Buck converter based switching power supply?

Question

^{simulate this circuit – Schematic created using CircuitLab}

I have designed this circuit based on a buck converter. I'm just having trouble finding out which values to plug in. I.E. C1, L1, and the frequency for the output of MCU.

Also, as this will be my first PSU, I wanted to get some circuit analysis before I plug this into the main.

Note that I realize the output of the buck converter will have to be turned back into AC, and run through a transformer to isolate the circuit, then turned back into DC.

Un-Specified Components:

This is a theoretical circuit, thus I have I have not committed the values and specificity of the components. My question is whether this circuit could theoretically work, and how to select components for certain voltage values on the output.

Intended Operation:

BR1: This bridge rectifier turns the main's AC into DC.

Voltage Divider R1-R2: The resistance values are undefined because I have not selected a microcontroller for MCU. The voltage here will be dependent on the needs of the micro-controller I eventually select as MCU.

MCU: A microcontroller which will turn the NPN 1 on and off to form the cycles for the buck converter. MCU's VCC is connected to a transistor (PNP 1). The function of PNP 1 can be defined as: IF there is no current coming from the buck converter, then the MCU is powered by the voltage divider R1-R2. This is because I would rather use the more stable buck converter's voltage to power the MCU than the unstable voltage through the voltage divider. The MCU will only be powered through the voltage divider until the first cycle begins.

NPN 1, L1, C1, D1, Load: This circuit constitutes a straight forward, run of the mill, buck circuit, with NPN 1 being used as the switch, fired by MCU.

Application of This Power Supply:

I'm building a 3-in-1 laser engraver, 3D printer, and CNC machine. I've built a prototype with a working area of 35mm x 35mm x 35mm. However, the prototype runs off of a Raspberry Pi, 3 dual-h-bridge modules , and a laser driver circuit. The end goal is to apply the code I wrote for the prototype to a completely scratch-built printer. The printer will require 9V at varying current loads. (I.E. the device driver needs more current to power a high torque CNC router than the 3D plastic extruder.)

Answer 1

You didn't say what the power supply will be powering, i.e. what the output voltage and current are. Thus, it's not clear that a switching converter is really the best solution to your problem. A simple linear regulator or even unregulated rectified AC might be good enough. Without knowing the voltage and current it's impossible to recommend a topology, much less component values.

As others have noted, your circuit has a lot of problems. Here's are the ones I noticed. There may be more.

1. The overall topology is that of an inverting buck-boost converter, not a non-inverting buck converter.

2. You're using the rectified mains voltage (~170V) directly to get an output that's probably in the range of 5V. With a buck converter, that gives a duty cycle of ~3%, which is probably too low to control effectively. Also, your rectified input supply does not have a capacitor. Finally, there's no isolation between the mains and the output, which means that a fault could be lethal. This is not something you should ignore. Imagine your family and friends standing around your grave in tears, and their only consolation is the smoking remains of a crappy power supply.

3. Powering an MCU via a voltage divider is questionable at best. Powering an MCU from the output side of the thing it's regulating is extremely questionable. MCUs like stable voltages, and misbehave if they don't get them.

4. Your MCU cannot drive the switching transistor directly, since its output signal will be at the same level as the output voltage, thus making \$V_{BE} = 0\$. Also, MOSFETs are usually better than BJTs for this application.

5. Feedback control is a pretty big subject. You can make a nice career out of implementing and tweaking control systems for power supplies. If you're not familiar with control theory, you probably won't want to deal with it as part of another project you're already having trouble with. Errors in the control algorithm for a high-voltage circuit can be lethal.

6. The MCUs feedback input and VCC are both grounded.

Here's what I recommend you do:

• Consider using a wall-wart instead of designing your own power supply. It will save you a lot of trouble.

• If you must design your own supply, put a transformer on the input to reduce the voltage and isolate the circuit from the mains. Be sure to use a fuse and insulate all exposed high-voltage conductors.

• Unless you need high efficiency (>80%), try using a linear regulator like a 7805 or LM317. An LDO will let you reduce your transformer output voltage for greater efficiency.

• If you really need a switch-mode power supply, get an integrated converter like the one Russell mentioned. The datasheet will tell you what component values to use. You can also find modular power supplies that include all components. Or you could even copy an existing circuit without modification.

• If you insist on designing your own switch-mode power supply from scratch, you will need to learn a lot more. For background knowledge, you'll need at least a bachelor's degree worth of EE education. This Coursera class can teach you about the basics of SMPS design and control systems. (I think you can do it even when there are no sessions scheduled.) Or you can find a textbook, or take a class at a local college.

This is a major project, not a stepping stone.

Answer 2

Your circuit shows a good attempt at trying to understand concepts of discrete offline switching power supply operation. It is however a very long way from the concepts that are being learned to implementing a practical design. It would not be impossible to keep hammering away at this design until all the misunderstandings bugs and errors were hammered out of it and necessary practical aspects added BUT this is not the best way to learn to do this and is costly on your time and effort, and on the time and effort of others, and may kill you along the way and will certainly kill various versions of the circuit if ever implemented.

If you look at the available circuits you will find that almost nobody implements direct off-line buck converters. It is sometimes stated that the conversion ratios required cannot be implemented with a single stage buck converter. While it is doable* it is usually not practical, sensible or safe.

I have noted and the end of this answer a few of the issues that this circuit has, as shown in the latest version. The new diagram is a vast improvement over the original but is still a long way from looking like a circuit that would even try to operate.

If you keep attempting this incremental path you will end up having to address issues which are not fundamental to the basics of what you are trying to do and you will miss out on learning about more basic issues that are important. Laying this circuit aside for now and starting with

• A low voltage input switching power supply

• Using a cheap readily available conceptually simple(but still very capable) IC

would be a much more productive approach.

I often recommend a very old & low cost control IC which is still very useful.
Unlike some potentially better performing modern ICs it can implement almost any topology and it has an internal output switch which means it can implement complete power supplies with minimal added components up to a certain power level. This is the olde but goode MC34063

Excellent ON Semi MC34063 data sheet - has basic circuits with PCB layouts and components lists.

AN920/D Superb ON Seminconductor application note 40+ pages of extremely good discussion, circuits, design information and more. They say -

• The goal of this application note is to convey the theory of operation of the MC34063 and μA78S40, and to show the derivation of the basic first order design equations. The circuits were chosen to explore a variety of cost effective and practical solutions in designing switching converters. Another major objective is to show the ease and simplicity in designing switching converters and to remove any mystical “black magic” fears.

TI MC34063 application note reasonable. 12 pages.

Basic design calculator for inverting buck-boosttopology

A zillion circuits using it

Related:

ON SemiSwitch-mode power supply reference manual

ON Semi LED circuit solutions

---

These comments relate to this version of the circuit:

You will note on inspection that your Vcc and ground are hard connected.
This is no doubt not what you intend.

While you say that " ... This circuit constitutes a straight forward, run of the mill, buck circuit, ", it doesn't. This is one version of what is known as a buck-boost circuit. This is because the output can assume a voltage above or below the input oltage in magnitude. It achieves this clever trick at the cost of inverting the output polarity - Vout is negative if Vin in positive, which is why D1 is connected as it is. Your pocessor will not operate on a -ve Vcc and if you rearranged it so it would (uC grtound = -Vcc, uC Vdd = ground) it will then not drive NPN1 correctly, even if NPN1 was crrectly configured, which it is not.

If you reverse D1 you do not get a +ve output - you get flaming ruin. Your basic circuit is not correct. D1 and L1 need to be interchanged with correct polarity on D1 for the new topology.

NPN1 need to be a PNP (or some other device) and driving it from a uC (above or below ground) takes substantially more than a drive resistor. While the diagram can be seen as conceptual there is not too much detail that is just plain wrong.

The intended operation of PNP1 is understood but you need much more "glue"components than are shown, and as shown, R1 & R2 will dissipate substantial ongoing energy when PNP1 is off (about 150 mW per mA of startup load so say around 1.5W for at 10 mA startup load.)

To convert this circuit into something that works safely, reliably or at all would require a long and unnecessary journey.

Answer 3

This is a simplified version of the circuit you have. Note the polarity is backwards from your circuit: