Nowadays, everyone is talking about renewable energies. People can't resist the cool, sleek looking solar panels and massive wind turbines that the government has placed all across the country in order to harvest "clean" energy, albeit only working in certain time or location (solar panels need sunlight and wind turbines need a location that's susceptible to wind) and takes up quite a lot of space (solar/wind farms. One single panel/turbine is not gonna power a country), renewables are still deemed the future of power generation, as fossil fuels are indeed turning earth into a massive rotating sweatbox.
But of course, I'm an engineer, not an environmental activist, so this post is not gonna be about climate change or how to save the polar bears. Instead, we'll look at how a boost converter enables more efficient energy harvest from solar panels by stepping up their voltage effectively (How it works, essentially).
 |
| Circuit Diagram of a Boost Converter with a MOSFET-based switch |
So, what's a boost converter anyway? A boost converter is a DC-DC conversion circuit that steps up voltage from its input to its output. It does so by switching the circuit on and off very quickly (typically thousands of times per second, with industry‑grade designs reaching into the millions), forcing an inductor within the circuit to store and then release that energy. Let's look at it in a bit more detail:
When the switch is closed, current flows through the inductor to the ground. During this time, the inductor stores energy by building up a strong magnetic field, and the diode is reverse biased because the voltage at the output (stored in the capacitor) is higher than the voltage at the closed switch, preventing any current flowing to the output.
When the switch is opened, the current is unable to flow to the ground through the switch anymore, so the inductor, bound by Lenz's Law, collapses its magnetic field to maintain current flow, generating a higher voltage that's combined with the input voltage. This combined potential is now higher than the output, making the diode forward biased and pushing current into the capacitor. Because the switch flips so fast, the capacitor stays charged, smoothing these spikes into a steady, high-voltage stream of DC voltage.
We can't, of course, use an actual switch and press it manually to boost the voltage, instead, we use transistors to help us do so. Most modern designs utilise MOSFETs for its high switching speeds and efficiency, but other transistors are used depending on the power requirements. For example, IGBTs or Thyristors are preferred for heavy-duty, very high-power industrial applications, while BJTs might be found in specialised low-cost or older designs. These transistors are usually driven by a PWM signal, which can precisely control the Duty Cycle to maintain a steady output. We'll discuss more about the duty cycle later.
 |
| Definition of Volt-Second Balance using calculus |
Now, we can work out the relationship between the input and the output by using the Volt-Second Balance principle, which states that over one full switching cycle (in steady-state), the average voltage across an inductor must be zero. Therefore, when the switch is closed, the inductor is connected directly across the input voltage. So, the input voltage must equal to the voltage across the inductor (let's call it Vl). When the switch opens, the inductor is now in series with the input and the output. The voltage across the inductor flips to push the current out. So, Vl = Vin - Vout. Since Vout > Vin, this value is negative, meaning the current is decreasing.
If we plug the variables in, we'll get the equations above after evaluating the integral. Note that tON is the time when the switch is closed (hence circuit is on), and T is the total period, which is the sum of the time the switch is closed and the time it is open.
If we rearrange the formulas, we get the ideal transfer function of a boost converter, which is Vin times 1/ (1-Duty Cycle). Duty Cycle (D) equals to tON/T because it represents the percentage of the total switching period that the switch remains closed. Now, suppose we want to boose a 1.5V standard AA battery to 5V to power an Atmel ATMega328P microcontroller (the one you find on the classic Arduino Uno). Based on the ideal transfer function, we would need a PWM signal with a duty cycle (D) of 70%. (It's not a very good idea in practice, as it'll drain the battery very quickly. Why do you think that is? Feel free to share your answers in the comment section below.)
 |
| Boost Converter Implementation on Breadboard |
 |
| Equivalent Circuit Diagram |
Now that we understand the theory, let's see if it works in practice. I've whipped up a boost converter circuit on breadboard, using my trusty favorite STM32G431KB microcontroller on a Nucleo development board to generate the PWM signal which will switch the MOSFET. And yes, you've rightly spotted that a 3.3V microcontroller pin is not going be enough to switch a regular MOSFET directly. So, I've opted for a special "Logic-Level" MOSFET, the classic IRLZ44N which is widely used across microcontroller-based designs. As for the inductor and the diode, I've chosen a 4.7mH inductor (bit too large for this experiment but it's the only one available in the lab) and a SB560 Schottky diode for fast switching speeds and lower voltage drop (Silicon diodes have around 0.7V, where Schottky diodes have around 0.3V). The Boosted voltage is then smoothed by a 1mF capacitor and is fed into a 1kΩ load resistor.
 |
| PWM signal from G431KB |
For this experiment, we'll be attempting to step up a 5V input to 20V. Using the ideal transfer function, we can work out that the duty cycle needs to be at 75%. Therefore, I programmed the G431KB microcontroller is programmed to output a 70kHz PWM signal (chosen for low current ripple within inductor and low switching loss) with 75% Duty Cycle using the timer peripheral, and the output observed from one of the GPIO pin headers programmed to output this wave using an oscilloscope, as shown in the image above. The 5V input voltage is also supplied by the Nucleo board, so no external power supplies are required.
 |
| Logic-Level MOSFET switch |
 |
| MOSFET Drain Voltage (C1, Blue) and PWM Signal at Gate (C2, Pink) |
The signal is connected to the gate of the IRLZ44N logic-level MOSFET (Logic-level simply means the MOSFET is designed to turn fully on with the low gate voltages produced by digital logic, typically around 3.3 V or 5 V. Whereas regular (standard) MOSFETs need around 10V), with a 100Ω gate resistor to reduce ringing within the signa, and a 10kΩ gate pull‑down resistor to keep the MOSFET remains securely in the off state by draining any stray charge from gate to ground, preventing unintended turn-ons or overheating during microcontroller startup or signal interruptions. If we connect an oscilloscope to the gate and the drain of the MOSFET, we can see that when the PWM signal is high (3.3V, switch closed), it conducts and pulls the drain down to 0V, which allows the inductor to store energy from the 5V input. When the PWM signal is low (0V, switch opened), The MOSFET stops conducting, and the inductor releases its stored energy out through the diode. Ideally, based on our calculations, it should reach around 20V, but the oscilloscope only observed around 17.8V, presumably due to losses from the inductor’s internal resistance.
 |
| Inductor's voltage waveform |
The inductor itself is observed to be switching between the input voltage (5V, switch closed) and the voltage across the inductor (Vl) (5V - 18V = -13V, switch opened). Therefore, the inductor is indeed effectively boosting the 5V input to around 18V total.
 |
| Input Voltage (C1, Blue) and Output Voltage (C2, Pink) |
To observe the final output of the boost converter, I connected the oscilloscope across the input voltage, and to the output voltage across the 1kΩ load resistor. We can see that the 5V input has been boosted to around 17.576V across the load. But wait! The formula did work out that the output voltage should be 20V? Where's that missing 2.4V?
To find out where the voltage has been lost (approximately, if we were to work out all the math for the losses, we'll be here all day), we'll have to start from the beginning. Let's start with the MOSFET.
 |
| IRLZ44N MOSFET Characteristics Tested on a Component Tester |
I've tested the IRLZ44N MOSFET used in this converter and noticed a rather unusually high gate capacitance of 4.03nF. While datasheet graphs suggest a maximum capacitance of 2.4nF at low voltages, the measured 4.03nF suggests that the MOSFET is out-of-spec. Nonetheless, the MOSFET remains thermally stable (cold) during operation. While the high gate capacitance certainly slows down switching transitions, the fact that the MOSFET remains cold indicates that switching losses are not the primary cause of the voltage drop. (There's also the possibility that the Component Tester is lacking in accuracy, as it's not a professional-grade tester.)
 |
| Datasheet of WE-TI Radial Leaded Wire Wound Inductor |
As for the inductor, the datasheet of the 4.7mH inductor states that it's got a DC resistance of 8.33Ω (typical), Which means that even at a low output current of around 17.5mA, the average current through the inductor is significantly higher. The energy lost as heat across the copper wire prevents the inductor from reaching the full potential calculated in the ideal formula (Which also means that the inductor isn't really designed for power conversions).
 |
| SB560 Schottky Diode Characteristics Tested on a Component Tester |
As for the SB560 Schottky Diode, unlike Silicon diodes that got a typical voltage drop of around 0.7V, Schottky diodes got a much lower voltage drop of around 0.3V, which is why we opted it for this experiment. For the one we used on the circuit, the tester measured a voltage drop of 0.262V, and a 0.86μA of reverse leakage current (which is normal for Schottky diodes and doesn't affect the output). So, the diode performed exceptionally well and its contribution to the overall voltage loss was only limited to the expected 0.262V from the
Schottky barrier and showed no unexpected losses.
So, where did most of the voltage go? It turns out if we look at the oscilloscope output of the Input Voltage, you'd realise that it's actually 4.68V, not an ideal 5V (presumably the circuit is loading the source from the Nucleo board), so the actual theoretical output voltage should be 18.72V, using the same formula. As observed from the MOSFET's drain voltage waveform peaking at 17.832V. This means that the inductor has lost around 0.89V, and after passing the diode of 0.262V drop, we ended up with the 17.576V (These values are based on automated measurements captured by the oscilloscope’s internal processing engine. While they provide a precise value of the voltage levels, they are treated here as practical estimates of the overall system losses to estimate the circuit's performance without requiring complex and tedious mathematical proofs.). As for the Capacitor, while it's Equivalent Series Resistance (ESR) and leakage current technically contribute to efficiency loss, these factors were considered negligible in this analysis as the component remained thermally stable and the low load current did not induce significant voltage ripple.
 |
| IC-based Boost Converter Module Board |
And there you have it. We have successfully recreated a boost converter circuit on breadboard. In practice, however, modern boost converters are typically Integrated Circuit (IC) based, such as the MT3608 module shown above. By incorporating the PWM source, control logic, and MOSFET into a single silicon chip, these modules simplify the overall design process while maintaining high efficiency. The module includes feedback, which allows the IC to monitor the output voltage in real-time, adjusting the PWM duty cycle for various loads. This "closed-loop" system is far more robust than the "open-loop" method we used in our experiment, as it can dynamically adjust and compensate for the very losses we identified earlier.
Anyway, I hope you've enjoyed this post as much as I'm writing it, as I do find DC-DC converters to be a very interesting topic to look into. I'd love to hear about your thoughts on DC-DC converters! Feel free to share your thoughts in the comments, and as always, I'll see you in the next post.
⚠️ FULL SAFETY DISCLAIMER
The project described in this post is for educational demonstration only. I have years of experience in electronics and have followed strict safety protocols. If you do not have a similar background in electronic and/or electrical engineering, do not attempt to recreate this circuit. It is easy to inadvertently generate dangerous voltages that are unsafe to touch or cause components or equipment significant and/or irreversible damage. Breadboarded circuits lack the thermal shutdowns and safety protections found in commercial products.
Therefore:
- If you are a student or hobbyist, do not build high-voltage switching circuits without the direct, in-person supervision of an industry professional.
- For testing and project use, it is cheaper, smaller, and significantly safer to purchase a pre-made IC module (like the MT3608). These are designed to be "plug-and-play" with safety standards already in place.
- By reading or attempting to replicate any part of this project, you do so at your own risk. I am not responsible for any damage to your equipment, personal injury, or damaged components resulting from the use of this information. Safety is your responsibility. Knowledge is no substitute for caution.
Click
here to read the blog's full terms and disclaimers.
This post draws on the following resources:
- Principles of Power Electronics by John G. Kassakian, David J. Perreault, George C. Verghese, and Martin F. Schlecht (Cambridge University Press, 2023)
- University of Exeter Lecture Materials by Prof. Mustafa Aziz
If we rearrange the formulas, we get the ideal transfer function of a boost converter, which is Vin times 1/ (1-Duty Cycle). Duty Cycle (D) equals to tON/T because it represents the percentage of the total switching period that the switch remains closed. Now, suppose we want to boose a 1.5V standard AA battery to 5V to power an Atmel ATMega328P microcontroller (the one you find on the classic Arduino Uno). Based on the ideal transfer function, we would need a PWM signal with a duty cycle (D) of 70%. (It's not a very good idea in practice, as it'll drain the battery very quickly. Why do you think that is? Feel free to share your answers in the comment section below.)
Comments
Post a Comment