Previously, we learned about the basic working principle of a Buck converter. It’s interesting, but it’s a bit difficult to understand. We therefore recommend this circuit: a 5V buck converter circuit using transistors.
We have an automatic Li-ion battery charger circuit that requires a 5V, 1A input. But we only have the old 12V, 0.5A power supply circuit (AC adapter).
So we will use the buck converter to reduce the voltage to 5V and increase the current to 1A.
Currently, my daughter is studying transistors, so we wanted to try out the basic transistor circuit to solve this problem. It may not be a complete switching regulator circuit.
But at least we can learn the basics of the buck converter along the way. We also like to use our old components, and they can also be good teachers for us.
How it works
It looks similar to the series voltage regulator of a linear power supply. However, the series voltage regulator works like a resistor that always reduces the current to the load, whereas a buck converter can increase the current.
So due to the similarity in both circuits, we are going to use the series voltage regulator as a base and build off from there.
The main change we made was adding the inductor L1. It works as an excellent electrical energy storage tank, similar to a capacitor. You can read more here.
Another component of note is Q1, which acts as a switching unit that works with L1, D1, and Q2 transistors. Both NPN (Q2) and PNP (Q1) transistors will feedback each other to continuously switch and generate a frequency.
When we apply electricity to the circuit, Q2 will work. Because a base current flows through R1 to bias Q2 into the conducting state. As a result, the positive current flows through pin E to pin B of Q1 and flows through C-E of Q2 to the GND.
In this case, it means that Q2 is used as Q1’s bias controller so that Q1 can supply a high current through pin C to coil L1.
The pin B of Q2 is also connected to pin C of Q3, which has ZD1 and R4 connected to pin B to control the voltage to be approximately 5.2 volts.
The way it can achieve this is when the output voltage level exceeds 5.2V, the current will flow through ZD1 to pin B of Q3, causing it to conduct current between its C and E pins. Functioning like a closed switch, giving the current a shorter path through pins B and E of Q2. As a result, Q2 stops conducting current, and so does Q1, and the output voltage drops.
Now coil L1 will start releasing the current, and D1, which acts as a rectifier, will rectify the negative potential and then charge up C2 through its negative polarity. Until L1 runs out of current, the cycle will start over again.
Testing this circuit
This circuit has a few components, and they are all easy to find.
0.25W or 0.5W Resistors, tolerance: 5%
- R1: 2.2K
- R2: 470Ω
- R3, R4: 220Ω
- C1: 1,000µF 25V
- C2: 1,000µF 16V
- Q1: TIP42 or equivalent, 100V 6A transistor
- Q2, Q3: S9013 or equivalent, 40V 0.5A Transistor
- D1: 1N5819, 40V 1A, Schottky Barrier Diode
- ZD1: 5.1V 500mA or 1W, Zener Diode
- L1: 220µH inductor coil
Recommended: If you are a beginner, read Electronic components list with images.
But there are some components that require extra consideration, including:
We made our own inductor L1 from a donut ferrite toroid core measuring 9.8 mm in diameter, 4.6 mm in ring width, and 5.9 mm in height. Then, wrap approximately 14 turns of 26AWG solid wire on the ferrite core. And we measured the inductance to be approximately 0.25mH or 250uH.
We picked the 1N5819 Schottky Barrier Diode as D1 because it has higher efficiency than using a regular 1N4007 silicon diode.
Then, we assemble the prototype circuit on a perforated PCB. Which is quite simple because there are not many components.
Once assembled, we test it out, as shown in the picture below. It worked quite well. The output voltage, when used with a Li-ion battery charger, drops to approximately 4.9V and can output approximately 1A. After nearly three hours, the battery was fully charged.
Testing the circuit with the Li-ion battery charger.
The voltage level of the Buck converter returned to 5.2V when there was no load.
Note: When there is a load, the input of the Buck converter is approximately 5.4V at 0.9A, and the output voltage is approximately 4.9V at 0.8A. It loses about 20% of the energy inside the Buck converter. It is not a complete switching regulator with about 90% or more efficiency.
This 5V buck converter circuit, although it is similar to a linear regulator, worked adequately. It can provide a constant voltage level of 5.2V when there is no load and up to 4.9V with a load that uses a current of approximately 1A. And it can be used continuously for a long time, with Q1 not getting hot either.
However, if you are looking for a circuit for day-to-day use, you should choose an IC, such as the MC34063, because we can determine the output voltage easily and accurately, and it has high efficiency with various protection systems.
Even though it cannot supply as many currents, we can still add transistors or MOSFETs to help drive up the current. We will be trying them out: How to increase more current for 34063 chip 5V 2A buck converter
I love electronics. I have learned them by creating the Electronic Circuits and Simple projects to teach my children. Most importantly I hope sharing our experience on this site will be helpful to you.