# Two 7805s on one source

This is a "made up" question in the sense it's not something I really need to know. It's not part of my daughter's school science project 8). But it's inspired by the 2x PSUs thread and the question on regulators thread....

Let's say I have a power supply at too high a voltage for my project (say 12VDC) and I decide to drop that to 5VDC with a 7805. But then I find, alas, that although my 12VDC supply can pump out the 2A my hypothetical project requires, the 7805 is only good for 1A.

So, can I put 2x 7805s in parallel, each supplying 1A @ 6V to my project for a 2A supply, and jointly drawing 12V @ 2A from my supply?

Would this work or is it a recipe for smoke release?

They won't share the load well. If you had a couple of ammeters you could see the result easily enough.

But you could divide the load: some stuff goes to "Reg#1" and other stuff goes to "Reg#2".

There is a discussion on that subject here

JimboZA: This is a "made up" question in the sense it's not something I really need to know. It's not part of my daughter's school science project 8). But it's inspired by the 2x PSUs thread and the question on regulators thread....

Let's say I have a power supply at too high a voltage for my project (say 12VDC) and I decide to drop that to 5VDC with a 7805. But then I find, alas, that although my 12VDC supply can pump out the 2A my hypothetical project requires, the 7805 is only good for 1A.

So, can I put 2x 7805s in parallel, each supplying 1A @ 6V to my project for a 2A supply, and jointly drawing 12V @ 2A from my supply?

Would this work or is it a recipe for smoke release?

You can do that provided you follow one or the other of the following:

1. Wire some of the components to use reg1 and others to use reg2 output, thus sharing the total load requirement between the two regulators. This however requires that you know the maximum current requirement for each different section you are powering and that either 'subload' requires less then 1 amp max.

2. Wire up two 1N4001 diodes, with one anode wired to reg1 and the other anode to reg2, and then wire the two diode cathodes together which becomes the shared common positive voltage to all the things requiring the voltage. The two regulators will effectively share the load when the total current draw is higher then one would have been able to handle. The downside is that the 'shared' output voltage will be less then +5vdc by a one forward voltage drop across a diode, so maybe 4.4 to 4.6 vdc output.

Lefty

Lefty's diode solution also works for multiple battery supplies, if you need more capacity or the ability to 'hot swap' a single battery set whilst your device continues running.

Where you do it with a regulator you could use a variable VR and set the VR output to be 5v PLUS the Vfd of the diode, say 5.7v total. That way you would have your 5 V common rail after the diodes.

Lefty option 2 sounds intriguing... would a couple of resistors do a similar thing having the resistors tied to give 1 output?

cjdelphi: Lefty option 2 sounds intriguing... would a couple of resistors do a similar thing having the resistors tied to give 1 output?

Diodes have a fairly constant voltage characteristic without allowing one regulator to back-feed the other - resistors are not going to do a great job here I think.

To be practical you'd use schottky diodes (forward voltage 0.4 to 0.5V max, not 1N4001's which have a forward voltage of 1.1V or so). These days schottky diodes are pretty much used for everything (unless very high voltage / high temperature, or if low reverse leakage matters). Schottky's have considerable reverse leakage at higher temperature...

MarkT:

cjdelphi: Lefty option 2 sounds intriguing... would a couple of resistors do a similar thing having the resistors tied to give 1 output?

Diodes have a fairly constant voltage characteristic without allowing one regulator to back-feed the other - resistors are not going to do a great job here I think.

To be practical you'd use schottky diodes (forward voltage 0.4 to 0.5V max, not 1N4001's which have a forward voltage of 1.1V or so). These days schottky diodes are pretty much used for everything (unless very high voltage / high temperature, or if low reverse leakage matters). Schottky's have considerable reverse leakage at higher temperature...

Agreed, schottky would be better. However adding a series diode in the ground pin line of the regulator chip is a very simple trick to 'compensate' for the output diode(s) voltage drop.

Lefty

Or one could up the 7805 to 7806....

But to be realistic all these tricks using linear regulator chips is kind of old hat, been done for decades, and still have the problem that linear regulators have, they suck because of wasting power via heat dissipation. In this day and age switching regulators is just the better solution unless your only goal is to try and use up your old spare parts supply. ;)

I would use these all the time over a typical linear regulator installation.

Lefty

The easiest solution is a .1 ohm resistor in each regulator output lead (called a ballast resistor). the second best has been mentioned but a diode cathode to ground and anode to the regulator ground will raise the output voltage by one diode drop. This will compensate for the series diode. There are also voltage regulators that are capable of more than 1 A. It is also possible to amplify the available current with a PNP 'wrapped' around the regulator. The benefit of this is that the regulator can both control the voltage and share in the load current. This can be done with a 78L05... Too. {Edit RKJ} But Lefty's idea is the best way to go.

Bob

So, can I put 2x 7805s in parallel, each supplying 1A @ 6V to my project for a 2A supply, and jointly drawing 12V @ 2A from my supply?

The load will draw what it wants (through the 7805) and you’re not really splitting the current here.

You can have (2) loads, each having it’s own 7805 regulator connected to the same unregulated source. But each load will draw whatever i wants. Just make sure they each load won’t draw more than the 7805 can safely supply.

Also, 78xx and 317/337 regulators are 1.5Amp rated, not 1Amp.

If you want more control over your output voltage, use an adjustable 317 with some trimmers, rather than a fixed voltage 7805 regulator.

Also, 78xx and 317/337 regulators are 1.5Amp rated, not 1Amp.

Well in my defense, see pic attached.

And since it’s a made up question, so I’ll just change the question to 3A 8)

Here’s a 1.5A part (assumption: adequate heatsinking required)

Now, for greater than 1.5Amp (3A, 5Amp), you’ll need external power transistors. Their datasheets offer example circuits.