A Boost converter, or parallel chopper, is a food with cutting which converts a continuous tension into another continuous tension of stronger value.

Applications

A converter boost is used when one wishes to increase the tension available of a continuous source. The systems supplied with accumulator battery often use several accumulators in series in order to have a sufficiently high level of tension. The place available being often limited, it is not always possible to have a sufficient number of elements. A converter boost makes it possible to increase the power provided by the batteries and thus to decrease the number of elements necessary to reach the desired level of tension. The hybrid vehicles as well as the lighting systems are two examples typical of use of the converters boost.

- The converters boost are used in applications of low power like the portable lighting systems. A white electroluminescent diode requires a tension of 2,7 V with 3,6 V approximately to function, a converter boost makes it possible to increase the power provided by a pile of 1,5 V in order to produce a flashlight low fuel consumption.
- The converters boost can also deliver tensions much higher in order to supply the tubes with cold cathode present in the retro-lighting of the screens at liquid crystals or the flash of the cameras for example.
- A hybrid car as Toyota Prius uses an electrical motor, requiring a tension of 500 V. Without converter boost, this car should embark 417 storage cells NiMH series-connected to supply this engine. However, Prius uses only 168 elements as well as a converter boost in order to pass the tension available of 202 to 500V.

Principle of operation

two configurations of a Boost converter following the state of the switch S

two configurations of a Boost converter following the state of the switch S

- The operation of a Boost converter can be divided into two phases distinct depending on the state from the switch S.
- A phase of accumulation of energy: when the switch S is closed, state passing, that thus involves the increase in the current in inductance the storage of a quantity of energy in the form of energy magnetic. The diode D is then blocked and it load is then disconnected from the food.
- When the switch is opened, inductance is then in series with the generator and its f.e.m. is added with that with the generator, booster effect. The current crossing inductance crosses then the diode D, the condenser C and load R. It results from it a transfer of the energy accumulated in inductance towards the capacity.

Conduction continue

formes d’ondes courant ⁄ tension dans un convertisseur Boost

formes d’ondes courant ⁄ tension dans un convertisseur Boost

When a Boost converter works in mode of continuous conduction, the current I_{L} crossing inductance is never cancelled. The figure shows the forms of waves of the current and the tension in a Boost converter.

- The output voltage is calculated in the following way.
- During the state passing, the switch S is closed, involving the increase in the current following the relation.
- V
_{i}= L * (dI_{L}) ⁄ dt - At the end of the state being on, the power I
_{L}increased: - ΔI
_{Lon}= ∫_{0}^{α*T}dI_{L}= ∫_{0}^{α * T}[(V_{i}* dt) ⁄ L] = V_{i}* α * T ⁄ L - α being the cyclic report ⁄ ratio. It represents the fraction of the period T during which the switch S led. α lies between 0, S never leads and 1, S leads all the time.
- During the blocked state, the switch S is opened, the current crossing inductance circulates through the load. If one considers a null voltage drop at the boundaries of the diode and a sufficiently large condenser to keep his constant tension, the evolution of IT is:
- V
_{i}- V_{0}= L * (dI_{L}⁄ dt) - Consequently, the variation of I
_{L}during the blocked state is: - ΔI
_{Loff}= ∫_{0}^{(1-α)T}dI_{L}= ∫_{0}^{(1-α)T}[(V_{i}- V_{0})dt ⁄ L] = [(V_{i}- V_{0})(1-α)T ⁄ L] - If it is considered that the converter reached its permanent mode, the quantity of energy stored in each one of these components is the same one at the beginning and the end of a cycle of operation. In particular, the energy stored in inductance is given by:
- E = ½L * I²
_{L} - Consequently, the current crossing inductance is the same one at the beginning and the end of each cycle of commutation. What can be written in the following way:
- ΔI
_{Lon}+ ΔI_{Loff}= 0 - While replacing ΔI
_{Lon}et ΔI_{Loff}by their expression, one obtains : - ΔI
_{Lon}+ ΔI_{Loff}= V_{i}* α * T ⁄ L + (V_{i}- V_{0})(1-α)T ⁄ L = 0 - What can be rewritten in the following way :
- V
_{0}⁄ V_{i}= 1 ⁄ 1-α - Thanks to this last expression, one can see that the output voltage is always higher than that of entry, the variable cyclic relationship between 0 and 1, qu' it increases with α and that theoretically it can be infinite when α approaches 1. Therefore one speaks about booster.

Discontinuous conduction

forms of waves running ⁄ tension in a Boost converter in discontinuous conduction.

forms of waves running ⁄ tension in a Boost converter in discontinuous conduction.

In certain cases, the quantity of energy required by the load is enough weak to be transferred in a time shorter than one period from commutation. In this case, the current crossing the inductor is cancelled during part of the period. The only difference with the principle in operation describes previously, is that inductance is completely discharged at the beginning of cycle. Although weak, the difference between continuous and discontinuous conduction has a strong impact on the formula of the output voltage.

- The output voltage can be calculated in the following way:
- As the current of inductance is null at the beginning of cycle, its maximum I
_{LMax}(at = α*T) vaut : - I
_{LMax}= (V_{i}* α * T) ⁄ L - During the blocked state, I
_{L}cancels after δ * T: - I
_{LMax}+ [(V_{i}- V_{0}) * δ * T] ⁄ L = 0 - By using the two last equations, δ is worth:
- δ = V
_{i}* α ⁄ V_{0}- V_{i} - The current in the I
_{0}load is equal to the average current crossing diode (I_{D}). The current crossing the diode is equal to that in inductance during the blocked state. - Consequently, the current crossing the diode can be written in the following way:
- I
_{0}= I^{-}_{D}= I_{LMax}⁄ 2 * δ - By replacing I
_{LMax}and δ by their respective expressions, one obtains: - I
_{0}= (V_{i}* α * T) ⁄ 2L * (V_{i}* α) ⁄ (V_{0}- V_{i}) = (V_{i}² * α² * T) ⁄ [2L(V_{0}- V_{i})] - Consequently, the profit of output voltage can be written in the following way :
- V
_{0}⁄ V_{i}= 1 + (V_{i}* α² * T) ⁄ 2L * I_{0} - This expression is much more complex than that obtained at the time of the study in continuous conduction. In discontinuous conduction, the profit in tension depends on the cyclic report/ratio but also on the tension of entry, of the value of the inductance and the output current.

Limit between conduction continues and discontinuous

evolution of the output voltage standardized of a Boost converter with an output current standardized

evolution of the output voltage standardized of a Boost converter with an output current standardized

As explained in the preceding paragraph, the converter functions in discontinuous conduction when the current required by the load is weak and it functions in continuous conduction for the more important currents. The limit between conduction continues and discontinuous conduction is reached when the current in inductance is cancelled just at the time of commutation.

- With the notations, that corresponds to :
- α * T + δ * T = T
- α + δ = 1
- In this case, the output current, current I
_{0lim}output in extreme cases of continuous and discontinuous conduction is given by the relation: - I
_{0lim}= I^{-}_{D}= I_{Lmax}⁄ 2 * (1 - α) - By replacing I
_{Lmax}by its expression in discontinuous conduction: - I
_{0lim}= (V_{i}* α * T) ⁄ 2L * (1 - α) - In extreme cases between the two modes of conduction, the output voltage obeys the expressions of the two modes. One will use that given for the mode of continuous conduction:
- V
_{0}⁄ V_{i}= 1 ⁄ 1 - α - One can thus rewrite I
_{0lim}in the following way : - I
_{0lim}= (V_{i}* T) ⁄ 2L * V_{i}⁄ V_{0}* [1 - (V_{i}⁄ V_{0})] - Let us introduce two new notations :
- The standardized tension, defined by |V
_{0}| = V_{0}⁄ V_{i} - Who corresponds to the profit in tension of the converter.
- The current standard, set by |I
_{0}| = L ⁄ (T * V_{i}) * I_{0} - The term T * V
_{i}⁄ L represents the maximum increase in current that can be achieved during a cycle, the current variation in the inductance achieved for α = 1 - We obtain, in steady state, |I
_{0}| equals 0 when the output current is zero and 1 for the maximum current that can provide the converter .
- Using these notations, we have:
- In continuous conduction, |V
_{0}| = 1 ⁄ 1 - α - In discontinuous conduction, |V
_{0}| = 1 + (V_{i}* α² * T) ⁄ 2L * I_{0}= 1 + α² ⁄ 2|I_{0}| - The current boundary between continuous and discontinuous conduction is :
- I
_{0lim}= V_{i}* T ⁄ 2L α (1 - α) = I_{0lim}⁄ 2|I_{0}| α (1 - α) - Therefore, the border between continuous and discontinuous conduction is described by :
- 1 ⁄ 2|I
_{0}| α (1 - α) = 1

The difference in behavior between continuous and discontinuous conduction is very clear. This may lead to servo output voltage problems.

Analysis average

A mean value template is a method for calculating the average period of the waveforms. It is to write the corresponding equations in each state of the system and multiply by the proportion of time spent in each state the converter.

- In the case of a boost converter in the conducting state, the change of current in the inductance is given by :
- L * dI
_{L}⁄ dt = V_{i} - During the off state, the voltage across the switch is equal to the output voltage :
- L * dI
_{L}⁄ dt = V_{i}- V_{0} - Therefore, the average change in inductance is obtained by multiplying by the two equations preceding the time spent in the state corresponding :
- αT for the state and (1- α)T for the off state) then dividing by the switching period :
- L * d
^{-}I_{L}⁄ dt = (α * T * V_{i}+ (1 - α)T * (V_{i}- V_{0})) 1 ⁄ T = α * V_{i}+ (1 - α) * V_{i}- V_{0}) - d
^{-}I_{L}⁄ dt represents the variations in the inductance at a slower than the switching frequency scale. - For a converter steady, d
^{-}I_{L}⁄ dt = 0 - The above equations thus become :
- α * V
_{i}+ (1 - α) * (V_{i}- V_{0}= 0 - Which can be written in the form :
- V
_{0}⁄ V_{i}= 1 ⁄ 1 - α - We find the equation obtained by the previous study

The interest of this method is that it masks the existence of the switches of the converter, thus authorizing a study of the converter with the traditional techniques of modeling uninterrupted ⁄ alternate.

Case of the circuit not-ideal

evolution of the output voltage of a Boost converter according to the cyclic report ⁄ ratio when the parasitic resistance of inductance increases.

evolution of the output voltage of a Boost converter according to the cyclic report ⁄ ratio when the parasitic resistance of inductance increases.

- The preceding study was made with the following assumptions:
- The condenser of exit has a sufficient capacity to provide a constant power, during a cycle of operation.
- The voltage drop at the boundaries of the diode is null.
- No losses by commutation in the semiconductors.
- No losses in the components generally.

These assumptions can be very far away from reality, the imperfections of the real components being able to have important effects on operation of the converter.

Taking into account of parasitic resistances

In the preceding study, the internal resistance of the components was not taken into account. That means that all the power is transmitted without loss of the source of tension towards the load. There exist however parasitic resistances in all the circuit because of the resistivity of materials used for its construction. Consequently, a fraction of the power transmitted by the source of tension is dissipated in these parasitic resistances.

For reasons of simplicity, one will consider here only the defects of inductance by modelling it by an inductance in series with a resistance. This assumption is acceptable because an inductance consists of a long wire which can thus have a clean resistance R_{L}. Moreover, the current crosses the reel in the two states of the converter.

- By using the method of the study in median value, one can write:
- V
_{i}= V^{-}_{L}+ V^{-}_{S} - With V
^{-}_{L}et V^{-}_{S}average tensions, on a cycle of operation, at the boundaries respectively of inductance and switch. - If it is considered that the converter is in permanent mode, the average current through inductance is constant.
- The average tension at the boundaries of inductance thus becomes :
- V
^{-}_{L}= L * (dI^{-}_{L}) ⁄ dt + R_{L}I^{-}_{L}= R_{L}I^{-}_{L} - When the switch is busy, V
_{S}=0. When it is blocked, the diode thus becomes busy V_{S}=V_{0}. - Consequently, the average tension through the switch is :
- V
^{-}_{S}= α * 0 + (1 - α)V_{0}= (1 - α)V_{0} - The output current is equal to that in inductance during the blocked state. The average current in inductance is thus written :
- I
^{-}_{L}= I_{0}⁄ 1 - α - If one regards the undulations of tension and current at exit as negligible, the load can be regarded as purely resistive.
- If one notes R the resistance of the load, the preceding expression becomes :
- I
^{-}_{L}= V_{0}⁄ (1 - α)R - By using the preceding equations, the tension of entry is written :
- V
_{i}= R_{L}* V_{0}⁄ (1 - α)R + (1 - α)V_{0} - This expression can be put in the form :
- V
_{0}⁄ V_{i}= 1 ⁄ (R_{L}⁄ R(1 - α)) + 1 - α - If the resistance of inductance is null, one finds the equation obtained in the ideal case.
- But the more R
_{L}increases, the more the profit in tension of the converter decreases compared to the ideal case. Moreover the influence of R_{L}increases with the cyclic report ⁄ ratio.

A Buck converter, or chopper series, is a food with cutting which converts a continuous tension into another continuous tension of low value.

- This type of converter is used for applications which one can classify in two categories :
- The applications aiming at obtaining a continuous tension fixes starting from a generator of tension continues higher.
- Conversion of the 12-24V provided by a battery of laptop towards the few Volts necessary to the processor.
- Conversion of the tension of the sector rectified uninterrupted fixes.
- The applications making it possible to obtain a tension adjustable but always lower than that presents to the entry.
- Variator of continuous tension.

For these two categories of application, one also wishes whom internal impedance of the system thus created either weak - same order of magnitude as that of the generator which feeds the device - what prohibits the use of a tension divider dissipating in the form of heat the excess of tension and having poor yield what is crippling for the applications of electronics of power.

Well conceived a Buck converter has a strong output and makes it possible to control the output voltage.

Principle of operation

### Buck-Boost converter

### Ćuk converter

### Converter sepic

Basic diagram of a converter sepic

### Converter flyback

basic diagram of a converter flyback.

Basic diagram of a Buck converter

Two configurations of a Buck converter following the state of the switch S

Two configurations of a Buck converter following the state of the switch S

- The operation of a Buck converter can be divided into two configurations according to the state of the switch S :
- In the state passing, the switch S is closed, the terminal voltage of inductance is worth V
_{L}= V_{i}- V_{0} - The current crossing inductance increases linearly. The terminal voltage of the diode being negative, no current crosses it.
- In the blocked state, the switch is open. The diode becomes busy in order to ensure the continuity of the current in inductance. The terminal voltage of inductance is worth V
_{L}= - V_{0}. The current crossing inductance decrease.

Continuous conduction

Forms of waves running ⁄ tension in a Buck converter

Forms of waves running ⁄ tension in a Buck converter

When a Buck converter works in mode of continuous conduction, the current I_{L} crossing inductance is never cancelled. The figure shows the forms of waves of the current and the tension in a Buck converter.

- The rate of increase in I
_{L}is given by: - V
_{L}= L * (dI_{L}) ⁄ dt - With V
_{L}equal V_{i}- V_{0}during the state passing and - V_{0}during the state blocked. - Consequently, the increase in I
_{L}during on-state is given by: - ΔI
_{lon}= ∫_{0}^{α*T}dI_{L}= ∫_{0}^{α*T}V_{L}⁄ L dt = (V_{i}- V_{0}) * α * T ⁄ L - In the same way, the fall of the current in inductance during the blocked state is given by :
- ΔI
_{loff}= ∫_{α*T}^{T}dI_{L}= ∫_{α*T}^{T}V_{L}⁄ L dt = - V_{0}(T - α * T) ⁄ L - If it is considered that the converter is in permanent mode, the energy stored in each component is the same one at the beginning and the end of each cycle of commutation. In particular, the energy stored in inductance is given by :
- E = ½ L * I²
_{L} - Consequently, the current I
_{L}crossing inductance is the same one at the beginning and the end of each cycle of commutation. What can be written in the following way : - ΔI
_{Lon}+ ΔI_{Loff}= 0 - While replacing ΔI
_{Lon}and ΔI_{Loff}by their expression, one obtains : - [(V
_{i}- V_{0}) * α * T] ⁄ L - [V_{0}* (T - α * T)] ⁄ L = 0 - What can be rewritten in the following way:
- V
_{0}= α * V_{i}

Thanks to this equation, one can see that the output voltage varies linearly with the cyclic report ⁄ ratio. The cyclic report ⁄ ratio lying between 0 and 1, the output voltage V_{0} is always lower than that of input. Therefore one speaks sometimes about reducing transformer.

Discontinuous conduction

Forms of waves running ⁄ tension in a Buck converter in discontinuous conduction.

Forms of waves running ⁄ tension in a Buck converter in discontinuous conduction.

In certain cases, the amount of power required by the load is enough weak to be transferred in a time shorter than one period from commutation. In this case, the current crossing the inductor is cancelled during part of the period. The only difference with the principle in operation describes previously is that inductance is completely discharged at the beginning of cycle. Although weak, the difference between continuous and discontinuous conduction has a strong impact on the formula of the output voltage.

- The output voltage can be calculated in the following way:
- It is always considered that the converter reached the permanent mode.
- The energy stored in the components is thus the same one at the beginning and the end of the cycle.
- That also generates that the average tension at the boundaries of inductance (V
_{L}) is null. - (V
_{i}- V_{0}) α * T - V_{0}* δ * T = 0 - One can deduce the value from it from δ
- δ = (V
_{i}- V_{0}⁄ V_{0}) * α - By supposing that the capacitor of output is sufficiently important to maintain the output voltage constant during a cycle of commutation, the output current delivered to the I
_{0}load is constant. - That implies that the current crossing the capacitor is of null median value. Consequently, we have:
- I
^{-}_{L}= I_{0} - With I
_{L}the average current crossing inductance. - The current I
_{L}crossing inductance is of form triangular. - Consequently, the median value of I
_{L}can thus be calculated geometrically in the following way : - I
^{-}_{L}= (½I_{Lmax}* α * T + ½I_{Lmax}* δ * T) 1 ⁄ T = [I_{Lmax}* (α + δ)] ⁄ 2 = I_{0} - The current in inductance is null at the beginning of cycle then increases during α * T until reaching I
_{Lmax}. - That wants to say that I
_{Lmax}is equal à : - I
_{Lmax}= V_{i - V0 ⁄ L α * T} - while replacing I
_{Lmax}in the preceding relations by his expression, one obtains : - I
_{0}= [(V_{i}- V_{0}) α * T (α + δ)] ⁄ 2L - One replaces δ by his expression :
- I
_{0}= [(V_{i}- V_{0}) α * T (α + (V_{i}- V_{0}⁄ V_{0}) α] ⁄ 2L - Who can put himself in the form :
- V
_{0}= V_{i}* [1 ⁄ (2L * I_{0}⁄ α² * V_{i}* T) + 1] - One can see that the expression of the output voltage is much more complicated than that obtained for conduction continues.
- Indeed, the output voltage depends not only on the tension of input (V
_{i}) and on the cyclic report ⁄ ratio α but also on the value of inductance (L), the period of commutation (T) and of the output current (I_{0}).

Limit between conduction continues and discontinuous

Evolution of the output voltage standardized of a Buck converter with an output current standardized.

Evolution of the output voltage standardized of a Buck converter with an output current standardized.

As explained in the preceding paragraph, the converter functions in discontinuous conduction when the current required by the load is weak and it functions in continuous conduction for the more important currents. The limit between conduction continues and discontinuous conduction is reached when the current in inductance is cancelled just at the time of commutation.

- With the notations of the figure, that corresponds to:
- α * T + δ * T = T
- α + δ = 1
- In this case, the output current I
_{0lim}is equal to the average current crossing inductance : - I
_{0lim}= I_{Lmax}(α + δ) ⁄ 2 = I_{Lmax} - While replacing I
_{Lmax}by its expression in discontinuous conduction : - I
_{0lim}= V_{i}- V_{0}⁄ 2L α * T - In extreme cases between the two modes of conduction, the output voltage obeys the expressions of the two modes. One will use that given for the mode of continuous conduction :
- V
_{0}= α * V_{i} - One can thus rewrite I
_{0lim}in the following way : - I
_{0lim}= V_{i}(1 - α) ⁄ 2L α * T - Let us introduce two new notations :
- The standardized tension, defined by |V
_{0}| = V_{0}⁄ V_{i}, which corresponds to the profit in tension of the converter. - The standardized current, defined by |I
_{0}| = L ⁄ T * V_{i}* I_{0} - The term T * V
_{i}⁄ L corresponds to the maximum increase in current which one can reach at the time of a cycle. - One thus obtains, in permanent mode, |I
_{0}| equalize 0 when the output current null east and 1 for the maximum current which the converter can provide. - By using these notations, one obtains:
- In continuous conduction, |V
_{0}| = α - In discontinuous conduction, |V
_{0}| = 1 ⁄ [(2L * I_{0}⁄ α² * V_{i}* T) + 1] = 1 ⁄ [(2|I_{0}| ⁄ α²) + 1] = α² ⁄ 2|I_{0}| + α² - The limiting current between conduction continues and discontinuous is :
- I
_{0lim}= [V_{i}(1 - α) ⁄ 2L] α * T = I_{0lim}⁄ |I_{0}| * [(1 - α)α ⁄ 2] - Consequently, the border between continuous and discontinuous conduction is described by :
- (1 - α)α ⁄ 2|I
_{0}| = 1 - The difference in behavior between continuous and discontinuous conduction is very clear. That can generate problems of control of the output voltage.

Case of the circuit not-ideal

Evolution of the output voltage of a Buck converter according to the cyclic report/ratio when the parasitic resistance of inductance increases.

Evolution of the output voltage of a Buck converter according to the cyclic report/ratio when the parasitic resistance of inductance increases.

- The preceding study was made with the following assumptions:
- The capacitor of output has a sufficient capacity to provide a constant power, during a cycle of operation.
- The voltage drop at the boundaries of the diode is null.
- No losses by commutation in the semiconductors.
- No losses in the components generally.
- These assumptions can be very far away from reality, the imperfections of the real components being able to have important effects on operation of the converter.

Undulation of the output voltage

The undulation of output voltage is the name given to the increase in the output voltage to the state passing and the reduction in the tension at the time of the blocked state. Several factors influence this undulation: the frequency of operation of the converter, the capacitor of output, inductance, the load but also characteristics of the control circuitry of the converter. At first approximation, one can consider that the undulation of tension results from the load and the discharge of the capacitor of output : dV_{0} = i * dT ⁄ C

During the blocked state, the current mentioned in this equation is the current one crossing the load. During the state being on, this power indicates the difference between the current delivered by the source in tension and the beam the load. The duration (dT) is defined by the cyclic report ⁄ ratio α and the quench frequency.

For on-state : dT_{on} = α * T = α ⁄ ƒ

For the blocked state : dT_{off} = (1 - α) * T = 1 - α ⁄ ƒ

The undulation of tension thus decreases with the increase in the capacitor of output or the quench frequency. The undulation of tension is generally fixed by the schedule of conditions of the feeding to realize. The choice of the capacitor is done on criteria of costs, volume available and also of the characteristics of different technology from capacitor. The choice of the quench frequency is done according to criteria of output, the output tending to drop with the increase in the quench frequency. The increase in the quench frequency can also pause of the compatibility issues electromagnetic.

The undulation of the output voltage east one of the disadvantages of the feedings with cutting and thus forms part of the measuring criteria of quality.

Specific structures

Synchronous rectification

Schematic diagram of a Buck converter to synchronous rectification. The diode D of standard the Buck converter was replaced by a second switch S_{2}.

Schematic diagram of a Buck converter to synchronous rectification. The diode D of standard the Buck converter was replaced by a second switch S

A synchronous Buck converter is a modified version of the traditional Buck converter in which one replaced the diode D by a second S_{2} switch. This modification makes it possible to increase the output of the converter because the voltage drop at the boundaries of a switch is weaker than that at the boundaries of a diode. It is also possible to increase the output further by keeping the diode in parallel of the second S_{2} switch. The diode then makes it possible to ensure the transfer of energy at the time of the short period or the switches are open. The use of a switch alone is a compromise between increase in the cost and the output.

The Buck converters are usually used in the computers in order to reduce the power provided by the feeding towards a weaker tension necessary to feed the CPU. These feedings must provide a fort running with a weak undulation of tension while remaining in a reduced volume.

In order to reduce the constraints on the semiconductors, these feedings use several converters connected in parallel. One speaks then about interlaced choppers because the converters lead in turn towards the same capacitor of output. The majority of the feedings of mother boards use 3 or 4 branches in parallel, but the manufacturers of semiconductor propose components being able to manage to 6 branches in parallel. In order to increase the output, these feedings use synchronous correction there too.

Interlaced Buck converter

Schematic diagram of a Buck converter interlaced with N branches with synchronous correction.

Schematic diagram of a Buck converter interlaced with N branches with synchronous correction.

For a feeding having N converters connected in parallel, the current will be distributed on N phases, thus limiting the constraints on each switch and increasing the thermal heat-transferring surface. Moreover, the frequency of the currents and tensions seen by the load will be N time higher than that of a simple converter, dividing of as much the undulation of output voltage

The interlacing of the converters brings also another advantage: the dynamic response of the system to the current fluctuations can be improved. Indeed, an important increase in the current required by the load can be satisfied by making lead several branches of the feeding simultaneously if it were planned for that.

Principle of implementation

One of the problems inherent in the interlaced choppers is to make sure that the current is equitably distributes on N branches of the feeding. The measurement of the current in the branches can be done without losses by measurement of the terminal voltage of the reel or the second switch when it is busy. These measurements are described as without losses because they use the internal losses with the components to take their measurement without generating additional losses. It is as possible to make this measurement on the terminals of a small resistance as one would have inserted in the circuit. This method with the advantage of being more precise than the two preceding ones but it poses problems in terms of costs, output and space.

The output voltage can it also be measured lossless through the higher switch. This method is more complex to set up than a resistive measurement because it is necessary to filter the noise generated by commutations but it with the advantage of being cheaper.

A Buck-Boost converter is a feeding with cutting which converts a continuous tension into another continuous tension of lower or greater value but of opposite polarity. A disadvantage of this converter is that its switch does not have a terminal connected to the zero, thus complicating its command.

Principle of operation

Basic diagram of a Buck-Boost converter

Two configurations of a Buck-Boost converter following the state of the switch S

Basic diagram of a Buck-Boost converter

Two configurations of a Buck-Boost converter following the state of the switch S

- The operation of a Buck-Boost converter can be divided into two configurations according to the state of the switch S :
- In the state passing, the switch S is closed, thus leading to an increase in the energy stored in inductance.
- In the blocked state, the switch S is open. Inductance is connected to the load and the capacity. It results from it a transfer of the energy accumulated in inductance towards the capacity and the load.

- Compared with the converters Buck and Boost, the main differences are:
- The output voltage east of polarity reverses that of input
- The output voltage can vary from 0 with - ∞ for an ideal converter.

Continuous conduction

Forms of waves running ⁄ tension in a Buck-Boost converter

Forms of waves running ⁄ tension in a Buck-Boost converter

When a Buck-Boost converter works in mode of continuous conduction, the current I_{L} crossing inductance is never cancelled. The figure shows the forms of waves of the current and the tension in a Boost converter.

- The output voltage is calculated in the following way:
- During the state passing, the switch S is closed, involving the increase in the current following the relation:
- V
_{i}= L * (dI_{L}⁄ dt) - At the end of the state being on, the power I
_{L}increased: - ΔI
_{Lon}= ∫_{0}^{α*T}dI_{L}= ∫_{0}^{α*T}[(V_{i}* dt) ⁄ L] = (V_{i}* α * T) ⁄ L - α being the cyclic report ⁄ ratio. It represents the duration of the period T during which the switch S led.
- α lies between 0 and 1.
- During the blocked state, the switch S is opened, the current crossing inductance circulates through the load.
- If one considers a null voltage drop at the boundaries of the diode and a sufficiently large capacitor to keep his constant tension, the evolution of IT is :
- dI
_{L}⁄ dt = V_{0}⁄ L - Consequently, variation of I
_{L}during the blocked state is : - ΔI
_{Loff}= ∫_{0}^{(1-α)*T}dI_{L}= ∫_{0}^{(1-α)*T}(V_{0}* dt) ⁄ L = [V_{0}(1 - α)T] ⁄ L - If it is considered that the converter is in permanent mode, the energy stored in each component is the same one at the beginning and the end of each cycle of commutation.
- E = ½ L * I²
_{L} - Consequently, the current I
_{L}crossing inductance is the same one at the beginning and the end of each cycle of commutation. What can be written in the following way : - ΔI
_{Lon}+ ΔI_{Loff}= 0 - While replacing ΔI
_{Lon}et ΔI_{Loff}by their expression, one obtains : - ΔI
_{Lon}+ ΔI_{Loff}= (V_{i}* α * T) ⁄ L + [V_{0}(1 - α)T] ⁄ L = 0 - What can be rewritten in the following way:
- V
_{0}⁄ V_{i}= -(α ⁄ 1 - α) - Thanks to this last expression, one can see that the output voltage is always negative.
- Its absolute value increases with α, theoretically until the infinite one when α approach 1.
- If the polarity is omitted, this converter is at the same time reducing transformer and booster.
- Therefore one describes it as Buck-Boost.

Discontinuous conduction

Forms of waves running ⁄ tension in a Buck-Boost converter in discontinuous conduction.

Forms of waves running ⁄ tension in a Buck-Boost converter in discontinuous conduction.

In certain cases, the amount of power required by the load is enough weak to be transferred in a time shorter than one period from commutation. In this case, the current crossing inductance is cancelled during part of the period. The only difference with the principle in operation describes previously, is that inductance is completely discharged at the beginning of cycle. Although weak, the difference between continuous and discontinuous conduction has a strong impact on the formula of the output voltage.

- The output voltage can be calculated in the following way :
- As the current of inductance is null at the beginning of cycle, its maximum I
_{Lmax}(at = α * T)) is worth : - I
_{Lmax}= (V_{i}* α * T) ⁄ L - During the blocked state, I
_{L}cancel yourself afterwards δ * T : - I
_{Lmax}+ [(V_{0}* δ * T) ⁄ L] = 0 - By using the two last equations, δ is worth :
- δ = -(V
_{i}* α) ⁄ V_{0}

The current in the I_{0} load is equal to the average current crossing the diode (I_{D}). As one can see it on the figure, the current crossing the diode is equal to that in inductance during the blocked state.

- Consequently, the current crossing the diode can be written in the following way:
- I
_{0}= I^{-}_{D}= (I_{Lmax}⁄)δ - By replacing I
_{Lmax}and δ by their respective expressions, one obtains : - I
_{0}= -{[(V_{i}* α * T) ⁄ 2L][(V_{i}* α) ⁄ V_{0}]} = -(V_{i}² * α² * T) ⁄ (2L * V_{0}) - Consequently, the profit of output voltage can be written in the following way:
- V
_{0}⁄ V_{i}= -(V_{i}* α² * T) ⁄ 2L * I_{0}

This expression is much more complex than that obtained at the time of the study in continuous conduction. In discontinuous conduction, the profit in tension depends on the cyclic report ⁄ ratio but also on the tension of input, of the value of the inductance and the output current.

Limit between conduction continues and discontinuous

Evolution of the output voltage standardized of a Buck-Boost converter with an output current standardized.

Evolution of the output voltage standardized of a Buck-Boost converter with an output current standardized.

As explained in the preceding paragraph, the converter functions in discontinuous conduction when the current required by the load is weak, and it functions in continuous conduction for the more important currents. The limit between conduction continues and discontinuous conduction is reached when the current in inductance is cancelled just at the time of commutation.

- With the notations of the figure, that corresponds to :
- α * T + δ * T = T
- α + δ = 1
- In this case, the output current I
_{0lim}is given by the relation : - I
_{0lim}= I^{-}_{D}= I_{Lmax}⁄ 2 (1 - α) - While replacing I
_{Lmax}by its expression in discontinuous conduction : - I
_{0lim}= (V_{i}* α * T) ⁄ 2L(1 - α) - In extreme cases between the two modes of conduction, the output voltage obeys the expressions of the two modes.
- One will use that given for the mode of continuous conduction :
- V
_{0}⁄ V_{i}= -(α ⁄ 1 - α) - One can thus rewrite I
_{0lim}in the following way: - I
_{0lim}= [(V_{i}* α * T) ⁄ 2L][V_{i}⁄ V_{0}(1 - α) - Let us introduce two new notations :
- The standardized tension, defined by |V
_{0}| = V_{0}⁄ V_{i}which corresponds to the profit in tension of the converter. - The standardized current, defined by |I
_{0}| = L ⁄ (T * V_{i})I_{0} - The term (T * V
_{i}) ⁄ L corresponds to the maximum increase in current which one can reach at the time of a cycle - One thus obtains, in permanent mode, |I
_{0}| equalize 0 when the output current null east, and 1 for the maximum current which the converter can provide. - By using these notations, one obtains:
- In continuous conduction, |V
_{0}| = -(α ⁄ 1 - α) - In discontinuous conduction, |V
_{0}= -(α² ⁄ 2|I_{0}|) - The limiting current between conduction continues and discontinuous is :
- I
_{0lim}= (V_{i}* T ⁄ 2L) α (1 - α) = I_{0lim}⁄ (2|I_{0}|) α (1 - α) - Consequently, the border between continuous and discontinuous conduction is described by :
- 1 ⁄ (2|I
_{0}|) α (1 - α) = 1

This curve was plotted on the figure. The difference in behavior between continuous and discontinuous conduction is very clear. That can generate problems of control of the output voltage.

Case of the circuit not-ideal

Evolution of the output voltage of a Buck-Boost converter according to the cyclic report ⁄ ratio when the parasitic resistance of inductance increases.

Evolution of the output voltage of a Buck-Boost converter according to the cyclic report ⁄ ratio when the parasitic resistance of inductance increases.

- The preceding study was made with the following assumptions:
- The capacitor of output has a sufficient capacity to provide a constant power, during a cycle of operation.
- The voltage drop at the boundaries of the diode is null
- No losses by commutation in the semiconductors
- No losses in the components generally

These assumptions can be very far away from reality, the imperfections of the real components being able to have important effects on operation of the converter.

Taking into account of parasitic resistances

In the preceding study, the internal resistance of the components was not taken into account. That means that all the power is transmitted lossless source of tension towards the load. There exist however parasitic resistances in all the circuit because of the resistivity of materials used for its construction. Consequently, a fraction of the power transmitted by the source of tension is dissipated in these parasitic resistances.

For reasons of simplicity, one will consider here only the defects of inductance by modelling it by an inductance in series with a resistance. This assumption is acceptable because an inductance consists of a long wire which can thus have a clean resistance R_{L}. Moreover, the current crosses the reel in the two states of the converter.

- By using the method of the study in median value, one can write:
- V
_{i}= V^{-}_{L}+ V^{-}_{S} - With V
^{-}_{L}et V^{-}_{S}average tensions, on a cycle of operation, at the boundaries respectively of inductance and interrupteur. - Si one considers that the converter is in permanent mode, the average current through inductance is constant.
- La average tension at the boundaries of inductance thus becomes:
- V
^{-}_{L}= L [(dI^{-}_{L}) ⁄ dt] + R_{LI-L = RLI-L} - When the switch is busy, V
_{S}= 0. - When it is blocked, the diode thus becomes busy V
_{S}= V_{i}-V_{0}. - Consequently, the average tension through the switch is :
- V
^{-}_{S}= α * 0 + (1 - α)(V_{i}- V_{0}) = (1 - α)(V_{i}- V_{0}) - The output current east opposed to that in inductance during the blocked state.
- The average current in inductance is thus written :
- I
^{-}_{L}= -I_{0}⁄ 1 - α - If one regards the undulations of tension and current at output as negligible, the load can be regarded as purely resistive.
- If one notes R the resistance of the load, the preceding expression becomes:
- I
^{-}_{L}= -V_{0}⁄ (1 - α)R - By using the preceding equations, the tension of input is written :
- V
_{i}= R_{L}* {[-V_{0}⁄ (1 - α)R] + (1 - α)(V_{i}- V_{0})} - This expression can be put in the form :
- V
_{0}⁄ V_{i}= -α ⁄ {[R_{L}⁄ R(1 - α)] + 1 - α}

If the resistance of inductance is null, one finds the equation obtained in the ideal case. But the more R_{L} increases, the more the profit in tension of the converter decreases compared to the ideal case. Moreover the influence of R_{L} increases with the cyclic report ⁄ ratio.

A Ćuk converter is a feeding with cutting which converts a continuous tension into another continuous tension of lower or greater value but of opposite polarity. Contrary to the other types of converters, which use an inductance, a Ćuk converter uses a capacitor to store energy. The Ćuk converter owes its name with its inventor, Slobodan Ćuk of California Institute off Technology, which was the first to describe this topology in an article.

Basic rule

It is a converter of the type "Boost-Buck" which allows energy exchanges controlled between a power source and another power source through a stage buffer made up of a source of tension. It is the dual circuit of the Buck-Boost converter

Ćuk converter without galvanic insulation

Principle of operation

Basic diagram of a Ćuk converter without galvanic insulation.

Two configurations of a Ćuk converter without galvanic insulation according to the state of the switch S.

Two configurations of a Ćuk converter without galvanic insulation according to the state of the switch S.

On this figure, the switches were replaced by a wire when they busy and are removed diagram when they are blocked.

Principle of operation

Basic diagram of a Ćuk converter without galvanic insulation.

Two configurations of a Ćuk converter without galvanic insulation according to the state of the switch S.

Two configurations of a Ćuk converter without galvanic insulation according to the state of the switch S.

On this figure, the switches were replaced by a wire when they busy and are removed diagram when they are blocked.

A Ćuk converter without galvanic insulation consists of two inductances, two capacitors, a switch and a diode.

The capacitor C is used to transfer energy between the source from tension of input (V_{i}) and that of output (V_{0}). For that, it is connected alternatively to the input or the outlet side of the converter thanks to the switch S and to diode D.

Two inductances L_{1} and L_{2} are used respectively to convert the source of tension of input and the source of output voltage (C_{o}) into power sources. Indeed a reel can be considered, over one short period, as a power source as it maintains this one constant. These conversions are necessary in order to limit the current when one connects the capacitor C to a source of tension (V_{0} or V_{i}).

the Ćuk converter can function with a continuous or discontinuous conduction while running. However, contrary to the other converters, it can also function with a discontinuous conduction in tension, the terminal voltage of the capacitor is cancelled during part of the cycle of commutation.

Continuous conduction

- If it is considered that the converter reached its permanent mode, the amount of power stored in each one of its components is the same one at the beginning and the end of a cycle of operation.
- In particular, the energy stored in inductance is given by:
- E = ½L * I²
_{L} - Consequently, the current crossing inductance is the same one at the beginning and the end of each cycle of commutation.
- Evolution of the current in an inductance being related to the tension on its terminals:
- V
_{L}= L * (dI ⁄ dt) - This relation enables us to see that the average tension at the boundaries of an inductance must be null in order to satisfy the conditions of permanent mode.
- If it is considered that the voltage drop at the boundaries of the diode is null and that the capacitors C and C
_{0}have a sufficiently large capacity compared to the durations of load and of discharge so that one can consider that the tensions on their terminals are constant, the terminal voltages of two inductances L_{1}and L_{2}must satisfy the following relations: - During the state passing, V
_{L1}= V_{i}because L_{1}inductance is connected directly to the source of tension of input. - V
_{L2}= V_{0}- V_{C}because inductance L_{2}is, as for it, connected in series with the capacitors C and C_{0}. - During the state blocked, V
_{L1}= V_{i}- V_{C}, inductance L_{1}being connected in series with the source of tension of input and the capacitor C. - V
_{L2}= V_{0}because, the diode D being busy, L_{2}is directly connected to the capacitor of output. - The converter being in a busy state of t = 0 à t = αT, (α being the cyclic report ⁄ ratio) then in a state blocked of t = αT à t = T.
- Median values of V
_{L1}and V_{L2}are written : - V
^{-}_{L1}= α * V_{i}+ (1 - α) * (V_{i}- V_{C}) = (V_{i}+ (1 - α) * V_{C}) - V
^{-}_{L2}= α * (V_{0}- V_{C}) + (1 - α) * V_{0}= (V_{0}- α * V_{C}) - As the two tensions are null in order to satisfy the conditions of permanent mode, one can deduce some by using the second equation :
- V
_{C}= V_{0}⁄ α - By replacing V
_{C}in the equation of V^{-}_{L1}by his expression, one obtains: - V
^{-}_{L1}= (V_{i}+ (1 - α) * V_{0}⁄ α) = 0 - What can be rewritten in the following way :
- V
_{0}⁄ V_{i}= -[α ⁄ (1 - α)] - One realizes that this expression is the same one as that obtained for the converter Buck-Boost.

Basic diagram of a converter sepic

A converter sepic of the acronym of is a feeding with cutting converting a continuous tension into another continuous tension, of different value. The value of the output voltage depends on the cyclic report ⁄ ratio of closing of the switch. This relation can be expressed in the following way : V_{0} ⁄ V_{i} = α ⁄ (1 - α).

This assembly was developed by Slobodan Ćuk at the end of the years 1970. It is usually used for the load of the accumulators.

With the difference of the Ćuk converter which is supplied by a power source and which feeds a power source, the converter sepic is fed by a source of tension but, thanks to the inversion of inductance and the diode, can feed a source of tension.

It is also possible to replace two inductances L1 and L2 by two inductances coupled on the same magnetic circuit.

The assembly forward consists in carrying out cutting primary side and including in the same way the bridge of diodes to the primary. Let us note that if the sector is of 240V, at the boundaries of the C_{1} capacitor one will have 340V.

- When the transistor is conducting: this tension of 340V is applied to the primary of the transformer with a primary current I
_{on1}. - It results from it in the principal secondary and inductance a I
_{on2}current - When the transistor is blocked: it is the same process as in the preceding assembly.
- The D
_{2}diode has as a role to avoid shorting-circuit the secondary winding. - The third rolling up known as of demagnetization makes it possible to return the magnetic energy, stored during the phase of conduction, towards C
_{1}(through D_{3}).

This assembly has various advantages: the currents in the bridge of diodes and the transistor are obviously weaker from where a loss of reduced energy. However it requires a rolling up of demagnetization, some additional diodes, a transistor having to support a tension reverses 680V when it is blocked, moreover one needs a transformer functioning for 70kHz.

The output reaches 83%, one can imagine other solutions using two transistors assembled into push-pull or to half-bridge facilitating filtering and simplifying the transformer. In against part such a device implies a perfect symmetry and a separate control of the two transistors.

- Operation is the following :
- V
_{0}= V_{i}* [n_{s}⁄ n_{p})(T_{on}⁄ T)] = V_{i}δ ⁄ n - With δ = t
_{on}⁄ T, cyclic report ⁄ ratio. - n = n
_{s}⁄ n_{p}, report ⁄ ratio of transformation. - When the transistor is conducting its current is the sum of three components.
- The current of magnetizing which at the end of the conduction is worth i
_{m}= V_{i}t_{on}⁄ L_{t} - L
_{t}is the inductance of magnetizing of the transformer. - The secondary current brought back to the primary summons current in the I
_{0}load and of the undulatory current in inductance dI_{L}. - At the boundaries of inductance:
- e = L * (di ⁄ dt) avec e = V
_{i}⁄ n - V_{0}= V_{i}⁄ n - V_{i}⁄ n δ = V_{i}⁄ n (1 - δ) - dI
_{L}= V_{i}⁄ n [1 - δ]t_{on}⁄ L - At the end of the period of conduction in inductance one has I
_{Lmax}= I_{0}+ dI_{L}⁄ 2 - Maybe with the primary I
_{Lmax}⁄ n - And thus I
_{Cmax}= i_{m}+ I_{Lmax}⁄ n

The transistor is blocked

The current magnetizing i_{m} is evacuated through D_{3} and two rollings up must be identical under penalty of seeing the magnetic core saturating itself at the end of a few periods. δ = 0.5 should then be made, if rollings up are different one will have δ different from 0.5.

In any rigor one needs V_{i}t_{on} = V_{m}(T - t_{on}) where V_{m} is the tension with the primary during the blocking of transistor V_{m} is different from V_{i} if the number of whorls of two rollings up are different, and δ = V_{m}/(V_{i } + V_{m}).

But attention if for example V_{m }= 2V_{i} the terminal voltage of the blocked transistor becomes V_{m }+ V_{i } = 3V_{i} is 1020V.

The figure above watch, on the curve top, evolution of the current which is on average equal to I_{0}. During the phase of conduction of the transistor it is equal to I_{0} + I_{D1} and during the phase of blocking it is I_{0 } - I_{D2}. The forms of variations I_{D1} and I_{D2} in inductance are given on the following diagrams.

basic diagram of a converter flyback.

A converter flyback is a feeding with cutting, generally with a galvanic insulation between the input and the output. Its basic diagram is the same one as that of a Buck-Boost converter in which one would have replaced inductance by a transformer. The converter flyback is probably the structure most used in electronics industry. It is generally reserved for the applications of reduced power.

Principle of operation

two configurations of a converter flyback according to the state of the switch S.

two configurations of a converter flyback according to the state of the switch S.

The basic diagram of a converter flyback is represented on the figure. It is the equivalent of a Buck-Boost converter in which one would have replaced inductance by two coupled inductances playing the part of transformer. Consequently the principle of operation of the two converters is very close. In both cases one distinguishes a phase from storage of energy in the magnetic circuit and a phase of restitution of this energy. The dimensioning of the magnetic circuit defines the amount of power which one can store but also the speed with which one can carry out storage and the destocking of it. It is an important parameter which determines the power that the feeding flyback can provide.

- The operation of a converter flyback can be divided into two stages according to the state of the switch T:
- In the state passing, the switch T is closed, the primary of the transformer is connected directly to the source of tension of input. It results an increase from it from the magnetic flux in the transformer. The terminal voltage of the secondary is negative, thus blocking the diode. It is the capacitor of output which provides the energy required by the load.
- In the blocked state, the switch is open. The energy stored in the transformer is transferred to the load.
- In the continuation of this article one will note :
- ℜ reluctance of the magnetic circuit of the transformer :
- φ flow in the magnetic circuit :
- n
_{1}the number of whorls of the transformer to the primary. - n
_{2}the number of whorls of the transformer to the secondary. - α the cyclic report ⁄ ratio.

Continuous conduction

forms of waves running ⁄ tension in a converter flyback.

forms of waves running ⁄ tension in a converter flyback.

When a converter flyback works in mode of continuous conduction, flow in the transformer is never cancelled. The figure shows the forms of waves of the current and the tension in the converter.

- The output voltage is calculated in the following way :
- On-state.
- Running to the primary.
- During the state passing, the switch T is closed, involving the increase in the current following the relation :
- V
_{e}= V_{1}= L_{1}* (dI_{1}⁄ dt) - One thus obtains :
- I
_{1}= I_{1min}+ V_{e}⁄ L_{1}* t - With I
_{1min}the value of the current with t = 0_{-}^{I1min}. - Also corresponds to the minimal value of the current I
_{1}. - Its exact value will be thereafter given. at the end of the state passing, I
_{1}reached its I_{1max}maximum value - I
_{1mx}= I_{1min}+ (V_{e}* α * T) ⁄ L_{1} - α being the cyclic report ⁄ ratio. It represents the duration of the period T during which the switch T led.
- α lies between 0 (T leads never) and 1 (T leads all the time).
- As for I
_{1min}, the value of I_{1max}will be given after the study of the blocked state. - Stored energy
- At the end of the state passing, W
_{e}energy stored in the transformer is worth : - W
_{e}= ½ L_{1}I²_{1max} - At the end of the state passing, the switch T opens thus preventing the I
_{1}current from continuing to circulate. - The conservation of energy stored in the transformer causes the appearance of a I
_{2}current in the secondary of the transformer, whose initial value I_{1max}can be calculated thanks to the conservation of energy stored in the transformer at the time of its passage of the primary towards the secondary: - W
_{e}= ½ L_{1}I²_{1max}= ½ L_{1}I²_{1max} - While replacing L
_{1}and L_{2}by their expression according to the ℜ magnetic circuit and number of whorls of rollings up of the transformer, one obtains : - W
_{e}= ½ (n²_{1}⁄ ℜ) * I²_{1max}= ½ (n²_{2}⁄ ℜ) * I²_{2max} - That is to say :
- ½ (n
_{1}⁄ n_{2}) * I_{1max} - Tensions
- The calculation of the tension V
_{2}can be done thanks to the relations flow ⁄ tension. The relative direction of windings being reversed, one a : - V
_{1}= n_{1}* (dφ ⁄ dt) et V_{2}= -n_{2}* (dφ ⁄ dt) - That is to say :
- V
_{2}= -(n_{2}⁄ n_{1}) * V_{1} - State blocked
- Running to the secondary, lasting the blocked state, the energy stored in the magnetic circuit at the time of on-state is transferred to the capacitor.
- V
_{s}= V_{2}= -L2 * (dI_{2}⁄ dt) - I
_{2}= I_{2max}-(V_{s}⁄ L_{2}* (t - αT) - At the end of the blocked state, I
_{2}reached its minimal value I_{2min} - Stored energy
- At the end of the blocked state, there is, as for the end of the state passing, conservation of energy stored in the transformer. One can thus write :
- W
_{e}= ½ L_{1}I²_{1min}= ½ L_{2}I²_{2min} - While replacing L
_{1}and L_{2}by their expression according to the reluctance ℜ magnetic circuit and number of whorls of rollings up of the transformer, one obtains : - W
_{e}= ½ (n²_{1}⁄ ℜ) * I²_{1min}= ½ * (n²_{2}⁄ ℜ) * I²_{2min} - That is to say :
- I
_{2min}= (n_{1}⁄ n_{2}) * I_{1min} - Tensions
- The calculation of the V
_{1}tension can be done thanks to the relations flow ⁄ tension. The relative direction of windings being reversed, one a : - V
_{1}= n_{1}* (dφ ⁄ dt) et V_{s}= V_{2}= -n_{2}* (dφ ⁄ dt) - That is to say :
- V
_{1}= -(n_{1}⁄ n_{2}) * V_{s} - The V
_{t}tension at the boundaries of the switch T is worth : - v
_{t}= V_{e}- V_{1}= V_{e}+ (n_{1}⁄ n_{2}) * V_{s} - Entered relation ⁄ left
- Tension
- If it is considered that the converter reached its permanent mode, the average tension at the boundaries of rollings up of the transformer is null.
- If one considers in particular the average tension V
^{-}_{2} - At the boundaries of the secondary winding:
- V
^{-}_{2}= 1 ⁄ T * [-(n_{2}⁄ n_{1}) * V_{e}αT + V_{s}* (T - αT)] = 0 - That is to say :
- V
_{s}= n_{2}⁄ n_{1}* α ⁄ 1 - α * V_{e} - One obtains the same relation as for the Buck-Boost converter with the report ⁄ ratio of transformation n
_{2}⁄ n_{1}near. - That is due to the basic diagram of a converter flyback which is the same one as that of a Buck-Boost converter in which one would have replaced inductance by a transformer of report ⁄ ratio n
_{2}⁄ n_{1}. - The output voltage does not depend on the output current, but only on the cyclic report ⁄ ratio and the tension of input.
- Running
- If it is considered that the converter is perfect, one finds at output the consumption in input:
- V
^{-}_{e}I^{-}_{1}= V^{-}_{s}I^{-}_{s} - That is to say :
- I
^{-}_{1}= V^{-}_{s}⁄ V^{-}_{e}* I^{-}_{s} - Finally :
- I
^{-}_{1}= n_{2}⁄ n_{1}* α ⁄ 1 - α * I^{-}_{s} - One can find the values of I
_{1min}and I_{1max}by calculating the median value of I_{1}: - I
^{-}_{1}= 1 ⁄ T ∫_{T}I_{1}(t) = 1 ⁄ T [I_{1min}α T + (α T(I_{1max}- I_{1min}) ⁄ 2] = α [I_{1min}+ (I_{1mx}- I_{1min}) ⁄ 2] - While replacing I
_{1max}- I_{1min}by its expression according to V_{e}, α , T et L_{1}: - I
^{-}_{1}= α [I_{1min}+ (V_{e}* α * T) ⁄ 2L_{1}] - Maybe with final while replacing I
^{-}_{1}by its expression according to the output current : - I
_{1min}= n_{2}⁄ n_{1}* 1 ⁄ 1 - α * I^{-}_{s}- (V_{e}* α) ⁄ (2L_{1}ƒ) - I
_{1max}= n_{2}⁄ n_{1}* 1 ⁄ 1 - α * I^{-}_{s}+ (V_{e}* α) ⁄ (2L_{1}ƒ) - Thanks to the report ⁄ ratio of transformation one obtains easily I
_{2min}et I_{2mx} - I
_{2min}= 1 ⁄ 1 - α * I^{-}_{s}- n_{2}⁄ n_{1}* (V_{e}* α) ⁄ (2L_{1}ƒ) - I
_{2max}= 1 ⁄ 1 - α * I^{-}_{s}+ n_{2}⁄ n_{1}* (V_{e}* α) ⁄ (2L_{1}ƒ)

Discontinuous conduction

forms of waves running ⁄ tension in a converter flyback in discontinuous conduction.

forms of waves running ⁄ tension in a converter flyback in discontinuous conduction.

In certain cases, the amount of power required by the load is enough weak to be transferred in a time shorter than one period from commutation. In this case, flow circulating in the transformer is cancelled during part of the period. The only difference with the principle in operation describes previously, is that the energy stored in the magnetic circuit is null at the beginning of cycle. Although weak, the difference between continuous and discontinuous conduction has a strong impact on the formula of the output voltage.

- The output voltage can be calculated in the following way :
- On-state
- With the state passing, the only difference between continuous and discontinuous conduction is that current I
_{1min}is null. - By taking again the equations obtained in continuous conduction and cancelling I
_{1min}, one thus obtains: - I
_{1}= V_{e}⁄ L_{1}* t - I
_{1max}= (V_{e}* α T) ⁄ L_{1} - I
_{2max}= n_{1}⁄ n_{2}* I_{1max}= n_{1}⁄ n_{2}* (V_{e}* α T) ⁄ L_{1} - and finally :
- V
_{2}= -(n_{2}⁄ n_{1}) * V_{e} - Blocked state
- During the blocked state, the energy stored in the magnetic circuit during on-state is transferred to the capacitor.
- V
_{s}= V_{2}= -L2 * dI_{2}⁄ dt - I
_{2}= I_{2max}-V_{s}⁄ L_{2}* (t - α T) - During the blocked state, I
_{2}cancels after δ * T : - I
_{2max}- V_{s}⁄ L_{2}δ * T = 0 - While replacing I
_{2max}by his expression, one obtains : - δ = V
_{e}⁄ V_{s}* L_{2}⁄ L_{1}* n_{1}⁄ n_{2}α - By replacing L
_{1}and L_{2}by their expression according to the reluctance ℜ of the magnetic circuit and the number of whorls of rollings up of the transformer, one obtains: - δ = V
_{e}⁄ V_{s}* n_{2}⁄ n_{1}α - Entered relation ⁄ left
- The current in the I
_{s}load is equal to the average current crossing the I_{2}diode. - The current crossing the diode is equal to that in the secondary during the blocked state.
- Consequently, the current crossing the diode can be written in the following way :
- I
_{s}= I^{-}_{2}= I_{2max}⁄ 2 δ - While replacing I
_{2max}and δ by their respective expressions, one obtains : - I
_{s}= n_{1}⁄ n_{2}* [(V_{e}* α T) ⁄ 2L_{1}] * V_{e}⁄ V_{s}* n_{2}⁄ n_{1}α = (V_{e}² * α² T) ⁄ 2L_{1}V_{s} - Consequently, the profit of output voltage can be written in the following way :
- V
_{s}⁄ V_{e}= (V_{e}* α² T) ⁄ 2L_{1}I_{s}

Limit between conduction continues and discontinuous

Evolution of the output voltage standardized of a converter flyback with an output current standardized.

Evolution of the output voltage standardized of a converter flyback with an output current standardized.

As explained in the preceding paragraph, the converter functions in discontinuous conduction when the current required by the load is weak, and it functions in continuous conduction for the more important currents. The limit between conduction continues and discontinuous conduction is reached when the current in inductance is cancelled just at the time of commutation.

- With the notations of the figure, that corresponds to :
- α * T + δ * T = T
- α + δ = 1
- In this case, the output current I
_{slim}is given by the relation : - I
_{slim}= I^{-}_{2}= I_{2max}⁄ 2 (1 - α) - While replacing I
_{2max}by its expression in discontinuous conduction : - I
_{slim}= n_{1}⁄ n_{2}* (V_{e}* α T) ⁄ 2L_{1}* (1 - α) - In extreme cases between the two modes of conduction, the output voltage obeys the expressions of the two modes. One will use that given for the mode of continuous conduction :
- V
_{s}⁄ V_{e}= n_{2}⁄ n_{1}* (α ⁄ 1 - α) - One can thus rewrite I
_{olim}in the following way: - I
_{slim}= n_{1}⁄ n_{2}* (V_{e}* α T) ⁄ 2L_{1}* n_{2}⁄ n_{1}* V_{e}⁄ V_{s}α = (V_{e}* α T) ⁄ 2L_{1}* V_{e}⁄ V_{s}α - Let us introduce two new notations :
- The standardized tension, defined by |V
_{s}| = V_{s}⁄ V_{e}, who corresponds to the profit in tension of the converter. - The standardized current, defined by |I
_{s}| = n_{2}⁄ n_{1}* L_{1}⁄ (T * V_{e}) * I_{s} - The term n
_{1}⁄ n_{2}* (T * V_{e}) ⁄ L_{1}corresponds to the maximum secondary current which one can theoretically reach at the time of a cycle. - One thus obtains, in permanent mode, |I
_{s}| equalize 0 when the output current null east, and 1 for the maximum current which the converter can provide. - By using these notations, one obtains:
- In continuous conduction, |V
_{s}| = n_{2}⁄ n_{1}* (α ⁄ 1 - α) - In discontinuous conduction, |V
_{s}| = n_{2}⁄ n_{1}* (α² ⁄ 2|I_{s}|) - The limiting current between conduction continues and discontinuous is :
- I
_{slim}= n_{1}⁄ n_{2}* V_{e}T ⁄ 2L_{1}α (1 - α) = (I_{slim}⁄ 2|I_{s}|) α(1 - α) - Consequently, the border between continuous and discontinuous conduction is described by :
- 1 ⁄ 2|I
_{s}| α (1 - α) = 1 - This curve was plotted on the figure for n
_{2}⁄ n_{1}= 1 - The difference in behavior between continuous and discontinuous conduction is very clear. That can generate problems of control of the output voltage.

Influence inductances of escapes

Diagram of a converter flyback with the primary inductance of escape.

Diagram of a converter flyback with the primary inductance of escape.

The forms of waves described previously are valid only if all the components are regarded as perfect. Actually, one can observe an overpressure at the boundaries of the switch ordered at the time of his opening. This overpressure comes from the energy stored in the inductance of L_{ƒ1} escape to the primary of the transformer.

The inductance of escape not being directly connected to the primary of the transformer, the energy which it contains at the time of the opening of the switch cannot be transferred to the secondary. The evacuation of the energy stored in this parasitic inductance will create an overpressure at the boundaries of the switch. Moreover, the cancellation of the current crossing the switch not being done under a null tension, L_{ƒ1} also will generate losses by commutations. These losses can be reduced by the addition of circuits of assistance to commutation.

There exists also an inductance of escape to the secondary. This inductance goes, it also, to generate losses and to decrease the energy provided by the feeding to the load. In the case of feeding having of multiple outputs, inductances of escapes to the secondaries will create different losses on each output.

Specific structures

Feeding with sinusoidal absorption

Feeding with sinusoidal absorption

In the case of a converter supplied with a bridge of diodes whose output is connected to a capacitor, the power-factor is not unit, mainly because of the shape of the absorptive current. This assembly, which does not comply with the rules of interconnections of the electronics of power, connects a source of tension, the sector, with another source of tension, the capacitor. It results from it that the current is limited only by the imperfections of the assembly. If the load of the bridge of diodes is a converter of the flyback type, then the rules of interconnections of the sources are complied with and it is possible to control the absorptive current. With an adapted control, one can force the converter to absorb a quasi sinusoidal current in phase with the tension of the sector and thus of unit power-factor.

Interlaced choppers

Feeding flyback with two transistors

Car-oscillating mode

Feeding flyback with two transistors

Car-oscillating mode

A converter flyback in car-oscillating mode varies its quench frequency in order to always function in extreme cases of continuous conduction and discontinuous conduction. Such a device makes it possible to reduce the size of the transformer and to limit the losses of the recoveries in the diode, on the other hand, it increases the constraints on the switch.

Applications

- The converters flyback are used to carry out feedings:
- Of low costs at multiple exits.
- With high voltage and low power.

Operation at constant power

While choosing to control constant the I_{1max} current, the flyback then delivers a constant power with the load. This is particularly well adapted to the gas-discharge lamp supply, such as for example the lamps with the metal halides, whose power must be maintained constant during all the lifespan, the tension of arc increasing according to the consumption of the electrodes (the cyclic report ⁄ ratio of the chopper evolves ⁄ moves consequently naturally). The control of such a converter is then very simple because it does not require the recourse to any regulation of power. So there is no risk of instabilities of regulation related to the dynamic characteristics of the lamp. In the case of an portable unit, supplied with battery, the compensation of the variation of tension of this one is obtained very easily by controlling the instruction of current to this variation. Dimmage is also simplified by the direct tuning of the current of instruction of the chopper.

execution time customer :

runtime server : 0.027 seconds