Is it possible (and advisable) to connect two or more 7805 regulators in parallel to obtain an increased current output?

I found several sources saying that it was not recommended, due to imbalances caused by small differences between devices. But then I found this circuit:

​

And since I found this circuit in the instructables website (which is not exactly known for being a infallible source of technical wisdom) I was wondering if anyone here could clarify things for me.

 

My National Semiconductor Voltage Regulator Handbook, 1982, shows plenty of circuits with 0.1 ohm on the output of several LM338 regulators. If NS says it can be done with resistors, it can be done.

But I'm not buying that diode set-up from Instructables until I do some math.

Throw some numbers at me about current and diode numbers, and I'll play with them.

 

Ugh. The voltage drop across D3 is not going to be the same as the voltage drops across D1 and D2, and they aren't going to be identical to one another, either. As a result, both output voltage accuracy and equality of current sharing between the two regulators are going to be only approximate.

 

I guess OBWO did that for me.

 

Ugh. The voltage drop across D3 is not going to be the same as the voltage drops across D1 and D2, and they aren't going to be identical to one another, either. As a result, both output voltage accuracy and equality of current sharing between the two regulators are going to be only approximate.

So what would you recommend? Simply put a 0.1 ohm resistor at the output of each 7805, as #12 suggested?

 

you could rather use a larger pass transistor and for instance a TL431 isnt there a circuit shown with pass transistor in some of the datasheets?

 

you could rather use a larger pass transistor

When you do that, you lose the self limiting/self protecting features of the regulator chip.
The first time a new-grad offered me a 3 pin regulator design, I cranked the load up to, "self protect" and the rectifiers exploded.
Naughty new-grad! You can't violate the specs on that model just because you don't understand the circuit.

ps, National Semiconductor suggested resistors. I just pointed to that fact and named the book.

 

So what would you recommend? Simply put a 0.1 ohm resistor at the output of each 7805, as #12 suggested?

Yup!

 

Yup!

Interesting ... so essentially, the 7805 in parallel behave similar to BJT connected in parallel? To counteract the "thermal runaway" phenomenon, right?

​

I cranked the load up to, "self protect"

You sure you didn't crank it up to "self destruct" instead?

​

 

Interesting ... so essentially, the 7805 in parallel behave similar to BJT connected in parallel? To counteract the "thermal runaway" phenomenon, right?

No, it doesn't have anything to do with thermal runaway; 7805's have internal current limiting, thermal shut down and safe operating area protection so they're pretty robust. The resistors on the 7805 outputs are there to enforce [approximately] equal current sharing.

 

No, it doesn't have anything to do with thermal runaway; 7805's have internal current limiting, thermal shut down and safe operating area protection so they're pretty robust. The resistors on the 7805 outputs are there to enforce [approximately] equal current sharing.

Ok then, so here's what I think I more or less understand until now:

The transistors are used because if the first 7805 is delivering more current than the other, then this current has to go through the resistor at its output. And since this resistor will dissipate more power than the other one (connected to the second 7805) then its resistance will also increase due to its increase in temperature. And this in turn will restrict the current flowing through the resistor, until both resistors are dissipating more or less the same amount of power.

This is a method to roughly equalize the current being delivered by both 7805's ... am I more or less right? or am I so off the mark that I'm not even wrong?

 

Is it possible (and advisable) to connect two or more 7805 regulators in parallel to obtain an increased current output?

I found several sources saying that it was not recommended, due to imbalances caused by small differences between devices. But then I found this circuit:

View attachment 113758​

And since I found this circuit in the instructables website (which is not exactly known for being a infallible source of technical wisdom) I was wondering if anyone here could clarify things for me.

Forget the diodes and select two regulators that have the same output under the expected load.

 

Ok then, so here's what I think I more or less understand until now:

The transistors are used because if the first 7805 is delivering more current than the other, then this current has to go through the resistor at its output. And since this resistor will dissipate more power than the other one (connected to the second 7805) then its resistance will also increase due to its increase in temperature. And this in turn will restrict the current flowing through the resistor, until both resistors are dissipating more or less the same amount of power.

This is a method to roughly equalize the current being delivered by both 7805's ... am I more or less right? or am I so off the mark that I'm not even wrong?

You're correct about the aim of this scheme-- to approximately equalize the currents being delivered by the two regulators-- but totally out in left field regarding how this is achieved. It has absolutely nothing whatsoever to do with resistor power dissipation, or changes in resistance with temperature (most have very low tempcos, BTW).

You've made it WAYYYYY more complicated than it really is. To see what's going on, try this simple back-of-the-envelope analysis:

Let's say you have two 7805's with slightly different output voltages, such as 4.98 volts and 5.01 volts. For analysis, model them as simple voltage sources with their negative ends grounded. Now connect their positive ends with a pair of 0.1 ohm resistors wired in series. And finally, connect a 5 ohm resistor between the junction of the two 0.1 ohm resistors and ground (to imitate a nominally 1 amp load).

Solve for voltages and currents, and you'll see how the sharing works.

 

When you do that, you lose the self limiting/self protecting features of the regulator chip.
The first time a new-grad offered me a 3 pin regulator design, I cranked the load up to, "self protect" and the rectifiers exploded.
Naughty new-grad! You can't violate the specs on that model just because you don't understand the circuit.

ps, National Semiconductor suggested resistors. I just pointed to that fact and named the book.

never heard of a plain old fuse perhaps.

 

You sure you didn't crank it up to "self destruct" instead?

You're lacking background information. The New-grad was competing with an existing design based on the LM723 foldback circuit. You can't blow them up under any condition I know of. The new-grad put the helper transistors on the 3 pin circuit and thus, forfeited the self-protection feature.

 

never heard of a plain old fuse perhaps.

Never heard of a foldback circuit. (LM723)
Never figured out that adding extra pass transistors would forfeit the self protecting features of the 3 pin regulators.

For his next act, the new-grad submitted a design for a torroid power transformer where the windings couldn't fit inside the doughnut, and got paid more than I did (because he had a college degree). I quit like a rat leaving a sinking ship for a job working on Lasers at a 43% pay increase. After the owner was replaced by an M.B.A. and I was replaced by a B.S.E.E. the company went out of business within a year.

 

You're correct about the aim of this scheme-- to approximately equalize the currents being delivered by the two regulators-- but totally out in left field regarding how this is achieved. It has absolutely nothing whatsoever to do with resistor power dissipation, or changes in resistance with temperature (most have very low tempcos, BTW).

You've made it WAYYYYY more complicated than it really is. To see what's going on, try this simple back-of-the-envelope analysis:

Let's say you have two 7805's with slightly different output voltages, such as 4.98 volts and 5.01 volts. For analysis, model them as simple voltage sources with their negative ends grounded. Now connect their positive ends with a pair of 0.1 ohm resistors wired in series. And finally, connect a 5 ohm resistor between the junction of the two 0.1 ohm resistors and ground (to imitate a nominally 1 amp load).

Solve for voltages and currents, and you'll see how the sharing works.

Excellent explanation, I'm going to model it in LTspice, and see what happens. Thanks!

EDIT: I guess I was not even wrong!

 

for a job working on Lasers

What type of lasers? CO2?

 

Never heard of a foldback circuit. (LM723)
Never figured out that adding extra pass transistors would forfeit the self protecting features of the 3 pin regulators.

For his next act, the new-grad submitted a design for a torroid power transformer where the windings couldn't fit inside the doughnut, and got paid more than I did (because he had a college degree). I quit like a rat leaving a sinking ship for a job working on Lasers at a 43% pay increase. After the owner was replaced by an M.B.A. and I was replaced by a B.S.E.E. the company went out of business within a year.

Well if you feel that way you have to follow some straitjacket means if theres a protection feature built into the IC you have to use this by all means?

DO you have a guarantee the IC will cut off and will not be destroyed?

2 or 3 Amps are not of much concern but if you have a battery or battery bank youll get nasty problems witohut proper fuses why not even use two fuses.

Ive seen SX1308 tiny SMD chip actually bursting into flames from a battery setup theres always such a possibility for power regulating IC either through some setup features like very low internal resistance, or an EMP event. Built in protection or not, Id never rely on that.

Sure there are nutcase people with degree who never built a circuit or never did some circuit work on their own not from a teacher sheet.

Ive learned a lot recently from industrial waste PCBs, professional, from factory production lines, ive taken a look how theyre built, what components, what brands.

All of these have fuses on the PCBs.

Ive desoldered an Intel 80188 + some parts, sold recently, got out the money I paid for the ewaste boards.

I often have my own opinion but really I consider all what has been done before professionally.
You get a feeling for what is right and what not after a while.

 

Is it possible (and advisable) to connect two or more 7805 regulators in parallel to obtain an increased current output?

I found several sources saying that it was not recommended, due to imbalances caused by small differences between devices. But then I found this circuit:

View attachment 113758​

And since I found this circuit in the instructables website (which is not exactly known for being a infallible source of technical wisdom) I was wondering if anyone here could clarify things for me.

How about making the regulators adjustable by a volt or so?