Introduction: Change 9 Volts to 5 Volts

By David Connolly
About: Electronics have always been fascinating to me. Things that get my attention are clocks, lamps, motion activated devices, light activated devices, laser intruder alert systems,solar and wind power, voltage inv…

Greetings to all my Instructable buddies. Today I'll show you how to convert 9 volts into 5 volts using the LM7805 fixed linear voltage regulator integrated circuit. If you would like more information about the LM7805 than I will be covering in this short Ible you should visit the Wikipedia page on this particular component.

Wikipedia's LM7805 page (actually titled 78XX)

Step 1: BOM (bill of Materials)

For this Instructable you will need:

  • 1 - LM7805 voltage regulator
  • 1 - Capacitor with the value of 0.33uF
  • 1 - Capacitor with the value of 0.1uF
  • 1 - 9 volt battery
  • 1 - 2.2 volt Red LED (you can use a different colored LED if you'd like, but then your final results may differ from what is displayed in the photos, ha!)
  • 1 - 150 ohm resistor

Step 2: Assemble and Use

I don't think any further explanation is needed at this point. But just in case. The positive lead from your battery goes to the left pin (input) on the LM7805. Attach one lead of the 0.33uF to the input pin and the other lead to ground. The output pin (right pin on the LM7805) will be your 5 volt source. Place the 0.1uF capacitor between the output pin and ground. In the example I'm using an LED as my 5 volt load. You could use whatever 5 volt device you need to use. There is a thing called "Dropout Value" which means your LM7805 will become unstable at about 2.5 volts above 5 volts on your input pin. In other words if you apply 7.5 volts or less to your input pin you may not get 5 volts from the output. One more thing I'd like to mention. You can apply more than 9 volts to the input pin. But (from the Wiki) the total power (voltage multiplied by current) going into the 7805 will be more than the output power provided. The difference is dissipated as heat. This means both that for some applications an adequate heatsink must be provided, and also that a (often substantial) portion of the input power is wasted during the process, rendering them less efficient than some other types of power supplies.

I hope this helps everyone who needed this information. Questions are welcomed.