5V - 2A Regulator

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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.

Supplies:

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....

2 People Made This Project!

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22 Discussions

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CharlesA126

Question 6 months ago on Introduction

This is dc-dc. Can is do it ac-dc regulated power supply?

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JohlyJ

7 months ago on Step 1

How many volts of the capacitor

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SaifA74

1 year ago

i am made smart pawerbank 5v 2m

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SaifA74

1 year ago

sir send me this vedio my watsup no......7210537221 ..... 5v 2m ...i hope you send .

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KentC16

1 year ago

sir what can i use as an alternative for LM78S05? Thanks hope you reply

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zachw44

1 year ago

Okay so I have a Aukey 5v 21Watt solar panel.
Problem-
Under full load the output voltage drops significantly, even while under full sun.
Solution-
Figure out how to regulate the voltage to within .1 volts.
I need to be able to boost 3-4v to 5v while also being able to pass through 5v when it's able to do so.
the panel has two 5v 2a USB 2.0 outputs.

suggestions?

1 reply
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sooraj619zachw44

Reply 1 year ago

this circuit works only with minimum 9V input and it cannot be used for boosing 3-4 v to 5V

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ketan11293

2 years ago

can i use to regulate current from two 18650 batteries to 5 v ?

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saidimed

3 years ago

I have a 6v 2A battery and i want to supply a raspberry pi can this regulator give me a 5v output?

1 reply
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baffourisaidimed

Reply 2 years ago

no , the input must be 7.5 volt or more

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viktor.bozicevic.9

2 years ago

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?

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verence

3 years ago on Introduction

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.

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 (lower than 5V, higher than 5V, about 5V with a lot of high frequency on top, it may change when you tap it with an oscilloscope probe...).

Add the two ceramics (100nF to 330nF) and you'll be fine.

6 replies
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verenceMuhammad BilalA

Reply 3 years ago

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!

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Muhammad BilalAverence

Reply 3 years ago

and the current supplied will be equal to current output ???

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verenceMuhammad BilalA

Reply 3 years ago

Yes. The input current is equal to the output current (plus the very small current the regulator needs for itself - but that is only µA to mA)

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Muhammad BilalAverence

Reply 3 years ago

i need to amplify the current output of a solar panel cud u plz help me ?

the output from it is 24 V and 0.1 A i need to increase the current up to 1A

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verenceMuhammad BilalA

Reply 3 years ago

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.