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This is a regulator circuit using LM78S05 IC. it is a 5v regulator IC with 2A current output . this circuit is very useful for those who need more current output . this is an easy circuit and consists of very few components.

Step 1: Parts List

1. 1 x LM78S05 IC

2. 1 x 220uf

3. 1 x 47uf

4. 1 x 1N5400

5. 1 x dot matrix board

Step 2: Circuit Diagram

Complete the circuit according to the circuit diagram and solder them according to the circuit diagram. use a heat sink for the IC use a large heatsink than the heatsink shown in picture since the heat was too high when I used this heatsink

Step 3: Conclusion

hope this circuit gets useful to you guys... give me feedbacks and rate this instructable anplease rate this instructable....

<p>I think we over do it, but why waste such a nice box ?</p>
<p>can i use to regulate current from two 18650 batteries to 5 v ?</p>
<p>I have a 6v 2A battery and i want to supply a raspberry pi can this regulator give me a 5v output?</p>
<p>no , the input must be 7.5 volt or more </p>
<p>Sir, </p><p>Can i use IC7805 here?</p>
<p>Hi, I made this circuit in the form of a USB charger that is powered with a 3S LiPo battery. Instead of 1N5400 diode I used 1N5408 diode (1000V max.), since I could only find it in my local electronics store, and I soldered the output + and - terminals to corresponding female USB socket. When I plug in my mobile phone, after a couple of secs the regulator gets very hot to touch, although phone draws 850mA max. when charging. Could you please help me how should I eliminate an overheating issue? Should I perhaps add ceramic caps in parallel with both input and output sides?</p>
Thanks for this! Was able to assemble from schematic and it powers my Pi2 with Edimax wifi great.
I'm using a 9v power supply (for guitar pedals) by the way
<p>This circuit will work perfectly in 99.9% of use cases. And will fail with the most unpredictable effects in the rest. Put two 100nF ceramic capacitors between input and ground resp. ground and output. Put them as close as possible to the IC. </p><p>As I said, otherwise the thing will work very nicely MOST of the time. With MOST of loads. And it may start oscillating and producing any kind of output voltage in other cases (<em>lower </em>than 5V, <strong><em>higher </em></strong>than 5V, about 5V with a lot of high frequency on top, it may change when you tap it with an oscilloscope probe...). </p><p>Add the two ceramics (100nF to 330nF) and you'll be fine.</p>
<p>cud u plz tell where to add the 330 nf capacitor ?</p>
<p>One from pin 1 to pin 2 (in to com) and one from pin 2 to pin 3 (com to out). As close as possible to the IC!</p>
<p>and the current supplied will be equal to current output ???</p>
<p>Yes. The input current is equal to the output current (plus the very small current the regulator needs for itself - but that is only &micro;A to mA) </p>
<p>i need to amplify the current output of a solar panel cud u plz help me ?</p><p>the output from it is 24 V and 0.1 A i need to increase the current up to 1A </p>
<p>Put ten of those panels in parallel. Or use a switching power regulator. But to get ten times the current, you can only get a tenth of the voltage (1A @ 2.4V). And as nothing is perfect, there will be losses in the regulator.</p>
<p>I always like tight little circuits such as this one, minimal parts, maximum performance.</p>
<p>For high power output you may need a bit more heatsink. You might also want to consider a reverse protection diode on the output too. It protects the regulator from the discharge of the output capacitor. You don't put the diode in series, like you did with the input though. It goes in parallel with the regulator, but backwards. Like this:</p>

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