000011.png,HDA0000467476040000021.png"\; state="\{\{state\}\}">V</figure-callout>}}_m\sqrt{\frac{1}{\mathrm{\&pi;}}{\&Integral;}_0^{\mathrm{\&pi;}}{\left(i_{\mathrm{in}}\left(t\right)\right)}^2\mathrm{d\&omega;t}}}=\frac{\sqrt{\frac{2}{\mathrm{\&pi;}}}{\&Integral;}_0^{\mathrm{\&pi;}}\frac{{\left(\sin \mathrm{\&omega;t}\right)}^2}{1+\frac{V_m}{{\mathrm{nV}}_o}\left|\sin \mathrm{\&omega;t}\right|}\mathrm{d\&omega;t}}{\sqrt{{\&Integral;}_0^{\mathrm{\&pi;}}{\left(\frac{\sin \mathrm{\&omega;<figure-callout\; label="t"\; filenames="HDA0000467476040000011.png,HDA0000467476040000021.png"\; state="\{\{state\}\}">t</figure-callout>}}{1+\frac{V_m}{{\mathrm{nV}}_o}\left|\sin \mathrm{\&omega;t}\right|}\right)}^2}}---\left(12\right)$

By formula (12), in conjunction with 3.1 joint design objectives, power factor PF is with V_m/ nV_ovariation rule curve as shown in Figure 3.2 reduce the control strategy of output current peak-to-average force ratio

3.1 traditional approach

By formula (8), formula (9) and formula (11), determine under ON time control mode output current instantaneous value i_owith mean value I_oratio

$\frac{i_o\left(t\right)}{I_o}=\frac{\mathrm{\&pi;}{\sin }^2\mathrm{\&omega;t}}{{\&Integral;}_0^{\mathrm{\&pi;}}\frac{{\sin }^2\mathrm{\&omega;<figure-callout\; label="t"\; filenames="HDA0000467476040000011.png,HDA0000467476040000021.png"\; state="\{\{state\}\}">t</figure-callout>}}{1+\frac{V_m}{{\mathrm{nV}}_o}\left|\sin \mathrm{\&omega;t}\right|}\mathrm{d\&omega;t}}\&CenterDot;\frac{1}{1+\frac{V_m}{{\mathrm{nV}}_o}\left|\sin \mathrm{\&omega;t}\right|}---\left(13\right)$

In the time of ω t=pi/2, formula (13) is got maximum, and peak-to-average force ratio is

$\frac{i_o\left(t\right)}{I_o}|\max =\frac{\mathrm{\&pi;}}{{\&Integral;}_0^{\mathrm{\&pi;}}\frac{{\sin }^2\mathrm{\&omega;<figure-callout\; label="t"\; filenames="HDA0000467476040000011.png,HDA0000467476040000021.png"\; state="\{\{state\}\}">t</figure-callout>}}{1+\frac{V_m}{{\mathrm{nV}}_o}\left|\sin \mathrm{\&omega;t}\right|}\mathrm{d\&omega;t}}\&CenterDot;\frac{1}{\left(1+\frac{V_m}{{\mathrm{nV}}_o}\right)}---\left(14\right)$

Make Fig. 4 according to formula (14), can find out and adopt while determining ON time control, peak-to-average force ratio is larger.

3.2 become ON time control method

For reducing output current peak-to-average force ratio, can in input current, inject 3 times and 5 subharmonic, establish input current expression formula and be

$i_{\mathrm{in}1+3+5}\left(t\right)=I_1(\sin \mathrm{\&omega;t}+I_3^{*}\sin 3\mathrm{\&omega;t}+I_5^{*}\sin 5\mathrm{\&omega;t})=I_1\sin \mathrm{\&omega;t}\&CenterDot;g\left(\mathrm{\&omega;t}\right)---\left(15\right)$

Wherein

i₁for fundamental voltage amplitude, I₃^*that 3 subharmonic amplitudes are based on I₁per unit value, I₅^*that 5 subharmonic amplitudes are based on I₁per unit value.

By power-balance, output current instantaneous value i_o1+3+5(t) be

$i_{o1+3+5}\left(t\right)=\frac{V_m\sin \mathrm{\&omega;t}}{V_o}{i}_{\mathrm{in}}\left(t\right)=\frac{V_m{I}_1}{V_{\text{o}}}\sin \mathrm{\&omega;t}\&CenterDot;(\sin \mathrm{\&omega;t}+I_3^{*}\mathrm{<figure-callout\; label="sin"\; filenames="HDA0000467476040000021.png,HDA0000467476040000032.png"\; state="\{\{state\}\}">sin</figure-callout>}3\mathrm{\&omega;t}+I_5^{*}\mathrm{<figure-callout\; label="sin"\; filenames="HDA0000467476040000041.png"\; state="\{\{state\}\}">sin</figure-callout>}5\mathrm{\&omega;t})=\frac{V_m{I}_1}{{\mathrm{<figure-callout\; label="V\; o\; sin"\; filenames="HDA0000467476040000021.png,HDA0000467476040000032.png"\; state="\{\{state\}\}">V</figure-callout>}}_o}{\sin }^{}{}^2\mathrm{\&omega;t}\&CenterDot;g\left(\mathrm{\&omega;t}\right)---\left(16\right)$

In half power frequency period, to formula (16) integration, can obtain output current mean value I in half power frequency period_o1+3+5

${\mathrm{<figure-callout\; label="I\; o"\; filenames="HDA0000467476040000011.png,HDA0000467476040000021.png"\; state="\{\{state\}\}">I</figure-callout>}}_o=\frac{2}{T_{\mathrm{line}}}{\&Integral;}_0^{\frac{T_{\mathrm{line}}}{2}}{\mathrm{<figure-callout\; label="i\; o"\; filenames="HDA0000467476040000011.png,HDA0000467476040000021.png"\; state="\{\{state\}\}">i</figure-callout>}}_o\mathrm{dt}=\frac{I_1{V}_m}{2V_o}---\left(17\right)$

Can be found out by formula (17), in output current mean value and input current, inject 3 times, 5 subharmonic amounts irrelevant.Guarantee to equate with the output current of determining under ON time control, can obtain

$I_1=\frac{{2V}_{\mathrm{<figure-callout\; label="o\; I\; o"\; filenames="HDA0000467476040000011.png,HDA0000467476040000021.png"\; state="\{\{state\}\}">o</figure-callout>}}{I}_o{1+3+5}_{}}{V_m}=\frac{{2V}_o{I}_o}{V_m}=\frac{{2P}_o}{V_m}---\left(18\right)$

According to formula (6), formula (10) and formula (15), ON time t_on1+3+5for

$t_{\mathrm{on}1+3+5}=\frac{{4P}_o{L}_p({\mathrm{nV}}_o+V_m|\sin \mathrm{\&omega;t}\left|\right)}{V_m^{2}n{V}_o}g\left(\mathrm{\&omega;t}\right)---\left(19\right)$

By formula (16) and formula (18), output current with the ratio of its mean value is

$i_{o1+3+5}^{*}\left(t\right)=\frac{i_{\mathrm{<figure-callout\; label="o"\; filenames="HDA0000467476040000011.png,HDA0000467476040000021.png"\; state="\{\{state\}\}">o</figure-callout>}1+3+5}\left(t\right)}{{\mathrm{<figure-callout\; label="I\; o"\; filenames="HDA0000467476040000011.png,HDA0000467476040000021.png"\; state="\{\{state\}\}">I</figure-callout>}}_o}---\left(20\right)$

After inputcurrent injection 3 times, 5 subharmonic, meet under the prerequisite of PF>=0.9 I₃^*, I₅^*must just can make peak-to-average force ratio minimum for particular value, get I₃^*=0.382, I₅^*=0.081, now peak-to-average force ratio is 1.40, PF=0.93.

3.3 matchings become ON time control method

ON time shown in formula (19) change function analog circuit realize more difficult, seek below relatively simple function come matching it.Can find out from formula (19), fitting function core is that (ω t) for matching g wherein.

Based on Taylor series

$f\left(x\right)=f\left(x_0\right)+f^{\&prime;}\left(x_0\right)(x-x_0)+\frac{f^{\&Prime;}\left(x_0\right)}{2!}{(x-x_0)}^2+\&CenterDot;\&CenterDot;\&CenterDot;+\frac{f^n\left(x_0\right)}{n!}{(x-x_0)}^n+\&CenterDot;\&CenterDot;\&CenterDot;---\left(21\right)$

Only retain first derivative item, (ω can approximate expression be t) g

$g\left(\mathrm{\&omega;t}\right)\&ap;g\left(\sin {\mathrm{\&omega;t}}_0\right)+\frac{d\left(g\left(\mathrm{\&omega;t}\right)\right)}{d\left(\sin \mathrm{\&omega;t}\right)}{|}_{t=t_0}(\sin \mathrm{\&omega;t}-\sin {\mathrm{\&omega;t}}_0)=a(1+k|\sin \mathrm{\&omega;t}\left|\right)---\left(22\right)$

According to formula (15) and formula (16), can obtain

$i_{\mathrm{in}}\left(t\right)=\frac{{2\mathrm{aP}}_o}{V_m}\sin \mathrm{\&omega;t}(1+k|\sin \mathrm{\&omega;t}\left|\right)---\left(23\right)$

$i_o\left(t\right)=\frac{V_m\sin \mathrm{\&omega;t}}{V_o}{i}_{\mathrm{in}}\left(t\right)=\frac{{2\mathrm{aP}}_o}{{\mathrm{<figure-callout\; label="V\; o\; sin"\; filenames="HDA0000467476040000021.png,HDA0000467476040000032.png"\; state="\{\{state\}\}">V</figure-callout>}}_o}{\sin }^{}{}^2\mathrm{\&omega;t}(1+k|\sin \mathrm{\&omega;t}\left|\right)---\left(24\right)$

Can be obtained by formula (1) and formula (23), adopt matching to become after ON time control, power factor PF is

$\frac{\frac{1}{T_{\mathrm{line}}/2}{\&Integral;}_0^{T_{\mathrm{line}}/2}{v}_{\mathrm{in}}\left(t\right)i_{\mathrm{in}}\left(t\right)\mathrm{<figure-callout\; label="dt"\; filenames="HDA0000467476040000011.png,HDA0000467476040000021.png"\; state="\{\{state\}\}">dt</figure-callout>}}{\frac{1}{\sqrt{2}}{V}_m{I}_{\mathrm{in}\_\mathrm{rms}}}=\frac{\frac{1}{\sqrt{2}}(1+\frac{8}{3\mathrm{\&pi;}}k)}{\sqrt{\frac{\mathrm{\&pi;}}{2}+\frac{8}{2}k+\frac{3\mathrm{\&pi;}}{8}{\mathrm{<figure-callout\; label="k"\; filenames="HDA0000467476040000021.png,HDA0000467476040000032.png"\; state="\{\{state\}\}">k</figure-callout>}}^2}}$

Get PF=0.93, can obtain k=-0.78.

According to formula (24), in half power frequency period, output current mean value is

$I_o=\frac{1}{\mathrm{\&pi;}}{\&Integral;}_0^{\mathrm{\&pi;}}{i}_o\left(\mathrm{\&omega;t}\right)\mathrm{d\&omega;t}=\frac{P_o}{V_o}(a+\frac{8\mathrm{<figure-callout\; label="ak"\; filenames="HDA0000467476040000021.png,HDA0000467476040000032.png"\; state="\{\{state\}\}">ak</figure-callout>}}{3\mathrm{\&pi;}})---\left(26\right)$

Guarantee to become after ON time control in matching, average output driving current remains unchanged,

$a+\frac{8\mathrm{<figure-callout\; label="ak"\; filenames="HDA0000467476040000021.png,HDA0000467476040000032.png"\; state="\{\{state\}\}">ak</figure-callout>}}{3\mathrm{\&pi;}}=1---\left(27\right)$

By k=-0.78 substitution formula (27), obtain a=3.086.(ideal curve t) of ω and matched curve are as shown in Figure 5 for g.

Can be in the hope of by formula (6), formula (10) and formula (23), after matching, become ON time into

$t_{\mathrm{on}}={4P}_o{L}_p\frac{({\mathrm{nV}}_o+V_m|\sin \mathrm{\&omega;t}\left|\right)a(1+k|\sin \mathrm{\&omega;t}\left|\right)}{{\mathrm{<figure-callout\; label="V\; m"\; filenames="HDA0000467476040000021.png,HDA0000467476040000032.png"\; state="\{\{state\}\}">V</figure-callout>}}_m^{}\&CenterDot;n{V}_o}---\left(28\right)$

By k=-0.78, a=3.086 substitution formula (24), can obtain the ratio i of output current and its mean value_o^*for

$i_o^{*}\left(t\right)=\frac{i_o\left(t\right)}{I_o}={6.16\mathrm{<figure-callout\; label="sin"\; filenames="HDA0000467476040000021.png,HDA0000467476040000032.png"\; state="\{\{state\}\}">sin</figure-callout>}}^2\mathrm{\&omega;t}(1-0.78|\sin \mathrm{\&omega;t}\left|\right)---\left(29\right)$

To above formula differentiate in half power frequency period, making derivative is zero to ask extreme value, can be in the hope of adopting matching to become after ON time control, and output current peak-to-average force ratio is 1.44.

4 performance comparison

4.1 former limit inductance and switching frequencies

For ease of analyzing, design parameter is as follows: v_in=85～265VAC, P_o=60W, V_o=24V, n=4.

By formula (4) and formula (11), determine switching frequency f under ON time_sfor

$f_s=\frac{1}{t_{\mathrm{on}}+t_{\mathrm{off}}}=\frac{{\mathrm{nV}}_{\mathrm{<figure-callout\; label="o\; V\; m"\; filenames="HDA0000467476040000021.png,HDA0000467476040000032.png"\; state="\{\{state\}\}">o</figure-callout>}}{V}_m^{}{}_2^{}{\&Integral;}_0^{\mathrm{\&pi;}}\frac{{\left(\sin \mathrm{\&piv;t}\right)}^2}{{\mathrm{nV}}_o+V_m\left|\sin \mathrm{\&omega;t}\right|}\mathrm{d\&omega;<figure-callout\; label="t"\; filenames="HDA0000467476040000021.png,HDA0000467476040000032.png"\; state="\{\{state\}\}">t</figure-callout>}}{2\mathrm{\&pi;}{L}_p{P}_o(1+\frac{V_m\left|\sin \mathrm{\&omega;t}\right|}{{\mathrm{nV}}_o})}---\left(31\right)$

Can be obtained by formula (4) and (28), become switching frequency f under ON time_s' be

$f_s^{\&prime;}=\frac{V_m^2{\left({\mathrm{nV}}_o\right)}^2}{12.34P_o{L}_p^{\&prime;}}\&CenterDot;\frac{1}{{({\mathrm{nV}}_o+V_m|\sin \mathrm{\&omega;t}\left|\right)}^2(1-0.78|\sin \mathrm{\&omega;t}\left|\right)}---\left(32\right)$

Adopt and determine ON time control, input voltage V_mone timing, f in half power frequency period_sconstantly change, when ω t=pi/2, f_sminimum.At different input voltage V_munder, f_sminimum value is also different.Adopt and become ON time control, input voltage V_mone timing, f in half power frequency period_sconstantly change, when ω t gets [0, pi/2] interior a certain value, f_sminimum.At different input voltage V_munder, f_sminimum value difference, its corresponding ω t is also different.Consider human auditory system frequency range, get f_smin=20kHz, can make the threshold inductance value under different input voltages, as shown in Figure 6.

By the threshold inductance value L under wide input range_pmax=656 μ H and L '_pmax=598 μ H are substitution formula (31) and formula (32) respectively, and the switching frequency that can obtain switching tube in two kinds of situations changes function

$f_s=\frac{{\mathrm{nV}}_{\mathrm{<figure-callout\; label="o\; V\; m"\; filenames="HDA0000467476040000021.png,HDA0000467476040000032.png"\; state="\{\{state\}\}">o</figure-callout>}}{V}_m^{}{}_2^{}{\&Integral;}_0^{\mathrm{\&pi;}}\frac{{\left(\sin \mathrm{\&omega;t}\right)}^2}{{\mathrm{nV}}_o+V_m\left|\sin \mathrm{\&omega;t}\right|}\mathrm{d\&omega;<figure-callout\; label="t"\; filenames="HDA0000467476040000021.png,HDA0000467476040000032.png"\; state="\{\{state\}\}">t</figure-callout>}}{2\mathrm{\&pi;}{L}_{p\max }{P}_o(1+\frac{V_m\left|\sin \mathrm{\&omega;t}\right|}{{\mathrm{nV}}_o})}---\left(33\right)$

$f_s^{\&prime;}=\frac{V_m^2{\left({\mathrm{nV}}_o\right)}^2}{12.34P_o{L}_{p\max }^{\&prime;}}\&CenterDot;\frac{1}{{({\mathrm{nV}}_o+V_m|\sin \mathrm{\&omega;t}\left|\right)}^2(1-0.72|\sin \mathrm{\&omega;t}\left|\right)}---\left(34\right)$

Fig. 7 is for being respectively (a) 85V, (b) 175V and (c) excursion of switching frequency under half power frequency period ON time decided at the higher level but not officially announced and the control of change ON time when 265V when input voltage.On the whole, within the scope of 85V～175V, under the control of change ON time, switching tube switching frequency is less; Within the scope of 175V～265V, determine under ON time control switching tube switching frequency less.

4.2 current peaks and effective value

According to formula (3) and formula (11), can be in half power frequency period, determine primary current peak I under ON time control_{lp_pk}with effective value I_{lP_rms}be respectively

$I_{\mathrm{Lp}\_\mathrm{pk}}=\frac{2\mathrm{\&pi;}{P}_o}{v_m{\&Integral;}_0^{\mathrm{\&pi;}}\frac{{\mathrm{<figure-callout\; label="sin"\; filenames="HDA0000467476040000021.png,HDA0000467476040000032.png"\; state="\{\{state\}\}">sin</figure-callout>}}^2\mathrm{\&omega;<figure-callout\; label="t"\; filenames="HDA0000467476040000011.png,HDA0000467476040000021.png"\; state="\{\{state\}\}">t</figure-callout>}}{1+\frac{V_m}{{\mathrm{nV}}_o}|\sin \mathrm{\&omega;t}}\mathrm{d\&omega;t}}---\left(35\right)$

$\begin{array}{c}{I}_{\mathrm{LP}\_\mathrm{rms}}=\sqrt{\frac{2}{T_{\mathrm{line}}}{\&Integral;}_0^{\frac{T_{\mathrm{line}}}{2}}{\left[i_{\mathrm{in}\_\mathrm{rms}}\left(t\right)\right]}^2\mathrm{dt}}=\sqrt{\frac{2}{T_{\mathrm{line}}}{\&Integral;}_0^{\frac{T_{\mathrm{line}}}{2}}{\left(\frac{V_m\left|\sin \mathrm{\&omega;t}\right|}{L_p}\right)}^2\text{\&CenterDot;}\frac{t_{\mathrm{on}}^3}{{3t}_s}\mathrm{dt}}\\ =\frac{{2P}_o}{V_m}\sqrt{\frac{\mathrm{\&pi;}}{3{\&Integral;}_0^{\mathrm{\&pi;}}\frac{{\mathrm{<figure-callout\; label="sin"\; filenames="HDA0000467476040000021.png,HDA0000467476040000032.png"\; state="\{\{state\}\}">sin</figure-callout>}}^2\mathrm{\&omega;<figure-callout\; label="t"\; filenames="HDA0000467476040000011.png,HDA0000467476040000021.png"\; state="\{\{state\}\}">t</figure-callout>}}{1+\frac{V_m}{{\mathrm{nV}}_o}\left|\sin \mathrm{\&omega;t}\right|}\mathrm{d\&omega;t}}}\end{array}---\left(36\right)$

In like manner, according to formula (3), formula (23) and formula (28), can be in half power frequency period, primary current peak I under the control of change ON time '_{lp_pk}with effective value I'_{lP_rms}be respectively

$I_{\mathrm{Lp}\_\mathrm{pk}}^{\&prime;}=\frac{12.34P_o}{V_m}\left|\sin \mathrm{\&omega;t}\right|(1+\frac{V_m\left|\sin \mathrm{\&omega;t}\right|}{{\mathrm{nV}}_o})(1-0.78|\sin \mathrm{\&omega;t}\left|\right)---\left(37\right)$

$I_{\mathrm{LP}\_\mathrm{rms}}^{\&prime;}=\frac{{4P}_o}{V_m}\sqrt{\frac{3.086}{3\mathrm{\&pi;}}{\&Integral;}_0^{\mathrm{\&pi;}}(1+\frac{V_m\left|\sin \mathrm{\&omega;t}\right|}{{\mathrm{nV}}_o}){\mathrm{<figure-callout\; label="sin"\; filenames="HDA0000467476040000021.png,HDA0000467476040000032.png"\; state="\{\{state\}\}">sin</figure-callout>}}^2\mathrm{\&omega;t}{(1-0.78|\sin \mathrm{\&omega;t}\left|\right)}^2\mathrm{dt}}---\left(38\right)$

Fig. 8 is that wide region is inputted lower two kinds of control mode primary current peak values and effective value waveform.As can be seen from the figure, than the control of fixing time, adopt and become conducting control, converter primary current peak value is smaller, and effective value is bigger.

The CRM Flyback LED driver of the low output current peak-to-average force ratio of 5 the present invention

In conjunction with Fig. 9, input voltage v_gthrough resistance the first resistance R₁with the second resistance R₂dividing potential drop obtains v_a=k_vgv_m| sin ω t|, k here_vgit is dividing potential drop coefficient.The 3rd resistance R₃, the 4th resistance R₄, the 5th resistance R₅, the 6th resistance R₆, the first diode D₁, the first capacitor C₁composition dividing potential drop peak sample circuit, v_c=1.28k_vgv_m, wherein R₃/ R₄=1.28R₁/ R₂.Output voltage V_othrough the 9th resistance R₉with the tenth resistance R₁₀dividing potential drop is again through the 7th resistance R₇, the 8th resistance R₈, the 4th operational amplification circuit A₄the amplifier of composition obtains v_d=k_vgnV_o, wherein R₉/ R₁₀=R₁/ R₂, R₈/ R₇=n-1.V_awith v_caccess subtraction circuit, wherein R₁₁=R₁₄, R₁₂=R₁₃, R₁₄/ R₁₃=0.78, be output as v_f=k_vgv_m(1-0.78|sin ω t|).V_awith v_daccess add circuit, wherein R₁₅=R₁₆=R₁₇=R₁₉=2R₁₈, be output as v_e=k_vg(nV_o+ V_m| sin ω t|).V_d, v_ewith v_faccess the first multiplier, it exports v_a=[k_vgv_m(1-0.78|sin ω t|) (nV_o+ V_m| sin ω t|)]/nV_o.Output current i_oobtain error signal v by output current feedback circuit_eA, v_eAwith v_aaccess the second multiplier, it exports v_c=v_eAv_a=[v_eAk_vgv_m(1-0.78|sin ω t|) (nV_o+ V_m| sin ω t|)]/nV_o, by v_chand over and cut the ON time that can obtain suc as formula Changing Pattern shown in (30) with sawtooth waveforms.Wherein v_a, v_c, v_d, v_e, v_f, v_a, v_c, v_bbe respectively the first dividing potential drop follow circuit 3, the second dividing potential drop peak sample circuit 4, the 3rd dividing potential drop amplifying circuit 5, add circuit 6, subtraction circuit 7, the 1st multiplier the 8, the 2nd multiplier 9, output voltage output current feedback circuit 10.Physical circuit is as follows:

The CRM Flyback LED driver of low output current peak-to-average force ratio of the present invention, comprises main power circuit 1 and control circuit, and described main power circuit 1 comprises input voltage source v_in, electromagnetic interface filter, diode rectifier circuit RB, transformer T₁, switching tube Q_b, diode D_b, filter capacitor C_o, filter inductance L_owith load LED, wherein input voltage source v_inbe connected with the input port of electromagnetic interface filter, the output port of electromagnetic interface filter is connected with the input port of diode rectifier circuit RB, and the output negative pole of diode rectifier circuit RB is reference potential zero point, the output cathode of diode rectifier circuit RB and transformer T₁the first winding N_pdifferent name end connect, transformer T₁the first winding N_psame Name of Ends access switching tube Q_bdrain electrode, switching tube Q_bsource electrode be connected zero point with reference potential, transformer T₁the second winding N_zdifferent name end be connected zero point with reference potential, transformer T₁tertiary winding N_ssame Name of Ends and diode D_banodic bonding, diode D_bnegative electrode access respectively filter capacitor C_oone end and filter inductance L_oone end, filter capacitor C_oother end access reference potential zero point, filter inductance L_othe other end be connected with the anode tap of load LED, the both end voltage of load LED is output voltage V_o; It is V that described control circuit adopts Changing Pattern_m(nV_o+ V_m| sin ω t|) (1-0.78|sin ω t)/nV_othe output signal driving switch pipe Q of ON time_bcomprise sawtooth waveforms comparison and switch tube driving circuit 2, the first dividing potential drop follow circuit 3, the second dividing potential drop peak sample circuit 4, the 3rd dividing potential drop amplifying circuit 5, add circuit 6, subtraction circuit 7, the first multiplier 8, the second multiplier 9, output current feedback circuit 10, the wherein output of sawtooth waveforms comparison and switch tube driving circuit 2 and switching tube Q_bgate pole connect; The input of the first dividing potential drop follow circuit 3 and input voltage sampled point V_gthe output cathode that is diode rectifier circuit RB connects, and the output terminals A of the first dividing potential drop follow circuit 3 is connected with input of add circuit 6 and an input of subtraction circuit 7 respectively; The input of the second dividing potential drop peak sample circuit 4 and input voltage sampled point V_gthe output cathode that is diode rectifier circuit RB connects, and the second output C of dividing potential drop peak sample circuit 4 and another input of subtraction circuit 7 are connected; The output D of the 3rd dividing potential drop amplifying circuit 5 respectively with another input of add circuit 6 and the 3rd input v of the first multiplier 8_zconnect; The second input v of the output E of add circuit 6 and the first multiplier 8_yconnect the first input end v of the output F of subtraction circuit 7 and the first multiplier 8_xconnect the first input end v of the output of the first multiplier 8 and the second multiplier 9_aconnect the output access sawtooth waveforms comparison of the second multiplier 9 and the input of switch tube driving circuit 2, the second input v of the output of output current feedback circuit 10 and the second multiplier 9_bconnect.

Described sawtooth waveforms comparison and switchtube driving circuit 2 comprise zero passage detection, rest-set flip-flop, driving, saw-toothed wave generator, the first operational amplifier A₁; The input of zero passage detection and transformer T₁the second winding N_zsame Name of Ends connect, the output of zero passage detection and the S of rest-set flip-flop end is connected, the R of rest-set flip-flop holds and the first operational amplifier A₁output connect, the Q of rest-set flip-flop end is connected with the input driving and the input of saw-toothed wave generator respectively, the output of saw-toothed wave generator and the first operational amplifier A₁positive input connect, the output of driving is output and the switching tube Q of sawtooth waveforms comparison and switchtube driving circuit 2_bgate pole connect, the output of the second multiplier 9 accesses the first operational amplifier A₁reverse input end be the input of sawtooth waveforms comparison and switchtube driving circuit 2.

The first described dividing potentialdrop follow circuit 3 comprises the second operational amplifier A₂, the first resistance R₁, the second resistance R₂; Wherein the first resistance R₁one end and input voltage sampled point V_gthe output cathode that is diode rectifier circuit RB connects, the first resistance R₁the other end and the second resistance R₂one end connects, and the first resistance R₁with the second resistance R₂common port access the first operational amplifier A₂positive input, the second resistance R₂the other end be connected zero point with reference potential, the second operational amplifier A₂reverse input end be directly connected with output terminals A, form in-phase voltage follower.

The second described dividing potential droppeak sample circuit 4 comprises the 3rd resistance R₃, the 4th resistance R₄, the 5th resistance R₅, the first diode D₁, the first capacitor C₁, the 6th resistance R₆, the 3rd operational amplifier A₃; Wherein the 3rd resistance R₃one end and input voltage sampled point V_gthe output cathode that is diode rectifier circuit RB connects, the 3rd resistance R₃the other end and the 4th resistance R₄one end connects, and the 3rd resistance R₃with the 4th resistance R₄common port access the 5th resistance R₅one end, the 4th resistance R₄the other end be connected zero point with reference potential, the 5th resistance R₅the other end and the first diode D₁after anodal series connection through the first diode D₁negative pole access the 3rd operational amplifier A₃normal phase input end, the first capacitor C₁with the 6th resistance R₆one end and the 3rd operational amplifier A after in parallel₃normal phase input end be connected, another termination reference potential zero point, the 3rd operational amplifier A₃inverting input be directly connected with output C.

The 3rd described dividing potentialdrop amplifying circuit 5 comprises the 7th resistance R₇, the 8th resistance R₈, the 9th resistance R₉, the tenth resistance R₁₀, four-operational amplifier A₄; Wherein the 7th resistance R₇termination reference potential zero point, the 7th resistance R₇the other end and the 8th resistance R₈one end connect, and the 7th resistance R₇with the 8th resistance R₈common port access four-operational amplifier A₄inverting input, the 8th resistance R₈the other end and four-operational amplifier A₄output D connect, the 9th resistance R₉one end and the output voltage V ofmain power circuit 1_oanodal connection, the 9th resistance R₉the other end and the tenth resistance R₁₀one end connect, and the 9th resistance R₉with the tenth resistance R₁₀common port access four-operational amplifier A₄normal phase input end.

Described addcircuit 6 comprises the 15 resistance R₁₅, the 16 resistance R₁₆, the 17 resistance R₁₇, the 18 resistance R₁₈, the 19 resistance R₁₉, the 5th operational amplifier A₅; Wherein the 15 resistance R₁₅one end is connected with the output terminals A of the first dividing potentialdrop follow circuit 3, other end access the 5th operational amplifier A₅positive input, the 16 resistance R₁₆one end is connected with the 3rd dividing potentialdrop amplifying circuit 5 output D, other end access the 5th operational amplifier A₅positive input, the 17 resistance R₁₇one end and the 5th operational amplifier A₅positive input connect, other end access reference point position zero point, the 18 resistance R₁₈one end access the 5th operational amplifier A₅reverse input end, other end access reference point position zero point, the 19 resistance R₁₉access the 5th operational amplifier A₅reverse input end and output E between, the second input v ofadd circuit 6 output E and the first multiplier 8_yconnect.

Described subtraction circuit 7 comprises the 11 resistance R₁₁, the 12 resistance R₁₂, the 13 resistance R₁₃, the 14 resistance R₁₄, the 6th operational amplifier A₆; Wherein the 11 resistance R₁₁one end is connected with the output terminals A of the first dividing potentialdrop follow circuit 3, and the other end is connected to the 6th operational amplifier A₆reverse input end, the 12 resistance R₁₂be connected to the 6th operational amplifier A₆reverse input end and output F between, the 13 resistance R₁₃one end is connected to the output C of the second dividing potential droppeak sample circuit 4, the 13 resistance R₁₃other end access the 6th operational amplifier A₆positive input, the 14 resistance R₁₄one end access the 6th operational amplifier A₆positive input, the 14 resistance R₁₄the other end be connected zero point with reference potential, the 6th operational amplifier A₆output be the first input end v that the output F of subtraction circuit 7 accesses the first multiplier 8_x.

Described output current feedback circuit 10 comprises the second transformer T₂, the second diode D₂, the 20 resistance R₂₀, the 21 resistance R₂₁, the second capacitor C₂, the 3rd capacitor C₃, the 22 resistance R₂₂, the 7th operational amplifier A₇; Wherein the second transformer T₂former limit Same Name of Ends and the transformer T of main power circuit 1₁tertiary winding N_sdifferent name end connect, the second transformer T₂former limit different name end be connected zero point with reference potential, the second transformer T₂secondary Same Name of Ends and the second diode D₂anodal connection, the second transformer T₂secondary different name end and the 20 resistance R₂₀rear and the second diode D connect₂negative pole connects, the 20 resistance R₂₀with the second diode D₂common port access the 21 resistance R₂₁one end, the 21 resistance R₂₁the other end and the second capacitor C₂one end connect, and the 21 resistance R₂₁with the second capacitor C₂common port access the 7th operational amplifier A₇inverting input, the second capacitor C₂the other end be connected zero point with reference potential, the 22 resistance R₂₂with the 3rd capacitor C₃access the 7th operational amplifier A after series connection₇reverse input end and output between, the positive input of the 7th operational amplifier A 7 and input voltage reference point V_ogconnect the 7th operational amplifier A₇output be the second input v that the output of output current feedback circuit 10 accesses the second multiplier 9_b.

In sum, the CRM Flyback LED driver of low output current peak-to-average force ratio of the present invention, in the situation that keeping power factor PF value to meet application requirements, adopt and become ON time control and realize and in input current, only contain a certain amount of identical with first-harmonic initial phase three times, quintuple harmonics, the switching tube ON time of driver is changed according to certain rule in a power frequency period, within the scope of whole 85V～265V ac input voltage, output current peak-to-average force ratio is reduced near 1.44, reduce output current peak-to-average force ratio, simultaneously little to other performance impacts.

Claims (8)

1. a CRM Flyback LED driver for low output current peak-to-average force ratio, is characterized in that, comprises main power circuit (1) and control circuit, and described main power circuit (1) comprises input voltage source v_in, electromagnetic interface filter, diode rectifier circuit RB, transformer T₁, switching tube Q_b, diode D_b, filter capacitor C_o, filter inductance L_owith load LED, wherein input voltage source v_inbe connected with the input port of electromagnetic interface filter, the output port of electromagnetic interface filter is connected with the input port of diode rectifier circuit RB, and the output negative pole of diode rectifier circuit RB is reference potential zero point, the output cathode of diode rectifier circuit RB and transformer T₁the first winding N_pdifferent name end connect, transformer T₁the first winding N_psame Name of Ends access switching tube Q_bdrain electrode, switching tube Q_bsource electrode be connected zero point with reference potential, transformer T₁the second winding N_zdifferent name end be connected zero point with reference potential, transformer T₁tertiary winding N_ssame Name of Ends and diode D_banodic bonding, diode D_bnegative electrode access respectively filter capacitor C_oone end and filter inductance L_oone end, filter capacitor C_oother end access reference potential zero point, filter inductance L_othe other end be connected with the anode tap of load LED, the both end voltage of load LED is output voltage V_o;

Described control circuit comprises sawtooth waveforms comparison and switch tube driving circuit (2), the first dividing potential drop follow circuit (3), the second dividing potential drop peak sample circuit (4), the 3rd dividing potential drop amplifying circuit (5), add circuit (6), subtraction circuit (7), the first multiplier (8), the second multiplier (9), output current feedback circuit (10); The wherein output of sawtooth waveforms comparison and switch tube driving circuit (2) and switching tube Q_bgate pole connect; The input of the first dividing potential drop follow circuit (3) and input voltage sampled point V_gthe output cathode that is diode rectifier circuit RB connects, and the output terminals A of the first dividing potential drop follow circuit (3) is connected with input of add circuit (6) and an input of subtraction circuit (7) respectively; The input of the second dividing potential drop peak sample circuit (4) and input voltage sampled point V_gthe output cathode that is diode rectifier circuit RB connects, and the output C of the second dividing potential drop peak sample circuit (4) is connected with another input of subtraction circuit (7); The output voltage V of the input of the 3rd dividing potential drop amplifying circuit (5) and main power circuit (1)_oanodal connection, the output D of the 3rd dividing potential drop amplifying circuit (5) respectively with another input of add circuit (6) and the 3rd input v of the first multiplier (8)_zconnect; The second input v of the output E of add circuit (6) and the first multiplier (8)_yconnect the first input end v of the output F of subtraction circuit (7) and the first multiplier (8)_xconnect the first input end v of the output of the first multiplier (8) and the second multiplier (9)_aconnect, the output access sawtooth waveforms comparison of the second multiplier (9) and the input of switch tube driving circuit (2), the second input v of the output of output current feedback circuit (10) and the second multiplier (9)_bconnect.

2. the CRM Flyback LED driver of low output current peak-to-average force ratio according to claim 1, it is characterized in that, described sawtooth waveforms comparison and switch tube driving circuit (2) comprise zero passage detection, rest-set flip-flop, driving, saw-toothed wave generator, the first operational amplifier A₁; The input of zero passage detection and transformer T₁the second winding N_zsame Name of Ends connect, the output of zero passage detection and the S of rest-set flip-flop end is connected, the R of rest-set flip-flop holds and the first operational amplifier A₁output connect, the Q of rest-set flip-flop end is connected with the input driving and the input of saw-toothed wave generator respectively, the output of saw-toothed wave generator and the first operational amplifier A₁positive input connect, the output of driving is output and the switching tube Q of sawtooth waveforms comparison and switch tube driving circuit (2)_bgate pole connect, the output of the second multiplier (9) accesses the first operational amplifier A₁reverse input end is the input of sawtooth waveforms comparison and switch tube driving circuit (2).

3. the CRM Flyback LED driver of low output current peak-to-average force ratio according to claim 1, is characterized in that, the first described dividing potential drop follow circuit (3) comprises the second operational amplifier A₂, the first resistance R₁, the second resistance R₂; Wherein the first resistance R₁one end and input voltage sampled point V_gthe output cathode that is diode rectifier circuit RB connects, the first resistance R₁the other end and the second resistance R₂one end connects, and the first resistance R₁with the second resistance R₂common port access the second operational amplifier A₂positive input, the second resistance R₂the other end be connected zero point with reference potential, the first operational amplifier A₁reverse input end be directly connected with output terminals A, form in-phase voltage follower.

4. the CRM Flyback LED driver of low output current peak-to-average force ratio according to claim 1, is characterized in that, described the second dividing potential drop peak sample circuit (4) comprises the 3rd resistance R₃, the 4th resistance R₄, the 5th resistance R₅, the first diode D₁, the first capacitor C₁, the 6th resistance R₆, the 3rd operational amplifier A₃; Wherein the 3rd resistance R₃one end and input voltage sampled point V_gthe output cathode that is diode rectifier circuit RB connects, the 3rd resistance R₃the other end and the 4th resistance R₄one end connects, and the 3rd resistance R₃with the 4th resistance R₄common port access the 5th resistance R₅one end, the 4th resistance R₄the other end be connected zero point with reference potential, the 5th resistance R₅the other end and the first diode D₁after anodal series connection through the first diode D₁negative pole access the 3rd operational amplifier A₃normal phase input end, the first capacitor C₁with the 6th resistance R₆one end and the 3rd operational amplifier A after in parallel₃normal phase input end be connected, another termination reference potential zero point, the 3rd operational amplifier A₃inverting input be directly connected with output C.

5. the CRM Flyback LED driver of low output current peak-to-average force ratio according to claim 1, is characterized in that, described the 3rd dividing potential drop amplifying circuit (5) comprises the 7th resistance R₇, the 8th resistance R₈, the 9th resistance R₉, the tenth resistance R₁₀, four-operational amplifier A₄; Wherein the 7th resistance R₇termination reference potential zero point, the 7th resistance R₇the other end and the 8th resistance R₈one end connect, and the 7th resistance R₇with the 8th resistance R₈common port access four-operational amplifier A₄inverting input, the 8th resistance R₈the other end and four-operational amplifier A₄output D connect, the 9th resistance R₉one end and the output voltage V of main power circuit (1)_oanodal connection, the 9th resistance R₉the other end and the tenth resistance R₁₀one end connect, and the 9th resistance R₉with the tenth resistance R₁₀common port access four-operational amplifier A₄normal phase input end.

6. the CRM Flyback LED driver of low output current peak-to-average force ratio according to claim 1, is characterized in that, described add circuit (6) comprises the 15 resistance R₁₅, the 16 resistance R₁₆, the 17 resistance R₁₇, the 18 resistance R₁₈, the 19 resistance R₁₉, the 5th operational amplifier A₅; Wherein the 15 resistance R₁₅one end is connected with the output terminals A of the first dividing potential drop follow circuit (3), other end access the 5th operational amplifier A₅positive input, the 16 resistance R₁₆one end is connected with the 3rd dividing potential drop amplifying circuit (5) output D, other end access the 5th operational amplifier A₅positive input, the 17 resistance R₁₇one end and the 5th operational amplifier A₅positive input connect, other end access reference point position zero point, the 18 resistance R₁₈one end access the 5th operational amplifier A₅reverse input end, other end access reference point position zero point, the 19 resistance R₁₉access the 5th operational amplifier A₅reverse input end and output E between, the second input v of add circuit (6) output E and the first multiplier (8)_yconnect.

7. the CRM Flyback LED driver of low output current peak-to-average force ratio according to claim 1, is characterized in that, described subtraction circuit (7) comprises the 11 resistance R₁₁, the 12 resistance R₁₂, the 13 resistance R₁₃, the 14 resistance R₁₄, the 6th operational amplifier A₆; Wherein the 11 resistance R₁₁one end is connected with the output terminals A of the first dividing potential drop follow circuit (3), and the other end is connected to the 6th operational amplifier A₆reverse input end, the 12 resistance R₁₂be connected to the 6th operational amplifier A₆reverse input end and output F between, the 13 resistance R₁₃one end is connected to the output C of the second dividing potential drop peak sample circuit (4), the 13 resistance R₁₃other end access the 6th operational amplifier A₆positive input, the 14 resistance R₁₄one end access the 6th operational amplifier A₆positive input, the 14 resistance R₁₄the other end be connected zero point with reference potential, the 6th operational amplifier A₆output be the first input end v that the output F of subtraction circuit (7) accesses the first multiplier (8)_x.

8. the CRM Flyback LED driver of low output current peak-to-average force ratio according to claim 1, is characterized in that, described output current feedback circuit (10) comprises the second transformer T₂, the second diode D₂, the 20 resistance R₂₀, the 21 resistance R₂₁, the second capacitor C₂, the 3rd capacitor C₃, the 22 resistance R₂₂, the 7th operational amplifier A₇; Wherein the second transformer T₂former limit Same Name of Ends and the transformer T of main power circuit (1)₁tertiary winding N_sdifferent name end connect, the second transformer T₂former limit different name end be connected zero point with reference potential, the second transformer T₂secondary Same Name of Ends and the second diode D₂anodal connection, the second transformer T₂secondary different name end and the 20 resistance R₂₀rear and the second diode D connect₂negative pole connects, the 20 resistance R₂₀with the second diode D₂common port access the 21 resistance R₂₁one end, the 21 resistance R₂₁the other end and the second capacitor C₂one end connect, and the 21 resistance R₂₁with the second capacitor C₂common port access the 7th operational amplifier A₇inverting input, the second capacitor C₂the other end be connected zero point with reference potential, the 22 resistance R₂₂with the 3rd capacitor C₃access the 7th operational amplifier A after series connection₇reverse input end and output between, the positive input of the 7th operational amplifier A 7 and input voltage reference point V_ogconnect the 7th operational amplifier A₇output be the second input v that the output of output current feedback circuit (10) accesses the second multiplier (9)_b.

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