Need help with LM317 based high current voltage regulator circuit

Thread Starter

sribasu

Joined May 10, 2017
18
I am new in Electronics, and spend my off time ib electronics as a hobbyist. I was trying to make a LM317 based high current voltage regulator. I came across the following link.
http://www.reuk.co.uk/wordpress/electric-circuit/lm317-adjustable-power-supply/

I am confused between circuit diagram 2 and 3 on this page. These 2 circuits shows the power transistor bases are connected to 2 different points. I already made a circuit using the 2nd diagram, which didn't work. I could only regulate between 18v - 27v for a 32v DC input. Although I understand that the 2 circuits use different number of transistors to support different amount of output current, but still in doubt which one is practically correct. I even simulated the 2nd circuit which worked per expectation. But when I built it, it didn't work. Could you please confirm which of these diagrams are correct with a little explanation?
 

dendad

Joined Feb 20, 2016
3,180
I must be missing something, both circuits look to be the same to me.
That is an easy way to do it, but not a very good way!
Also, maybe in that circuit, add a 1K resistor between the base and emitter pins of the 2N3055 to make sure it turns off.

Have a look at page 17 of the data sheet.
http://www.ti.com/lit/ds/symlink/lm317.pdf
This is a much better way to do it.
What current rating are you after?
Remember if you are running high currents through an analog regulator, particularly at low output voltages, there will be a lot of heat generated. Make sure you have a big heat sink, and maybe a cooling fan.
 

dl324

Joined Mar 30, 2015
9,335
Welcome to AAC! Please post schematic on this site to avoid future issues with broken links; or requiring people to go to untrusted sites.
I already made a circuit using the 2nd diagram, which didn't work. I could only regulate between 18v - 27v for a 32v DC input.
Is this the circuit you're referring to?
upload_2017-5-10_16-1-43.png

It's a horrible power supply. It places two diode junctions in the output where the regulator can't compensate for the voltage drops that are dependent on current.

It should be adjustable from about 0-27V.
 
Why would you want to use a 317?
Dont get me wrong, I like them, they are much more flexible than most monolithic regulators and make a passable current source quickly and easily but as was stated earlier they are analouge which means any voltage drop is going to be converted to heat, either in the reg or its output transistor.

Newer buck regulators, PWM on sterroids with a relativly large inductor are cheap and easy to build provide you stick to recommended values.
You can also buy preassembled boards and kits...

If you are trying to build something simply to try and understand it and learn then that is a whole different ball game and you need to think about the approach and how you ask the question.

It is a good idea to break circuits down i to functional blocks/systems...
An LM317 attempts to maintain a fixed voltage on its Ref pin and requires a minimum load current to do it.
They also have a tendency to ring, oscillate, at some output capacitences, which is worth checking for.

Your circuit has two elements to it...
A regulator, which should be linked to the output voltage in such a way as to avoid any semiconducting junctions and an output gevice to handle high current and power dissipation.

You could also look at voltage follower/buffer circuits which work by causing the output to match the input.

What is your goal something that works well or a learning opportunity?
Both are equally valid but require different questions to get you where you want to be.

Have fun,
Al
 

Thread Starter

sribasu

Joined May 10, 2017
18
Did you have a load of some sort connected when you were measuring the voltage. If not the leakage through the 2N3055 might be enough to give that reading on the voltmeter.
You are right, it was no load voltage. So should I attach a load (how much is reasoable here?) and then measure?
 

Thread Starter

sribasu

Joined May 10, 2017
18
A regulator, which should be linked to the output voltage in such a way as to avoid any semiconducting junctions and an output gevice to handle high current and power dissipation.
Thanks for your response. Being aware of the drawbacks of these old linear regulators, I am making it just for learning purpose, not for real use. Could you please help me understand the quoted part of your reply?
 

Thread Starter

sribasu

Joined May 10, 2017
18
I must be missing something, both circuits look to be the same to me.
That is an easy way to do it, but not a very good way!
Also, maybe in that circuit, add a 1K resistor between the base and emitter pins of the 2N3055 to make sure it turns off.

Have a look at page 17 of the data sheet.
http://www.ti.com/lit/ds/symlink/lm317.pdf
This is a much better way to do it.
What current rating are you after?
Remember if you are running high currents through an analog regulator, particularly at low output voltages, there will be a lot of heat generated. Make sure you have a big heat sink, and maybe a cooling fan.
I am looking for 5 amp current. There is a difference on the base pin connections of the transistors and the Vref feedback of LM317. I am not sure if a pull down resistor is required, as you said to make it working or not.
 

Thread Starter

sribasu

Joined May 10, 2017
18
Welcome to AAC! Please post schematic on this site to avoid future issues with broken links; or requiring people to go to untrusted sites.
Is this the circuit you're referring to?
View attachment 126441

It's a horrible power supply. It places two diode junctions in the output where the regulator can't compensate for the voltage drops that are dependent on current.

It should be adjustable from about 0-27V.
Thanks for responding. I am new to the forum and I will make sure that I avoid remote linking.

I do not understand your response fully though. Did you mean to say that the circuit is wrong, or its performance is going to be horrible? Is there any clue why I cqn't reach below 18v, if its expected to regulate between 0-27v? I understand that I would loose roughly 5v from the upper limit due to the semiconductors and LM317 Vdrop. Is that what you referred horrible?
 

dendad

Joined Feb 20, 2016
3,180
A regulated power supply should keep the output voltage fairly constant at the set voltage over the full range of current.
This circuit you have found will not do that well. It will probably do for many applications but as a regulated power supply, it is not the best practice.
Also, you loose the over current control that a properly designed circuit will provide.
Try putting 5 x 47R 600mW resistors in series to give you 240 ohms across the output.
 
you need to read the datasheet/s. I put a link to them in another thread earlier about limiting current with an LM338, which is thew same fammily and has very simmilar functionality.

What I was trying to suggest is that yhou build a stable regulator, low current, first and then buffer the output and move the feedback reference.

A 317 is always trying to maintain a fixed voltage on the REF pin which I think is 1.25V bur you should check.
The REF is relative to out so the whole things works differently to a 'standard' regulator like the 78nn/79nn series which are still extensivly used in low power applications. ( a go to unless you need something different)

To work a317 must have a minimum load current, as a drain, although it isnt much. However if you set it small at the maximum output using a simple resistor it will be to small at a low output. so either get clever or set it just big enough at blow output and accept the losses at higher voltages.

Functionally...
The unit drives a current into Adjust and your circuit manipulates this into a voltage.
The voltage on Adjust is compared to Vout...
If the voltage difference is less than 1.25 the regulator delivers more current to Vout, increasing the voltage across the load
If the voltage difference is less greater 1.25 the regulator delivers less current to Vout, decreasing the voltage across the load
Pulling Adjust to ground, in fact anything less than Vout - 1.5, will shut the regulator down.
The current on Adjust is limited, by the regulator, to less than 100uA.

In practice the simplest circuit only needs 2 resistors, one of which is likely to be variable, and a couple of capacitors to stabilise things.
A third resistor is required in paralell with the load to meet the minimum current requirement, which is required for open circuit regulation, IE no load.
Again look at the datasheet

To buffer the output, to get a larger current capability, you would need a 'follower circui't / 'voltage mirror'
Have a look at this...
https://www.allaboutcircuits.com/technical-articles/how-to-buffer-an-op-amp-output-for-higher-current-part-1/

Opamps are relativly easy to use and will give you a well controlled wsay to invert the voltage/current signal you are going to need on your output transistor/FET.
They are also a 'basic building block' of more complex circuits, so good learning. There are good tutorials on this site.

Have a play with this...
http://www.falstad.com/circuit/circuitjs.html?cct=$+1+0.000005+10.20027730826997+43+10+42 t+256+192+320+192+0+1+-10.35701397507593+0.7656568664521153+100 t+256+288+320+288+0+-1+18.183874940229607+-0.6934542182423478+100 w+320+208+320+240+0 w+320+240+320+272+0 w+320+304+320+352+0 r+256+288+256+352+0+220 r+256+192+256+128+0+220 d+256+240+256+288+1+0.805904783 R+256+240+192+240+0+1+40+10+10+0+0.5 w+256+352+320+352+0 w+256+128+320+128+0 w+320+128+320+176+0 R+256+128+256+80+0+0+40+30+0+0+0.5 d+256+192+256+240+1+0.805904783 w+320+240+400+240+0 r+400+240+400+352+0+10 g+400+352+400+384+0 O+400+240+448+240+0 w+256+352+400+352+0 o+8+64+0+4098+48+0.025+0+2+8+3 o+17+64+0+4098+24+0.00009765625+1+1

(And look at the simple push pull follower without the diodes which has a pronounced deadband, under the circuits menu)

This simple simulator, isnt, and dessnt claim to be the most complete solution out there and has some significant limitations but I find it quite useful to do a quick proof of concept and model simple tutorial circuits. You can download it to use locally.

If you had your regulator output where the AC source is, with its feeback network connected to the output I expect it would be quite stable.
( Waiting for the storm of folk telling me I am wrong now... :) )

Have fun,
Al
 

dl324

Joined Mar 30, 2015
9,335
Did you mean to say that the circuit is wrong, or its performance is going to be horrible?
Voltage regulation will be affected by the two PN junctions on the output of the regulator. If you set the voltage at a low current, voltage will drop when load increases. If you set the voltage with a high load, voltage will increase when the load decreases.
Is there any clue why I cqn't reach below 18v, if its expected to regulate between 0-27v?
I think it has been established that you measured voltage range with no load. You need to have a load on the output.
 

DickCappels

Joined Aug 21, 2008
6,012
That is a very poor way to boost the output current of an LM317 (did somebody say that before?) As sribasu said, he is just doing this for the learning experience. After this he might want to try other approaches, but for now let's help him do what he wants to do.

I just had a look at the 2N3055 datasheet -the leakage current can approach 1 milliamp. That load resistor (anything less than 1k should help a lot) is necessary for you to see what your circuit would actually do under load because without a load the circuit cannot work correctly. (I think that has been said a few times earlier in this thread.
 

Thread Starter

sribasu

Joined May 10, 2017
18
That is a very poor way to boost the output current of an LM317 (did somebody say that before?) As sribasu said, he is just doing this for the learning experience. After this he might want to try other approaches, but for now let's help him do what he wants to do.

I just had a look at the 2N3055 datasheet -the leakage current can approach 1 milliamp. That load resistor (anything less than 1k should help a lot) is necessary for you to see what your circuit would actually do under load because without a load the circuit cannot work correctly. (I think that has been said a few times earlier in this thread.
Thank you sir for your response.
 
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