Linear power supply project

[Therm Mr.]

I re-made some of the readings in my last post the changes are in red. I am not sure how I got them wrong the first time.

[Therm Mr.]

I have re-double checked everything on my circuit. If there is a wiring flaw I don't recognize it as such.

I did try a larger cap at c12 even though the light is out now. It did make the light not go out as quickly when I remove the test probe from pin 7. Just a bit more delay. I'm glad I did, it helps me to understand what that cap is doing.

https://tinyurl.com/23j2679m
The above link is the simulation @pqass made. If I am understanding it correctly Pin 3 of the LF411 should move between 2.25mv and 1.25v when the adjust pot moves from stop to stop? I Is this correct? Because I have 0 v at that pin. My five volt reference stops after the 14.3k (15K in the Sim) resistor between the 5V Ref and Pin 3. Am I understanding this correctly?

Thanks

Therm

[pqass]

I'm glad you're focusing on the core part of the circuit.  There's no point in adding more (current limiting, CC/CV indicators) until you build a working foundation.

Review the golden rules of op amps; especially the 3rd where the difference between +/- inputs should be zero in a steady-state negative feedback arrangement.

Below, I walk-through how the voltage regulation works via negative feedback (refer to the simulation in your last message and your latest schematic).    Lets start...

a. The default position of the darlington Q1 is fully on via R3 (1.2K) to +12V (ref+bias supply).
b. If the darlington is fully on then the unregulated source voltage (to its left) is let through to the output (right side) with minimal voltage drop.
c. Notice the GND symbol on the +OUTPUT that ties the GND of the ref+bias supplies (+12V, -12V, +5VREF) to this point.
d. If the full unregulated source voltage is let through, then the -OUTPUT is the negative of the unregulated source voltage (in our case the battery symbol) relative to the same GND.
e. We have a resistor divider between the lower -12V (fully unreg. source voltage), the voltage adj. pot R37, R15 (15K), and +5VREF.
f. Given that +5VREF being relative to the same GND point on +OUTPUT,
   if the fully unregulated source is let through, that makes it 12V (full unreg. source voltage) + 5VREF = 17V across the pot R37 and R15.
g. Lets say the pot is half-way (25K), then we have 17V across 25K+15K=40K.
   Q: What's the voltage at the mid-point between them? 
   A: 25K/40K=0.625*17V=10.6V (lower) and 17V-10.6V=6.4V (upper)
h. The +5VREF is like an anchor; it's always +5V from GND (via TL431 regulator action).
   Given the upper part of the divider being 6.4V to the mid-point, that places the mid-point below GND (+5VREF-6.4V=-1.4V).
i. The mid-point is connected to +Vin of the op amp, the -Vin is tied to GND.
   So now +Vin is lower than -Vin!  This causes the output of the op amp to swing negative.
j. The negative-going op amp output flows through the diode to turn-on Q2 (PNP),
   which lowers the current to the darlington (via Q2 winning against R3),
   which (Q1) is no longer fully-on (acts like a growing resistor with a growing voltage drop lowering the voltage going from left to right side),
   which lowers the voltage between R15+pot divider,
   which lowers the voltage on the upper and lower legs of the divider to the point where the mid-point is no longer negative WRT -Vin of the op amp.
k. The op amp will closely watch any difference between +Vin and -Vin and give a negative kick or back-off (output a positive voltage) to let R5 turn-off Q2, letting R3 turn-on the darlington a bit more.
l. Until you change the the adj. pot and the op amp needs to seek a new equilibrium.
m. Nothing is instantaneous.  The unregulated source voltage takes time to ramp up, the op amp also takes time to swing positive or negative, etc.
   The actions above will kick into regulation as soon as +Vin is lower than -Vin and won't let a -1.4V difference happen.

[Therm Mr.]

 @pqass.

That explanation will take some time for me to fully digest. I was totally misunderstanding the function of the op amp. You have provided me with a good sequence of operation to logically follow to try and trouble shoot my issues. Knowing what is supposed to happen, will help me try and find out what is not happening.

Thanks

Therm

[Therm Mr.]

I have tested and checked and re-checked. I am only getting adjustment on my output from 0-2.7 V. This is low range and I should be getting 0-8 V. It hits 2.7 at about 3-4 turns and goes no higher. I did all the math and tried to match the sample @pqass provided and Vin - And Vin + do remain at 0. Pin 6 will vary from -.81 to -.24 from stop to stop on the main Pot. Using a TL071 had no effect readings were pretty much the same. So I am back with the LF411.

I even walked away from it for two days just to clear my mind in case I was too focused and could not see the obvious. On the positive side I did already receive my custom transformer. It came much faster than I thought it would. I have yet wire it in to my circuit yet but it looks like a very good quality build.

I am posting a table of my readings comparing it to pqass's simulation. I am thinking about trying a spare TIP 142 thinking maybe I damaged the one in use. I would have done that already but it requires me soldering jumper wires onto it so it will plug into the Bread board. I am not really that hopeful replacing it will make a difference because I am getting 2.7 V of adjustment. I figure if it were bad it would not do that. But who knows.

Any suggestions for trouble shooting would be appreciated. I may just have to pull everything off the board and try again.

Thanks
Therm

[xavier60]

The zero volts between the opamp inputs indicates that it has control of the loop. Does it stay at zero when the pot is turned passed to point where the output stops increasing passed 2.7V?
Although I dont think it is the cause, check for oscillation at the opamp's output.
The output should go to full voltage when the loop is broken by removing the diode or disconnecting the pot.

[pqass]

Just to confirm,
the op amp power supply MUST be +12V (pin 7) and -12V (pin 4, not GND/+OUTPUT).   
And a multimeter (set to DCV) between pin 7 (red lead) and pin 4 (black lead) should be 24V.
The TL071 is not a rail-to-rail op amp and the 0.299V output (pin 6) smells like it's the max. lower excursion.
The output (according to the simulation) should be -158mV (below GND/+OUTPUT).

Also, verify
that if you were to disconnect the emitter of Q2 (2N4036) from the base of Q1 (darlington), leaving the 1.2K pull-up between darlington base and +12V, that the output (DCV between +OUTPUT and -OUTPUT) should be just a bit below the +17V (unreg.) source voltage.

[xavier60]

I mocked up the circuit in the simulation, minus the diodes. Works fine, just some ringing on load transients due to having no frequency compensation.
This is of little direct help except to prove that the design is ok.

[Therm Mr.]

I mocked up the circuit in the simulation, minus the diodes. Works fine, just some ringing on load transients due to having no frequency compensation.
This is of little direct help except to prove that the design is ok.

Loop always stays at 0. If I remove the diode Reading is 10.6 V. Removing the Pot has no effect.

I am not mocking up the sim exactly. I am mocking up my circuit which is very similar to it. (see pic)

Thanks

Therm

[xavier60]

With the pot removed, the opamp's + input must pull to REF, causing the opamp's output to swing hard positive, assuming the - input is tied to S+.

Edit: remove CR7.

[Therm Mr.]

Just to confirm,
the op amp power supply MUST be +12V (pin 7) and -12V (pin 4, not GND/+OUTPUT).

Yes Pin 4 and 7 are regulated power. 7) +12.04 and 4) -11.96

And a multimeter (set to DCV) between pin 7 (red lead) and pin 4 (black lead) should be 24V.

Yes it is 24.05 V

The TL071 is not a rail-to-rail op amp and the 0.299V output (pin 6) smells like it's the max. lower excursion.
The output (according to the simulation) should be -158mV (below GND/+OUTPUT).

I am using the LF411 I only tried the TL071 briefly. Pin 6 is -.318 v

Also, verify
that if you were to disconnect the emitter of Q2 (2N4036) from the base of Q1 (darlington), leaving the 1.2K pull-up between darlington base and +12V, that the output (DCV between +OUTPUT and -OUTPUT) should be just a bit below the +17V (unreg.) source voltage.

If I pull the emitter of Q2 I get 15.52 V at the output between +/-.


Thanks

Therm

[andrewtaylor]

I don't know the details of the HP transformer switching. It is common to have transformer tap switching with just using more or less windings in series, not no extra parallel connection. This does help with reducing the power loss, but does not allow for a higher current at low voltags (at least not much and not higher current limit used). T

In the original HP supply,
the wire for the lower voltage range has higher diameter.
That's why with simple tap switching (not the more complicated series-paralle scheme I usually use) HP has higher current at he lower voltage range.

Or to keep a long story short:
it's a specialls manufactured/wound transformer .-)

Knowing this, the trick is easily done 

[andrewtaylor]

I have tested and checked and re-checked. I am only getting adjustment on my output from 0-2.7 V. This is low range and I should be getting 0-8 V. It hits 2.7 at about 3-4 turns and goes no higher.

Ttry to remove the LM358 that you planned for curent control.When the voltage control then works as planned, remember the old trick for HW debugging:

If control loops interact, try to seprate them.

Of course, be careful NOT to short the output when there is no CC loop due to removed LM358 

[xavier60]

CR6, if damaged, is also capable of causing the present fault state.

[Therm Mr.]

I have tested and checked and re-checked. I am only getting adjustment on my output from 0-2.7 V. This is low range and I should be getting 0-8 V. It hits 2.7 at about 3-4 turns and goes no higher.

Ttry to remove the LM358 that you planned for curent control.When the voltage control then works as planned, remember the old trick for HW debugging:

If control loops interact, try to seprate them.

Of course, be careful NOT to short the output when there is no CC loop due to removed LM358 

I did this at your suggestion and it had and it no effect.

Thanks

Therm

[Therm Mr.]

CR6, if damaged, is also capable of causing the present fault state.

I removed CR6 and CR7 one at a time with no change to pin 3. It stayed at 0.

Thanks

Therm

[xavier60]

Either the current path from REF to pin 3 is open circuited or some low resistance path from pin 3 to S+

[Therm Mr.]

Either the current path from REF to pin 3 is open circuited or some low resistance path from pin 3 to S+

I have 12k Ohms resistance between Pin 3 and 5v ref.

Pin 3 to S+ Junction 13.8k Ohms.

Thanks

Therm

Edit: I did discover an intermittent connection on the anode side of CR7 and I corrected the fault. Now my output only adjusts to 1.6V. Pin 2 and 3 are still zero, Pin 6 is .98V when Pot is 0 ohms and .54v at 50k ohms.   

[xavier60]

Although my mock-up uses an LM358, I will state the pins for a single opamp.
With nothing else but a 10K resistor connecting my 5V REF to pin 3, I measure 5V on pin 3. Pin 6 measures 10.7V.
The main output measures 14.7V with a 1K load with 15V unregulated input.
This condition must be able to be replicated.

[pqass]

See attached for a minimal breadboard version of the voltage regulator only schematic from reply#176.
I left out the Q2 (PNP) base protection diode and made the following substitutions (what I had on-hand):

• TL081 op amp,
• Q1=2SC1983 darlington,
• Q2=2N3906,
• op amp output diode is some random signal diode,
• 12K vs 10K load resistor, and
• 68K trimmer pot vs. 50K for voltage adjustment.

A isolated 3-output power supply provided +12, -12, +5; all tied to the same GND.  I didn't have a TL431 to generate +5VREF from the +12V.   The "unregulated" source supply was provided by another entirely separate isolated +20V supply.

I was able to vary the output voltage from below 1V to 19V, although, I only had a voltmeter attached with no significant load.