Author Topic: Help with how this PSU works. (With revised design)  (Read 7502 times)

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Offline Evangelopoulos PanagiotisTopic starter

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Help with how this PSU works. (With revised design)
« on: February 01, 2020, 03:45:58 pm »
[EDIT: Check out the revised design at the bottom]

Greetings everyone,

I found this linear power supply in a dumpster a while back and only now I decided to reverse engineer it.
I did this mainly because I wanted to understand how it works and later maybe build my own.
I created the schematic of the thing but I can't understand how it really works. Can someone provide me with an explanation on how it works?
I'm familiar with all its components but I can't piece them together right.

Moreover, there is a 10uF that goes from the positive to mains earth (not shown in the schematic). What's its use?

Finally, what changes should I make if I need it to supply more power? The poor TIP3055 gets quite hot at 4A even with that huge ass heatsink.

EDIT: I forgot to mention that I have already repaired it a few times. The TIP3055 seems to get burned really easily. I just fixed it again and I was about to test a DC-DC converter when a weird spike caused the thing to get destroyed again... Is it a good idea to parallel a couple of TIP3055s?

Thank you in advance for your time and effort,
E.P.
 
« Last Edit: May 12, 2020, 05:37:06 pm by Evangelopoulos Panagiotis »
 

Offline duak

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Re: Help with how this PSU works.
« Reply #1 on: February 02, 2020, 02:50:44 am »
Greetings EP,

Yes, you will need more TIP3055 transistors in parallel.  In the best case, each can dissipate 90 W.  Under short circuit conditions they will have to dissipate 4 A x 45 V = 180 W.  Two is the minimum and I would consider 3 or 4.  That heat sink looks small for the possible power dissipation but it might be possible to mount four TIP3055s by using both sides. 

There should be a 1K0 resistor between the emitter of the BD243C and the right hand side of the 0R33 resistors.  This helps reduce collector to base leakage current at high temperatures.

Each TIP3055 will require a 0R33 resistor between its emitter and the ammeter terminal to help share current between them.  Each TIP3055 should also have a 10R resistor in series with its base to prevent oscillation.

The current sense voltage developed across each 0R33 resistor should connect to the - input of op-amp A with its own 4K7 resistor.  The 4K7 connected to the + input of the op-amp can be replaced with a 1K0 to equalize the resistances the op-amp sees.   The C ADJ variable resistor will have to be adjusted to maintain a safe maximum current.
« Last Edit: February 02, 2020, 02:52:39 am by duak »
 

Offline Evangelopoulos PanagiotisTopic starter

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Re: Help with how this PSU works.
« Reply #2 on: February 02, 2020, 08:09:47 am »
Greetings EP,

Yes, you will need more TIP3055 transistors in parallel.  In the best case, each can dissipate 90 W.  Under short circuit conditions they will have to dissipate 4 A x 45 V = 180 W.  Two is the minimum and I would consider 3 or 4.  That heat sink looks small for the possible power dissipation but it might be possible to mount four TIP3055s by using both sides. 

There should be a 1K0 resistor between the emitter of the BD243C and the right hand side of the 0R33 resistors.  This helps reduce collector to base leakage current at high temperatures.

Each TIP3055 will require a 0R33 resistor between its emitter and the ammeter terminal to help share current between them.  Each TIP3055 should also have a 10R resistor in series with its base to prevent oscillation.

The current sense voltage developed across each 0R33 resistor should connect to the - input of op-amp A with its own 4K7 resistor.  The 4K7 connected to the + input of the op-amp can be replaced with a 1K0 to equalize the resistances the op-amp sees.   The C ADJ variable resistor will have to be adjusted to maintain a safe maximum current.

Thank you duak for the tips. I shall try them as soon as I get my hands on some more TIP3055s. I ran out of them... Hopefully sharing the load across 4 of them will make it a bit more durable. I'm guessing the extra ballast resistors of 0.33Ω should be rated at 5W as well right?
 

Offline aheid

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Re: Help with how this PSU works.
« Reply #3 on: February 02, 2020, 12:26:26 pm »
I drew the most important lines for the control circuit in the attached image.

The core regulation is provided by the pass transistor, used as a variable resistor. In order for the NPN pass transistor to conduct, the base must have a positive voltage compared to the emitter. Since the emitter is hooked directly to the positive output of the supply, that means the control circuit must have access to voltage above the positive output rail.

This has been solved by using the secondary windings on the transformer to provide an independent set of power rails, and then by connecting the "ground" of the secondary rails to the positive primary rail (purple in drawing). The LM7808 then provides a stable 8V above the positive output for the control circuit.

The control circuit consists of two main elements:

1) voltage control (red tones in drawing)
2) current control (cyan tones in drawing)

They both use opamps to drive the pass transistor through a diode-OR configuration. The opamp which has the lowest output voltage will be in control. Thus the control opamps can ever only actively reduce the output voltage. This means that even if the voltage controller raises its output voltage, in an effort to increase the output voltage, the current controller can override this by keeping its output low. It also means that without either opamp "putting the breaks on", the pass transistor will be saturated, giving maximum output.

The current controller is relatively easy to figure out. It compares the voltage across a current sense resistor on the positive output with an adjustable reference voltage derived from the control circuit rails. If the current sense voltage (- input ) is higher than the reference voltage (+ input ), the opamp will pull its output down. This will reduce the voltage of the pass transistor base (in darlington configuration), increasing the resistance of the transistor and thus reducing the voltage on the positive output. This setup works because the control circuit ground is tied to the low side of the current sense resistor.

The voltage controller works similar except it compares the output voltage, but does so in a slightly convoluted way. The key here is that the negative output terminal has a negative voltage compared to the control circuit ground. The negative output voltage is mixed with the 8V rail (used as a voltage reference), such that the sum (red blob in drawing) is zero volts with respect to the control circuit ground (~= positive output terminal).

Note that the control circuit ground is buffered by the 220uF capacitor (far right), which means small changes in the positive output terminal will not affect the control circuit ground. Thus for small changes of the output voltage, the 8V and the control circuit ground will essentially stay fixed, thus only the feedback from the negative output terminal will affect the opamp.

If the negative output terminal is not negative enough (ie output voltage is too low), then the sum at the blob will be above zero volts, and thus input to the + input on the opamp will be higher than the - input and the opamp output voltage will go up. This "releases the breaks" of the voltage controller and, unless the current controller overrides it, will drive the pass transistor harder, reducing the resistance in the transistor, thus increasing output voltage. Similarly if the output voltage is too high, the negative output terminal will be too negative, causing the + input to be lower than the - input, driving the opamp output low.

At least that's my take on it, modulo typos and brainfarts :)
« Last Edit: February 02, 2020, 12:30:31 pm by aheid »
 
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Offline xavier60

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Re: Help with how this PSU works.
« Reply #4 on: February 02, 2020, 12:44:54 pm »
Or use TIP35C instead. Beware of fakes.
A problem with the manual tap changing is the very high dissipation when the output is shorted while in high range.
And also the hazard of having  secondary wiring near live wiring at the switch.
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Offline Evangelopoulos PanagiotisTopic starter

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Re: Help with how this PSU works.
« Reply #5 on: February 02, 2020, 12:50:32 pm »
That helped a lot! It seems easy now that you explained it. Thank you so much aheid for your help! Though I'm a bit confused about that two diodes that face each other at the input of the voltage control opamp.

The core regulation is provided by the pass transistor, used as a variable resistor.

Cant a MOSFET be used in such application for extra durability since they can be used as variable resistors as well?  But I'm guessing the control circuit would have to be altered beyond recognition...
 

Offline Evangelopoulos PanagiotisTopic starter

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Re: Help with how this PSU works.
« Reply #6 on: February 02, 2020, 12:55:46 pm »
A problem with the manual tap changing is the very high dissipation when the output is shorted while in high range.
And also the hazard of having  secondary wiring near live wiring at the switch.

Honestly I've been willing to change that switch and have a second one for the range selection at the back. That switch handling both live and secondary wires gives me the  heebie jeebies... I'll do a rewiring as well. Lots of exposed contacts in there.
 

Offline aheid

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Re: Help with how this PSU works.
« Reply #7 on: February 02, 2020, 01:09:32 pm »
That helped a lot! It seems easy now that you explained it. Thank you so much aheid for your help! Though I'm a bit confused about that two diodes that face each other at the input of the voltage control opamp.

You're welcome! The diodes there is just protection, it prevents the difference between the two inputs to be larger than a diode drop (~0.6V) in case of sudden, large load changes (ie load suddenly disconnected etc).

The core regulation is provided by the pass transistor, used as a variable resistor.

Cant a MOSFET be used in such application for extra durability since they can be used as variable resistors as well?  But I'm guessing the control circuit would have to be altered beyond recognition...

You could use a MOSFET, though most MOSFETs these days are made for switching applications, while the pass transistor (be it a BJT or a MOSFET) will operate in the linear regime thus you need one designed for that. I find it more difficult to find MOSFETs that are designed for linear workloads compared to BJTs, but they exist. If you go this route, stay far away from anything optimized for switching, don't use trench-style devices but rather planar MOSFETs, and make sure to check that the SOA chart has a DC line. Also the downwards slope on the right hand side of the SOA chart should have a kink, if it's a straight line don't trust it.

Though I'm no expert so maybe someone here has better suggestions.

As for using a MOSFET, I don't think you need to change the control circuitry too much in principle, since it's already primarily voltage based. You'd drop the darlington configuration, you'll probably need to adjust the 2.2 something Ohm resistor between the 7808 and base (gate now), and probably also add a 10-kohm resistor or something like that between the gate and source (emitter). Since you only have 8V to play with, you'll need a MOSFET that can saturate at that voltage. If not you could seemingly change the 7808 for a 7812, but then you'll have to adjust some of the other resistors in the control circuit, though maybe you have enough leeway in the Vadj and Cadj potentiometers to get away with that.
 

Offline Evangelopoulos PanagiotisTopic starter

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Re: Help with how this PSU works.
« Reply #8 on: February 02, 2020, 01:19:55 pm »
That helped a lot! It seems easy now that you explained it. Thank you so much aheid for your help! Though I'm a bit confused about that two diodes that face each other at the input of the voltage control opamp.

You're welcome! The diodes there is just protection, it prevents the difference between the two inputs to be larger than a diode drop (~0.6V) in case of sudden, large load changes (ie load suddenly disconnected etc).

Simple yet effective and interesting.

The core regulation is provided by the pass transistor, used as a variable resistor.

Cant a MOSFET be used in such application for extra durability since they can be used as variable resistors as well?  But I'm guessing the control circuit would have to be altered beyond recognition...

You could use a MOSFET, though most MOSFETs these days are made for switching applications, while the pass transistor (be it a BJT or a MOSFET) will operate in the linear regime thus you need one designed for that. I find it more difficult to find MOSFETs that are designed for linear workloads compared to BJTs, but they exist. If you go this route, stay far away from anything optimized for switching, don't use trench-style devices but rather planar MOSFETs, and make sure to check that the SOA chart has a DC line. Also the downwards slope on the right hand side of the SOA chart should have a kink, if it's a straight line don't trust it.

Though I'm no expert so maybe someone here has better suggestions.

As for using a MOSFET, I don't think you need to change the control circuitry too much in principle, since it's already primarily voltage based. You'd drop the darlington configuration, you'll probably need to adjust the 2.2 something Ohm resistor between the 7808 and base (gate now), and probably also add a 10-kohm resistor or something like that between the gate and source (emitter). Since you only have 8V to play with, you'll need a MOSFET that can saturate at that voltage. If not you could seemingly change the 7808 for a 7812, but then you'll have to adjust some of the other resistors in the control circuit, though maybe you have enough leeway in the Vadj and Cadj potentiometers to get away with that.

I'll stick with the TIP3055 or TIP35C as Xavier recommended and maybe I'll try out a MOSFET version in a breadboard or something and see its performance.

Thank you again for everything!
Have a nice day,
E.P.
 

Offline aheid

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Re: Help with how this PSU works.
« Reply #9 on: February 02, 2020, 03:15:07 pm »
The diode input protection is a common trick, typically used between ground and input, and between input and supply voltage. Again to ensure the inputs don't stray too far away. This can cause some suprise, as Dave himself showed quite nicely here: https://www.eevblog.com/2015/12/18/eevblog-831-power-a-micro-with-no-power-pin/

Though usually the internal protection are not designed to handle any significant load, hence why they added the external diodes which can handle a lot more.
« Last Edit: February 02, 2020, 03:17:27 pm by aheid »
 

Offline xavier60

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Re: Help with how this PSU works.
« Reply #10 on: February 02, 2020, 07:06:58 pm »
Also the CV and CC loops will likely need frequency compensation if instability is noticed during testing.
If purchasing the TIP35C BJT from riskier sources, the TO-247 variants are less likely to be fakes.
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Offline duak

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Re: Help with how this PSU works.
« Reply #11 on: February 02, 2020, 07:42:15 pm »
I built a similar power supply years ago using MOSFETs.  While the MOSFETs were rated for greater current and power dissipation, I went back to bipolar devices because they were easier to stabilize and didn't fail under some conditions.  Since then, something called "Electro-thermal Instability" has been discovered to explain why MOSFETs can fail in linear power applications.  However, don't let me stop you.  Many people have used MOSFETs in linear power supplies.  The important thing to look at is the Safe Operating area of the device, not just the maximum ratings.

With four TIP3055 transistors, I would consider larger emitter resistors, say 0R47 or 0R68.  If each carries 1 A, they will dissipate 0.47 W or 0.68 W respectively.  Each should be rated for 1 W although I would use 2 W so there would be less drift in the current limit as they heat up.  It would also allow for an increase in ouput current if the TIP3055 transistors can handle the extra power.
« Last Edit: February 02, 2020, 10:51:16 pm by duak »
 

Offline aheid

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Re: Help with how this PSU works.
« Reply #12 on: February 02, 2020, 09:46:36 pm »
For fun I tried to recreate it in LTSpice. Made my own homebrew LM7808 implementation, probably doesn't behave exactly like the real deal but don't think the difference matters too much.

The sim as configured will simulate the voltage setpoint being changed between 3.3V and 10V, but the current limiter is configured so it kicks in at about 830mA, thus limiting the 10 Ohm load to 8.3V.

To change the current limit, change RcsetH and RcsetL while maintaining the constraint RcsetH + RcsetL = 1kOhm. To change the voltage limit, change Rvset.

In the simulation at least, the 220uF capacitor that buffers the V0 line (C7 in sim) limits the dynamic response. Lowering it to 22uF makes for a much more responsive output. However I guess in real life there might be a reason why you'd want to stick to 220uF.
 

Offline Evangelopoulos PanagiotisTopic starter

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Re: Help with how this PSU works.
« Reply #13 on: February 02, 2020, 10:31:57 pm »
For fun I tried to recreate it in LTSpice.

Wow you went the extra mile for this! Currently I'm away from my main PC but as soon as I get back I'll play around with it a bit to see how it behaves!
 

Offline Evangelopoulos PanagiotisTopic starter

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Re: Help with how this PSU works.
« Reply #14 on: February 02, 2020, 10:39:30 pm »
If purchasing the TIP35C BJT from riskier sources, the TO-247 variants are less likely to be fakes.

I'll probably be buying from LCSC so I think I'm good. Once I bought a few 2N3055s from AliExpress for like 5€ and when I opened one, the silicon was tiny. No wonder they got destroyed in no time...

With four TIP3055 transistors, I would consider larger emitter resistors, say 0R47 or 0R68.  If each carries 1 A, they each have to rated for 1 W although I would use 2 W.

Thank you for the wattage info. I was about to go overboard with 5W cement resistors.
 

Offline Evangelopoulos PanagiotisTopic starter

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Re: Help with how this PSU works.
« Reply #15 on: May 12, 2020, 05:32:31 pm »
Greetings everyone again,

It's been a while since I've done anything with this power supply. My desire to create a somewhat decent linear power supply was postponed due to recent events.

I integrated some of the changes that were proposed:
Added: 4 parallel TIP35C
Added: Ballast resistors
Added: Output protection diodes
Added: 10R base resistors

I have a few questions before I go on and create the PCB.
-I changed the 1k pot that controls the current to a 10k one because that's whats mounted on my project box. I'd rather not change it because it has an integrated switch in it. Can I do that?

-I want to add a green LED that will turn on when the output voltage is at the desired level. Will adding a LED on the output of the voltage Op-Amp, same as the current one, do the trick?

-Have I messed up the output protection?  :P

Once I'm confident enough with this design I'll make the project public and paste the link here incase someone would like to build one themselves!  ;D

Thank you all in advance,
E.P.
 

Offline xavier60

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Re: Help with how this PSU works. (With revised design)
« Reply #16 on: May 12, 2020, 09:56:18 pm »
Don't make a PCB yet. The way I see it, the CC is sensing output voltage and the VC is sensing nothing.
Ill look for example schematics.
Most designs put D10 between the output and the Collectors of the output transistors to protect them from an external voltage source being connected while the PSU is powered off.
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Offline aheid

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Re: Help with how this PSU works. (With revised design)
« Reply #17 on: May 12, 2020, 09:58:28 pm »
Don't make a PCB yet. The way I see it, the CC is sensing output voltage and the VC is sensing nothing.

Looks like the LTSpice sim I made above, unless I missed something?
 

Offline Evangelopoulos PanagiotisTopic starter

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Re: Help with how this PSU works. (With revised design)
« Reply #18 on: May 12, 2020, 10:03:48 pm »
Don't make a PCB yet. The way I see it, the CC is sensing output voltage and the VC is sensing nothing.

The output positive is connected to the 0V label which is also connected to the inverting input of the VC Op-Amp.

Most designs put D10 between the output and the Collectors of the output transistors to protect them from an external voltage source being connected while the PSU is powered off.


I didn't quite get that... :-[
 

Offline xavier60

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Re: Help with how this PSU works. (With revised design)
« Reply #19 on: May 12, 2020, 10:05:13 pm »
Don't make a PCB yet. The way I see it, the CC is sensing output voltage and the VC is sensing nothing.

Looks like the LTSpice sim I made above, unless I missed something?
I can't see the labels clearly. Can you post it with higher res?
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Offline xavier60

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Re: Help with how this PSU works. (With revised design)
« Reply #20 on: May 12, 2020, 10:12:59 pm »
In this example, D6 mainly  prevents the output transistors' B-E junctions from being reversed biased if an external source is connected while the PSU is not being powered. https://www.eevblog.com/forum/projects/linear-lab-power-supply/?action=dlattach;attach=751023
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Offline Evangelopoulos PanagiotisTopic starter

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Re: Help with how this PSU works. (With revised design)
« Reply #21 on: May 12, 2020, 10:18:53 pm »
Oh ok now I get it. I shall change the D10s position then.
 

Offline xavier60

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Re: Help with how this PSU works. (With revised design)
« Reply #22 on: May 12, 2020, 10:24:47 pm »
Your design's topology has been around since the sixties.
This is a useful reference, https://www.hpl.hp.com/hpjournal/pdfs/IssuePDFs/1962-07.pdf
Back in those days, the germanium power transistors were just about all PNP, explaining why the pass transistors are in the negative rail.
The design I linked in my previous post has comparable performance, but does not need auxiliary control rails.
« Last Edit: May 12, 2020, 10:26:59 pm by xavier60 »
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Offline Evangelopoulos PanagiotisTopic starter

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Re: Help with how this PSU works. (With revised design)
« Reply #23 on: May 12, 2020, 10:35:34 pm »
So am I good to go? Any feedback on the other questions I asked?
 

Offline xavier60

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Re: Help with how this PSU works. (With revised design)
« Reply #24 on: May 12, 2020, 10:41:27 pm »
So am I good to go? Any feedback on the other questions I asked?
About the Pot I can't say until the other problems are corrected in the schematic.
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