Simple DIY Linear PSU

[panoss]

Also it has no BAT85, so I put BAT42. It 's the D5.

[panoss]

Ok, I managed to simulate it!

I use TL072 as for LM358 Proteus has no simulation model.
I 've also replaced D45H11(Q2) with a BD912 because D45H11 has no simulation model.

I power it with a voltage source, 30V.
I put the CV Pot in the middle, CC Pot on the top (max current).
These are the results:
RL=30 Ohm, 0.47A, 14.2V
RL=25 Ohm, 0.57A, 14.2V
RL=20 Ohm, 0.71A, 14.2V
RL=15 Ohm, 0.94A, 14.2V
RL=10 Ohm, 1.40A, 14.0V
RL=5  Ohm, 1.56A   7.8V

Conclusion: It cannot give enough current.

I have uploaded in .pdf (change it to .pdsprj) and in zip format.

[xavier60]

The BD912 likely has low current gain. Try a BD138-16 for now.  Most of the BJTs I tested with  had gains of 200 and more.
The D45H11 has unusually high gain for a large BJT.
The voltage at the Base of Q1 should be checked during testing to get an idea of how much total gain the output path has.

[panoss]

.Try a BD138-16 for now.

It 's not included in Proteus' libraries.

[xavier60]

The design on  mike_mike's thread uses a BD244a. Check what the modeled current gain is.

[panoss]

The design on  mike_mike's thread uses a BD244a. Check what the modeled current gain is.

I don't have the simulation file of mike_mike's design.
I 'm expecting someone to send it to me.

[xavier60]

The design on  mike_mike's thread uses a BD244a. Check what the modeled current gain is.

I don't have the simulation file of mike_mike's design.
I 'm expecting someone to send it to me.

You mainly need to find a driver transistor with reasonable gain. What did you use in the simulation of LM723 design?
Keep in mind to regularly check Q1's Base voltage.
The .pdf you posted seems corrupted.

[panoss]

The design on  mike_mike's thread uses a BD244a. Check what the modeled current gain is.

I don't have the simulation file of mike_mike's design.
I 'm expecting someone to send it to me.

You mainly need to find a driver transistor with reasonable gain. What did you use in the simulation of LM723 design?
Keep in mind to regularly check Q1's Base voltage.
The .pdf you posted seems corrupted.

Ok, I imported the D45H11 's model from a PSpice model for this transistor. (it 's doable)
We do have an improvement but it's far from proper working:
RL=10 Ohm, 1.42A, 14.2V
RL=5 Ohm, 11.5V,  2.31A

I reuploaded both .zip and .pdf files.

[xavier60]

Did you note Q1's Base voltage? Can you obtain Q1's Collector current from the simulation?

[panoss]

RL=10 Ohm, Q1 Vb= 7.12V   Ic=0A
RL=5   Ohm, Q1 Vb= 7.60V   Ic=0A

No current through Q1?? 

[xavier60]

RL=10 Ohm, Q1 Vb= 7.12V   Ic=0A
RL=5   Ohm, Q1 Vb= 7.60V   Ic=0A

No current through Q1??

Q1's Collector current should be the same as the current through R10 witch can be calculated from its voltage drop.
I would also expect the op-amps to be allowing Q1's Base voltage to go up much closer to 9V. Check this.
Without knowing the current gain of the output transistors, I can't work out the expected maximum output current.
You can try lower values for R10 as a temporary fix.
Watch for stability problems.

[kallek]

That .pdf file does not open. At least base current go through because voltage drop. Are you sure you are measuring dc current?

[panoss]

kallek, the pdf file is not actually a pdf file but a pdsprj file: you have to change the .pdf to .pdsprj.
Yes, I'm measuring DC.

I changed the Amp Meter and now it shows current flow! (the other amp meter I used has a low resollution!!!):

RL=10 Ohm, Q1 Vb= 7.12V   Ic=0.00143A  Ve=6.45V
RL=5   Ohm, Q1 Vb= 7.60V   Ic=0.00189A  Ve=6.9V

[panoss]

Q1's Collector current should be the same as the current through R10

RL=5 Ohm, Q1: Ic=0A 0.00189A   IR10=0.00190A

[kallek]

No I did not. I assume that then I need Proteus too. Currents looks like ok, the difference is probably base current. Next measure Q2 current to check if pass transistors get enough base current.

[panoss]

I would also expect the op-amps to be allowing Q1's Base voltage to go up much closer to 9V. Check this.
Without knowing the current gain of the output transistors, I can't work out the expected maximum output current.
You can try lower values for R10 as a temporary fix.
Watch for stability problems.

With R10=1k:
RL=5 Ohm, Q1: Vb=7.60V
RL=10 Ohm, Q1: Vb=7.15V

With R10=300 Ohm:
RL=5 Ohm, Q1: Vb=6.43V
RL=10 Ohm, Q1: Vb=6.13V

With R10=300 Ohm it seems to work a lot better (it can give more current and voltage remains stable).

For knowing the current gain of the output transistors I should simulate my first design? (the kakopa design from my first post?)

[panoss]

Next measure Q2 current to check if pass transistors get enough base current.

With RL=5 Ohms and R10=300 Ohms:
Q2 Ie=0.147A
Q3 Ib=0.0691A
Q4 Ib=0.0691A

[xavier60]

Next measure Q2 current to check if pass transistors get enough base current.

With RL=5 Ohms and R10=300 Ohms:
Q2 Ie=0.147A
Q3 Ib=0.0691A
Q4 Ib=0.0691A

That works out to 20 for the output transistors. The simulator must be using the minimum spec from the data sheet.

[xavier60]

150 for the driver transistor which is reasonable.
A suitable value for R10 can be chosen after the design is built with real parts.

[panoss]

So you 're saying the circuit is ok?
I should built it and after this experiment with the value of R10?

[xavier60]

So you 're saying the circuit is ok?
I should built it and after this experiment with the value of R10?

That's right. Aim for a gm of 10 to 15. That is the amount of output current change to change of Q1's Base voltage.
A gm of 10 would mean that the output current changes by 1A for a change of 0.1V at Q1's Base.

[panoss]

            Output current change
gm = -----------------------------------
           Q1's Base voltage change
Is this correct?

e.g if Q1's Base voltage increases by 10% and Output current increases by 100%, this is a gm of 10?

[xavier60]

            Output current change
gm = -----------------------------------
           Q1's Base voltage change
Is this correct?

e.g if Q1's Base voltage increases by 10% and Output current increases by 100%, this is a gm of 10?

You  have the right idea but use absolute changes, not percentages.
Use 2 resistors in parallel for R17 so that the LM317 output can be easily set to the desired voltage.
It is important that the unreg voltage stays high enough to prevent the LM317 from dropping out of regulation.
I don't expect a problem with a 15VAC secondary.
The regulation stability can only be as good as that of the LM317 which I think is reasonably good.

[panoss]

I don't expect a problem with a 15VAC secondary.

I 'll try with 15VAC and 30VAC as these are my transformer 's output voltages.

Another question:
Which resistors should be power resistors?
I think R13, R14 and R18 should be 5W rated?

[xavier60]

At 5A the dissipation is 0.625W for each of the 100mΩ resistors and 1.25W for the 50mΩ resistor which can be 2x 100mΩ resistors in parallel.
So 2 or 3 watt resistors will be fine. 5W resistor on ebay are likely to be cheaper.
I like to place larger areas of copper print under power resistors.`