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seven voltsss

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Tom Selier 2024-02-04 15:54:47 +01:00
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\section{Introduction}
Students of the Electrical Engineering department of the University of Applied Sciences Utrecht are to design a DC-DC Boost converter in the fifth semester during the Hardware Design course.
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The goal of the project is to design a DC-DC Boost converter on breadboard that can step up a voltage of around $3V$ to any voltage between $3.3V$ and $7.5V$. The converter should be able to deliver a current of $50mA$ at any given voltage. The entire system is to be controlled by a STM32F4 series microcontroller which should provide a way to control the output voltage through software.
The goal of the project is to design a DC-DC Boost converter on breadboard that can step up a voltage of around $3V$ to any voltage between $3.3V$ and $7V$. The converter should be able to deliver a current of $50mA$ at any given voltage. The entire system is to be controlled by a STM32F4 series microcontroller which should provide a way to control the output voltage through software.
An example circuit and a couple of components are provided and mandatory to use, such as the inductor, the capacitor and the MOSFETs. Apart from these components, the circuit can be altered and or extended to the student's liking to improve its function as long as the final circuit is able to deliver the required current and voltage.
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\item The MOSFETs, BS170 series
\end{list}
In the example circuit, there was only one MOSFET and it was directly controlled by the MCU. This circuit was able to reach a voltage of $7.5V$, but not at the required current. This is due to the fact that the BS170 series MOSFET has a fairly high $R_{DS(on)}$ of $1.2\Omega$ \cite{Onsemi_2022}. To increase the output current, a second MOSFET was added to the circuit. This way, the $R_{DS(on)}$ is halved, effectively doubling the current at which the inductor can be charged if we neglect the resistance of the inductor itself. The downside of adding a second MOSFET is that it also doubles its input capacitance, however, with an additional circuit modification this can be overcome.
In the example circuit, there was only one MOSFET and it was directly controlled by the MCU. This circuit was able to reach a voltage of $7V$, but not at the required current. This is due to the fact that the BS170 series MOSFET has a fairly high $R_{DS(on)}$ of $1.2\Omega$ \cite{Onsemi_2022}. To increase the output current, a second MOSFET was added to the circuit. This way, the $R_{DS(on)}$ is halved, effectively doubling the current at which the inductor can be charged if we neglect the resistance of the inductor itself. The downside of adding a second MOSFET is that it also doubles its input capacitance, however, with an additional circuit modification this can be overcome.
A second addition to the circuit is the push-pull driver on the gate of the MOSFETs. The driver consists of two NPN transistors that are controlled by inverted signals from the MCU, this way the MOSFET gates are quickly switched between $0V$ and $3.3V$. The push-pull driver was chosen over a pull-down resistor because it would cause several issues; A low resistor value would quickly discharge the gate capacitance, but it would also decrease the driving current when turning the MOSFET on. A high resistor value would have little effect on the gate capacitance. The push-pull driver solves both of these issues by quickly discharging the gate capacitance by connecting them to ground while not effecting the driving current when turning the MOSFET on. This way, the MOSFETs are switched faster, increasing the efficiency of the circuit.