more output current with LM317 ?

Hi,
for some time I have been wondering if I can make LM317 to output more than 1.5Amps. After many hours of searching I found that I can achieve this in 2 ways. The circuits are in LM317 datasheet which can be downloaded from here :
http://www.ti.com/lit/ds/symlink/lm317.pdf
1- By using a transistor with LM317 (page 16) .
2-By using several LM317 regulators in parallel and using TL080 (page 15).
Second one seems better IF load is equally divided between regulators, it also has the benefit of producing less heat when  high currents are drawn.
My question is: will the second circuit (LM317s in parallel) work and output more than 2Amps? is load equally divided between regulators in this design ? can I replace TL080 with another IC ?
By the way, there is a circuit in page 12 called "tracking preregulator" , what is the purpose of this design ?

PS. I know there are switching regulators which output more current, but I'd like to experiment with LM317.

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-max-1 year ago

You can use a 2N3055 or other NPN BJT transistor as an "emitter follower." The 2N3055 can dissipate more power and supports a higher current output (make sure you use a proper heatsink). If I'm not mistaken, the collector connects to Vcc, the base connects to the output of the regulator (Not sure, but I think there should be a small pulldown resistor) and the emitter is the new output, and the feedback resistors should then be connected to the emitter to allow the regulator to regulate the output of the emitter follower stage.

The disadvantage is that adding the additional transistor output stage will further increase the dropout voltage by at least an additional 0.7 volts.

ehsan_zt (author)  -max-1 year ago

yes, this I've come across this design, but the only problem is lm317 getting hot. The circuit that I have mentioned earlier with several regulators in parallel seems a better idea ( it produces high current and reduces heat) but of course IF it works...

-max- ehsan_zt1 year ago
Adding more regulators in "parallel" will not reduce the heat produced. in fact, it will increase, (but only ever so slightly, not nearly enough to make a difference) due to quiescent current. It really will just spread the heat produced across the 2 regulators, as each one delivers (ideally) half the total current.

Heat produced in a linear regulator will be equal to difference input vs output voltage, times the current through it.* (Input current and output current are equal*)

* ignoring quiescent currents
ehsan_zt (author)  -max-1 year ago

Thanks for detailed explanation . as you said the heating problem still exist in either design, I may have to use a better regulator.

-max- ehsan_zt1 year ago

If you want to use the LM317 in a way to prevent it from producing too much heat, or if you want like it to be very efficient, you just need to make sure that the voltage dropped away by the regulator is small.

-max- ehsan_zt1 year ago

A simple switch mode regulator may be the way to go. They are not easy to build, requiring an inductor, some good capacitors, a scotchy diode, some passives, a switching regulator chip of some sort, and careful design and layout.

You can either buy one of those little DC to DC power converter boards, or build one from scratch. I actually built a few from scrap electronics pulled off of old boards.

-max- ehsan_zt1 year ago
When using the additional power transistor output stage, most of the heat produced should be dissipated in it, not the lm317. Emitter followers have a voltage gain of 1, and a pretty high current gain. This means that the voltage at the emitter output will 'follow' the base voltage, ignoring a constant 0.6v drop.

Because of the high current gain, only a small base current is needed to get the desired voltage, and the gate looks like a high impedance, while the output is a low impedance (the voltage on the emitter is constant and does not change as you draw more current)

Yes, the 2nd circuit will work. Yes, it will equally divide the current, as far as its able. That's why the 0.2 Ohm resistors are there. The TL080 can be replaced with any op amp, that can work single supply, and with rail to rail output so a 741 isn't a great choice.

A tracking pre-regulator stabilises the input to the main regulator, and makes the output of the main even quieter.

There are other clever pre-regulator schemes where the tracking pre- stage is a switcher, and the main regulator is a linear one. This makes the linear regulator dissipate much less heat, while giving a quiet output.

ehsan_zt (author)  steveastrouk1 year ago

The pre-regulator scheme was interesting Thank you for clearing that out.

how is that 0.2 ohm resistor calculated ? for example if I want each lm317 to output 0.5 Amps, what resistor should I use ?
should I use 5w resistors ? (or less maybe ?)
Isn't it better to use resistors at input of each regulator?

Can you please suggest a replacement for TL080, I'm not really familiar with op-amps for power supplies.

Sorry for asking too much,I get curious sometimes :-D

There is no fixed current limit in the circuit shown, each regulator could provide up to 1.5A, so that circuit has a limit of 4.5A.

The resistors are called sharing resistors, and their value doesn't affect the output current. The resistor is sized to take up a few mV of difference between the output voltages of the regulators, so here, even flat out, they are dissipating only .25 Watts. Putting the resistors at the input won't work.

Try an amp like the LM358.

ehsan_zt (author)  steveastrouk1 year ago

I've looked at LM358 datasheet to figure out which legs should I use.Here is what I think, I'd appreciate if you could check it out to see if there are any mistakes:
pin 8 should be connected to 100 ohm resistor,
pin 2 should be connected to 5K resistors,
pin 3 should be connected to 150 ohm resistor,
pin 4 should be connected to ground.
I'm not sure what pins should be connected to 200pF capacitor and base of 2n2905 ...

Don't think of it as "pins", look at the circuit function.

V+ goes to pin 7 (that's where the 100 Ohm is connected.

V- (ground, here) goes to pin 4

Output, pin 6 goes to the base of the transistor.

In - goes to pin 2

In_ goes to pin 3

With the lm358, the 200pf isn't needed.

ehsan_zt (author)  steveastrouk1 year ago

shouldn't pin 8 be connected to V+ (100 ohm resistor) ?

I've made corrections to the original circuit according to your instructions although I wasn't sure what you meant with "In -" and "In__",is it right ?

LM317 in Parallel.jpgLM358.JPG

TL080 is a single OPamp where pin 1 is N1/comp and pin 8 is comp pin 7 is vcc pin 6 is out and pin 5 is offset N2 pins pin 2 is - in pin 3 is + in and pin 4 is ground.

That diagram is of a TL082.

You can use it connecting pin 8 as you said and only using one of the OPamps.

You can get the datasheet here:

http://www.maxim4u.com/

ehsan_zt (author)  Josehf Murchison1 year ago

you meant like this ?
by the way, lm358 doesn't have comp and n2,aren't they important in this application ?

LM317 in Parallel'.jpg

Yes, that's right. Modern amps like the 358 don't need the comp pin.

ehsan_zt (author)  steveastrouk1 year ago

thanks , I will try it.

I can't read the page you linked.

You can get hundreds of circuits just by Googling "lm317 high current regulator circuit".

What circuit you use is more what output you want.

ehsan_zt (author)  Josehf Murchison1 year ago

Sry, it was a link to lm317 texas instruments datasheet. here's the original link:

http://www.ti.com/lit/ds/symlink/lm317.pdf

I had googled that, but most of the circuits are using a transistor for higher current, it works but it still produces so much heat. I'd like to divide load between several regulators so that each of them are running colder .

That's better now I know what you are saying.

That circuit on page 12, I don't get why you would regulate the voltage across a voltage regulator when the voltage regulator regulates the voltage output.

The circuit on page 15 regulates the voltage up to 4 amps and as long as the Opamp you use is the same basic device it should run the circuit. If you are trying to keep it cool something a little more powerful might do you.

I used the search filter here to find 17 OPamps that might do you.

http://ca.mouser.com/Electronic-Components/

You can get datasheets here also.

ehsan_zt (author)  Josehf Murchison1 year ago

thank you, I will look it up .
"If you are trying to keep it cool something a little more powerful might do you."
you mean like another regulator or another Opamp ?

I was referring to the OPamp, the TL080 is about 0.625 watt it will heat up less with a 1 watt Opamp.

So will transistors if you need 10 watt and you use a 20 watt they don't get as hot.

You might also want to look into a BA00DDOWxx series or some other regulator that can carry 2 amps or more. The LM150, LM250, or the LM350, 3 amp adjustable voltage regulator.

ehsan_zt (author)  Josehf Murchison1 year ago

thanks for your suggestion. LM350 seems interesting, I may end up using it.