Hello there I am making RFID reader and I need 12 v to power up the reader I am using 12v wall adapter I need 12v to 5v dc converter . Currently I am using 7805 voltage regulator to give 5 v to Nodemcu and relay but I don't know is it efficient for a long use . I need my device to work for 24*7, I don't know 7805 is suitable for that could any one recommend what kind of converter should I use for this purpose.
If your 7805 doesn't get too warm it should be ok. Did you attach
a heatsink to it?
yes i did ,I am using T0220 heat sink for that I need to build a case for the project and i wish to use a closed one so i don't know 7805 will last that long or not
If it doesn't get too warm inside of your case it should be fine.
These 12V to 5V switching converters, providing 0.5A or 1A are inexpensive, very efficient, reliable and dissipate very little heat.
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The 7805 is an old, reliable and excellent friend many times. Know it can discipate max 1 Watt being naked. Using a heatsink, more is possible.
What is the current used out of it? That current times 5 is the Watts...
Can you post a link to the relay and any other parts that will be driven from the 5V line?
And what is the rating of the
Alternatively why dont you use a 5V rfid reader, and a 5V supply?
This is the relay Module and level shifter and the Nodemcu v3
Currently i am using Wiegand rfid reader it operates on 12v
A 7805 is not "efficient", electrically speaking, but it should be plenty reliable for 24x7 operation if it's not getting too warm (as others have said.) You'd want to look for one of the fancier switching regulators if overall power consumption was an issue (battery operation) , or if the power demands of your application were higher (which WOULD cause the 7805 to get warm.)
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i need less than 500 mA in 5v output it only required to drive nodemcu a single channel relay and level shifter. My other power consuming devices are in 12v and i am currently using 12v 2amp power adaptor. For 12v i need 500mA power
Sry if my power calculations are off . i am new to electronic stuff and currently learning
The wattage dissipated by the regulator is simply the voltage droped across the regulator times the current. So 12V - 5V = 7V. 500mA = 0.5A. 7 * .5 = 3.5 Watts your regulator will need to dissipate. You need to use a heat sink. Be sure to use the appropriate capacitors or it will oscillate and get real hot. Currently you can get Buck regulators for less then a buck fro china.
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That's a different story then the one I replied to. Designing small computers since 1989, using 7805, those things are well known.
I don't know how to use appropriate capacitor for 7805 currently i am currently using 0.33uf and 0.1uf
There is a big secret on sizing them, it is the data sheet. Be careful as not all parts are the same so look at the data sheet from the manufacturer for the part you have. I like to put a bulk 47uF cap on the input side of the regulator. Keep the leads as short as possible.
Did you mean by placing capacitor near to the 7805
I place SMD capacitors at the points the TO220 package entrees the board. Some 7805s get goofy if the lead length between them and the cap is over 1"/2.5cm. That is why if you take things apart with that regulator the caps are right next to it with reasonably heavy conductors.
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ok currently i am using THT ones but i keep the distance with in 2.5cm
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You have it!
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Just Gut Out a Car charger for Mobile Phones... 
A small board by Adafruit: