Half-Bridge Push-Pull Converter

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- The values of all input fields can be changed.
- If an input field is left empty, a default value is chosen. This value is displayed after leaving the input field in question.
- The switch mode power supply operates within a certain input range i.e. between
*V*_{in_min}and*V*_{in_max}.

**Note:**- For the european mains of 230V +/-10% and behind the rectifier and the smoothing (with a voltage ripple of 10%) the input voltage range is between
*V*_{in_min}= 250V and*V*_{in_max}= 360V. - For wide range Switch Mode Power Supplies the input voltage range of the mains is from 100Vac -10% (Japan) to 240Vac +6% (Great Britain). In this case, the DC input range of the power supply is from
*V*_{in_min}=110V to*V*_{in_max}=360V. - For use of a power factor pre-regulator the input voltage range is normally from
*V*_{in_min}=360V to*V*_{in_max}=400V.

- For the european mains of 230V +/-10% and behind the rectifier and the smoothing (with a voltage ripple of 10%) the input voltage range is between
- The program needs the output values
*V*_{out}and*I*_{out}. - The switching frequency
*f*is the operating frequency of the transistor. - If the field "proposal" is activated for the inductor
*L*, a value for*L*and the corresponding current ripple Δ*I*_{L}is proposed. These values are laid out such that Δ*I*_{L}= 0.4*I*_{out}with*V*_{in_max}as the input voltage. - If the field "proposal" for the input field "
" is activated, the turns ratio*N*_{1}/*N*_{2}*N*_{1}/*N*_{2}is proposed. This suggestion is chosen such that the required output voltage can be achieved using*V*_{in_min}as an input voltage. - If you do not agree with our proposals, you can change
*N*_{1}/*N*_{2}or*L*as well as Δ*I*_{L}. The field "proposal" is then deactivated automatically. - The value
*V*_{in}is the value for the calculation of the current and voltage diagrams on the right side of the display.*V*_{in}must lie between*V*_{in_min}and*V*_{in_max}.

Illustration 1: Half-Bridge Push-Pull Converter |

The Push-Pull converter drives the high-frequency transformer with an AC voltage, where the negative as well as the positive half swing transfers energy. The capacitor-bridge generates, in its centre point, a voltage of ½ *V*_{in}.

The primary transformer voltage *V*_{1} can be +½ *V*_{in}, -½ *V*_{in} or *zero* depending on whether the upper transistor, the lower transistor or neither is on.

On the secondary side, the AC voltage is rectified, so that *V*_{3} is a pulse-width-modulated voltage which switches between ½·*V*_{in}·(*N*_{2}/*N*_{1}) and zero. Due to the rectification, the pulse-frequency of *V*_{3} is equal to 2· *f* .

The Low-Pass filter, formed by the inductor *L* and the output capacitor *C*_{out}, produces the average value of *V*_{3}. For continuous mode (*I*_{L} never becomes zero) this leads to:

The Duty cycle of this converter may theoretically increase to 100%. In practice this is not possible because the serial connected transistors, *T*_{1} and *T*_{2}, have to be switched with a time difference to avoid a short circuit of the input supply.

Due to the fact that the duty cycle *t*_{1}/*T* can theoretically increase to 100%, it follows for the turns ratio that:

In the program, this value is multiplied by a factor of 0.95, so that the proposed value for *N*_{1}/*N*_{2} includes a small margin which guarantees the demagnetisation of the core, when the input voltage is minimal, (remember: at minimum input voltage the duty cycle reaches its maximum).

For the allocation of the inductor *L*, the same rules as for the
Buck Converter can be used. One also distinguishes between **discontinuous** and **continuous mode**, depending on whether or not the inductor current falls to zero during the on-time of the transistor.

During continuous operation:

- In continuous mode the output voltage depends only on the duty cycle and the input voltage, it is load independent.

During continuous mode, with

- The change in inductor current is load independent. The output current
*I*_{out}is taken to be the average value of the inductor current*I*_{L}.

In that moment, when the inductor current becomes zero, the voltage

Continuous Mode | Discontinuous Mode |

Illustration 2: Operating modes of the Half-Bridge Push-Pull Converter

- The larger the chosen value of the inductor
*L*, the smaller the current ripple Δ*I*_{L}. However this results in a physically larger and heavier inductor. - The higher the chosen value of the switching frequency
*f*, the smaller the size of the inductor. However the switching losses of the transistor also become larger as*f*increases. - The smallest possible physical size for the inductor is achieved when Δ
*I*_{L}= 2*I*_{out}at*V*_{in_max}. However, the switching losses at the transistors are at their highest in this state. - Choose Δ
*I*_{L}so that it is not too big. The suggestions proposed by us have adequately small current ripple along with physically small inductor size. With a larger current ripple, the voltage ripple of the output voltage*V*_{out}becomes clearly bigger while the physical size of the inductor decreases marginally. - It is best not to alter the turns ratio
*N*_{1}/*N*_{2}proposed by us.

** V_{in_min}**,

Using these parameters, the program produces a **proposal for N_{1}/N_{2} and L**:

(the factor of 0.95 is taken into account to allow for the fact that the duty cycle

Δ

From this it follows that:

- For
**Δ**the converter is in continuous mode and it follows that:*I*_{L}< 2*I*_{out}

- For
**Δ**the converter is in discontinuous mode and it follows that:*I*_{L}> 2*I*_{out}

Main page | | How to use the program | | Function principals | | Mathematics used in the program | | Help for HF transformer |

Top of page | | Application | | Tips | | Literature Notes | | Help for choking coils |