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.
<p>How did you choose the PCB? I mean how much current can it withstand? how to choose the proper one?</p>
<p>Sorry my BAD(: </p><p>Thank you good brother, BTW if i connect the 78xx regulator in series, then i will get more voltage YES...</p>
<p>excuse me, how can i increase the maximum input voltage ?<br>in your case, the maximum input voltage is 35v.<br>how can i increase it to 60v ?</p>
<p>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?</p>
<p>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.</p><p>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)</p><p>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.</p>
Really nice and useful project. I'll be using it in my 8*8*8 LED cube for a huge current.! ;)
<p>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. </p>
<p>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.</p>
<p>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.</p>
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). <br> <br>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.
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.
D1 &amp; 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 ?
Yes, D3 is for compensation. <br>It should drop the same as D1 and D2.
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.
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.
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.
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.<br><br>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).<br><br>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.<br><br>To learn more, get the part numbers off of them and look up their datasheets online.
the circuit i wanted back then in 2009 was for testing you know learning. <br>i found many circuits to use these diodes and for anyone else interested in using psu schottky diodes yes they are very high current. <br>Example. Car amps use in a home. <br> <br>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 <br> <br>these diodes work great. as AC-DC said there in parallel and work fine like that. <br> <br>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.
If I were to use a car amp in a house (doubtful I would because I build my own amps), I'd use a switching computer PSU as that is the most cost effective way to get 30A+ @12V, then I'd probably throw an LC filter after it and liberally place film decoupling capacitors on the amp board to decrease the high frequency noise.<br><br>The primary problem with those diodes is their low voltage tolerance, for most applications when you get into high current you are also at a higher voltage than the typical schottky diodes found in consumer electronics. Car amps are just an exception since they are made to work within the 12V automotive electrical system. Otherwise, for amp PSU the more ideal diodes are &quot;soft recovery&quot; type to decrease power rail noise.
you can use UA7805MK regulator 5V ,5Amp. instead of this circuit <br>Check the Datasheet below: <br>http://www.alldatasheet.com/datasheet-pdf/pdf/130557/FAIRCHILD/UA7805.html
Unfortunately, you can't rely on this circuit to work every time. If the regulators are very well matched (same batch) it might work. If they have significantly different output voltages, one will volunteer more current than the other and you will find it shuts down well before the circuit gives the X2 maximum current. It is a perennial problem with regulated supplies, if you want to parallel them, you need some sort of active current sharing circuitry as well. The diode isolation circuit you've shown works well to provide X1 current with redundency if one regulator should fail, another common technique, but it can't reliably ensure load sharing between regulators.
Linear regulators have increasing forward voltage drop as current increases, if they are the same spec and the circuit is sound (no bad solder joints, low trace impedance, etc) they will come close enough to sharing current equally because they won't ever be THAT far off from each other in output... the voltage tolerances for a regulator aren't that loose.<br><br>You will not necessarily need an active current sharing circuit even if what you suggested were true. Simply putting a low ohm resistor on each output would suffice.
you mean the input so the Regulators do not go over there rated current. <br>running the Regulators @ 1AMP each by putting a Resistor on each input of each regulator to get the current you need best be a little under the rating of each regulator for cool operation will more then do the job the rest is fine. <br> <br>(for this exact project it should be ok as most wifi and routers dont go over 2amps.)
No I mean, look at the regulator datasheet... for example I refer to this one:<br>http://www.fairchildsemi.com/ds/LM%2FLM7805.pdf<br><br>Now suppose you need 1.5A output, so using only one of them would not work. If you place two in parallel without a resistor on the input it could still work.<br><br>On the datasheet, see page 21 figure 5. If one regulator is handling more current it will get hotter than the other (assuming same heatsinking) and the normalized output voltage will start dropping a little, not a lot but remember these are regulators with a fair amount of precision so they won't be too far apart in output in the first place in the same circuit/conditions. That drop brings them a little closer to equal current.<br><br>Now look at the dropout voltage spec on page 12. 2.0V at 1A. If one regulator starts handling more current, the forward voltage drop rises from a little under 1.0V towards 2.0V, so the other regulator starts using more and more current till they reach a near equilibrium. Not &quot;near&quot; by precision measurement standards, but near enough if you leave some margin in the circuit (not trying to get 2.0A from 2 x 1A regulators but rather a more conservative amount of current like 1.5A.<br><br>It is true that most (consumer grade and small office equipment) wifi and routers don't consume over 2A, but I would not use a linear regulator for this purpose because of the heat level, the wasted power, and the availability of low cost ready made switching PSU. Plus, there might be another option. Many wifi/routers don't even need regulated power, right after the power jack on the board the next stage is a power regulation circuit that drops the input voltage down to 5V or lower and all your external PSU needs to do is stay below the voltage rating of the capacitors on that router regulation subcircuit at the lightest load the router causes. Of course the builder would have to examine the router PCB and have a basic familiarity with regulation subcircuits to identify the presence of one instead of just assuming it, but it is almost always the case that any router with over 5.0V power input does have this subcircuit regulating down to one or more lower voltages.<br><br><br>
Normally rather than using a second 7812, you would use a power bipolar transistor like the 2N3055A (NPN) <a rel="nofollow" href="http://www.onsemi.com/PowerSolutions/product.do?id=2N3055A">http://www.onsemi.com/PowerSolutions/product.do?id=2N3055A</a> and 2N2955 (PNP) and related. <br/><br/>See Figure 2 at &lt;http://sound.westhost.com/project77.htm&gt;<br/><br/>&quot;LM317T Voltage Regulator with Pass Transistor&quot; at &lt;http://www.qsl.net/yo5ofh/hobby%20circuits/power_sourse_circuits.htm&gt; <br/><br/><a rel="nofollow" href="http://www.users.qwest.net/%7Eptaylor/Electronics/DC-PCsupply/PC-DC-AT-supply.jpg">http://www.users.qwest.net/%7Eptaylor/Electronics/DC-PCsupply/PC-DC-AT-supply.jpg</a><br/><br/><a rel="nofollow" href="http://www.zen22142.zen.co.uk/Circuits/Power/1230psu.htm">http://www.zen22142.zen.co.uk/Circuits/Power/1230psu.htm</a><br/>
Yes, those are alternative circuits if you really want high current outputs.
colinbeeforth is correct, the parallel regulators circuit is an <strong>hack</strong> (hey it's neat, but it has caveats), and while it may of worked for you, you cannot be certain that all individual 78xx regulators will perform identically. <br/><br/>Using a series pass transistor is no more complicated that wiring up this parallel regulator circuit. See <strong>High Current Voltage Regulator</strong> sample schematic on page 13 of the LM340 / LM78xx datasheet from National. That circuit only needs 5 components versus 7 required (<em>not counting either LED &amp; their current limiting resistors</em>) in your circuit. <br/><br/>You can substitute the pass transistor with a different NPN transistor with adequate current and power ratings. For 2-3A possible easy to find transistors would include 2N3055, 2N3771, 2N3773, 2N5301, TIP41, TIP120, and many other NPN power transistors.<br/>
It is true it is a hack but you don't need to worry much about identical performance regulators unless you are really pushing the design limits instead of conservative engineering like designing for less than 70% of expected average current. The reason is that as output current rises the voltage drop across the regulator increases so they will essentially equalize out close enough in current per each regulator.<br><br>I agree a regulator plus pass transistor is the best option IF someone needs a simple linear regulated PSU instead of a switching PSU.
Oops, that should be <a rel="nofollow" href="http://www.national.com/ds/LM/LM340.pdf">LM340 / 78xx datasheet</a>.<br/><br/>
Just remember that whatever voltage you "Lose" in a 78xx Reg will come out as heat. E.G. Say you pop 12V through a 7805. 12v IN; 5v OUT; 7V disipates as heat. The 7V will turn the 7805 Regulator into a toaster. A rather large heatsink is mandatory and I have found that one twice the presumed size is a minimum. I'll upload a nifty project I did using the above example, so you can have some fun. BTW, makes a nice hand warmer on a cold day. This is my 1st post here, so in advance, sorry if I pissed anyone off. That is/was not my intent. :p
Actually that is not quite correct. People usually use linear regulators for low current applications so 7V isn't necessarily much of a drop, won't necessarliy heat up the regulator much. Yes it is too much if someone is trying to parallel two of them to get 3A out, but it is a bad design in the first place to use 2 x linear regulators for a 3A load, very wasteful of power and then you need additional materials, size and weight for the heatsink and maybe a fan too.<br><br>There's not much point to making a PSU like this though except for certain boutique applications that need very clean power like an analog audio amp. Otherwise you can pick up a 3A switching PSU off eBay/etc for very little money that is ready to use, lighter, less heat and energy waste, etc. In fact it will probably cost less than an empty project case to put the homemade PSU in!
It is not the 7 volts that dissipates as heat, it is the <em>power </em> (Watts) that is converting into &quot;waste&quot; heat. <br/><br/>So that is 7V times the current draw (up to 1A for TO-220 package or 0,1 A for the TO-92 package), which is 7W or 0.7W depending on which 7805 regulator you are using.<br/><br/>At 7 watts you would certainly need to attach a heatsink for the supply to be reliable and stable.<br/>
Indeed I do hear where you are coming from. Of course I understand the relationship between Voltage Amperage and Wattage.<br/><br/>My point here was that the differential of 7 Volts goes somewhere and that somewhere... is in heat.<br/>I acknowledge that heat is of course noted in terms of watts and Volts x Amps = Watts.<br/><br/>I happen to use 220's all the time, so I did not differentiate the power handling of a 92.<br/><br/>Merry Hannukah and Happy Christmas !<br/>
No Harm done, opinions are welcome here.
Hi all i posted about schotkey diodes if anyone needs to know how they work just pm me i have alot of info about them and they can be used as a normal diode or as a fullwave bridge just a little tricky to place on pcb, regarding this project of the LM78XX its great but if you need more current its best to use a PNP Transistor as a Current pass instead of using two LM78XX and this config the Current is only limited by the Transistors and its Voltage source.
If you use a shallow glass dish (like a cassarole dish) for etching your PCB, you can shine a flashlight up through the bottom to check the progress of the etching easily.
woow cool trick :P<br /> gonna remeber that one :P<br /> maybe i'll make a box with a glass plate and brithe white leds :P
Hakkı abi, bu projeye destek için çok sağ ol. Derin Erkan
anyone know a way of putting 7812's to make about 4A output? I have a LCD I need to make a PSU for. Also a "Sharpie" permanent marker is good as an etch-resist pen, we get them in canada at walmart, though I have recently found a pack of 5 at the dollar-store (I will not let you know how much they cost though).
You could simply use a LM338 which can handle up to 5A continuous output with adequate heatsinking.<br/><br/>See <a rel="nofollow" href="http://www.national.com/mpf/LM/LM338.html">LM338 from National</a> for datasheet and example schematic. <br/><br/>Or you could use a 7812 with a series pass transistor that can handle the higher current. Examples are given in any manufacturers' datasheet. <br/><br/>See the <strong>High Current Voltage Regulator</strong> example schematic on page 13 of the National's <a rel="nofollow" href="http://www.national.com/ds/LM/LM340.pdf">LM340 / LM7812T datasheet</a>. <br/><br/>For a 4A power supply I would suggest using a single 2N3771, 2N5301, or 2N5302 for the pass transistor with a large heatsink attached.<br/> <br/>Personally I don't recommend using this parallel 7812 regulator <strong>trick</strong>, as the 7812's have an risk of thermal run-away, if the current handling is not evenly shared between them. It may work, but it is a hack, and may not work with all 7812s, and has a higher risk of failure than using a just as easy alternative solution. <br/>
Thanks, I will look into them
Nice one

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