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I new to electronic engineering and am trying to learn electronic engineering by creating little projects. I have a few questions about powering my Raspberry Pi (5 V DC @1000 mA) from a 12 V DC 1000 mA power supply. The 12 V power supply will provide power to the TFT monitor, and I want to also power my Raspberry Pi from the same source.

After doing some research, I can make a 5 V regulator using a 7805 voltage regulator and two 100 µf electrolytic capacitors. The plan is to put a heat sink on the 7805 to help dissipate the heat. My goal is to have one source of power input for my project.

My schematic:

5 V Regulator

And a Fritzing diagram on the hook up:

Enter image description here

  1. Is this the correct approach?
  2. Is my schematic correct?
  3. Is my hookup correct?
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    \$\begingroup\$ schematic is wrongly wired...5V name and GND are same. Output terminal pin3 is 5V. \$\endgroup\$
    – user19579
    Commented Apr 12, 2014 at 14:46
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    \$\begingroup\$ You can purchase USB chargers meant to be used in vehicles very in expensively. Look for one meant to charge an iPad, and you'll get 5V @ 2A out of it, more than enough for the raspberry pi, and as a bonus it'll be a switching regulator, putting out less heat and using less current than the linear regulator you are talking about in your question. \$\endgroup\$
    – Adam Davis
    Commented Apr 12, 2014 at 15:00
  • \$\begingroup\$ A 1 Watt supply will not do. Power is Voltage times Current, so if the 5V Supply needs at least 1 Amp, power is 5 * 1 = 5W. Extra margin implies you want 7-10W (i.e. 2 Amps as posted above). You may even have an old 5V wall wart lying around - I have a number in a drawer. It's funny that from all the sophisticated (and expensive) electronics I have bought over the years the only parts that persist as useful are these. \$\endgroup\$
    – user40286
    Commented Apr 12, 2014 at 15:14
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    \$\begingroup\$ Get a car USB charger. They are typically switching regulators designed for ~12V DC to 5V, exactly what the Raspberry Pi (and you) need. \$\endgroup\$
    – Passerby
    Commented Apr 12, 2014 at 19:37

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Knowing that many people have issues with a 5V/1A power supply, it being on the weak side, I expect RPi to draw a fair amount of current, quite near to the maximum current rating of the regulator. In theory your circuit is largely correct, but you will need a huge heat sink dissipate somewhere near 7W (more than half the total power used). A good heat sink is probably more expensive than a capable switching converter. I'd check Internet for a cheap buck or DC/DC converter.

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  • \$\begingroup\$ Thanks - will this DC/DC convertor do? Bothhand SM1-0512 LF 1W Single Output SMD DC-DC Converter radioshack.com/product/index.jsp?productId=20016706 \$\endgroup\$
    – PhillyNJ
    Commented Apr 12, 2014 at 11:37
  • \$\begingroup\$ Quit searching RadioShack. They're bizarrely expensive and the price to quality ratio is at least equally bad. Apart from that the spec of the device does not meet your requirements. Try amazon or ebay eg.: amazon.co.uk/niceeshop-LM2596-Converter-Module-1-23V-30V/dp/… Try to find qouple alternatives yourself. You need > 1A output current, 5V output voltage and at least 12V input voltage. \$\endgroup\$
    – jippie
    Commented Apr 12, 2014 at 15:15
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Your schematic is wrong and will not work, because there is no DC current path through the capacitor into the 5V rail. Confusingly the Fritzing diagram seems to be different and gets it right.

Stylistic criticism: normally you would put the positive power rails at the top and ground at the bottom, with the regulator as a T-shape in the middle.

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Start by estimating current consumption of the devices that you will be powering. You don't have to be very precise at this, just a ballpark figures with err on the higher side. When you know your current requirements, you can estimate power wasted in the 7805 regulator.

Raspberry Pi alone can use up to 0.5 A. Let's say your display draws another 0.5 A, 1 A in total. Your input voltage is 12 V, and the output voltage 5 V. This means the linear regulator will have to burn (12 V - 5 V) * 1 A = 7 W of power.

You have to check the datasheet for your particular flavor of LM7805, but that is dangerously close or even above the absolute maximum ratings of LM7805.

I'd suggest getting a cheap DC-DC buck converter module off eBay. I've tested this one - pretty low noise and much better efficiency than LM7805 - Details about Mini DC-DC Buck Converter Step Down Module Power Supply For aeromodelling.

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  • \$\begingroup\$ Just don't forget to set the output voltage correctly before hooking up the RPi ;o) \$\endgroup\$
    – jippie
    Commented Apr 12, 2014 at 15:32

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