Simple DIY Linear PSU

[panoss]

To change LM317 's Vout I must change R16? Or R17? Or something else?

[xavier60]

To change LM317 's Vout I must change R16? Or R17? Or something else?

Try R17 first. ll try to work it out here.

[xavier60]

680Ω for R17 will set it to a bit under 7V.

[panoss]

Ok, I did the mods:
R17 = 680Ω
Two 1N4148 in place of D3, voltage across them is 1.22V.

Also I changed Vin to 18V.

RL=no load, Vout=16.93V
RL=20Ω,      Vout=14.61V
RL=10Ω,      Vout=13.34V

I lowered it to 12V.
RL=no load, Vout=12.04V
RL=20Ω,      Vout=11.99V
RL=10Ω,      Vout=11.96V

So at lower voltages it 's more stable.
Maybe if I put a second transistor it 'll be more stable at higher voltages?

[xavier60]

Ok, I did the mods:
R17 = 680Ω
Two 1N4148 in place of D3, voltage across them is 1.22V.

Also I changed Vin to 18V.

RL=no load, Vout=16.93V
RL=10Ω,      Vout=14.61V
RL=20Ω,      Vout=13.34V

I lowered it to 12V.
RL=no load, Vout=12.04V
RL=10Ω, Vout=11.99V
RL=20Ω, Vout=11.96V

So at lower voltages it 's more stable.
Maybe if I put a second transistor it 'll be more stable at higher voltages?

As the amount of current approaches a transistors maximum rating, the gain usually drops a lot.
What maximum current  do you need?
The   TIP35C transistors in my bench supply have a combined current rating of 50A to handle 5A. This isn't unusual.
What is Q1's Base voltage with the output voltage set to 9V unloaded?

[panoss]

What maximum current  do you need?
The   TIP35C transistors in my bench supply have a combined current rating of 50A to handle 5A. This isn't unusual.

I 'll make it for 5A too.
So two TIP35Cs (they have maximum Collector current 25A) will be ok, right?

What is Q1's Base voltage with the output voltage set to 9V unloaded?

It's 2.55V.

[Kleinstein]

With higher current the limiting factor is usually the SOA (at higher voltage) and maybe the heat sink. So the current rating of the output transistors is often not that important. The other point to watch for with higher power is the heat dissipation in the shunt.

[panoss]

shunt.

??
Which is the shunt? R13 & R14?

[kallek]

shunt.

??
Which is the shunt? R13 & R14?

Shunt means low value resistor what is used to measure current by its voltage drop. So it is R18.

[panoss]

R18 is actually two resistors, R18 and R18-B in parallel, both rated at 2W.
I hope this will be proved enough for heat dissipation.

[kallek]

Sure it is enough. R18 is total 0.05 ohm so you can calculate power dissipation with 5 amp current by formula P=RI^2 = 1.25W. But as mentioned earlier, make sure that traces next to these resistors are quite wide. As well as other high current traces too.

[Kleinstein]

With the usual voltage drop of some 200-500 mV one already has a power of 400 mW to 1 W at 2 A.
To get a stable resistance the temperature rise should be low. So the 2x2 W may be just OK for 2 A if the voltage drop is low (some limitations when the set current is low, may want a good OP for the current limit than).

For the resistor the permissible temperature rise is also different, depending on the type - some of the power wire wounds are allowed to go very high, so there rating is high, but the useful power as a shunt is still relatively low.

[xavier60]

An important test for a power supply is how the output voltage responds to mains power loss.
Without an oscilloscope, this test will need to be done with a multimeter while input voltage(18V) is gradually decreased to zero.
One way is by temporarily connecting a very large capacitor across the 18V so that the voltage falls gradually when the mains is switched off.
The output should be set to 1V, then monitored as the 18V falls.

[panoss]

Which value is 'very large'?
4700uF?
10000uF?
More?

And I should measure the time that it takes for Vout to go from 1V to 0V?

[xavier60]

Which value is 'very large'?
4700uF?
10000uF?
More?

And I should measure the time that it takes for Vout to go from 1V to 0V?

10,000uF will make it about 1v/s. You are looking for any increase of output voltage which is what we don't want to see.

[panoss]

I did the test and everything seems normal.
I saw no increase of the output voltage.

I plug it in, set the Vout=1V, unplug it. It stays for a few seconds at 1V and then gradually, in a few seconds, decreases till 0.

Till now (I have tried a few things) it works VERY WELL!

[xavier60]

That's good to know.

[panoss]

This is thanks to you Xavier, thank you very much.
Have you build it or I 'm the first?

[xavier60]

I built the design into a working bench supply some time ago. The circuitry around Q1(Q2) is a little different and it relies on a micro-controller  to shut off the output when the mains is switched off. It uses a higher performance TLC072 op-amp which does not have much benefit.
Actually, using a TLC072 or other types in the current design might cause a mild voltage spike during power down. It would need testing.
https://www.eevblog.com/forum/projects/linear-lab-power-supply/msg2388873/#msg2388873

As far as I can tell, mike_mike 's implementation of the design worked out well also.

[panoss]

Hi again.
I have a problem with the PSU: I short circuited the output and something was burnt (I saw some (very little) smoke).
Since then it only outputs the maximum voltage (20V).

I checked every component one by one: I desoldered one pin from every diode and checked them, I desoldered  Q1 and checked it, I desoldered Q2, Q3, Q4 (they are in a separate schematic & pcb, not seen in the attached schematic) and checked them.
I measured every resistor, they all are ok...
I even checked the LM358 as per these suggestions, it 's ok.
These are the voltage measurements, input voltage is 20V, any help is more than welcome:

[xavier60]

Because there is no drive current flowing through Q1, there must be a problem in the output block.
If a BJT seems ok on diode test, also check for junction leakage on resistance range, mainly C-B.
Are you still using undersized transistors? 

[panoss]

If a BJT seems ok on diode test, also check for junction leakage on resistance range, mainly C-B.

So I measure resistance between C-B? Which measurement is ok and which implies a damaged transistor?

Are you still using undersized transistors?

I went back to D209L  (for Q3 and Q4).
Q2 is a BC557.
They worked fine (before short circuiting the PSU 's output).

[xavier60]

If a BJT seems ok on diode test, also check for junction leakage on resistance range, mainly C-B.

So I measure resistance between C-B? Which measurement is ok and which implies a damaged transistor?

Are you still using undersized transistors?

I went back to D209L  (for Q3 and Q4).
Q2 is a BC557.
They worked fine (before short circuting them).

Any good silicon PN junction should measure totally open circuit, no leakage at all with the probes connected in the direction that will reverse bias the junction.
You need to get suitable parts some day.

[panoss]

I measured Q2, Q3 & Q3.
Q2: C-E, resistance: 32 ohms! Both directions...
So it 's like a resistor of 32Ω, it 's bypassing Q3 & Q4, so this is why it 's outputing the input voltage!

Btw, in diode mode it looks fine!

You need to get suitable parts some day.

This will happen some day...

[WattsThat]

I measured Q2, Q3 & Q3.
Q2: C-E, resistance: 32 ohms! Both directions...
So it 's like a resistor of 32Ω, it 's bypassing Q3 & Q4, so this is why it 's outputing the input voltage!

Btw, in diode mode it looks fine!

You need to get suitable parts some day.

This will happen some day...

I suspect meter operator error. Chances are about 99.999% that Q2 transistor is shorted collector to emitter. Checking on DMM diode setting should show less than 0.1 volt C to E regardless of lead polarity. A good transistor should show INF.

Measure a 27 ohm resistor on your meters diode check function - what does it show?