For an electronics enthusiast, having a 5 volt DC (direct current) power supply available can be very useful. Many operational amplifiers, micro controllers and other integrated circuits (ICs) require a 5 volt power supply (although most can operate in the 3-15 volt range). This article describes how to build a simple 5V DC power supply, capable of supplying current up to 1.5 Amps. You will need to weld various components together.
Steps
Step 1. Consider one of the AC adapter cables as the positive terminal
Consider the other as the grounded terminal. It does not matter which of the two you have selected as the positive terminal, but, from now on, you must remember your choice.
Step 2. Connect the positive terminal of the AC adapter to the end of the diode not marked with a strip
You are connecting the positive terminal to the anode of the diode; current will flow into the diode through this end and only in this direction, to charge the capacitor you will connect later.
Step 3. Locate the capacitor terminal located at a marked stripe on its body
Usually this strip is white and shows the identification of the minus sign. This is the negative terminal, which you will need to connect to the ground terminal of the AC adapter.
Step 4. Connect the other terminal of the capacitor to the terminal of the diode marked with the strip
That is, connect the positive terminal of the capacitor to the cathode of the diode. The diode allows the transformer current to charge the capacitor, and prevents the capacitor from discharging through the transformer during the negative cycle.
Step 5. Connect Pin 1 of the integrated circuit for voltage regulation to the connection node between the positive terminal of the capacitor and the terminal on the marked end of the diode
Pin 2 is the reference for grounding, also called the "Common" terminal, and should be connected to the ground terminal of the AC adapter. Pin 3 is the output. The voltage difference between Pin 3 and ground will be 5 volts.
Advice
- The terminal next to the marked stripe on the diode body is always the cathode (negative terminal) of the diode.
- You can get all the parts you need from a common electronics distributor, such as Digikey and Mouser.
- Add a capacitor between the output and the ground to improve the transient response.
- The TL780-05 5V regulator is capable of delivering 1.5A of current, but if the AC adapter you use is unable to do the same, then the output current of your power supply will be limited by the maximum current of your adapter B. C.
- Try using a 12 volt or lower AC adapter. Higher voltages will cause the regulator to dissipate more power, causing it to overheat.
- Your new regulator will be able to deliver up to 1.5A of current, at 5 Volts. The integrated circuit for voltage regulation may overheat in case of higher current values; therefore consider adding a heatsink, in the case of high power applications.
- Assemble the circuit on a prototyping board to make building easier.
- Even higher voltage values will require precautions and the use of a secondary regulator, to lower the voltage down to a value manageable by the 5V regulator.
- To enhance the design, add a full-wave rectifier in place of the current design's half-wave rectifier.
- This is a very simplified project; you can find improvements in the technical data sheet of the TL780-05 component.
Warnings
- None of the voltage values used in this project are dangerous. The AC adapter is the only element in contact with the mains voltage. If you want to open the plastic casing of the adapter, make sure you have disconnected it from the power supply first.
- The 5V regulator can get very hot when asked to supply a lot of current. Be careful as it may overheat to the point of causing sunburn.
- By overheating the 5V regulator too much, you may soon burn it out.
- If you connect an electrolytic capacitor inversely you could cause it to explode. Make sure that the negative terminal of the capacitor (the one tapped with the strip) is always at a lower voltage than the positive terminal, and that the voltage across the capacitor does not exceed the maximum voltage limit of the capacitor itself.
