3.3v and 5v Power Supply

This is my very very very first attempt at any real electronics and someone suggested starting with a simple power supply. So I was hoping that SE could sanity check my diagram.

What I want is deadly simple, a supply that can accept 7.2V DC via a battery or 9V DC from a wall wart and puts out both 3.3V and 5V.

I decided I would follow SparkFun's power supply tutorial, but attempt to "add" a 3.3V supply to the circuit and I honestly have no idea if I did this right, but here it is...

I plan on using the following regulators (3.3V & 5V).

Pretty much any feedback is welcome, but I mostly want to know if this will accomplish what I want and if any improvements can be made (and if so some explaination would be great).

• Consider using the LM1117-5.0 LDO regulator for the 5 Volt regulation, instead of the 7805 - to work with the low headroom you are providing it. Apr 9, 2013 at 3:59
• It's recommended to not immediately accept an answer, to stimulate more answer activity on the question. In future, please give a question a day or two before accepting. Apr 9, 2013 at 12:25
• Sorry about that, will do in the future. I'm used the StackOverflow where answers on programming things tend to come in quickly. Apr 9, 2013 at 18:26

The input range of your 5V regulator is 7 -25V. In case you want to use 7.2V as input voltage, there is only 0.2V margin. This is not OK, because if the input voltage drops let's say 500mV over the diode, to 6.7V, and you can be sure that the 5V output rail would be dropping also (at ~ 4.7V). This kind of unstable voltage on the supply rail can disturb the function of digital circuits and has to be avoided.

Also, you can improve the design with some small ceramic capacitors (e.g. 100nF) at each input, for your regulators. Place them close to the pin (a few mm). Have a look on the datasheet recommended schematic (page 3 in the 3V3 spec).

Other than that, your design seems OK. You can keep the 3V3 supplied directly from battery. It may heat up, but, it is specified up to 15V input. Just make sure you dissipate the heat properly. Make a little testing. It will not blow up from the first trial.

• Good catch on the Diode voltage drop. Apr 9, 2013 at 7:57

While everything shown is correct, you need to take in mind what type of regulators they are. The 5v is a common LM7805 1.5A regulator, and the 3.3v is a LD1117V33 800mA low-drop out regulator, similar to the LM78xx family. Both are Linear Regulators.

First, the LM7805 is not a low-drop out regulator. It needs at least 2v above it's output to work properly. At 7.2v Battery input, that will quickly drop, depending on the current draw and battery discharge rate. A low dropout version would be better. And then I noticed the Diode. Any common diode will have 0.6 to 0.8v drop across it, as current increases beyond anything significant (10~100ma). This would make the Battery 7.2v way below the LM7805's Minimum Input Voltage, and it will not be happy. A lower dropout version would be mandatory for proper operation in this case.

Second, as linear regulators, they will turn any voltage above it's output, into heat. If you power them from 9V, and draw say 250mA each, that is (9V - 5V) * 250mA = 1 Watt of power wasted into heat for the LM7805, and (9V - 3.3V) * 250mA = 1.43 Watts on the LD1117V33. Heat will start becoming a concern, especially for the 3.3v regulator. A better option would be to have the 3.3v Regulator's input coming from the 5v regulator's output. While this means the 5v Regulator's load will increase by whatever load the 3.3V regulator consumes, the overall affect is that the 3.3V regulator will produce less heat.

Be sure to use a heatsink on the 7805, and watch your current draw.

• This is pretty spot on. I would just add that the size of the capacitors needed really depends on what you are powering, but the values shown should be fine for most things. Apr 9, 2013 at 3:05