Author Topic: Lab Power Supply - The Lost Current  (Read 19890 times)

0 Members and 1 Guest are viewing this topic.

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 2817
  • Country: au
Re: Lab Power Supply - The Lost Current
« Reply #25 on: February 03, 2018, 06:25:49 am »
Also, does the discrepancy in current readings exist in CV mode? The loop should be stable in CV mode.
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 

Offline radoczi94Topic starter

  • Contributor
  • Posts: 44
  • Country: hu
  • Electronics lover since 12y/o
Re: Lab Power Supply - The Lost Current
« Reply #26 on: February 03, 2018, 08:36:44 am »
Don't modify it until you have found a definite reason for the problems. Have you checked for oscillations?
When you checked the regulator PCB for a possible short bypassing the shunt, did you have all of the wires connected?
 
Try a "diode-OR" to switch between the CV and CC control loops. This is a common and effective way to implement constant current limiting on a "normally CV" power supply.

Check out the attached schematic of an HP6214A lab power supply.

Also check out this Keysight article: http://powersupply.blogs.keysight.com/2012/07/how-does-power-supply-regulate-its.html





Thanks, will try it.

Edit: One thing came into my mind. I am not shure, that I can put a diode in series with the CV opamp. What I could do is to move the diode on the CV opamp output (D7) and tie it's anode to the driver transistors' (Q4) base.

There is a 5W 0.68 ohm resistor on the regulator PCB. I can't see the marking of the resistor next to it? Are these the shunt resistors?
Yes, it was in there, when I took the picture. When I put those 0,68R resistors,then I realised, that the "missing" 2 amps is more or less linear with the output load current.

 The wires were connected, just removed the shunts, when I was checking for shorts. I'm not sure, how to check for oscillations without a scope. My idea is to give this thing a try without th CC opamp installed.


Also, does the discrepancy in current readings exist in CV mode? The loop should be stable in CV mode.
Yes, it is stable in CV mode.

The CC mode output current is pretty stable too. There are some smaller deviations, like 200mA or so, typically downwards. But it is stable, the meter does not moving at all, the current value just deviating a little bit every time the CC kicks in.
 

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 2817
  • Country: au
Re: Lab Power Supply - The Lost Current
« Reply #27 on: February 03, 2018, 09:32:29 am »
I was manly curious to know if there is a difference between output current and shunt current in CV mode also.
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 

Offline radoczi94Topic starter

  • Contributor
  • Posts: 44
  • Country: hu
  • Electronics lover since 12y/o
Re: Lab Power Supply - The Lost Current
« Reply #28 on: February 03, 2018, 10:08:33 am »
That is one thing I did not measured, at least I don't remember. Will do that too, if I have time.

You may probably noticed, my english is not the best, sorry for that.I'm trying.
« Last Edit: February 03, 2018, 10:20:08 am by radoczi94 »
 

Offline C

  • Super Contributor
  • ***
  • Posts: 1346
  • Country: us
Re: Lab Power Supply - The Lost Current
« Reply #29 on: February 03, 2018, 04:21:45 pm »
You may probably noticed, my english is not the best, sorry for that.I'm trying.
An doing good, Very simple, If you have problems to or from english just say so.  :-+

oscillations without a scope
With no scope the best you can do is trying to measure AC with meter.
On DC mode with meter, only last digit value should change.

Keep in mind that a Lab Power Supply is to help you when building new circuits & protect the circuit.

Think of what happens when you are building a PWM dimmed LED circuit. At some point in this design, the power supply will see a load changing from almost no current to very high current very quickly. With volt meter, you can not see total effect on lab power supply.
If you had a scope, a good test of power supply is a test load is one that changes a lot.

Could help you a lot if you created a spreadsheet to record your measurements in. You should measure circuit many times with different loads.

All these measurements will help you understand how the circuit is working and spot where it is not working as you think it should..

You are having problems with CC mode, Would be good to have all the measurements so you can compare before change to after a change.

It might help you a lot if you make your schematic a lot better.

You have
Power paths to output.

Power to run control circuit.

Reference block.

Output voltage sense.

Output current sense.

Voltage control circuit.

Current control circuit.

Output control circuit.

Some quick cut & paste moves could make circuit easer to under stand.

One way is to make top of schematic output & control with bottom power for control & reference..


Make a box around circuit that supplies power to control circuit.
Box top is between D3 & X-1
Box right is between x2-2 & LED1
Move this box down below where X1-3 goes across  schematic.

Move Reference block. down also where X1-3 goes across  schematic.

Box left is between IC1 & X3-3
Box right is between X4-2 & R19

What I am calling Output Control is area above R26.

Voltage Sense
  R21 & R27 could be vertical next to C17

Think these changes and a few more will change the schematic from cram it in to a more logical schematic.

A more logical schematic can let you see problem areas.
===






   


 

Offline radoczi94Topic starter

  • Contributor
  • Posts: 44
  • Country: hu
  • Electronics lover since 12y/o
Re: Lab Power Supply - The Lost Current
« Reply #30 on: February 05, 2018, 09:43:49 am »
Quote
oscillations without a scope
With no scope the best you can do is trying to measure AC with meter.
On DC mode with meter, only last digit value should change.
Thanks for the idea, will try that. My meter has freq counter function, if the amplitude is high enough, it is able to estimate the osc. freq. Will try that too. If I have time.

About the schematic.
I think I understand the schematic as it is, but yeah, it could be a lot easier to read, definitely needs some optimizing. I put in those regulating circuits and the reference. http://www.electronics-lab.com/project/0-30-vdc-stabilized-power-supply-with-current-control-0-002-3-a/ Here is the original schematic.

I want to redesign the PCB, it had some problems, the circuit needs some fusing and a handful of protection diodes. It went trough a lot of soldering and desoldering, now it looks like it's made by some blind one handed apprentice.
« Last Edit: February 05, 2018, 10:26:47 am by radoczi94 »
 

Offline C

  • Super Contributor
  • ***
  • Posts: 1346
  • Country: us
Re: Lab Power Supply - The Lost Current
« Reply #31 on: February 05, 2018, 12:36:13 pm »

Before you start changing thinks, would be good to understand the circuit.
Most times each circuit has problem areas. There are normally many ways to work around a problem, some better then others. Some fixes help in one area while making other areas worse.

In a lot of ways, the Electronics-lab link of last post is much better. 

Look at something simple.

Last schematic
Transistor labeled Q1
  With Q1 acting like a switch, why short out the output of U2? Not good for U2 especially when there is another way to do same thing.
Q1 could pull down on junction of R13 and Q2!

About the schematic.
I think I understand the schematic as it is, but yeah, it could be a lot easier to read, definitely needs some optimizing. I put in those regulating circuits and the reference. http://www.electronics-lab.com/project/0-30-vdc-stabilized-power-supply-with-current-control-0-002-3-a/ Here is the original schematic.

So if you really understand current schematic and compare it to this schematic some of the bad changes should really stand out.

These days it is not uncommon to have a load that is doing high current PWM. Not uncommon to have many different PWM loads. During testing/design a Lab Power Supply is the power supply for this new circuit. Until you know exactly what value of series resistor to use for that PWM dimmed LED the law power supply CC mode is the protector of that high dollar LED, would be a good idea for it to function properly.
With this being a simple cheap supply, you may not have the best but still want the best it can be.

So think about a load that changes
Initially with Max current of CC mode the output of this supply will be rapidly changing from VC mode at some voltage to some current value in CC mode. How does the lab power supply handle this? What changes in the lab power supply circuit.

To make it easy the negative output of supply is connected to ground reference

With respect to ground reference
what is point labeled 7 on last schematic doing?
What is cathode of D5 doing?
What is anode of D7 doing?
What is voltage across C5 doing?
Is C5 a good idea or is it causing problems?
What is base of Q2 doing? How does it change with low/high output voltage, low/high output current?

Each little change can effect output of the supply.

A change that makes one part of circuit better that at same time makes supply output worse is not a great change. Some changes that could effect output can be hidden by other parts of circuit, but there is a limit to how much.

Try to understand last post schematic and then look at your first posted schematic. What does each change do?

What is changes to circuit when load changes such that same CC mode current is two different output voltages?

I want to redesign the PCB, it had some problems, the circuit needs some fusing and a handful of protection diodes. It went trough a lot of soldering and desoldering, now it looks like it's made by some blind one handed apprentice.

If you are going to redesign the PCB, I would strongly suggest a redesign of circuit first to fix the many problems both versions have.

 

Offline radoczi94Topic starter

  • Contributor
  • Posts: 44
  • Country: hu
  • Electronics lover since 12y/o
Re: Lab Power Supply - The Lost Current
« Reply #32 on: February 05, 2018, 02:37:45 pm »
Quote
Last schematic
Transistor labeled Q1
  With Q1 acting like a switch, why short out the output of U2? Not good for U2 especially when there is another way to do same thing.
Q1 could pull down on junction of R13 and Q2!
You meant R15? Because if I pull down the R13-Q2 juntion, Q1 shorts C1 to ground, causing rather large currents and it would not be so good for the cap and the transistor too.
The opamp should not care about that short, it has "infinite" output short circuit protection to each power rail and ground. But still, it's better to do that trough a resistor (if you meant to pull down the base of Q2).

Quote
So think about a load that changes
Initially with Max current of CC mode the output of this supply will be rapidly changing from VC mode at some voltage to some current value in CC mode. How does the lab power supply handle this? What changes in the lab power supply circuit.

To make it easy the negative output of supply is connected to ground reference

With respect to ground reference
what is point labeled 7 on last schematic doing?
What is cathode of D5 doing?
What is anode of D7 doing?
What is voltage across C5 doing?
Is C5 a good idea or is it causing problems?
What is base of Q2 doing? How does it change with low/high output voltage, low/high output current?

On the last, linked schematic:
All with the respect of the output negative
-Point 7
It is the "high current" power rail. With a varying load it is also varying, depending on the main transformator and the capacity of the filtering cap.
-Cathode of D5
The cathode of D5 is on the ground.If we measure that point in the respect of the negative output, there should be the voltage drop across the shunt resistors. A square wave if the varying load is on the output.(If the load is pwm)
-Anode of D7
That's the negative voltage rail, there sould be the zener voltage+the shunt voltage drop.A square vawe with a dc offset.
-C5
should be same as the point 7. I think it is a deoupling cap, but not shure. Therefore, I do not really know if it is necessary or it is bad.
-Base of Q2
There should be the output voltage + Q4 and Q2 B-E voltage drop. At maximum voltage, there should be 31-32, at minimum voltage there should be the between 1-2 volts, depending on the transistors. At low output current there should be output voltage+ the B-E voltage drops, at max. current, there should be 0 volts. Or even negative voltages.

I wanted to make a stabilised supply for the IC-s, because my transformer is 28Vac and capable of 5amps. The rectified voltage would be much higher, than the opamps would tolerate, that is why tose parts were put in. Changed the voltage reference also, because I did not liked that circuit at all, never understood, actually. Why are these changes so critical?

Quote
If you are going to redesign the PCB, I would strongly suggest a redesign of circuit first to fix the many problems both versions have.


I was intended to do so, but I can not really found the problems with this circuit on my own. If you guys could help me find those probelms, explaining, why are they a problem and how can it be solved, I will correct them. I'm just not at the level of do this on my own. The more I learn about electronics, the more dumb I feel myself.
 

Offline C

  • Super Contributor
  • ***
  • Posts: 1346
  • Country: us
Re: Lab Power Supply - The Lost Current
« Reply #33 on: February 05, 2018, 09:22:57 pm »

Yes  R15

Q1 controls base of Q2

Think this through a step at a time.

What happens if I short Q2 base to the positive output?

U2 supplies current to other side of R15.  Assume for a moment that U2 output is shorted to point labeled circle 7, In practice U2 will be limited to less then this. What is max current possible.

After you think you have some answers look at end of this post.

Quote
-Base of Q2

If you think about it, Base of Q2 is output control. It will change based on voltage in VC mode and try to change by current in CC mode.

So  all of the circuit is changing based on current drop across R7 except for what is directly connected to circle 4 ( negative output.

C5
C5 is trying to prevent voltage changes between circle 7 & circle 4.
You stated that circle 7 changes based on current demand of output. So C5 is also trying to prevent fast CC mode response to a current change.
My initial thoughts is that C5 is bad, makes supply worse in both CC & VC modes.


Quote
I wanted to make a stabilised supply for the IC-s, because my transformer is 28Vac and capable of 5amps. The rectified voltage would be much higher, than the opamps would tolerate, that is why tose parts were put in. Changed the voltage reference also, because I did not liked that circuit at all, never understood, actually. Why are these changes so critical?

look at last schematic at U1
Think of how an op amp works, Output is like a variable resistor to +V and a variable resistor to -V.
Effects of Changes to +V are reduced by power supply rejection ratio of op amp

Your first posted schematic has R13 changing by load current & filtered some by C10
You have power supply rejection ratio of U1 vs rejection of TL431.

But by creating the stabilised supply  for IC's you now have all IC's having to fight against Load current changes on their supply lines vs IC output.
You have a bunch of capacitors connected to what is noise for what you want on output. Examples are C3, C4 C5,C7,C9,C6.

So you are starting from last schematic that has a poor CC Mode and then increasing source voltage creating more problems.

Might be a good idea to fix CC Mode in such a way that more source voltage becomes less of a problem, instead of hack the hack the hack.

There are many ways to do this.
With no scope and limited testing capability you will get a "I think it is good lab power supply". With changes you have made from last schematic, your limited testing is saying "I know it is broke/bad"

A simple circuit for a power supply that has a CV mode and a CC mode is as follows.
A pull up resistor that turns on the output to MAX.
Your Voltage reg & Current reg then limit the max.
Your current regulator's diode is setup to do this. Needs to control output directly not via voltage regulator as this adds a second control loop that is talked about.
Output of Voltage regulator needs to connect to diode with anode of two diodes common and controlling output.

So you have one diode for CC Mode & one diode for VC mode & you could add a third diode or more diodes for shutdown or other limit conditions.

With low side current sense you have a problem that any control circuit current changes connected to negative output also effects your current sense.

If this is a 0V to 30V supply, You could think of it as a 5V to 35V supply.
The negative output regulates the -5V supply you have.
With -5v Controlled by output negative, you could then use this to control +V to op amps.  You have just removed supply bounce by current.
This leads to control circuit power around negative output.

You could think of floating control circuit power with positive output as the reference. This can make it easy to have higher voltages and make high side current sense easer. The down side is that today most think of negative ground.

What I am trying to show is how you think about a circuit can change what you see. A small change can make a huge difference.

So step one is understanding both circuits better and comparing what the differences do.

Any change needs you to understand the effects of the change.

Answer
Q1 controls base of Q2 question.
You missed that when Q1 turns off Q2, collector of Q2 can rise and will not short out C1

D10 keeps Q2 base from going below  circle 3 (positive output ) by diode drop. At a Diode drop below Q2 is off, This in tern means that Q4 is off. The only thing left trying to hold base of Q2 high is R15.
Max current is R15 with voltage of circle 7 - circle 4


Now Think of what U2 will do.  Say output was set at 1 V.
Pulling Q2 base low will cause output voltage drop which will cause U2 to turn on harder trying to make U2 output more positive.
So at start very little current through R15, but will increase to U2 max output voltage current with Q1 having to conduct this current.


 
The following users thanked this post: radoczi94

Offline radoczi94Topic starter

  • Contributor
  • Posts: 44
  • Country: hu
  • Electronics lover since 12y/o
Re: Lab Power Supply - The Lost Current
« Reply #34 on: February 05, 2018, 11:17:29 pm »
Huh, that is a lot of information in one go. Now I see, why this circuit is a total failure. Maybe with extra, stabilised supply voltages for the opamps,with a decent voltage reference and with the OR diode gate, this thing will be able to work like as it should.  :horse:

Or maybe I should just look for another schematic. Designing for another 4-5 days, taking away from my precious time, building it, for 2 days, experimenting for weeks, because why would it work for the first time. Or just getting an old, well proven circuit from somewhere, understanding how it works and building it as it is. I need to think this trough.

I really wanted to build an indestructible, foolproof power supply, that is why I used that massive transformer and heatsink. With the case I built, it's weight is around 20 kgs, and not even used the full 5 amps of the transformer.

Anyway, thanks for all help, it's really appreciated!
« Last Edit: February 06, 2018, 12:59:40 am by radoczi94 »
 

Offline C

  • Super Contributor
  • ***
  • Posts: 1346
  • Country: us
Re: Lab Power Supply - The Lost Current
« Reply #35 on: February 06, 2018, 07:05:35 pm »
I really wanted to build an indestructible, foolproof power supply, that is why I used that massive transformer and heatsink. With the case I built, it's weight is around 20 kgs, and not even used the full 5 amps of the transformer.
No such thing exists. All have limits and is up to user to keep connected load in limits.
Example a 30v power supply and idiot connects a 48v battery to output.
For power supply to protect against this is very hard to do, best to think of limiting the damage.
Example 1000F cap bank connected to output. The power supply would be at current limit for days. How long can power supply survive with output at 0V & Max current.
Sure you can add to this list.
So the best you can do is the limit damage and user has to supply remaining protection.

One thing I would not pass on is using what you have to learn. A lot could be learned.
Back in the 80's HP was doing a lot of internal research. A lot of things HP sold were first designed,built & used to cure a research problem. Often a power supply consisted of many modules or boards. This makes upgrades or new designs easer.

So before you look for something better, would be good to know what to look for.

The separate parts have to work together to make the whole thing good.

You can have very good blocks in schematic, but they have to be connected together in a way that makes total good.

You also have to keep in mind that nothing is perfect.

You have a cap across output terminals.
For CC mode you want no cap or small cap. The reason is that cap charge gets dumped into a short on output.
For CC mode any cap on output slows rate of current change to hold desired current.

For VC mode to hold constant output you want that cap huge.
For VC mode to detect a load change on output, smaller lets circuit see same change sooner.

The right cap or caps can stop supply from oscillations.
There are more things

So output cap is war of sizes with an option of making CC or VC better.
Then it could be that expected loads will survive the dump of output cap on short such that a little larger cap is ok.
Or load is not effected badly with a voltage change.

For a lab power supply you need to think of all the nasty loads that could be connected.
Step one for lab power supply is to protect connected load.
If connected load is real bad then step two is for lab power supply to try to protect it's self.

Then you have option of big load power supply & small load power supply where you pick best for load

So at cost of some of your time you could learn a lot from what you have.

Keep in mind the many ways to do things.
One change from last schematic to first is power and reference.

Last schematic  used op amp's power supply rejection to lower noise.
First schematic I see creating problems by coupling noise into control power.

Again bad schematic hides details.

Take last  schematic and redraw it.
Power section
 on left is D1-D4, R1 & C1
on Right is
   current sensor R7
   Output control Q2,Q4 & R16
   ?  C3
   Some protection by D10, D11
   Output filter C7


What is left is Control circuit.

Now if you do not have ability to easily do this, use your versions.

Now the Control circuit needs inputs to function.
Voltage and Current inputs should be differential.
Voltage sense is connection to  Circle 3 and Circle 4
Current sense is connection to each side of R7

For control circuit to control output you could need connections to circle 7, circle 3, circle 4, left side of R7
 
Note that control and sense should be separate wires or paths, A Kelvin connection.

Time to learn
Does current PC board do this separation?
Does your first do this?
It's not shown on any schematic posted, yet is simple and cheap to do.
If you look for new design, something to look for.
Note also this separates the high heat parts, parts that fail on heat.

With the schematic redone, You should be able to see some problem areas and have some ideas on how to improve circuit

So a great schematic is very important.
Makes it easy to understand the circuit quickly. Can show problem areas & and many more things.

Now for control section separate out it's parts.
Reference section
   U1, D8, R4,R5,f6
it;s outputs are connection to R18 & circle 6

Should note a few things here
Power is via U1 V+ & V-
Part of differential sense is by R4 & R6 being connected to circle 4.
Should note that V+ and current set pot is main source current changes.

Should note that last schematic makes great use of op amp's power supply rejection to lower noise. A change to less is not an improvement.

The V+ connection of U1,U2 & U3 is connection to circle 7 for output control.
Adding a second line across to circle 7 could show this.

So I see a good first step is to learn from what you have now & fix schematic to show details and make understanding easer.
Take a huge number of measurements with your meter with different settings & loads. Save this to look back to to see a change is good or bad.
Each change a new spreadsheet of measurements.

Unknown is how well the last schematic functioned before changes.
Could be original has problems or you created problem for CC mode. Only measuring will tell.

One common thing is to have separate power for control & load supply.

So you have a massive transformer, would be a shame not to use it if you see a need for the power supply it can be a part of.
The way I see a single PCB is a way to mass produce something.
If you want something good that can upgrade then many PCB's or circuits is better. A good circuit design also makes this easy.

If you look at old HP supplies you often see it is made of many modules. A module is often used in many different supplies.

Think of future some, today you build best power supply you can. Next year you could think of it as junk. Easy to change junk could become new supply better then first.
 
The following users thanked this post: radoczi94

Offline radoczi94Topic starter

  • Contributor
  • Posts: 44
  • Country: hu
  • Electronics lover since 12y/o
Re: Lab Power Supply - The Lost Current
« Reply #36 on: February 06, 2018, 09:21:47 pm »
Quote
No such thing exists. All have limits and is up to user to keep connected load in limits.
Example a 30v power supply and idiot connects a 48v battery to output.
For power supply to protect against this is very hard to do, best to think of limiting the damage.
Example 1000F cap bank connected to output. The power supply would be at current limit for days. How long can power supply survive with output at 0V & Max current.
Sure you can add to this list.
So the best you can do is the limit damage and user has to supply remaining protection.

Yes, I know that, i did not really meant it as it is, just basic protection and really good cooling. I have some desperate ideas, like having a huge parallel diode reverse biased and a fuse on the output in series (may be fast blow), it could prevent a reverse polarity battery for example. A huge truck battery will kill the whole thing anyway, so not much protection, but better than nothing. A series diode with proper cooling could help also, but the voltage drop can be an issue.

I see now, I can not really build an ok-ish psu without external voltages. So I have to go to the store, and buy a pair of small transformers, build a good secondary supply for the control circuit. Even if I hate wiring work.

I was not aware of the Kelvin-connection thing.I did not knew, that such thing exist. Saw in a bunch of EEVBlog vids, but never knew, if it is intended or not. Therefore my design has absolutely noting like that. Thank you for that!

Otherwise my pcb design is a bit beta-ish, tried to maintain wide traces for the 3A current paths, so the contol gone into a tiny little space between huge polygons.Never again.

I think I will do some further investigation on what is happening around the circuit, to see if I can make it work, but I don't really feel the motivation to stick to this schematic and make it an acceptable psu.

My buddy sent me an old schematic he built two of this. It has separate power supplies, opamps controlling trough an OR gate etc. A lot of things I seen in older Tektronix schematics. The schematic itself is a perfect example of how not to draw a schematic. My only concern is the old parts used. If it works, I can replace the 741s with better opamps, like OP07, which has excellent DC characteristics ( the BW is a bit low, so certainly not the fastest opamp on the market). The other thing is the 723. It can be used as a switching regulator so it's not hard to make it oscillate. https://drive.google.com/file/d/0B7yeYY6jyIVVaHZpTlhTQVl2b0wzVHpEUXAtS1VtU1E0b2o0/view?usp=sharing Here it is.I kinda like it, have not found any better yet. The description is pretty good too, explains the principle.
« Last Edit: February 06, 2018, 11:21:10 pm by radoczi94 »
 

Offline C

  • Super Contributor
  • ***
  • Posts: 1346
  • Country: us
Re: Lab Power Supply - The Lost Current
« Reply #37 on: February 06, 2018, 11:58:21 pm »

In my past I do not remember any Tektronix supplies.
Used a lot of HP and a few other brands.

If you spend some time looking at old, look at HP power supplies

Protection
  As a last resort protection the simple thing is have control circuit blow a fuse.
On input side you have many choices. 
Blow line fuse, blow a fuse on secondary
On output side you want to get disconnected from load.
Think the best device is the SCR or triac to do this.

Quote
I see now, I can not really build an ok-ish psu without external voltages. So I have to go to the store, and buy a pair of small transformers, build a good secondary supply for the control circuit. Even if I hate wiring work.
You can use any type of supply to supply control circuit power. If you have any electronics junk you have something you could probably use.

The important part is the separation.
Think of this, you have a possible large current draw on input bridge diodes. Current = voltage drop, changing current = noise
A simple change is just using a second diode bridge and filter. You gain some isolation from the noise. Separate source is more noise isolation.
Actual power voltages is not that critical. Some designs are easer if control power exists below or above output terminals of supply.
Important thing here is op anps have a sweet spot in center between supply with problems at close to supply rails.

Kelvin-connection
  You may have seen some current sense resistors with a Kelvin-connection. For a lab power supply you want a Kelvin-connection to output terminals for voltage sense. Better supplies add remote voltage sense terminals to extend Kelvin-connection to load.
Can be important to use on a PCB as well.  Think of When this changes where do the currents go/flow. Think of this, a voltage based temp sensor and a current based temp sensor. On end of a 1000 feet of wire one is crap with other still great.

Quote
Otherwise my pcb design is a bit beta-ish, tried to maintain wide traces for the 3A current paths, so the contol gone into a tiny little space between huge polygons.Never again.
Guess why you see separate boards for high current or one small area for high current.

Quote
I think I will do some further investigation on what is happening around the circuit, to see if I can make it work, but I don't really feel the motivation to stick to this schematic and make it an acceptable psu.

Investigate to learn. If you chop up the schematic like I suggest, then future is replacing blocks on schematic.

The 723 is ok for a cheap simple power supply but is not great.
The link you post from my quick read is not great.

Keep in mind that you want the lesser value of Voltage or Current.
Link Instead of simple you have a mess.
In link voltage control A1 if not modified wants to turn on output more  in current limit mode while current control needs to turn it off more.
Any change in A1 output messes up good current control!

Add a pull up resistor to output transistor causing output to turn on full blast.

A2 is all ready, Too much current and it pulls down lowering current to set point.

A1 needs a diode and then will work same way and reduce to set point.

Just want you want for a supply that has CC & VC
Lower output level win.
Now you have two separate circuits controlling output. You do not have to modify one for other to work properly.

Now look at range for the change.
A diode has a voltage vs current curve. The curve is fairly flat but not flat. Change over goes from one output control, a current to the two outputs equal , 50% of one controlling. If it takes 20 ma to hold down output then you would look at two spots on diode curve, 20 ma & 10 ma. and get some voltage value.
Now due to feed back this becomes mostly hidden and change from CC to VC is fast with few if any problems.
 
Want something great just read a HP manual for a power supply. Every thing is in it, how it works,  schematic, parts list and often PC Board layout.

Gave you a hint above but you may have not caught it.
Current goes the distance, You can connect low side circuit to high side circuit with current. A current change then can control output.

 
 

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 2817
  • Country: au
Re: Lab Power Supply - The Lost Current
« Reply #38 on: February 07, 2018, 12:12:18 am »
Because of the way the voltage adjust potentiometer is configured as a VR in the Elektor design, it is not fail safe. If the potentiometer becomes scratchy, the PSU output could spike to full voltage. I think it is better to use the potentiometer as a potentiometer to directly vary the reference voltage. Including a wiper bleed resistor would make it fail safe. The reference voltage will need to be negative.
I don't like the remote sensing. It should be omitted.
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 

Offline radoczi94Topic starter

  • Contributor
  • Posts: 44
  • Country: hu
  • Electronics lover since 12y/o
Re: Lab Power Supply - The Lost Current
« Reply #39 on: February 07, 2018, 09:31:44 am »
One thing I do not really understand about this old schematic. 723 makes the reference. As far as I know emitter followers have a voltage gain of 1. How is 0-30 volts on the output, if the Vref is 7,15V? I'm shure, that some resistor makes this magic. Maybe R17 is talking back? Why don't just pull it up to the collectors? I've seen in some HP and Tektronix supplies, the base were just pulled up.

What I do not like about this, is the 723. Maybe a little bit too crusty. I'm shure there are much better solutions for a good reference.

Because of the way the voltage adjust potentiometer is configured as a VR in the Elektor design, it is not fail safe. If the potentiometer becomes scratchy, the PSU output could spike to full voltage. I think it is better to use the potentiometer as a potentiometer to directly vary the reference voltage. Including a wiper bleed resistor would make it fail safe. The reference voltage will need to be negative.
I don't like the remote sensing. It should be omitted.

You mean, like the current set pot? Dividing the vref with the pot itself?
 

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 2817
  • Country: au
Re: Lab Power Supply - The Lost Current
« Reply #40 on: February 07, 2018, 12:10:49 pm »
In your old design, The output Darlington functioned as an Emitter follower which buffered the output of the CV op-amp which was able to swing by the whole range required for 0-30 volts output.

In the Elektor design and many others, everything is referenced to the positive output. This includes the Darlington Emitter, 0 volt reference for the low voltage split rails(+/- 12 volts) and the voltage reference source, a TL431 possibly.

A good way to properly drive the Darlington's Base is to source current from a voltage source that is always a fixed amount higher than the Emitter so that there is enough available Base drive current even when rising Emitter voltage approaches Collector voltage. This can only be from the +12 volt rail.

 This makes it possible to fully turn on the Darlington. This is desirable when the PSU is operating at high output voltage and current so that as much as possible of the unregulated voltage on the Collector can be transfered to the output. The regulator will be said to have a low dropout voltage, a good thing.

In the Elektor design, the Darlington can no longer be thought of as an Emitter Follower. It is a series pass element conducting current from the unregulated rail to the output under control of either the CV op-amp or CC op-amp.

Google "Darlington pair"
 Do you understand what op-amps do?


One thing I do not really understand about this old schematic. 723 makes the reference. As far as I know emitter followers have a voltage gain of 1. How is 0-30 volts on the output, if the Vref is 7,15V? I'm shure, that some resistor makes this magic. Maybe R17 is talking back? Why don't just pull it up to the collectors? I've seen in some HP and Tektronix supplies, the base were just pulled up.

What I do not like about this, is the 723. Maybe a little bit too crusty. I'm shure there are much better solutions for a good reference.

Because of the way the voltage adjust potentiometer is configured as a VR in the Elektor design, it is not fail safe. If the potentiometer becomes scratchy, the PSU output could spike to full voltage. I think it is better to use the potentiometer as a potentiometer to directly vary the reference voltage. Including a wiper bleed resistor would make it fail safe. The reference voltage will need to be negative.
I don't like the remote sensing. It should be omitted.

You mean, like the current set pot? Dividing the vref with the pot itself?
Yes
« Last Edit: February 07, 2018, 12:12:22 pm 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: radoczi94

Offline radoczi94Topic starter

  • Contributor
  • Posts: 44
  • Country: hu
  • Electronics lover since 12y/o
Re: Lab Power Supply - The Lost Current
« Reply #41 on: February 07, 2018, 12:53:19 pm »
So the floating supply's ground is referenced to the output.This means, that the floating positive rail and the Vref is on higher potential than the output, and the floating negative rail is lower. This is because the floating supply'ss ground tied up to the output by R23 47R. Right?

This way the Vref is able to source current to the Darlington bases, because the Vref's potential is higher than the emitter(s).

As I understood, the opamps are sucking away current from the Darlington base, or the CV oamp tries to push it, if the output voltage is not high enough, but it can not be happen, because the diode is reverse biased,so the Vref opens the Darlington more.

If I'm right, the Vref needs to be capable of providing enough current to drive the transistors, and the voltage dividers at the opamps.In the same time, it needs to be stable with low drift. Maybe a bare TL431 is not enough for this, a lot depends on the beta of the Darlington. Edit: The total beta of the darlingtons in worst case 400, so 8-10mA is enough for 3-4A output.

A good way to properly drive the Darlington's Base is to source current from a voltage source that is always a fixed amount higher than the Emitter so that there is enough available Base drive current even when rising Emitter voltage approaches Collector voltage. This can only be from the +12 volt rail.

There are some violations to this, because there is some voltage drop across the shunt. But we want this, so the voltage drop is compensated by the control circuit. Right?

Because of the way the voltage adjust potentiometer is configured as a VR in the Elektor design, it is not fail safe. If the potentiometer becomes scratchy, the PSU output could spike to full voltage. I think it is better to use the potentiometer as a potentiometer to directly vary the reference voltage. Including a wiper bleed resistor would make it fail safe. The reference voltage will need to be negative.
I don't like the remote sensing. It should be omitted.
Why would be practical to have a negative reference voltage? Just in case of pots getting scratchy?
« Last Edit: February 07, 2018, 03:01:28 pm by radoczi94 »
 

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 2817
  • Country: au
Re: Lab Power Supply - The Lost Current
« Reply #42 on: February 07, 2018, 04:34:30 pm »
So the floating supply's ground is referenced to the output.This means, that the floating positive rail and the Vref is on higher potential than the output, and the floating negative rail is lower. This is because the floating supply'ss ground tied up to the output by R23 47R. Right?
That's right. That area of the circuit can be left the same but with the sense wire tied directly to the + output terminal.
This way the Vref is able to source current to the Darlington bases, because the Vref's potential is higher than the emitter(s).
As I understood, the opamps are sucking away current from the Darlington base, or the CV oamp tries to push it, if the output voltage is not high enough, but it can not be happen, because the diode is reverse biased,so the Vref opens the Darlington more.
If I'm right, the Vref needs to be capable of providing enough current to drive the transistors, and the voltage dividers at the opamps.In the same time, it needs to be stable with low drift. Maybe a bare TL431 is not enough for this, a lot depends on the beta of the Darlington. Edit: The total beta of the darlingtons in worst case 400, so 8-10mA is enough for 3-4A output.

Both low voltage rails should be regulated for best performance and the drive current for the Darlington should come from the +12v rail. You won't have a positive reference if you decide to go for a negative reference anyway.
The low voltage rails don't strictly have to be +/-12v. A bit lower would be ok if your small transformer doesn't output enough to regulate 12v.

A good way to properly drive the Darlington's Base is to source current from a voltage source that is always a fixed amount higher than the Emitter so that there is enough available Base drive current even when rising Emitter voltage approaches Collector voltage. This can only be from the +12 volt rail.
There are some violations to this, because there is some voltage drop across the shunt. But we want this, so the voltage drop is compensated by the control circuit. Right?
That won't be a problem.

Because of the way the voltage adjust potentiometer is configured as a VR in the Elektor design, it is not fail safe. If the potentiometer becomes scratchy, the PSU output could spike to full voltage. I think it is better to use the potentiometer as a potentiometer to directly vary the reference voltage. Including a wiper bleed resistor would make it fail safe. The reference voltage will need to be negative.
I don't like the remote sensing. It should be omitted.

Why would be practical to have a negative reference voltage? Just in case of pots getting scratchy?
Pots going scratchy is all too common. It is actually the wiper losing contact with the track. Have a think about what happens next if the pot is configured as a VR as in the Elektor design?
As far as I can see, the reference needs to be negative if the control is to be wired as a potentiometer rather than a VR.
Time to start sketching some circuits.
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 

Offline C

  • Super Contributor
  • ***
  • Posts: 1346
  • Country: us
Re: Lab Power Supply - The Lost Current
« Reply #43 on: February 07, 2018, 04:43:01 pm »
Simple question

The Elektor link

How much trust should you give to an article in a mag where the operating principles does not match the final design?

The article describes how it works in figure 2.
A classic design where CC mode is hacked on to a VC mode regulator.
Some of this hack shows in final circuit.

The final circuit figure 3 does not match the theory of operation in figure 2.
Here you have the 723 suppling power to R9 & have the analog OR for CC & CV ( D4, D5)

So Author could not describe a simple analog OR to make sense of circuit.

Then you have if it was truly a VC & CC regulator, why only one indicator of mode?  How much effect does driving that indicator have on mode switch?
Note that it is possible that a load is CC & VC at same time. With load noise a good supply can be changing fast between the three states.

With out a scope, trying to update this very old design I would say is a fail.
 It is what it is an attempt at a cheap precision power supply.

some thing to keep in mind, If you are going to be building analog circuits a Positive & Negative supply is nice to have during design.
 

Offline radoczi94Topic starter

  • Contributor
  • Posts: 44
  • Country: hu
  • Electronics lover since 12y/o
Re: Lab Power Supply - The Lost Current
« Reply #44 on: February 08, 2018, 01:41:48 am »
Quote
Both low voltage rails should be regulated for best performance and the drive current for the Darlington should come from the +12v rail. You won't have a positive reference if you decide to go for a negative reference anyway.
The low voltage rails don't strictly have to be +/-12v. A bit lower would be ok if your small transformer doesn't output enough to regulate 12v.
I think I will go with +-15V. I have a few salvaged 2*18V*0,15A PCB mains transformers, I want to use them. Then put around some 78-7915 regulators some filtercaps, and a preload resistor or whatnot.


Quote
Pots going scratchy is all too common. It is actually the wiper losing contact with the track. Have a think about what happens next if the pot is configured as a VR as in the Elektor design?
As far as I can see, the reference needs to be negative if the control is to be wired as a potentiometer rather than a VR.

If the pot works as just a voltage and the contact is failing in the wiper, the output goes low, because the voltage difference between the inputs is positive. If the VR fails, the goes to full chooch.Oops.

Simple question

The Elektor link

How much trust should you give to an article in a mag where the operating principles does not match the final design?

The article describes how it works in figure 2.
A classic design where CC mode is hacked on to a VC mode regulator.
Some of this hack shows in final circuit.

The final circuit figure 3 does not match the theory of operation in figure 2.
Here you have the 723 suppling power to R9 & have the analog OR for CC & CV ( D4, D5)

So Author could not describe a simple analog OR to make sense of circuit.

Then you have if it was truly a VC & CC regulator, why only one indicator of mode?  How much effect does driving that indicator have on mode switch?
Note that it is possible that a load is CC & VC at same time. With load noise a good supply can be changing fast between the three states.

With out a scope, trying to update this very old design I would say is a fail.
 It is what it is an attempt at a cheap precision power supply.

some thing to keep in mind, If you are going to be building analog circuits a Positive & Negative supply is nice to have during design.

Had a feeling about that, but only concentrated on that voltage setting pot. Which was basically the 90% of the theoretical part.As far as i know, this thing is able to work, my buddy made two of these and both worked first-second try.



Spent half a day googling around to see in search of a usable schematic, nothing. I will probably end up poking parts into breadboard. With only a multimeter, for now. I wanted to buy a scope since ages, so I guess, the time has come.
 

Offline C

  • Super Contributor
  • ***
  • Posts: 1346
  • Country: us
Re: Lab Power Supply - The Lost Current
« Reply #45 on: February 08, 2018, 04:06:14 am »

Google
  find a model number for an HP supply and find the manual using google.
As I said most are fantastic with everything. Was common in 80's to have everything in manual so user could repair it if necessary.

One thing you will find is that controlled power up and down is built in to design not a tack on.

Little things are included in a great design.   You might design the voltage control pot to change from 0 to Vref. A pot failure becomes 0 a normal setting.
The great design can have a 0 output of supply equal some %  not 0 of reference. A pot failure becomes something not normal and electronics can detect problem and shut down power supply with an error. That analog OR lets you have many circuits that can shutdown supply.

So google and study a HP supply and see what is missing in the simple hacks.  Manual not schematic as it's in manual.

The problem with a supply is often not the DC it's the load caused AC changes causing the power supply to fail. It really takes some fancy test equipment to check the AC part.

One thing some forget is what a power supply is. Most are crippled power op amps. Think it through If you had an op amp that could do X amps with a range of Y volts you have a power supply. Removing half of op amp's power stage is a cost & heat savings. The better lab supplies still have some sink capability.
Think of what a direct coupled audio amp is. Could be thought of as a bidirectional power supply.

So You have a transformer, what are the spec's, Just one winding?

Old school is using pot's to set things, New is using DAC's. Would be good to design so that ether can be used. Some were controlled by resistor decades.

Keep your eyes open. In first schematic you posted, you had Q4 as an NPN. A change to a PNP  would have let you have a supply to limits of the 2N3055 while having lower voltage control circuit.

6227B Power Supply
https://literature.cdn.keysight.com/litweb/pdf/06227-90001.pdf?id=734411
DC Power Supplies - Discontinued and Obsolete Products
https://www.keysight.com/en/pc-1000002054%3Aepsg%3Apgr/dc-power-supplies-discontinued-and-obsolete-products?nid=-536902299.0.00&cc=US&lc=eng


.
 
 

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 2817
  • Country: au
Re: Lab Power Supply - The Lost Current
« Reply #46 on: February 08, 2018, 07:36:58 am »
Most of us would agree that for this type of PSU project that the topology that has everything referenced to the +out is a good starting point. The rest is details, not all will be trivial ones.
radoczi94, at the heart of understanding what is going on, it is essential that you understand op-amps. Do you understand how op-amps function and how they are applied?
I'm planning on doing some related experiments shortly.
HP 54645A dso, Fluke 87V dmm,  Agilent U8002A psu,  FY6600 function gen,  Brymen BM857S, HAKKO FM-204, New! HAKKO FX-971.
 

Offline radoczi94Topic starter

  • Contributor
  • Posts: 44
  • Country: hu
  • Electronics lover since 12y/o
Re: Lab Power Supply - The Lost Current
« Reply #47 on: February 08, 2018, 06:58:52 pm »
The various protection circuits are neccessary for shure. One thing I already did, is a switch-on delay with a 555 and a relay switching the output. It works, but yeah, switch contacts doesn't really like switching DC. There could be a switch-off circuit too, like in the Electronics-Lab design, monitoring the AC, even on the mains side trough an optocoupler and then connecting to the OR gate. Or even a thermal shutdown (I don't think that this is absolutely necessary in my case.)

But I can throw all the protection to the trash can if the 2 main function doesn't work. Since I do not have a scope (yet), what I can do now is just sit down and learn as much as I can. And maybe search for more suitable parts, do some sketches, collect ideas.

That massive transformer is 250VA 2X(0-3-12-24V)*5A. For this thing, 4A output current would be comfortable at 25V output. Could have an automated tap selector circuit. At least for the 12V tap. I think the heatsink could handle it either way, but why piss away 100W?

What i know about opamps...well, basics. Understood the DC operation well, and some of the AC. I found a lot of informative stuff on the Analog Devices website, there is a lot of useful stuff. I'm reading those, also found some Linear stuff about undesirable oscillations.
« Last Edit: February 08, 2018, 07:11:51 pm by radoczi94 »
 

Offline xavier60

  • Super Contributor
  • ***
  • Posts: 2817
  • Country: au
Re: Lab Power Supply - The Lost Current
« Reply #48 on: February 09, 2018, 12:59:34 am »
Just because a PSU has CC, doesn't really mean that it's short circuit protected. Even my Agilent U8002A will supply over 20 amps into a sub ohm load for 100us before it begins to sluggishly current limit.
I mocked up a CV/CC regulator to experiment with the CC response. The cause of the delayed CC response was because the output of the CC op-amp normally sits at close to its full + rail voltage. When the PSU is suddenly overloaded, it then takes a long time for the CC op-amp's output to slew down to the point where it takes control of the Base.
One of the reasons is that the loop compensation capacitor is usually connected directly between the op-amp's output and inverting input. I have connected the capacitor to the other side of the ORing diode so that the op-amp's output can slew at its maximum rate until the ORing diode conducts.
Because the LM358 that I have used doesn't slew that fast anyway, 0.3V/us, I have put a diode and LED in its feedback path that keeps its output at 2.2v so that it doesn't have far to swing down to take control of the Base. I am ordering some faster NE5532 op-amps to see if I can omit this extra complication.

Extra: I have used an LED for the CC ORing diode to give CC indication.
« Last Edit: February 09, 2018, 01:12:27 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 xavier60

  • Super Contributor
  • ***
  • Posts: 2817
  • Country: au
Re: Lab Power Supply - The Lost Current
« Reply #49 on: February 09, 2018, 09:53:50 am »
The 13V rails in my regulator experiment are supplied from 2 regulated  plug packs. I noticed that the regulator's output would pulse to full voltage when I powered down the plug packs only. I have added protection to prevent this from happening. Q3 cuts Base drive current when the +/-13v rails drop by a few volts.
It was also an ideal place to add an on/off switch.
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