1
\$\begingroup\$

This question already has an answer here:

The L805CV voltage regulator is capable of producing a consistent 5v 1.5A draw. I am creating a small raspberry pi case and have a 12v 3a power supply to accomplish the project. I am powering a 12v fan and a few other peripherals at 5v including the pi. I have 10uF capacitors for both inputs and outputs, however I ran into a problem fairly quickly. I researched very thoroughly that nearly any heat sink I looked at would overheat the device. Being in Texas you can assume a maximum ambient temperature of around 40*c. the heatsinks have a dissipation rate of 23*c/w and here's the datasheet for the regulator: https://www.sparkfun.com/datasheets/Components/LM7805.pdf.

Now that you have the information let me ask my question. I'm attempting to run the 12v supply into the 5v regulator at 1A. however this very quickly overshoots the allowed temperatures as this is 7 watts to dissipate. Would it be easier to get a 9v supply and just adjust the voltage upwards for the 12v fan or is there a way in order that 12v 1A can be pushed through this without overheating?

If I did choose to go with 9v what circuit would you recommend in order to bump up the voltage 3v for the fan? The fan is an Antec TriCool 120mm LED fan with a maximum current draw of 300mA at 12v. Thank you for your time.

This is not a repeat subject as I am doing low voltage conversions at a maximum of 7w whereas the other topic addresses 24v 2A down to 5v .1A. not only does this not address the fact that he is using a pathetic amount of current but it's obvious why it was overheating. this is asking for a solution it's not asking why it's happening. I'm simply asking how to fix it

\$\endgroup\$

marked as duplicate by m.Alin, Ricardo, PeterJ, Adam Lawrence, Daniel Grillo Feb 18 '15 at 10:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 4
    \$\begingroup\$ Why not use a buck converter, they are more efficient and quite cheap these days. \$\endgroup\$ – PlasmaHH Feb 16 '15 at 20:31
  • 1
    \$\begingroup\$ I agree with the buck converter suggestion. You can get a drop-in replacement for your linear 7805. I used a murata-ps.com/data/power/oki-78sr.pdf recently for a small project... \$\endgroup\$ – brhans Feb 16 '15 at 20:38
  • \$\begingroup\$ You couldn't find a better heatsink? Digikey lists ~300 parts with 10°C/W or better thermal rating. Some of them are better then 1°C/W! \$\endgroup\$ – Connor Wolf Feb 18 '15 at 5:39
0
\$\begingroup\$

I found an alternative solution which may be of use to all those dealing with the same problem. I have ordered a pack of L7808CV IC's which will convert the 12v 1a down to 8v (4 watts is much easier to dissipate than 7) and then 8v down to 5v. I am using 10uf Caps on both inputs and outputs to keep a consistent draw on both ends. If you need a single project i would recommend the buck converter option, however since I already had a set of 7805's for 5$ I just added another 5$ leaving me with 10 7805's and 10 7808's to play with.

-- I would only recommend the following solution to those who are buying for multiple projects and need 10 sets of one item --

\$\endgroup\$
  • \$\begingroup\$ 4 watts from the first regulator, then 3 watts from the second. Total is still 7 watts, although you get to spread the heat around some, using two heatsinks. \$\endgroup\$ – gbarry Feb 18 '15 at 5:28
  • \$\begingroup\$ I have a bulk of heat sinks sitting around with all the proper thermal/mounting gear. as i stated its only ideal when you have the parts for a mass of projects \$\endgroup\$ – Darksun96 Feb 18 '15 at 16:07

Not the answer you're looking for? Browse other questions tagged or ask your own question.