Introduction: Increasing Current on 78xx Series Regulators
Typically 78xx series regulators have a maximum load current capacity of 1 to 1.5 Amperes. Using this design you can double the maximum current of your 78xx regulator. This design was posted on the Net by I Hakki Cavdar of Karadeniz Technical University, Trabzon, Turkey. I have revised some of the components values due to heating concerns and to suit my intended application. Picture #2 is the schematic diagram.
Step 1: Preparing the Components & Circuit Board
Component List:
IC1 and IC2 - 78xx series regulator IC ( 7805 for 5V, 7812 for 12V etc.)
D1,D2 & D3- 1N4003 ( 3 Amp Diodes )
D4 & D5 - Light Emitting Diodes (LED)**
R1 & R2 - 4.7 K , 1/2 watt resistor **
C1 & C2 - 4700 uF / 16V electrolytic capacitor
C3 - 47,000 uF / 35V electrolytic capacitor
Printed Circuit Board ( PCB )
Etching Solution
WaterProof marker
** - optional components
Cut the Circuit Board using a hacksaw, click the picture for a better view. Using the waterproof marker, draw this to the copper side of the circuit board- copy the RED diagram. Take note of the pin distances of the components so that placing them afterwards will be a breeze. Put the PCB in the Etching solution and wait until you'll see the copperless plate ( around 20 minutes ). Rinse the PCB with water. Clean the Marker Ink with Acetone to expose the copper. Drill the holes for the components and your PCB is ready to go.
Step 2: Drawing in the Copper Clad Board
Draw the circuit pattern on the copper side using a waterproof marker. The other picture is what it looks like in the other side.
Step 3: Etching
After proof reading your drawing,soak it in the Etching Solution.I am using Ferric Chloride to do it.
Step 4: After Etching
The copper drawn with the marker remains. Clean it with Acetone to get rid of the marker ink and expose the copper.
Step 5: Drilling
Drill the component holes and your done with the PCB.
Step 6: Placing and Soldering the Components
In placing the components, I always place the resistors first, in this case R1 and R2. Next are the capacitors C1, C2 and C3, please always check the polarities of their pins (you can check this by reading the plastic covering of the capacitor, usually its there) to avoid your capacitor blowing up. You might not want a hot liquid and lots of paper shreds in you face. Next, is to insert the LEDs D4 and D5, again take note of their pin polarities (anode and cathode), this won't blow if polarities are not correct, only that it won't light up. Lastly insert diodes D1,D2,D3 and the 2 regulators.
Once all the components are in place, double check their polarities again and your ready. Place the PCB upside down exposing the copper side with the component pins. In my experience, its better to solder the components first before cutting the excess pins but some people I know are more comfortable cutting the pins first before soldering, so this one depends on your liking. Clean all excess pins protruding from the soldered area and your ready for testing your project.
Step 7: Testing and Other Mods
This circuit is very easy to test, just connect a power supply to the input at C1. Note that input voltages should be higher than your desired output.For example, if you want a 12V output, your input voltage should be like 16 Volts or higher -- the 78xx regulator can handle input voltages up to 35V. If everything goes well, your 2 LEDs will light up. If it doesn't, check if there are any voltage coming from your output with a Multimeter then check the LED's pins. The output of this circuit is dependent on your 78xx series regulator, say you connected a 7812 regulator, output should be in the range of 11.3 to 11.5 Volts. I have added sufficient heatsink to prevent overheating of the regulator. I connected this to my wireless router and stayed stable after powering it up for 2 days straight. I found a small CPU fan and added it to reduce heat even further, although its not necessary, might as well make use of it.
Be the First to Share
Recommendations
Make It Bridge
For the Home Contest
Big and Small Contest
45 Comments
Question 4 years ago on Introduction
47,000 uf cap? Is that a super cap or something.
From the sizes on the picture I assume both cap rating have an extra 0 on each? Please clarify. Thanks.
5 years ago
Interesting. I was considering a similar approach, but read just about everywhere it was a bad idea as both chips were not balanced. I see someone else brought up the same concern and got a reply that spoke of balancing by the developed heat. Not sure if that would work..
As this is from 2006, would be interesting to hear the experience of the Author.
Having said that, there are now (11 years later) cheap LM2596 buck converters that can deliver 3 amps (if cooled), so that might be a better option
5 years ago
How did you choose the PCB? I mean how much current can it withstand? how to choose the proper one?
6 years ago
Sorry my BAD(:
Thank you good brother, BTW if i connect the 78xx regulator in series, then i will get more voltage YES...
6 years ago
excuse me, how can i increase the maximum input voltage ?
in your case, the maximum input voltage is 35v.
how can i increase it to 60v ?
7 years ago on Introduction
How do you ensure current is pulled evenly from both regulators? I've heard parralel regulators is a bad idea because slight defects can cause one of the regulators to supply all the current. Are the diodes doing this?
Reply 7 years ago
It's not only the diodes. Mounting both regulators to one big heatplate is also of importance. Once one regulator is overcharged, it will get too hot and heat up the other one too so that the output is regulated by the negative heat coefficient both regulators have.
the capacitors are preventing a swinging in the voltage (however, I don't know why they have to be that high in capacity, to my mind double the capacity of the datasheet should work well enoguh)
the diodes prevent the current from one regulator to leak back to the other. Max current per regulator is about 1.8A, so you really need AT LEAST 2A diodes. But they are really cheap (0.1USD per piece), so just get 3A ones. 50V is enough, but there are also 100V and 400V diodes i think.
8 years ago
Really nice and useful project. I'll be using it in my 8*8*8 LED cube for a huge current.! ;)
8 years ago on Introduction
This is a great instructable for increasing the output current of a 7805. The 78XX series is simple to wire up for novice/beginner electrobeginners (like me), one is positive, one is ground, the other is output. Sometimes, I must have that an item can handle the extra amperage. 7805 limitations if i remember right are at 1 amp. Having it at between 2 and 3 amps is even better. Now if there was a way to do 10 amps, lol.
8 years ago on Introduction
Great idea. I have a whole bunch of 7805s in my bits box, but at the moment need a PSU with 2A or more current. I will definitely make this.
9 years ago on Introduction
Interesting instructable. I just have one revision to your instructable and that is that the 1N4003 is not a 3A diode but is a 1A diode. This is true for the 1N4001 - 1N4007 with their only difference being that they can handle higher voltages.
9 years ago on Introduction
Hello Everyone, It is I, Matt from 5 years in the future. Old thread but google brought me here, I don't know why I didn't try here before google. But the web is filled with DON'T DO THIS!!!, sorry web, I just did, 5 LM7805's 3.5A vary warm but not burning yet (3 hours). I found another link for the list http://www.reuk.co.uk/High-Current-Voltage-Regulation.htm and as explained above with the resistors on the output, and this neat trick, both on that page. Not enough 1 ohm resistors on hand but I do say, this will work (at least for 3 hours).
My only hiccup was: My case is ground and ground is my case, my case is also my heatsink, so... metal on metal, I loose the 0.7 volts :( but [knock on wood] my 2 pi's 2 powered hubs, and usb 3.0 hard drive all run ok with 4.3 volts.
13 years ago on Introduction
D1 & D2 are dropping 0.7 volts. So you get output voltage that is less by 0.7 V than the desired value. The diode also has feeble internal resistance, which makes it drop more voltage when higher amount of current is drawn. I suggest the use of schottky diodes with very small internal resistance.
Reply 9 years ago on Introduction
D1 & D2 are dropping 0.7v yes...but is that why the designer included D3 which, as I understand it, would elevate the regulator voltage by 0.7V to compensate ?
Reply 9 years ago on Introduction
Yes, D3 is for compensation.
It should drop the same as D1 and D2.
Reply 13 years ago on Introduction
Hi i have some of these shottky Diodes from old computer psu do you know a simple circuit for them so i can learn more about them it looks like they have two ac in and one dc out i am thinking its for Transformers with Paired Windings but i am not sure but i tryed to look where people used Schottky i not found anything. anyway thanks for any input you could give me on them in advanced.
Reply 13 years ago on Introduction
Schottky diode has two terminals like an ordinary diode. What you have is most probably a bridge consisting of 4 diodes. Bridge produces full wave rectified output from an AC input.
Reply 13 years ago on Introduction
no this is three its in a Mosfet case as i said its out of a Computer psu here is the model number s30sc3m MTO-3P is the case and the pin 1 has a symbol of a diode that points to the middle pin 2 pin three has the same but facing the middle as far as i know they can have really high amp's like 50amp's hence i thought you mean this the 4 pin bridge diodes i have used for full wave before but these 3 pins is the ones i am not sure how to use.
Reply 11 years ago on Introduction
You don't design a circuit around the fact of having a diode, you use the diode when you have a circuit you wanted to build that needs one.
What you have is two schottky diodes in a single casing arranged in parallel for higher current and isolation of the two inputs. They are usually arranged with two anodes (input) and one cathode (output).
It is not necessarily for transformers with paired windings, it's for any use where you would have two schottky diodes in parallel in a circuit and need the high current capability (but low peak voltage) they can withstand. Frankly they are most suited for switching power supplies with an output under 20VDC, in most other circuits you can find a better diode for the purpose whether it be smaller or cheaper or capable of higher voltage.
To learn more, get the part numbers off of them and look up their datasheets online.
Reply 11 years ago on Introduction
the circuit i wanted back then in 2009 was for testing you know learning.
i found many circuits to use these diodes and for anyone else interested in using psu schottky diodes yes they are very high current.
Example. Car amps use in a home.
nearly all car amps need about 30amps+ any person who has tried to find transformers for something like knows that its not cheap fact is its cheaper to buy a nice house amp for the price of a high current DC tranformer for 12volts
these diodes work great. as AC-DC said there in parallel and work fine like that.
secondly for high current you cant find a better diode for high Current. especially for the price, most people can get diodes rated at 10 to 40amps and for many people that is pushing it.Thus Schottky Diodes work fine for 50/60hz AC to DC with some caps and inductors to make a nice clean DC you have a good high current PSU and one big Transformer. The Datasheets show they work well with 50/60hz.