Introduction: DIY Power Supply for 5V and 3.3V

I'm currently migrating my old and hopefully interesting projects from my blog www.tiny-labs.com to instructables. I use this power supply regularly to power my embedded projects in the lab. It is very handy and tiny. I've used an USB port as power supply for my projects before. Some ICs need 3.3V instead of 5V as Vcc. Therefore I decided to build this small power supply. It uses a range of 7-14V (I use 7V for less power loss) as input voltage and supplies 5V and 3.3V with 1A each. Both supplies can be switched individually.

I hope you can use this power supply for your projects. If you have any questions or suggestions please feel free to ask/post them in the comments or via PM.

Enjoy :)

Suggestions:

I got two suggestion by gm280 and rafununu concerning the heat production. They suggest to mount heatsinks to the ICs. They are right, especially for high current consumptions. Consider that the energy, in this case thermal energy, increases proportional to the voltage difference between input and output voltage. A heatsink can be easily attached with a screw or a clip directly on the LM1086. Heat paste would increase the thermal conductivity between the IC and the heatsink. There a lot of differently shaped heatsinks for the TO-220 package (this is how the housing design is called). Just ask your dealer of trust or do a quick google search. I must admit that I don't have overheat problems, but I'm mostly using the 5V line (which leads to less power loss) with a small load.

Rafununu suggests to replace the two diodes with only one Schottky diode as well.

JohnC430 suggests to limit the output current to a reasonable amount, like 600mA. That would make the power supply overheat-proofed.

Another hint by jhsa140467 concerning the heat problem would be placing another LM1086-5 right before the 3.3V circuit to reduce the voltage drop and therefor the heat generation at the LM1086-3.3

Step 1: Circuit Diagram

The concept is very clear. I want to have a small and handy power supply, which can provide 3.3V and 5V with at least 1A current. The output voltage should be constant even under heavy load and should have less noise.

I use to linear voltage regulator from Texas Instruments, the LM1086-5 and the LM1085-3.3. You can find typical application diagrams in the data sheet. I'll add the data sheet to the attachments. Please don't be confused, this data sheet is valid for all versions of LM1086-xx.

The two regulators are powered with the same source and the both circuits can be switched on/off individually.

This is the complete part list:

• Power plug, any kind you find and to which you have a wall-pluged power supply to
• 2 switches, make sure that they can handle your maximal current
• 2 1N4004 diodes to separate the two circuits
• LM1086-5 (for the 5V circuit)
• LM1086-3.3 (for the 3.3V circuit)
• 2 1k Ohm resistors for the status LEDs
• 2 small red LEDs for status on/off
• 4 Electrolytic capacitor (e-caps) of 100uF for smoothing the voltages and handle load switches
• 4 capacitors of 100nF for smoothing the voltages and handle load switches
• Some wires :D
• Some kind of plugs for the output voltages and GND. I use male pin headers, because I use cables in my lab, which have female pin headers soldered to them

You can use different or add more LM1086-xx with another output voltage to this project. But I think 5V and 3.3V is enough for the most embedded projects.

I use eagle cad to plan the circuit and later on the PCB.

Step 2: First Prototype on a Breadboard

Before I go on, I want to test my circuit. I set up everything on a breadboard and measure the voltages at the outputs. It works perfectly. Next step is to plan the PCB layout and to route the wires.

Step 3: Layout and Soldering

Now it's time to put the circuit into a layout for the PCB. You can easily switch in eagle cad to the layout view. It takes some time and experience to place everything in the right position. There are few rules I try to follow. The capacitors should be as close to the pins of the ICs as possible. The whole design should be compact but still easy to solder. I try to separate the two circuits visually and to apply some kind of symmetry. Another important rule is to minimize the crossing over of wires, which means to minimize the number of bridges you will need. Consider the heat, which is produced by your parts, especially by the ICs.

In this case you can see my result in the foto above. I need one wire bridge (the red line). You may achieve a different version, but that is okay. There is no right solution. Just try to apply as many features, I've mentioned before, as possible to the layout.

Next step is to solder everything on a PCB. PCB is not the right word here. PCB means printed circuit board. I don't print anything. I solder it by hand with wires and not traces. But you can send your design to a company, which provides such services. Or you can do it at home. There are some technics to this. Just check instructables.com, I guess you will find some examples.

My board has a width of 35 mm (1.38") and a length of 46 (1.81") mm.

Step 4: Measurements

Congratulations, you are done. The power supply is finished and can be used for your projects.

The last thing to do, but this is completely optional, is to measure the quality of the power supply. There should be no difference, if you use the same design and the same parts. I use a digital oscilloscope to measure the output voltage without and with a load. The third measurement shows to performance during the moment a load is applied to the output.

Step 5: 3.3V Output Without Load

There is no load applied to the 3.3V output port. The voltage has a small noise of +/-10mV. There is no voltage drop. This is more than enough for my purposes.

Step 6: 3.3V Output With a Load of 330mA

In this setup I want to measure the performance of the power supply under a load. The simplest way to apply a load is to use a resistor. I use a big 10 Ohm resistor. Make sure you have one, which can handle some power. I use a resistor which can handle 4 Watts. It's pretty big.

V = R*I --> I = V/R --> I = 3.30V / 10Ohm --> I = 0.33A

I apply a load of 330mA. The noise level is the same +/-10mV. There is a small voltage drop as well of 20mV. This is still enough for my purposes.

Step 7: 3.3V Output With a Load Switch to 330mA

The noise level and the voltage drops are important for a power supply. Another important criteria is the behavior during the switching process. I use a MOSFET to switch between two loads (from small to big). The MOSFET is triggered by a function generator with a square-wave signal. This signal is used to trigger the oscilloscope as well.

The noise level is the same +/-10mV. The voltage drops immediately a bit. There is no resonance or big amplitude.

Step 8: Conclusion

The power supply suites my requirements. The voltage is smooth with less noise. The small voltage drop under heavy load is negligible. The process of switching the loads does not interfere with the performance.

I hope you enjoyed the small instructables :)

Participated in the
Power Supply Contest