Boost 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}. - 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, the proposed choking coil
*L*and the current ripple Δ*I*_{L}are chosen such that Δ*I*_{L}= 0.4*I*_{in}, for*V*_{in_min}as the input voltage. (*I*_{in}=*I*_{out}·*V*_{out}/*V*_{in}) - If you should not be content with the proposed values, you can change
*L*or Δ*I*_{L}. If this is the case, the field "proposal" is 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: Boost Converter |

The transistor works as a switch which is turned on and off by a pulse-width-modulated control voltage. The ratio between on-time and the period *t*_{1}/*T* is called the **Duty Cycle**.

For the following analysis it will be assumed that the transistor is simplified as an ideal switch and the diode has no forward voltage drop. In the program itself, the diode will take into account a forward voltage drop *V*_{F} = 0.7V.

During the on-time of the transistor, the voltage across *L* is equal to *V*_{in} and the current *I*_{L} increases linearly. When the transistor is turned off, the current *I*_{L} flows through the diode and charges the output capacitor.

The function of the boost converter can also be described in terms of energy balance:

During the on-phase of the transistor, energy is loaded into the inductor.

This energy is then transferred to the output capacitor during the blocking phase of the transistor.

The output voltage is always larger than the input voltage. Even if the transistor is not switched on and off the output capacitor charges via the diode until *V*_{out} = *V*_{in}. When the transistor is switched the output voltage will increase to higher levels than the input voltage.

- The Boost Converter is not short circuit proof, because there is inherently no switch-off device in the short-circuit path.

With the help of Faraday's Law the continuous mode and steady state conditions can be established.

From this it follows that:

- For continuous mode the output voltage is dependent on the duty cycle and the input voltage, it is independent of the load.

Continuous Mode | Discontinuous Mode |

Illustration 2: Operating modes of the Boost 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. - 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. - 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*_{in}at*V*_{in_min}. However, the switching losses at the transistors are at their highest in this state.

** V_{in_min}**,

Using these parameters, the program produces a **proposal for L**:

where

From this it follows that:

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

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

Main page | | How to use the program | | Function principals | | Mathematics used in the program | | |

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