Author Topic: Simple DIY Linear PSU  (Read 17884 times)

0 Members and 1 Guest are viewing this topic.

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 3034
  • Country: au
Re: Simple DIY Linear PSU
« Reply #75 on: March 14, 2020, 05:11:17 pm »
When the Inverting input voltage is less than the Non-inverting input voltage, the output of the op-amp should go to its maximum voltage. About 1.4V less than supply voltage for the LM358.

RL=no load, Vpin7=8.52V   Vpin8=9.69V
RL=20 Ohm,  Vpin5=8.13    Vpin8=9.30V
RL=10 Ohm,  Vpin5=7.87    Vpin8=9.04V
  Vpin8 going from 9.69V to 9.04V  means that the input voltage is dropping by a lot.
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 

Offline panossTopic starter

  • Frequent Contributor
  • **
  • Posts: 327
  • Country: gr
Re: Simple DIY Linear PSU
« Reply #76 on: March 14, 2020, 05:46:22 pm »
No problem using a SMPS so long as the voltage is correct and holds up when loaded.
The Collector of Q1 should be close to 18V?


RL=no load, Q1 Vc = 17.33V
RL=20 Ohm, Q1 Vc = 15.22V
RL=10 Ohm, Q1 Vc = 14.04V

I also tried with an SMPS (12V):
RL=no load, Vout=11.4V
RL=20 Ohm, Vout=10.3V
RL=10 Ohm, Vout=9.9V

EDIT: I tried with a second SMPS (a PC power supply). The results are the same.
« Last Edit: March 14, 2020, 05:54:32 pm by panoss »
 

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 3034
  • Country: au
Re: Simple DIY Linear PSU
« Reply #77 on: March 14, 2020, 05:55:54 pm »
No problem using a SMPS so long as the voltage is correct and holds up when loaded.
The Collector of Q1 should be close to 18V?


RL=no load, Q1 Vc = 17.33V
RL=20 Ohm, Q1 Vc = 15.22V
RL=10 Ohm, Q1 Vc = 14.04V

I also tried with an SMPS (12V):
RL=no load, Vout=11.4V
RL=20 Ohm, Vout=10.3V
RL=10 Ohm, Vout=9.9V
RL=10 Ohm, Q1 Vc = 14.04V, means that 4.5V is being applied to B-E of Q2 which means that it's damaged. This is assuming that the supply is holding up at 18.5V.
But why was the 9V rail dropping before? Are you using an LM317?
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 

Offline panossTopic starter

  • Frequent Contributor
  • **
  • Posts: 327
  • Country: gr
Re: Simple DIY Linear PSU
« Reply #78 on: March 14, 2020, 06:00:13 pm »
RL=10 Ohm, Q1 Vc = 14.04V, means that 4.5V is being applied to B-E of Q2 which means that it's damaged. This is assuming that the supply is holding up at 18.5V.
But why was the 9V rail dropping before? Are you using an LM317?
Yes, I 'm using an LM317. (I just took one more look to make sure  :D)
I 'll replace it to be sure, maybe it 's damaged.
« Last Edit: March 14, 2020, 06:01:51 pm by panoss »
 

Offline panossTopic starter

  • Frequent Contributor
  • **
  • Posts: 327
  • Country: gr
Re: Simple DIY Linear PSU
« Reply #79 on: March 14, 2020, 09:12:03 pm »
Well, I replaced:
-the S8550 with the BC557
-the LM317 with another LM317

Results are the same.
 

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 3034
  • Country: au
Re: Simple DIY Linear PSU
« Reply #80 on: March 14, 2020, 09:15:35 pm »
Well, I replaced:
-the S8550 with the BC557
-the LM317 with another LM317

Results are the same.
Just with the 20 ohm load, measure all of the voltages in the output stage,
Q2 Emitter
Q2 Base
Q2 Collector

Q3 Emitter
Q3 Base
Q3 Collector

With respect to the top  end of the CS resistor.
« Last Edit: March 14, 2020, 09:22:56 pm by xavier60 »
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 

Offline panossTopic starter

  • Frequent Contributor
  • **
  • Posts: 327
  • Country: gr
Re: Simple DIY Linear PSU
« Reply #81 on: March 14, 2020, 09:27:14 pm »
With respect to the top  end of the CS resistor.
Which is the CS resistor?
 

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 3034
  • Country: au
Re: Simple DIY Linear PSU
« Reply #82 on: March 14, 2020, 09:30:23 pm »
With respect to the top  end of the CS resistor.
Which is the CS resistor?
The Current Sense resistor R18. Always take measurements with respect to  the end that connects to the negative output terminal,
What value is it?
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 

Offline panossTopic starter

  • Frequent Contributor
  • **
  • Posts: 327
  • Country: gr
Re: Simple DIY Linear PSU
« Reply #83 on: March 14, 2020, 09:35:13 pm »
R18 is two 100m resistors in parallel, so it's 50milliohm.
 

Offline panossTopic starter

  • Frequent Contributor
  • **
  • Posts: 327
  • Country: gr
Re: Simple DIY Linear PSU
« Reply #84 on: March 14, 2020, 09:48:48 pm »
Input voltage: 12V.
RL=20 Ohms.

Q2:
Ve=11.67V
Vb=10.83V
Vc=11.58V

Q3:
Ve=11.66V
Vb=11.59V
Vc=10.30V
« Last Edit: March 14, 2020, 09:50:19 pm by panoss »
 

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 3034
  • Country: au
Re: Simple DIY Linear PSU
« Reply #85 on: March 14, 2020, 10:14:10 pm »
Input voltage: 12V.
RL=20 Ohms.

Q2:
Ve=11.67V
Vb=10.83V
Vc=11.58V

Q3:
Ve=11.66V
Vb=11.59V
Vc=10.30V
The voltages for Q2 look about right. 0.84V B-E voltage means it is being driven hard for its size which is expected.
Voltages for Q3 only make sense if the Vc and Ve are swapped? B-E voltage of 1.29V means that it's being stressed also.
Looks like output voltage setting is higher than what's available. Try decreasing the setting.
You need to get the proper transistors for it. Don't test at more than 500mA for now.

When it's happily regulating, the Base of Q1 should drop to a lower voltage and the CV op-amp's input voltages will be the same as each other.
« Last Edit: March 15, 2020, 02:37:51 am by xavier60 »
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 
The following users thanked this post: panoss

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 3034
  • Country: au
Re: Simple DIY Linear PSU
« Reply #86 on: March 15, 2020, 12:20:10 am »
The D45H11 and 2x TIP35C combination in my bench supply project have a total current gain of about 10,000.
Q1 only needs to sink about 1mA for 5 amps of output.
There needs to be another 3 or so milliamps reserve to drive Q2 deep into saturation as the regulator approaches its dropout state of about 0.9V.
I have some minor changes in mind that will allow the regulator to work better with low input voltages.  I would like to see it working properly first of all.
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 

Offline panossTopic starter

  • Frequent Contributor
  • **
  • Posts: 327
  • Country: gr
Re: Simple DIY Linear PSU
« Reply #87 on: March 15, 2020, 08:13:24 am »
Input voltage: 12V.
RL=20 Ohms.

Q2:
Ve=11.67V
Vb=10.83V
Vc=11.58V

Q3:
Ve=11.66V
Vb=11.59V
Vc=10.30V
Voltages for Q3 only make sense if the Vc and Ve are swapped? B-E voltage of 1.29V means that it's being stressed also.
Vb=11.59V  Ve=11.66V
Vb-e=-0.07V   
That is, b is lower than e for 0.07V. (So Q3 is in cutoff?)

You made a mistake: you swapped Ve with Vc (Vb=11.59V  Vc=10.30V , this is a difference of 1.29V)
« Last Edit: March 15, 2020, 08:16:14 am by panoss »
 

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 3034
  • Country: au
Re: Simple DIY Linear PSU
« Reply #88 on: March 15, 2020, 08:41:40 am »
Q3 is NPN. Its Collector should connect to the input rail and should always be more positive than its Emitter which should connect to the output rail.
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 
The following users thanked this post: panoss

Offline panossTopic starter

  • Frequent Contributor
  • **
  • Posts: 327
  • Country: gr
Re: Simple DIY Linear PSU
« Reply #89 on: March 15, 2020, 09:33:43 am »
Indeed...you 're right Xavier...I had swapped C & E in Q3 and Q4... (omg... :palm:.. I 'm gonna hide in some cave...In Eagle, I had connected Q3 and Q4 to a PNP package...:palm: )


Ok, I corrected it:
RL=no load, Vout=11.39V
RL=20 Ohms, Vout=10.82V
RL=10 Ohms, Vout=10.55V
« Last Edit: March 15, 2020, 09:41:51 am by panoss »
 

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 3034
  • Country: au
Re: Simple DIY Linear PSU
« Reply #90 on: March 15, 2020, 09:41:36 am »
That's holding up much better. You still need to reduce the CV setting down to something like 9V to allow it to regulate.
If the voltage measurement is taken at the correct points on the PCB, the voltage should only drop by millivolts with increasing load.
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 

Offline panossTopic starter

  • Frequent Contributor
  • **
  • Posts: 327
  • Country: gr
Re: Simple DIY Linear PSU
« Reply #91 on: March 15, 2020, 09:46:45 am »
You still need to reduce the CV setting down to something like 9V
How will I do this?
 

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 3034
  • Country: au
Re: Simple DIY Linear PSU
« Reply #92 on: March 15, 2020, 09:51:39 am »
You still need to reduce the CV setting down to something like 9V
How will I do this?
Reduce the setting on the CV Pot. If the the output voltage doesn't reduce, there could be damage to the output transistors.
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 

Offline panossTopic starter

  • Frequent Contributor
  • **
  • Posts: 327
  • Country: gr
Re: Simple DIY Linear PSU
« Reply #93 on: March 15, 2020, 10:12:08 am »

The output voltage changes when I turn the CV Pot.
But not linearly:
starting from max (Vout=11.37V), I turn the pot and for almost the half of the route voltage does not change.

After this it starts lowering till 0V!  :-+

The potentiometer is a Piher 10k 426M.
« Last Edit: March 15, 2020, 10:43:30 am by panoss »
 

Offline panossTopic starter

  • Frequent Contributor
  • **
  • Posts: 327
  • Country: gr
Re: Simple DIY Linear PSU
« Reply #94 on: March 15, 2020, 10:40:03 am »
Ok, I got it!
I lowered Vout to 8.89V.
RL=no load,   Vout=8.89V
RL=20 Ohms, Vout=8.86V
RL=10 Ohms, Vout=8.83V

 :-+

Wow! This is rock solid!

 

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 3034
  • Country: au
Re: Simple DIY Linear PSU
« Reply #95 on: March 15, 2020, 11:04:44 am »
Ok, I got it!
I lowered Vout to 8.89V.
RL=no load,   Vout=8.89V
RL=20 Ohms, Vout=8.86V
RL=10 Ohms, Vout=8.83V

 :-+

Wow! This is rock solid!
Well done for now.
Depending on how much more experimenting you want to do.
I would like D3 replaced by 2 forward biased signal diodes in series and the control rail changed to 7V.
The zener diode does not need to be removed.
« Last Edit: March 15, 2020, 11:07:05 am by xavier60 »
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 

Offline panossTopic starter

  • Frequent Contributor
  • **
  • Posts: 327
  • Country: gr
Re: Simple DIY Linear PSU
« Reply #96 on: March 15, 2020, 11:15:11 am »
I would like D3 replaced by 2 forward biased signal diodes in series and the control rail changed to 7V.
The zener diode does not need to be removed.
What are the signal diodes? Zener? (ok, I found it here. A 1N4148 is a signal diode) What shall I achieve with the signal diodes? Greater stability in Zener 's Vout?

Which is the 'control rail'? The output of the LM317? And why should I change it to 7V? I'm gonna use it with Vin=15V or Vin=30V.
If D3 (which is the zener diode) gets replaced, how the zener diode will not be removed?
« Last Edit: March 15, 2020, 11:35:59 am by panoss »
 

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 3034
  • Country: au
Re: Simple DIY Linear PSU
« Reply #97 on: March 15, 2020, 11:36:19 am »
I would like D3 replaced by 2 forward biased signal diodes in series and the control rail changed to 7V.
The zener diode does not need to be removed.
What are the signal diodes? Zener? Something else? What shall I achieve with the signal diodes?
Which is the 'control rail'? The output of the LM317? And why should I change it to 7V? I'm gonna use it with Vin=15V or Vin=30V.
If D3 (which is the zener diode) gets replaced, how the zener diode will not be removed?
"Signal diodes" usually refers to small general purpose silicon diodes like 1N4148 or 1N914.
The 2 diodes will drop the voltage across D3's location to about 1.4V. The original 5.1V zener will have no effect and can be left in circuit in case the mod has to be undone.
 I'm trying to achieve a control range at the Base of Q1 of about 4V.  The output of the LM317 needs to be dropped to about 7V for it to work out.

This can be achieved by going to a lower voltage zener but low voltage zeners don't regulate very well.
« Last Edit: March 15, 2020, 11:38:58 am by xavier60 »
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 

Offline panossTopic starter

  • Frequent Contributor
  • **
  • Posts: 327
  • Country: gr
Re: Simple DIY Linear PSU
« Reply #98 on: March 15, 2020, 11:44:44 am »
You 're saying I should do this?:
 

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 3034
  • Country: au
Re: Simple DIY Linear PSU
« Reply #99 on: March 15, 2020, 11:47:22 am »
Yes, its a very simple mod including the LM317 voltage change. I would normally try it here.
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf