# How to step 12v to 5v and 3.3v

I'm still pretty new to electronics, and this is my first time on the EE StackExchange, so please bare with me :)

I'm currently working on designing a schematic for a quick prototyping board, in Eagle, for use with the Intel Edison. On the board there are a couple chips that require 5v to operate (an FE1.1s and a PCM2902E), and the Edison needs 3.3v to run.

In my schematic, I have a standard barrel jack and two screw terminal that I'm going to use to power the board. Now as I have a history of acidentally blowing stuff like LEDs, some chips, and speakers by acidenatlly applying to much voltage, or acidenatlly using a wall adapter set too high, I want to install some protection. (Especially since the Edsion don't have any type of voltage regulator, I don't want to fry a \$50 board!)

The idea is that the user can input any voltage (5v and above, say up to 12v) through the screw terminals or the barrel jack, and the board will shift to voltage to the levels it needs.

Right now, using this tutorial I have a 5v voltage regulator in my schematic to shift the input down to the 5v I need to run the chips. Now I need another circut to shift it down to 3.3v so I can power the Edison

simulate this circuit – Schematic created using CircuitLab

Whats the best way to go about shifting the voltage down to 3.3v for the Edison? and is there a better way to shift the unknown input voltage (max of 12v) to 5v?

• Hi, Keith M. Mkeith here. You must be the person everyone has always been mistaking me for my whole life when they ask me if my name is backwards. How much current do you need on 5V and 3.3V. There are two main types of regulators that would possibly make sense for you. There are linear regulators (such as your 78LXX) and buck converters (which use inductors and capacitors to provide lower voltages without wasting too much power). If you provide the current (amperage) estimates, it will help decide between the two. Nov 9, 2015 at 2:15
• @mkeith on the 5V (if I'm reading these datasheets right jfd-ic.com/Documents/FE1.1s%20Data%20Sheet%20(Rev.%201.0).pdf and ti.com/lit/ds/symlink/pcm2900.pdf) I need about 50-100ma, and I can't find the current on the datasheet for the edison, for the 3.3v adafruit.com/datasheets/EdisonDatasheet.pdf Nov 9, 2015 at 2:31
• @mkeith Just relized I made a mistake, the FE1.1s needs 100ma for 4 full-speed ports running off it, which is what I'm using it for, and the PCM2902E takes 10ma Nov 9, 2015 at 3:01
• The FE1.1 is a 4-port hub controller. Are you actually going to make 4 ports available to external devices? If so, you will need to supply (ideally) 500 mA at 5V for each port. If the USB connections are going to on-board devices, then you just need to include the power consumption of those devices in your budget. Nov 9, 2015 at 18:43
• Yep, 3 of the ports are connected to USB type A female connectors, and the 4th jack will have the option with a header to either be used as a usb soundcard through the PCM2902E, or as a 4th female connector Nov 9, 2015 at 19:44

## 2 Answers

LM78XX is a good option to solve your problem, but I strongly suggest you to investigate on LM2596 and mic29302. These regulators are much better than the old (but always effective) LM78XX series. If you are new to electronics, I recommend you to start with the newest available components. You will need some additional capacitors for these regulators but it's nothing hard, you will find all required information in the datasheets. Also, using LM2596 you can adjust the output voltage by changing the resistors values, so you can re-use these components in future projects. LM78XX series has defined output voltages that cant be changed.

• Those look like their exactly what I need, thanks! Nov 9, 2015 at 3:11
• m not too worried about being able to change the voltage, since my schematic is a design of a PCB, so I'll probably stick with a couple of defined output ones Nov 9, 2015 at 3:31
• Would you care to elaborate on why I should use the LM2596 over the LM78XX I already have in my schematic? I'm curious as to why the LM2596 is better. I see that the LM2596 has 5v and 3.3v versions, so I could get the same kind of regulator for both the parts of my circuit which would probably be good Nov 9, 2015 at 3:48
• Maybe you will see no difference on this project because LM2596 has many additional functions that you will not be using this time, but it is just much modern than LM7805. That is why I recommend you to start with it. In the future LM7805 will not be enough for you and you will need to migrate to 2596, why not to do it now!? Nov 9, 2015 at 13:12
• And of course you can use 2596 to step down to 3.3 or 5 or any other value since it has an adjustable output that you can set by changing the resistors value Nov 9, 2015 at 13:13

What you have will work very well for anything above about 6.3 V down to 5.0 -- Use an LM7805-type regulator.

You can get 3.3 either from the 5 V regulator (LM317 is suitable), or directly from the ~ 12 V input also. An LM317 is suitable for that also; there's little difference between these solutions, although powering the 3.3 V directly from the input may spread the power dissipation out a little better.

• Hey, @jp314, I think you accidentally gave some bad advice there. The LM317 datasheet recommends that Vin be at least 3V greater than Vout. It will work fine to generate 5V, but you cannot use it to drop 5V down to 3.3. Should use an LDO for that, or use 12V for Vin. But then you really need to worry about dissipation. Nov 9, 2015 at 2:21
• Yes, you are correct -- an ld1117v33 would be a better choice if chaining the output from the 5 V regulator. Nov 9, 2015 at 5:43