In this post we investigate how to connect popular voltage regulator ICs such as 7812, 7805 in parallel for acquiring high current output from the ICs.
Voltage regulator chips mostly have their maximum current output specs fixed to some predetermined levels. Increasing them to a higher level would normally call for external out board transistors and complicated associated circuitry which might be difficult to configure for the new hobbyists. Connecting a few of them in parallel, possibly solves the problem.
The idea was requested by Mr.Raja.
Technical Specifications
Sir,
could i use three L 7815 voltage regulator ic in parallel, to get 15 volt 4 amps dc current from about 20 volt 5 amp dc source?
Sir, as LM 338 and their equivalent ic s ( which gives 5 amps ) are not available in my town. I planned to use three 7815 in parallel. Is my idea works? If so please help me.
How can i connect them in parallel ? Could i connect input of all three 7815 ic by a common wire or i should separate mutually by a diode of 2 amp? And what about out put, should i separate them or use
a common wire? And i think, i can connect the negative terminal of ic with a common wire. Is it? Please guide me.
Solving the Circuit Request
Although not recommended by many, the issue can be handled simply by connecting the regulators in parallel, as shown in the following diagram.
Here we can see the terminals of the all the three ICs connected in parallel except the output pins which are terminated with individual diodes.
However the above connection might face a crucial drawback. Since all the ICs wouldn't have precisely identical characteristics and specs could vary with their current limits, and ultimately lead to one of them supplying greater amount of current than the other, and overheating in the course.
Although this won't pose a threat to the ICs as these are always thermally protected from inside, it's never a good idea to have a semiconductor device sizzle unnecessarily.
The issue can be very easily tackled by connecting the counterparts over a common heatsink, as shown in the diagram below.
Since the tab for the ICs connect with their identical common leads (ground lead), doesn't need any sort of isolation in the form of mica isolation kit etc.
Just get them on-board over a common aluminium plate and now you can relax as the heat dissipation across the plate would result in correct transition of heat enabling each one with an equal share of current at their respective outputs which in return would result in an optimally combined higher current outputs, as required.
Circuit Schematic
Kesava Raj says
Thank u sir….
Kesava Raj says
For mobile charging input voltage of 7805 I'm uing 6v 4.5ah battery …it will be okay sir….
Swagatam says
yes it will do!
Swagatam says
adding 5408 diodes will drop 0.6V from the 7805 which could create problems in charging. instead you could try adding an outboard transistor to raise the current as shown here
https://homemade-circuits.com/2012/05/dc-to-dc-double-cell-phone-charger.html
Kesava Raj says
Hai sir…
For smartphone mobile charging….shall i use 2 7805 in parallel….o/p using two 3 amps diode 5408….it will provide output 5v 2 amps sir….
Kesava Raj says
Thank u sir….
But i replace it new one sir ….stil same problem….when i disconnect the relay ..7812 o/p ..11.9 showing correctly….when i connect in relay it will show 11.7….when i activate the 7805 ic….the 7812 will varying voltage automatically…..till now i can't able to find solution….now I'm using zener as a regulator….pls make it any solution….sir
Swagatam says
Kesava, connect the IC 555 with 7812 and check the response….7805 is actually not needed.
Kesava Raj says
Hai sir…i have 3 doubts…
1.I use 7812 and 7805 both working good…7805 is using for 555…
7812 is using for Relay….
Sometimes relay working good…Sometimes not working when i check o/p volt of 7812 it showing 1.2…
What i want do….sir…
Pls help me sir
Swagatam says
Kesava, your 7812 could be faulty or not connected properly, check its connections or simply change it with a new good quality IC
Kesava Raj says
Hai sir….
1.Using 3 7812 it will provide 3 amps
2.3 ampere or 6 ampere Diode we want to use…
3. I need output 12v 5 amps..shall i use 5 ics…..
Pls help me sir
Swagatam says
for 3 amp you must use 6amp diode (6A4), for 5 amp output you may have to incorporate 6 to 7nos of 7812 in parallel.
Kesava Raj says
Hai genius…
The 3 7812 using…
It will provide 3 ampere or not
Swagatam says
Hi Kesava, yes it will allow you to get 3 amps max
Kaustubh Vaidya says
Hello Sir,
output of TWO different LM7812 can be connected in series, so that 24vdc output is possible? If that possible how do I connect?
Swagatam says
hello kaustubh,
that's not possible, you can try adding a 12V zener in series with the ground terminal of the IC and a 1K resistor between output and ground pins to achieve the same…or simply replace the 7812 with a 7824
Jusi Bayani says
Sir, the maximum input voltage of LM78XX series is 35v. In order to handle higher input voltage, say 80vdc/10amps, what circuit would you recommend using a simple transistor design?
Swagatam says
Jusi, presently I have the following design which could be modified for getting the specified outputs:
https://homemade-circuits.com/2012/01/how-to-make-versatile-variable-voltage.html
Aveface says
I'm curious to know if the spacing of, or rather the space between, the devices is crucial? Also, what's the draw of the devices themselves? By that I mean, Is there substantial current draw with not no load on the regulated volatge?
Swagatam says
Placing them closely would help more uniform heat sharing so it's recommended. Current consumption of each IC could be around 3mA
musa kizito says
Sir, is it possible to regulate a 12v 65Ah car battery to 6v dc and still have close to the 56Ah at the output by connecting 7806 regulators in parallel. If yes, then how many pieces of the regulators will be used. Thanks.
Swagatam says
Yes it would be possible but you will have to use as many as 60 nos of them, instead you can use 10nos LM196 and connect then in parallel by following the conditions provided in the above article.
VIJAYAKUMAR BURRI says
sir
here i am looking for solar inverter,but without battery for day time purposes ,no back up required, kindly give me some idea about this circuit
Swagatam says
Hi Vijayakumar,
What kind of load do you want to operate? please specify the load type and power, I'll try to help you with the right circuit.
VIJAYAKUMAR BURRI says
thank u sir
here i am asking about residential need purposes ,like fan,light,tv.
kind regards
vijay
Swagatam says
Hi Vijay,
I'll try to design it and post it soon in this blog.
Anonymous says
This configuration may decrease voltage by 0.5-0.7 volts.
The gnd of all the regulators be raised by employing a 1N400x diode. Do you agree??
Swagatam says
It may be done for 7812, but for 7815 as proposed above the issue won't matter much.
Anonymous says
I have 1 amp rated transformer, re-wind the secondary with thicker(0.4mm to 1mm) wire then i get more amps? If i need more current than volt, what changes should be done with 1amp xfmr?
Swagatam says
current can be increased by increasing the thickness of the secondary winding.
To decrease voltage reduce the number of turns, and to increase current use thicker wire.
mfkrzeminski says
It is actually not that simple. Thicker secondary will indeed allow greater current capacity, will lower the heat losses becaue of lower resistance but to actually get more juice out of that transformer the primary might need to be rewound; please take note that the current in the primary will increase accordingly. Also the limited crossection of the core might not allow more flux between the windings. If it's a classic E-I laminated core it also performs heatsinking function to some extent especially for primary that is deep burier.
J-boy says
Please is it possible to produce a joule thief circuit that can use a 4×1.5v battery in parallel to produce 1.5v and 6amp and upconvert it to 12v? Please if possible how can i get the circuit?
Swagatam says
Yes it is possible but it won't produce 6amps.
Rather its capacity will increase to 6 AH, AH refers to charge or discharge rate in terms of current in an hours time, it's does not suggest the current delivering capacity.