CN104753350A - Method used for prediction convergence control of inductive current in booster circuit - Google Patents

Abstract

The invention discloses a method used for prediction convergence control of inductive current in a booster circuit. The method includes: predicting an inductive current value of a next moment through an inductive current value iL(n-1) of a previous moment; predicting to acquire a value of a duty ratio of a switch tube at an nTS time moment according to power conservation; subjecting a calculation result to convergence processing according to a steady-state expected value; predicting to acquire an average value of the duty ratio of the switch tube and the duty ratio of the switch tube in a steady state in the process of convergence so as to acquire a final duty ratio Dx(n) of the switch tube and generate PWM (pulse width modulation) with the duty ratio of Dx(n) to control the switch tube to realize control on inductive current. By the method, the duty ratio of the switch tube is calculated quickly by taking the inductive current as controlled amount; when a given is suddenly increased or reduced, the given can be tracked quickly, so that an algorithm can acquire good dynamic-steady state characteristics.

Description

A kind of for inductive current prediction convergence control method in booster circuit

Technical field

The invention belongs to Power Electronic Circuit control technology field, be specifically related to a kind of for inductive current prediction convergence control method in booster circuit.

Background technology

In recent years, boosting (Boost) converter is a kind of electronic power switching circuit be widely adopted, through being usually used in the equipment to direct current energy transmission.Usually by output voltage or the output current of the modulation duty cycle control circuit of the power tube in regulating circuit, the input side of booster circuit is generally photovoltaic array, storage battery, super capacitor or other power supply, by regulating and controlling converter, it is made to export electric energy by oriented load.

Nowadays, Boost has different control methods, such as PI control, Repetitive controller, fuzzy control, synovial membrane control etc.The advantage that Repetitive controller has control precision high, shortcoming is that algorithm is complicated, and parameter determines comparatively difficulty.The advantage of fuzzy control has stronger robustness and vulnerability to jamming, and shortcoming is that the control precision of system reduces and bad dynamic performance.The advantage that synovial membrane controls improves dynamic, the steady-state behaviour of system, and controlled system can be made to have stronger robustness; Shortcoming is under this control can not make power switch component be operated in fixing frequency, and output voltage ripple is comparatively large, also higher to the designing requirement of filter.And traditional PI control algolithm, it has the advantages such as applicability is better, Control System Design is simple, but the design that its shortcoming is then output feedback ontrol is based target control errors instead of based on model cootrol, dynamic responding speed is comparatively slow, static difference is comparatively large, control effects is poor, can not optimal control be realized, the requirement that Switching Power Supply significantly improves dynamic response, control precision can not be met.

Summary of the invention

The object of this invention is to provide a kind of for inductive current prediction convergence control method in booster circuit, solve the problem that the dynamic response characteristic existed in prior art is slow, static difference is comparatively large, control precision is poor.

The technical solution adopted in the present invention is, a kind of for inductive current prediction convergence control method in booster circuit, specifically implements according to following steps:

The magnitude of voltage v of the low tension potential source in step 1, sampling booster circuit _l, high-tension electricity potential source magnitude of voltage v _h, (n-1) T _sthe inductor current value i in moment _l(n-1), (n-1) T _sthe duty ratio D of switching tube in cycle _bo(n-1),

Wherein, T _sfor a switch periods of switching tube;

Step 2, judge nT _sgiven value of current i on moment inductance _lrefn whether () be less than 0, if be not less than 0, forwards step 3 to; If be less than 0, then nT _sthe duty ratio D of switching tube in cycle _xn ()=0, forwards step 9 to;

Step 3, by formula (1) calculate nT _sthe inductive current predicted value in moment

${\hat{i}}_L\left(n\right)=\left\{\begin{array}{cc}w(n-1),& w(n-1)\&GreaterEqual;0\\ 0,& w(n-1)<0\end{array}\right.---\left(1\right)$

Wherein,

$w(n-1)=i_L(n-1)+\frac{1}{L}\&CenterDot;T_S\&CenterDot;[v_L-(1-D_{\mathrm{bo}}(n-1))\&CenterDot;v_H]---\left(2\right)$

L is inductance value;

Step 4, judgement whether set up, if set up, forward step 5 to; If be false, then nT _sthe duty ratio D of switching tube in cycle _xn ()=1, forwards step 9 to;

Step 5, by nT _sgiven value of current value i on moment inductance _lrefn () assignment is to nT _sthe mean value of cycle internal inductance electric current that is:

$\stackrel{\&OverBar;}{i_L}\left(n\right)=i_{\mathrm{Lref}}\left(n\right)---\left(3\right)$

By the inductive current predicted value obtained in formula (1) calculate nT _sthe duty ratio D of switching tube in cycle _bo(n):

$D_{\mathrm{bo}}\left(n\right)=1-\sqrt{D_{\mathrm{bt}}+\frac{2L}{v_H\&CenterDot;T_S}[{\hat{i}}_L\left(n\right)-\stackrel{\&OverBar;}{i_L}\left(n\right)]}---\left(4\right)$

Wherein, D _bt=v _l/ v _h;

Step 6, according to formula (5) calculate w (n):

$w\left(n\right)={\hat{i}}_L\left(n\right)+\frac{1}{L}\&CenterDot;T_S\&CenterDot;[v_L-(1-D_{\mathrm{bo}}\left(n\right))\&CenterDot;v_H]---\left(5\right)$

Judge whether w (n) >=0 sets up, if be false, then forward step 7 to; If set up, then forward step 8 to;

Step 7, according to the inductive current predicted value obtained in formula (1) calculate nT _sthe duty ratio D of switching tube in cycle _bo(n):

$D_{\mathrm{bo}}\left(n\right)=\frac{-L\&CenterDot;{\hat{i}}_L\left(n\right)}{v_L\&CenterDot;T_S}+\frac{\sqrt{(1-D_{\mathrm{bt}})\&CenterDot;[L^2\&CenterDot;{{\hat{i}}_L}^2\left(n\right)+2L\&CenterDot;v_L\&CenterDot;T_S\&CenterDot;\stackrel{\&OverBar;}{i_L}\left(n\right)]}}{v_L\&CenterDot;T_S}---\left(6\right)$

Step 8, calculate duty ratio D after intermediate value process according to formula (7) _x(n):

$D_x\left(n\right)=\frac{1}{2}(D_{\mathrm{bo}}\left(n\right)+D_z\left(n\right))---\left(7\right)$

Wherein, for nT _sthe duty ratio of switching tube during cycle homeostasis;

Step 9, generation duty ratio are D _xn the PWM ripple of () carrys out control switch pipe.

Feature of the present invention is also,

The detailed process obtaining formula in step 5 (4) is:

When switching tube is opened, inductance continues energy storage, and inductive current increases, nT _sthe increment Delta i of inductive current in switching tube opening process in cycle _l+for:

${\mathrm{\&Delta;i}}_{L+}=i_{L\max }\left(n\right)-i_L\left(n\right)=\frac{1}{L}{v}_L{D}_{\mathrm{bo}}\left(n\right)T_S---\left(8\right)$

Wherein, i _lmaxn () is nT _sthe maximum of cycle internal inductance electric current,

When switching tube turns off, the pad value Δ i of inductive current _l-for:

${\mathrm{\&Delta;i}}_{L-}=\frac{1}{L}(v_L-v_H)(1-D_{\mathrm{bo}}\left(n\right))T_S---\left(9\right)$

At nT _scycle is to (n+1) T _sin cycle, inductive current is by i _ln () changes i into _l(n+1), (n+1) T _sthe inductive current i in moment _l(n+1):

$i_L(n+1)=i_L\left(n\right)+{\mathrm{\&Delta;i}}_{L+}+{\mathrm{\&Delta;i}}_{L-}=i_L\left(n\right)+\frac{1}{L}\&CenterDot;T_S\&CenterDot;[v_L-(1-D_{\mathrm{bo}}\left(n\right))\&CenterDot;v_H]=w\left(n\right)---\left(10\right)$

Due to the reverse turn-off characteristic of diode, inductive current cannot decay to negative value continuously, then i _l(n+1) there is a discontinuity point, expression formula is:

$i_L(n+1)=\left\{\begin{array}{cc}w\left(n\right),& w\left(n\right)\&GreaterEqual;0\\ 0,& w\left(n\right)<0\end{array}\right.---\left(11\right)$

Formula (11) is converted to:

$i_L(n+1)=\left\{\begin{array}{cc}w\left(n\right),& D_{\mathrm{bo}}\left(n\right)\&GreaterEqual;1-D_{\mathrm{bt}}-\frac{L\&CenterDot;i_L\left(n\right)}{v_H\&CenterDot;T_S}\\ 0,& D_{\mathrm{bo}}\left(n\right)<1-D_{\mathrm{bt}}-\frac{L\&CenterDot;i_L\left(n\right)}{v_H\&CenterDot;T_S}\end{array}\right.---\left(12\right)$

Wherein D _bt=v _l/ v _h;

In the ideal situation, input v _lthe ENERGY E of release _vLfor:

$E_{\mathrm{vL}}=v_L\&CenterDot;\stackrel{\&OverBar;}{i_L}\left(n\right)\&CenterDot;T_S---\left(13\right)$

Input v _lthe ENERGY E of release _vLa part is stored on inductance, shows as the variable quantity of cycle internal inductance electric current, this part energy Δ E _lbe expressed as:

${\mathrm{\&Delta;E}}_L=\frac{1}{2}L[{i_L}^2(n+1)-{i_L}^2\left(n\right)]---\left(14\right)$

Another part shows as to output v for load consumption _hcharging, the ENERGY E that monocycle internal burden consumes _offfor:

$E_{\mathrm{off}}=v_H\&CenterDot;\stackrel{\&OverBar;}{i_{\mathrm{off}}}\left(n\right)\&CenterDot;(1-D_{\mathrm{bo}}\left(n\right))\&CenterDot;T_S---\left(15\right)$

Wherein, for v when switching tube turns off _haverage current:

$\stackrel{\&OverBar;}{i_{\mathrm{off}}}\left(n\right)=\left\{\begin{array}{cc}{i}_L\left(n\right)+{\mathrm{\&Delta;i}}_{L+}+\frac{1}{2}{\mathrm{\&Delta;i}}_{L-},& w\left(n\right)\&GreaterEqual;0\\ \frac{L\&CenterDot;{[i_L\left(n\right)+{\mathrm{\&Delta;i}}_{L+}]}^2}{2\&CenterDot;(v_H-v_L)(1-D_{\mathrm{bo}}\left(n\right))\&CenterDot;T_S},& w\left(n\right)<0\end{array}\right.---\left(16\right)$

In the ideal situation, and ignore various loss, obtain according to the conservation of energy:

E _vL＝ΔE _L+E _off(17)

When w (n) >=0:

$\frac{1}{2L}\&CenterDot;v_H\&CenterDot;T_S\&CenterDot;{D^2}_{\mathrm{bo}}\left(n\right)-\frac{1}{L}\&CenterDot;v_H\&CenterDot;T_S\&CenterDot;D_{\mathrm{bo}}\left(n\right)+\frac{1}{2L}\&CenterDot;T_S\&CenterDot;(v_H-v_L)-i_L\left(n\right)+\stackrel{\&OverBar;}{i_L}\left(n\right)=0---\left(18\right)$

When meeting

$\frac{1}{L^2}\&CenterDot;{T_S}^2\&CenterDot;v_H\&CenterDot;v_L-2\&CenterDot;\frac{1}{L}\&CenterDot;v_H\&CenterDot;T_S[\stackrel{\&OverBar;}{i_L}\left(n\right)-i_L\left(n\right)]\&GreaterEqual;0---\left(19\right)$

Namely $\stackrel{\&OverBar;}{i_L}\left(n\right)\&le;i_L\left(n\right)+\frac{1}{2L}\&CenterDot;v_L\&CenterDot;T_S---\left(20\right)$

Time formula (18) have solution, solution is:

$D_{\mathrm{bo}}\left(n\right)=1-\sqrt{D_{\mathrm{bt}}+\frac{2L}{v_H\&CenterDot;T_S}[i_L\left(n\right)-\stackrel{\&OverBar;}{i_L}\left(n\right)]}---\left(21\right).$

The detailed process obtaining formula in step 7 (6) is:

As w (n) < 0:

$\frac{1}{2L}\&CenterDot;v_H\&CenterDot;v_L\&CenterDot;{T_S}^2\&CenterDot;{D^2}_{\mathrm{bo}}\left(n\right)+v_H\&CenterDot;i_L\left(n\right)\&CenterDot;T_S\&CenterDot;D_{\mathrm{bo}}\left(n\right)+\frac{1}{2}\&CenterDot;L\&CenterDot;{i_L}^2\left(n\right)-(v_H-v_L)\&CenterDot;\stackrel{\&OverBar;}{i_L}\left(n\right)\&CenterDot;T_S=0---\left(22\right)$

When meeting

$v_H\&CenterDot;(v_H-v_L)\&CenterDot;{T_S}^2\&CenterDot;[{i_L}^2\left(n\right)+2\&CenterDot;\frac{1}{L}\&CenterDot;v_L\&CenterDot;T_S\&CenterDot;\stackrel{\&OverBar;}{i_L}\left(n\right)]\&GreaterEqual;0---\left(23\right)$

Namely $\stackrel{\&OverBar;}{i_L}\left(n\right)\&GreaterEqual;-\frac{L\&CenterDot;{i_L}^2\left(n\right)}{2\&CenterDot;v_L\&CenterDot;T_S}---\left(24\right)$

Time formula (22) have solution, solution is:

$D_{\mathrm{bo}}\left(n\right)=\frac{-L\&CenterDot;i_L\left(n\right)}{v_L\&CenterDot;T_S}+\frac{\sqrt{(1-D_{\mathrm{bt}})[L^2\&CenterDot;{i_L}^2\left(n\right)+2L\&CenterDot;v_L\&CenterDot;T_S\&CenterDot;\stackrel{\&OverBar;}{i_L}\left(n\right)]}}{v_L\&CenterDot;T_S}---\left(25\right).$

The invention has the beneficial effects as follows:

1. the present invention is a kind of different from general modeling analysis for the modeling method of inductive current prediction convergence control method in booster circuit, traditional modeling is all analyzed by DCM and CCM two kinds of mode of operations, and usually obtains duty ratio according to small-signal analysis.The present invention is by summing up DCM, CCM, stable state and the concrete state such as dynamic, and carry out entirety classification, according to the rule of the conservation of energy, obtain the relation between the duty ratio of switching tube and average inductor current, inductance parameters, input voltage, output voltage, instantaneous inductor electric current, switch periods, and then prediction obtains the value of the duty ratio of switching tube, again according to stable state desired value, convergence process is carried out to result of calculation, the mean value of the duty ratio of switching tube when convergence process employing is predicted and obtained duty ratio and the stable state of switching tube, thus obtain the duty ratio of final switching tube.

2. the present invention is according to the conservation of energy, calculates the duty ratio of switching tube using inductive current as controlled variable fast.If do not add convergence, 1 switch periods just energy tracing preset; If add convergence, 2 switch periods just energy tracing preset, dynamic response characteristic is fast.And in given unexpected increase or when reducing suddenly, can track given fast, describe the static and dynamic performance that this algorithm can reach good, static difference is little, control precision good.

Accompanying drawing explanation

Fig. 1 is booster circuit sketch in the present invention;

Fig. 2 is the inductive current dynamic waveform figure of booster circuit in the present invention;

Fig. 3 is seven kinds of working mode figures of booster circuit inductive current in the monocycle in the present invention;

Fig. 4 is switching tube current waveform schematic diagram in the Single switch cycle in the present invention;

Fig. 5 is the flow chart of inductive current forecast Control Algorithm of the present invention;

Fig. 6 is the asynchronous inductive current oscillogram of sampled value in the present invention;

Inductive current oscillogram when Fig. 7 is stable state in the present invention.

Embodiment

Below in conjunction with the drawings and specific embodiments, the present invention is described in detail.

Be illustrated in figure 1 the present invention's boosting (Boost) circuit diagram, Boost circuit comprises low tension potential source v _l(low tension potential source is generally the device of the energy such as super capacitor or storage battery energy storage), low tension potential source v _lnegative pole respectively with high-tension electricity potential source v _h(be generally the voltage of the Support Capacitor of DC bus parallel connection, because capacitance voltage can not suddenly change, so a voltage source can be regarded in the short time as) negative pole, switching tube S emitter connect, the collector electrode of switching tube S is connected with the positive pole of inductance L, diode D respectively, the negative pole of diode D and high-tension electricity potential source v _hpositive pole connect.

Switching tube S be carry anti-paralleled diode or there is anti-paralleled diode characteristic can switch-off power switching device.

Be illustrated in figure 2 the dynamic waveform of inductive current in Boost circuit, wherein, i _ln () opens action moment nT at switching tube for inductance _s(T _sa switch periods for switching tube) electric current, T _sfor a switch periods of switching tube, D _bofor the duty ratio of switch.

Boost describes the steady state operating mode of switching circuit usually with CCM, DCM, this division have ignored the dynamic duty process of circuit, be illustrated in figure 3 seven kinds of working mode figures of monocycle internal inductance electric current, characterize the mode of operation of Boost circuit by these seven kinds of mode of operations.

In switching tube opening process, flow through the electric current of switching tube as shown in Figure 4.When switching tube is opened, inductance continues energy storage, so inductive current increases, and the increment Delta i of the inductive current of this one-phase _l+as shown in Fig. 4 (a), increment Delta i _l+be expressed as by formula (8):

${\mathrm{\&Delta;i}}_{L+}=i_{L\max }\left(n\right)-i_L\left(n\right)=\frac{1}{L}{v}_L{D}_{\mathrm{bo}}\left(n\right)T_S---\left(8\right)$

Wherein, i _lmaxn () is nT _sthe maximum of cycle internal inductance electric current,

When switching tube turns off, input to induction charging, but inductance by diode to load discharge, the decay of the inductive current of this one-phase as shown in Fig. 4 (b), the pad value Δ i of inductive current _l-for:

${\mathrm{\&Delta;i}}_{L-}=\frac{1}{L}(v_L-v_H)(1-D_{\mathrm{bo}}\left(n\right))T_S---\left(9\right)$

At nT _scycle is to (n+1) T _sin cycle, inductive current is by i _ln () changes i into _l(n+1), then (n+1) T _sthe inductive current i in moment _l(n+1) be:

$i_L(n+1)=i_L\left(n\right)+{\mathrm{\&Delta;i}}_{L+}+{\mathrm{\&Delta;i}}_{L-}=i_L\left(n\right)+\frac{1}{L}\&CenterDot;T_S\&CenterDot;[v_L-(1-D_{\mathrm{bo}}\left(n\right))\&CenterDot;v_H]=w\left(n\right)---\left(10\right)$

Due to the reverse turn-off characteristic of diode, inductive current cannot decay to negative value continuously, so in this case for i _l(n+1) there is a discontinuity point, expression formula is:

$i_L(n+1)=\left\{\begin{array}{cc}w\left(n\right),& w\left(n\right)\&GreaterEqual;0\\ 0,& w\left(n\right)<0\end{array}\right.---\left(11\right)$

Formula (11) is converted to:

$i_L(n+1)=\left\{\begin{array}{cc}w\left(n\right),& D_{\mathrm{bo}}\left(n\right)\&GreaterEqual;1-D_{\mathrm{bt}}-\frac{L\&CenterDot;i_L\left(n\right)}{v_H\&CenterDot;T_S}\\ 0,& D_{\mathrm{bo}}\left(n\right)<1-D_{\mathrm{bt}}-\frac{L\&CenterDot;i_L\left(n\right)}{v_H\&CenterDot;T_S}\end{array}\right.---\left(12\right)$

Wherein D _bt=v _l/ v _h;

For Boost circuit, in the ideal situation, input v _lthe ENERGY E of release _vLfor:

$E_{\mathrm{vL}}=v_L\&CenterDot;\stackrel{\&OverBar;}{i_L}\left(n\right)\&CenterDot;T_S---\left(13\right)$

Input v _lthe ENERGY E of release _vLa part is stored on inductance, shows as the variable quantity of cycle internal inductance electric current, this part energy Δ E _lbe expressed as:

${\mathrm{\&Delta;E}}_L=\frac{1}{2}L[{i_L}^2(n+1)-{i_L}^2\left(n\right)]---\left(14\right)$

Another part shows as to output v for load consumption _hcharging, the ENERGY E that monocycle internal burden consumes _offfor:

$E_{\mathrm{off}}=v_H\&CenterDot;\stackrel{\&OverBar;}{i_{\mathrm{off}}}\left(n\right)\&CenterDot;(1-D_{\mathrm{bo}}\left(n\right))\&CenterDot;T_S---\left(15\right)$

Wherein, for v when switching tube turns off _haverage current:

$\stackrel{\&OverBar;}{i_{\mathrm{off}}}\left(n\right)=\left\{\begin{array}{cc}{i}_L\left(n\right)+{\mathrm{\&Delta;i}}_{L+}+\frac{1}{2}{\mathrm{\&Delta;i}}_{L-},& w\left(n\right)\&GreaterEqual;0\\ \frac{L\&CenterDot;{[i_L\left(n\right)+{\mathrm{\&Delta;i}}_{L+}]}^2}{2\&CenterDot;(v_H-v_L)(1-D_{\mathrm{bo}}\left(n\right))\&CenterDot;T_S},& w\left(n\right)<0\end{array}\right.---\left(16\right)$

In the ideal situation, and ignore various loss, obtain according to the conservation of energy:

E _vL＝ΔE _L+E _off(17)

When w (n) >=0, obtained by formula (8), (9), (12), (13), (14), (15), (16), (17):

$\frac{1}{2L}\&CenterDot;v_H\&CenterDot;T_S\&CenterDot;{D^2}_{\mathrm{bo}}\left(n\right)-\frac{1}{L}\&CenterDot;v_H\&CenterDot;T_S\&CenterDot;D_{\mathrm{bo}}\left(n\right)+\frac{1}{2L}\&CenterDot;T_S\&CenterDot;(v_H-v_L)-i_L\left(n\right)+\stackrel{\&OverBar;}{i_L}\left(n\right)=0---\left(18\right)$

When meeting

$\frac{1}{L^2}\&CenterDot;{T_S}^2\&CenterDot;v_H\&CenterDot;v_L-2\&CenterDot;\frac{1}{L}\&CenterDot;v_H\&CenterDot;T_S[\stackrel{\&OverBar;}{i_L}\left(n\right)-i_L\left(n\right)]\&GreaterEqual;0---\left(19\right)$

Namely $\stackrel{\&OverBar;}{i_L}\left(n\right)\&le;i_L\left(n\right)+\frac{1}{2L}\&CenterDot;v_L\&CenterDot;T_S---\left(20\right)$

Time formula (18) have solution, solution is:

$D_{\mathrm{bo}}\left(n\right)=1-\sqrt{D_{\mathrm{bt}}+\frac{2L}{v_H\&CenterDot;T_S}[i_L\left(n\right)-\stackrel{\&OverBar;}{i_L}\left(n\right)]}---\left(21\right).$

From formula (20), the equal sign when duty ratio is 1 in formula (20) is just set up, so under restriction duty ratio is less than the condition of 1, formula (18) must have solution; When formula (20) is false given current value cannot be realized in the instruction book cycle, need the multicycle to complete, directly given duty ratio is set to higher limit.

As w (n) < 0, obtained by formula (8), (9), (12), (13), (14), (15), (16), (17):

$\frac{1}{2L}\&CenterDot;v_H\&CenterDot;v_L\&CenterDot;{T_S}^2\&CenterDot;{D^2}_{\mathrm{bo}}\left(n\right)+v_H\&CenterDot;i_L\left(n\right)\&CenterDot;T_S\&CenterDot;D_{\mathrm{bo}}\left(n\right)+\frac{1}{2}\&CenterDot;L\&CenterDot;{i_L}^2\left(n\right)-(v_H-v_L)\&CenterDot;\stackrel{\&OverBar;}{i_L}\left(n\right)\&CenterDot;T_S=0---\left(22\right)$

When meeting

$v_H\&CenterDot;(v_H-v_L)\&CenterDot;{T_S}^2\&CenterDot;[{i_L}^2\left(n\right)+2\&CenterDot;\frac{1}{L}\&CenterDot;v_L\&CenterDot;T_S\&CenterDot;\stackrel{\&OverBar;}{i_L}\left(n\right)]\&GreaterEqual;0---\left(23\right)$

Namely $\stackrel{\&OverBar;}{i_L}\left(n\right)\&GreaterEqual;-\frac{L\&CenterDot;{i_L}^2\left(n\right)}{2\&CenterDot;v_L\&CenterDot;T_S}---\left(24\right)$

Time formula (22) have solution, solution is:

$D_{\mathrm{bo}}\left(n\right)=\frac{-L\&CenterDot;i_L\left(n\right)}{v_L\&CenterDot;T_S}+\frac{\sqrt{(1-D_{\mathrm{bt}})[L^2\&CenterDot;{i_L}^2\left(n\right)+2L\&CenterDot;v_L\&CenterDot;T_S\&CenterDot;\stackrel{\&OverBar;}{i_L}\left(n\right)]}}{v_L\&CenterDot;T_S}---\left(25\right).$

Because average inductor current is permanent be more than or equal to 0, therefore necessarily have solution by formula (24) known equation (22), then only need meet w (n) < 0, just can use formula (25) computed duty cycle.

Because existing Digital Signal Processing (DSP) unit delay cannot be avoided, so duty ratio must be calculated before the cycle starts, therefore need to predict nT _sthe inductive current in moment

The present invention is used for inductive current prediction convergence control method in booster circuit, specifically implements according to following steps, as shown in Figure 5:

The magnitude of voltage v of the low tension potential source in step 1, sampling booster circuit _l, high-tension electricity potential source magnitude of voltage v _h, (n-1) T _sthe inductor current value i in moment _l(n-1), (n-1) T _sthe duty ratio D of switching tube in cycle _bo(n-1);

Step 2, judge nT _sgiven value of current i on moment inductance _lrefn whether () be less than 0, if be not less than 0, forwards step 3 to; If be less than 0, then nT _sthe duty ratio D of switching tube in cycle _xn ()=0, forwards step 9 to;

Step 3, by formula (1) calculate nT _sthe inductive current predicted value in moment

${\hat{i}}_L\left(n\right)=\left\{\begin{array}{cc}w(n-1),& w(n-1)\&GreaterEqual;0\\ 0,& w(n-1)<0\end{array}\right.---\left(1\right)$

Wherein,

$w(n-1)=i_L(n-1)+\frac{1}{L}\&CenterDot;T_S\&CenterDot;[v_L-(1-D_{\mathrm{bo}}(n-1))\&CenterDot;v_H]---\left(2\right)$

L is inductance value, T _sfor a switch periods of switching tube;

Step 4, judgement whether set up, if set up, forward step 5 to; If be false, then nT _sthe duty ratio D of switching tube in cycle _xn ()=1, forwards step 9 to;

Step 5, by nT _sgiven value of current value i on moment inductance _lrefn () assignment is to nT _sthe mean value of cycle internal inductance electric current that is:

$\stackrel{\&OverBar;}{i_L}\left(n\right)=i_{\mathrm{Lref}}\left(n\right)---\left(3\right)$

By the inductive current predicted value obtained in formula (1) substitute in the solution formula (21) of formula (18), calculate nT _sthe duty ratio D of switching tube in cycle _bo(n):

$D_{\mathrm{bo}}\left(n\right)=1-\sqrt{D_{\mathrm{bt}}+\frac{2L}{v_H\&CenterDot;T_S}[{\hat{i}}_L\left(n\right)-\stackrel{\&OverBar;}{i_L}\left(n\right)]}---\left(4\right)$

Wherein, D _bt=v _l/ v _h;

Step 6, according to formula (5) calculate w (n):

$w\left(n\right)={\hat{i}}_L\left(n\right)+\frac{1}{L}\&CenterDot;T_S\&CenterDot;[v_L-(1-D_{\mathrm{bo}}\left(n\right))\&CenterDot;v_H]---\left(5\right)$

Judge whether w (n) >=0 sets up, if be false, then forward step 7 to; If set up, then forward step 8 to;

Step 7, according to the inductive current predicted value obtained in formula (1) substitute in the solution formula (25) of formula (22), calculate nT _sthe duty ratio D of switching tube in cycle _bo(n):

$D_{\mathrm{bo}}\left(n\right)=\frac{-L\&CenterDot;{\hat{i}}_L\left(n\right)}{v_L\&CenterDot;T_S}+\frac{\sqrt{(1-D_{\mathrm{bt}})\&CenterDot;[L^2\&CenterDot;{{\hat{i}}_L}^2\left(n\right)+2L\&CenterDot;v_L\&CenterDot;T_S\&CenterDot;\stackrel{\&OverBar;}{i_L}\left(n\right)]}}{v_L\&CenterDot;T_S}---\left(6\right)$

If the initial value of the inductive current of sampling is different, then may there is situation as shown in Figure 6, namely occur the problem that inductive current is not restrained.

Step 8, in order to solve the problem that inductive current is not restrained, only nT need be got _sthe duty ratio D of switching tube in cycle _buthe duty ratio D of (n) and stable state moment switching tube _zthe mean value D of (n) _xn (), namely calculates D according to formula (7) _x(n):

$D_x\left(n\right)=\frac{1}{2}(D_{\mathrm{bo}}\left(n\right)+D_z\left(n\right))---\left(7\right)$

When switching tube reaches stable state, the waveform of inductive current as shown in Figure 7, namely

Δi _L+＝-Δi _L-，

Can obtain: nT _sthe duty ratio D of switching tube during cycle homeostasis _z(n) be:

$D_z\left(n\right)=\frac{v_H-v_L}{v_H};$

Step 9, generation duty ratio are D _xn the PWM ripple of () carrys out control switch pipe.

The rule of the present invention's foundation conservation of energy obtains the duty ratio of switching tube, calculates the duty ratio of switching tube using inductive current as controlled variable fast.If do not add convergence, 1 switch periods just energy tracing preset; If add convergence, 2 switch periods just energy tracing preset, dynamic response characteristic is fast.And in given unexpected increase or when reducing suddenly, can track given fast, describe the static and dynamic performance that this algorithm can reach good, static difference is little, control precision good.

Claims (5)

1., for an inductive current prediction convergence control method in booster circuit, it is characterized in that, specifically implement according to following steps:

The magnitude of voltage v of the low tension potential source in step 1, sampling booster circuit _l, high-tension electricity potential source magnitude of voltage v _h, (n-1) T _sthe inductor current value i in moment _l(n-1), (n-1) T _sthe duty ratio D of switching tube in cycle _bo(n-1),

Wherein, T _sfor a switch periods of switching tube;

Step 2, judge nT _sgiven value of current i on moment inductance _lrefn whether () be less than 0, if be not less than 0, forwards step 3 to; If be less than 0, then nT _sthe duty ratio D of switching tube in cycle _xn ()=0, forwards step 9 to;

Step 3, by formula (1) calculate nT _sthe inductive current predicted value in moment

${\hat{i}}_L\left(n\right)=\left\{\begin{array}{cc}w(n-1),& w(n-1)\&GreaterEqual;0\\ 0,& w(n-1)<0\end{array}\right.---\left(1\right)$

Wherein,

$w(n-1)=i_L(n-1)+\frac{1}{L}\&CenterDot;T_S\&CenterDot;[v_L-(1-D_{\mathrm{bo}}(n-1))\&CenterDot;v_H]---\left(2\right)$

L is inductance value;

Step 4, judgement whether set up, if set up, forward step 5 to; If be false, then nT _sthe duty ratio D of switching tube in cycle _xn ()=1, forwards step 9 to;

Step 5, by nT _sgiven value of current value i on moment inductance _lrefn () assignment is to nT _sthe mean value of cycle internal inductance electric current that is:

${\stackrel{\&OverBar;}{i}}_L\left(n\right)=i_{\mathrm{Lref}}\left(n\right)---\left(3\right)$

By the inductive current predicted value obtained in formula (1) calculate nT _sthe duty ratio D of switching tube in cycle _bo(n):

$D_{\mathrm{bo}}\left(n\right)=1-\sqrt{D_{\mathrm{bt}}+\frac{2L}{v_H\&CenterDot;T_S}[{\hat{i}}_L\left(n\right)-{\stackrel{\&OverBar;}{i}}_L\left(n\right)]}---\left(4\right)$

Wherein, D _bt=v _l/ v _h;

Step 6, according to formula (5) calculate w (n):

$w\left(n\right)={\hat{i}}_L\left(n\right)+\frac{1}{L}\&CenterDot;T_S\&CenterDot;[v_L-(1-D_{\mathrm{bo}}\left(n\right))\&CenterDot;v_H]---\left(5\right)$

Judge whether w (n) >=0 sets up, if be false, then forward step 7 to; If set up, then forward step 8 to;

Step 7, according to the inductive current predicted value obtained in formula (1) calculate nT _sthe duty ratio D of switching tube in cycle _bo(n):

$D_{\mathrm{bo}}\left(n\right)=\frac{-L\&CenterDot;{\hat{i}}_L\left(n\right)}{v_L\&CenterDot;T_S}+\frac{\sqrt{(1-D_{\mathrm{bt}})\&CenterDot;[L^2\&CenterDot;{{\hat{i}}_L}^2\left(n\right)+2L\&CenterDot;v_L\&CenterDot;T_S\&CenterDot;{\stackrel{\&OverBar;}{i}}_L\left(n\right)]}}{v_L\&CenterDot;T_S}---\left(6\right)$

Step 8, calculate duty ratio D after intermediate value process according to formula (7) _x(n):

$D_x\left(n\right)=\frac{1}{2}(D_{\mathrm{bo}}\left(n\right)+D_z\left(n\right))---\left(7\right)$

Wherein, for nT _sthe duty ratio of switching tube during cycle homeostasis;

Step 9, generation duty ratio are D _xn the PWM ripple of () carrys out control switch pipe.

2. one according to claim 1 is used for inductive current prediction convergence control method in booster circuit, and it is characterized in that, the detailed process obtaining formula (4) in described step 5 is:

When switching tube is opened, inductance continues energy storage, and inductive current increases, nT _sthe increment Delta i of inductive current in switching tube opening process in cycle _l+for:

$\mathrm{\&Delta;}{i}_{L+}=i_{L\max }\left(n\right)-i_L\left(n\right)=\frac{1}{L}{v}_L{D}_{\mathrm{bo}}\left(n\right)T_S---\left(8\right)$

Wherein, i _lmaxn () is nT _sthe maximum of cycle internal inductance electric current,

When switching tube turns off, the pad value Δ i of inductive current _l-for:

$\mathrm{\&Delta;}{i}_{L-}=\frac{1}{L}(v_L-v_H)(1-D_{\mathrm{bo}}\left(n\right))T_S---\left(9\right)$

At nT _scycle is to (n+1) T _sin cycle, inductive current is by i _ln () changes i into _l(n+1), then (n+1) T _sthe inductive current i in moment _l(n+1) be:

$i_L(n+1)=i_L\left(n\right)+\mathrm{\&Delta;}{i}_{L+}+\mathrm{\&Delta;}{i}_{L-}=i_L\left(n\right)+\frac{1}{L}\&CenterDot;T_S\&CenterDot;[v_L-(1-D_{\mathrm{bo}}\left(n\right))\&CenterDot;v_H]=w\left(n\right)---\left(10\right)$

Due to the reverse turn-off characteristic of diode, inductive current cannot decay to negative value continuously, then i _l(n+1) there is a discontinuity point, expression formula is:

$i_L(n+1)=\left\{\begin{array}{cc}w\left(n\right),& w\left(n\right)\&GreaterEqual;0\\ 0,& w\left(n\right)<0\end{array}\right.---\left(11\right)$

Formula (11) is converted to:

$i_L(n+1)=\left\{\begin{array}{cc}w\left(n\right),& D_{\mathrm{bo}}\left(n\right)\&GreaterEqual;1-D_{\mathrm{bt}}-\frac{L\&CenterDot;i_L\left(n\right)}{v_H\&CenterDot;T_s}\\ 0,& D_{\mathrm{bo}}\left(n\right)<1-D_{\mathrm{bt}}-\frac{L\&CenterDot;i_L\left(n\right)}{v_H\&CenterDot;T_s}\end{array}\right.---\left(12\right)$

Wherein D _bt=v _l/ v _h;

In the ideal situation, input v _lthe ENERGY E of release _vLfor:

$E_{\mathrm{vL}}=v_L\&CenterDot;{\stackrel{\&OverBar;}{i}}_L\left(n\right)\&CenterDot;T_S---\left(13\right)$

Input v _lthe ENERGY E of release _vLa part is stored on inductance, shows as the variable quantity of cycle internal inductance electric current, this part energy Δ E _lbe expressed as:

$\mathrm{\&Delta;}{E}_L=\frac{1}{2}L[{i_L}^2(n+1)-{i_L}^2\left(n\right)]---\left(14\right)$

Another part shows as to output v for load consumption _hcharging, the ENERGY E that monocycle internal burden consumes _offfor:

$E_{\mathrm{off}}=v_H\&CenterDot;\stackrel{\&OverBar;}{i_{\mathrm{off}}}\left(n\right)\&CenterDot;(1-D_{\mathrm{bo}}\left(n\right))\&CenterDot;T_S---\left(15\right)$

Wherein, for v when switching tube turns off _haverage current:

$\stackrel{\&OverBar;}{i_{\mathrm{off}}}\left(n\right)=\left\{\begin{array}{cc}{i}_L\left(n\right)+\mathrm{\&Delta;}{i}_{K+}+\frac{1}{2}\mathrm{\&Delta;}{i}_{L-},& w\left(n\right)\&GreaterEqual;0\\ \frac{L\&CenterDot;{[i_L\left(n\right)+\mathrm{\&Delta;}{i}_{L+}]}^2}{2\&CenterDot;(v_H-v_L)(1-D_{\mathrm{bo}}\left(n\right))\&CenterDot;T_S},& w\left(n\right)<0\end{array}\right.---\left(16\right)$

In the ideal situation, and ignore various loss, obtain according to the conservation of energy:

E _vL＝ΔE _L+E _off(17)

When w (n) >=0:

$\frac{1}{2L}\&CenterDot;v_H\&CenterDot;T_S\&CenterDot;{D^2}_{\mathrm{bo}}\left(n\right)-\frac{1}{L}\&CenterDot;v_H\&CenterDot;T_S\&CenterDot;D_{\mathrm{bo}}\left(n\right)+\frac{1}{2L}\&CenterDot;T_S\&CenterDot;(v_H-v_L)-i_L\left(n\right)+{\stackrel{\&OverBar;}{i}}_L\left(n\right)=0---\left(18\right)$

When meeting

$\frac{1}{L^2}\&CenterDot;{T_S}^2\&CenterDot;v_H\&CenterDot;v_L-2\&CenterDot;\frac{1}{L}\&CenterDot;v_H\&CenterDot;T_S[{\stackrel{\&OverBar;}{i}}_L\left(n\right)-i_L\left(n\right)]\&GreaterEqual;0---\left(19\right)$

Namely ${\stackrel{\&OverBar;}{i}}_L\left(n\right)\&le;i_L\left(n\right)+\frac{1}{2L}\&CenterDot;v_L\&CenterDot;T_S---\left(20\right)$

Time formula (18) have solution, solution is:

$D_{\mathrm{bo}}\left(n\right)=1-\sqrt{D_{\mathrm{bt}}+\frac{2L}{v_H\&CenterDot;T_S}[i_L\left(n\right)-{\stackrel{\&OverBar;}{i}}_L\left(n\right)]}---\left(21\right).$

3. one according to claim 1 is used for inductive current prediction convergence control method in booster circuit, and it is characterized in that, the detailed process obtaining formula (6) in described step 7 is:

As w (n) < 0:

$\frac{1}{2L}\&CenterDot;v_H\&CenterDot;v_L\&CenterDot;{T_S}^2\&CenterDot;{D^2}_{\mathrm{bo}}\left(n\right)+v_H\&CenterDot;i_L\left(n\right)\&CenterDot;T_S\&CenterDot;D_{\mathrm{bo}}\left(n\right)+\frac{1}{2}\&CenterDot;L{i_L}^2\left(n\right)-(v_H-v_L)\&CenterDot;{\stackrel{\&OverBar;}{i}}_L\left(n\right)\&CenterDot;T_S=0---\left(22\right)$

When meeting

$v_H\&CenterDot;(v_H-v_L)\&CenterDot;{T_S}^2\&CenterDot;[{i_L}^2\left(n\right)+2\&CenterDot;\frac{1}{L}\&CenterDot;v_L\&CenterDot;T_S\&CenterDot;{\stackrel{\&OverBar;}{i}}_L\left(n\right)]\&GreaterEqual;0---\left(23\right)$

Namely ${\stackrel{\&OverBar;}{i}}_L\left(n\right)\&GreaterEqual;-\frac{L\&CenterDot;{i_L}^2}{2\&CenterDot;v_L\&CenterDot;T_S}---\left(24\right)$

Time formula (22) have solution, solution is:

$D_{\mathrm{bo}}\left(n\right)=\frac{-L\&CenterDot;i_L\left(n\right)}{v_L\&CenterDot;T_S}+\frac{\sqrt{(1-D_{\mathrm{bt}})[L^2\&CenterDot;{i_L}^2\left(n\right)+2L\&CenterDot;v_L\&CenterDot;T_S\&CenterDot;\stackrel{\&OverBar;}{i_L}\left(n\right)]}}{v_L\&CenterDot;T_S}---\left(25\right).$

4. the one according to claim 1-3 any one is used for inductive current prediction convergence control method in booster circuit, and it is characterized in that, in described step 1, booster circuit comprises low tension potential source v _l, low tension potential source v _lnegative pole respectively with high-tension electricity potential source v _hnegative pole, switching tube S emitter connect, the collector electrode of switching tube S is connected with the positive pole of inductance L, diode D respectively, the negative pole of diode D and high-tension electricity potential source v _hpositive pole connect.

5. one according to claim 4 is used for inductive current prediction convergence control method in booster circuit, it is characterized in that, described switching tube S be carry anti-paralleled diode or there is anti-paralleled diode characteristic can switch-off power switching device.