Linear power supply

[wakoko79]

Hi,

Can someone please give me what is happening in my circuit? It is a linear power supply (24V, ~6Amax)

Here is the schematic:
View attachment 66112


- V1 is an LM336- 5.0V voltage reference
- The network R2 an R4 is actual circuit is 3k - 1k pot - 12k network instead so that I can adjust the voltage a little
- R7 is a current sense resistor for the base of the pass transistor, Q8 turns on when around 0.7/.5=~1.4A flows through R7
- R3 is the pull-up resistor for the lm393 comparator, the value was just a guess with a little trial and error.
- CURRENT LIMIT:
- R14 and Q6 composes the *constant current limit. It samples the current through one of the pass transistors, it then pulls the comparator output to near ground so that Q2 would stop supplying base current to the pass transistors, thereby limiting the current through one pass transistor to about (0.7/.27)= ~2.6A
- R12, R13, R16, R17, Q7, R15 and Q1 comprise the current-fold limiting. It's too long to describe it so I'll leave it at that, but it does the same thing to the comparator output as the constant current limit

- Q2 is a darlington BJT. It supplies base current to the pass transistors
- Q3, Q4 and Q5 are the pass transistors. R9, R10 and R11 serves as current sense resistors for the pass transistors as well as equalizing resistors to them.
- R5 is a bleeder resistor.
- C2 is the output filter capacitor.
- R6 is the simulated load (3A - half load), my proposed full load is 6A.
- The bridge diodes are changed to a monolithic GBPC3510-W, it's better.


So there is my circuit. This is OK in simulations, but in real actual circuit, it fails.
It regulates to 24V at no load but output voltage drops to about 22V at half load (actually, the output starts at about 20V then climbs up to about 22V as te=he temperature of the components rise). The output waveform is also wavy. I use an electronic load for testing.

Care to give me pointers? I'm confused why my circuit's voltage goes down below 24V. As I think about it, a linear power supply at higher loads will show higher ripples right? But what my circuit is showing is not ripples where regular voltage dips may be seen. The whole waveform just goes down.

Look at these waveforms. The probe was taken at the output of the actual board (NOT at the e-load terminal).

(no load)
View attachment 66113


(3A load)
View attachment 66114


And here is the test set-up:
View attachment 66115
View attachment 66116


Thanks!

 

[ronsimpson]

What is the voltage across C1, C3? There is some 120hz ripple on it at load. The minimum voltage must be 2 volts (maybe 3 volts) more than the output voltage.

Is there ripple across V1? 5.1V Zener? Often there is a capacitor across the Zener.

 

[wakoko79]

What is the voltage across C1, C3? There is some 120hz ripple on it at load. The minimum voltage must be 2 volts (maybe 3 volts) more than the output voltage.

Is there ripple across V1? 5.1V Zener? Often there is a capacitor across the Zener.

That is supposed to be a rectified 24Vac coming from the tx. So that's about 29Vdc (loaded + rectifier drop).
I checked the unloaded voltage (not really unloaded since there is a bleeder resistor there), it's about 34.5Vdc.

BTW, C1 C3 and C4 are actually six 3.3mF caps.

I tried to place a 10nF cap across the zener before, here are the waveforms:

(without 10nF cap, 3A load)
View attachment 66118


(with 10nF, 3A load)
View attachment 66119

A bummer, right? Or should I increase the capacitance?

 

[ronv]

You probably don't have enough capacitance at the output of the bridge, Scope that supply and see if there isn't a lot of ripple. It needs to be about 30 volts at its lowest point under full load.

 

[atferrari]

Is the trafo up to the specs?

Maybe it can't cope with the load. Been there.

 

[wakoko79]

Before I used the GBPC 3510-W, I used 4 p600b diodes. I it works up to 6A, though the output is somehow noisy. Maybe the voltage drop across the rectifier bridge is bigger. The transformer is rated at 12A BTW.

But come on people, any other ideas?

 

[ronsimpson]

(RonV and I) agree you do not have enough input voltage.

Another way to test that: Change the feedback resistors so the supply will output 15 volts not 24 volts. (15 to 20V) Now what happens? Does it work at 1/2 power? Is the ripple less?

If you don't like that test: remove one of the input caps. C1 or C3 Is the problem worse?

Another test: Add more input caps. Is the problem better?

 

[bountyhunter]

So there is my circuit. This is OK in simulations, but in real actual circuit, it fails.

I propose that be put up as a sticky......

but what do I know.

The schematic above can't be accurate: it shows an LM393 being used as an error amplifier but that can't work because it's a comparator. That can't work in a linear regulator. An error amp has to be an op amp and it also requires external feedback compensation.

It also shows no current drive from power transistors Q3 - Q5 going to the output. Their emitters go to the point labeled ground 1?

schematic makes no sense at all.

 

[ronsimpson]

Bounty Hunter,

"gnd1" is the negative side of the bridge. (only that connection and has nothing to do with "gnd" at the output.) Most power supplies have a common ground and the transistors pull up on the load. This design has a common "+" supply and the transistors pull down. It does make sense if you stand on your head.

I agree that a LM393 voltage comparator is not an error amplifier. I have seen several of these internet designs recently and they seem to work. The comparator switches on/off/on/off very fast and the slow transistors filter that out. Strange design but works.

Q5 hearts my head. and R13 "current fold back" I think.

 

[ronv]

Yep, Kinda plays with you mind. The claim to fame is that you don't need isolation for the transistors.

 

[bountyhunter]

Bounty Hunter,

"gnd1" is the negative side of the bridge. (only that connection and has nothing to do with "gnd" at the output.) Most power supplies have a common ground and the transistors pull up on the load. This design has a common "+" supply and the transistors pull down. It does make sense if you stand on your head.

I agree that a LM393 voltage comparator is not an error amplifier. I have seen several of these internet designs recently and they seem to work. The comparator switches on/off/on/off very fast and the slow transistors filter that out. Strange design but works.

Q5 hearts my head. and R13 "current fold back" I think.

What node is the regulated output?

with respect to what ground point?

 

[ronv]

The ground symbol is regulated with respect to positive of the bulk supply.

 

[ronsimpson]

Output is across R6.
gnd1 is not connected to the output.

 

[JimB]

Ohh, my eyes!!!!

Before even attempting to analyse what could be wrong with that circuit, DRAW IT AGAIN, PROPERLY!!!

Remove all the twists and turns to try and fit into a drawing space which was too small in the first place.

How the hell anyone is supposed to make sense of that schematic, as drawn, is beyond me.

JimB

 

[wakoko79]

Bounty Hunter,

"gnd1" is the negative side of the bridge. (only that connection and has nothing to do with "gnd" at the output.) Most power supplies have a common ground and the transistors pull up on the load. This design has a common "+" supply and the transistors pull down. It does make sense if you stand on your head.

I agree that a LM393 voltage comparator is not an error amplifier. I have seen several of these internet designs recently and they seem to work. The comparator switches on/off/on/off very fast and the slow transistors filter that out. Strange design but works.

Q5 hearts my head. and R13 "current fold back" I think.

Exactly, thank you.
The original circuit is here:
ludens.cl/Electron/Ps20/Ps20.html

I posted before about this design (it is somewhere in this forum, lost). I tweaked the circuit to improve it a little. Well, I could change the op amp part and make a linear one (what I really intended at first) but I just can't find a starting circuit for the compensation. How much compensation do I need? Topologies? I had a hard time finding basic circuit for this even in the books I used (I mean for op amp linear supplies; suprisingly, or maybe I'm just bad with searching the books because it has no ctrl+f function). So ok, I think the rest of the circuit could be left intact, any basic circuit out there?

 

[bountyhunter]

Well, I could change the op amp part and make a linear one (what I really intended at first) but I just can't find a starting circuit for the compensation. How much compensation do I need? Topologies? I had a hard time finding basic circuit for this even in the books I used (I mean for op amp linear supplies; suprisingly, or maybe I'm just bad with searching the books because it has no ctrl+f function). So ok, I think the rest of the circuit could be left intact, any basic circuit out there?

I reviewed the original design and here is what it says:

No parts were added to control frequency response, loop damping, etc. All trust was placed on the 741's rather low frequency response and high stability, combined with a 1000µF capacitor across the output. In practice this has proven to work well enough, but purists may want to experiment with the loop response and add some compensation capacitor.

Good luck. using a comparator in place of an op amp can't work, using an uncompensated LM741 may work over a range of loading. I have no idea. Based on what I read in the original article, the "designer" did not do sufficient testing to see if the circuit works properly.... according to the designer's own words:

I originally did not have R11 in the design. A fair number of readers built the power supply and had good results. But then I got a mail from Carl Ressel, who built it and tried it out feeding a variable voltage onto C1. Very reasonable, only that without R11, Q2 took a very high current while the input voltage was too low to allow the supply to produce normal output! This situation could be fatal for Q2 in the event of a brownout. With the added R11, the problem is solved. This shows how easy it is to overlook a situation that normally does not happen, but sometimes can! A good design must be able to handle any situation. Thanks, Carl, for this test!

BTW, based on part selection, I would wager this design is at least 20 years old. I would not use it. It has no loop compensation and the current limiting is archaic, and will vary by about a 2:1 ratio based on the temp of the power transistors. There are much better designs available even using an old chip like the LM723.

 

[ronv]

Compare the size of C1, C2, and C3 in your circuit compared to the original. ( muuuuch bigger) Try something like a LT1013 it can run at your supply voltage.

 

[wakoko79]

To put your eyes and minds at ease, here's a picture:
View attachment 66159

There are basically 3 nodes:
1) Vout - this is the regulated out put in reference to OUTPUT GROUND (which is shown having the ground symbol). This is also the rectified input(now, WRT the input ground which is labeled as "gnd1").
2) gnd1 - the input ground
3) GND (symbol) - the output ground

In most linear power supply, the output ground is tied with the input ground, then you regulate at the output pin. THIS IS NOT THE CASE for this circuit. Here, you regulate at the output ground. The reason why the symbol for ground is there is so that I can in simulation, the output voltage waveform will be automatically drawn, nothing more.
To put it simply, Vout has the highest potential, and the point of lowest potential is the input ground "gnd1". The output ground with the ground symbol is supposedly 24V lower than the Vout.

I reviewed the original design and here is what it says:



Good luck. using a comparator in place of an op amp can't work, using an uncompensated LM741 may work over a range of loading. I have no idea. Based on what I read in the original article, the "designer" did not do sufficient testing to see if the circuit works properly.... according to the designer's own words:



BTW, based on part selection, I would wager this design is at least 20 years old. I would not use it. It has no loop compensation and the current limiting is archaic, and will vary by about a 2:1 ratio based on the temp of the power transistors. There are much better designs available even using an old chip like the LM723.

Better designs? Any examples?

Compare the size of C1, C2, and C3 in your circuit compared to the original. ( muuuuch bigger) Try something like a LT1013 it can run at your supply voltage.

The original circuit was meant for a 13.8V 25A power supply. Mine is just 6A.

I'll test the input voltage as advised by ronsimpson, updates later.=)

 

[ronv]

The original circuit was meant for a 13.8V 25A power supply. Mine is just 6A.

I see. So you reduce the current by a factor of 4 and reduce the filter cap by 2 orders of magnitude. Good luck with that.

The bigger one at the output is the only thing keeping it stable.

 

[wakoko79]

I see. So you reduce the current by a factor of 4 and reduce the filter cap by 2 orders of magnitude. Good luck with that.

The bigger one at the output is the only thing keeping it stable.

I wanted bigger caps, but there is no more space. Plus bigger caps means bigger $$$. I read from somewhere, that as a rule of thumb, add 2.2mF(I really can't remember the value) for every ampere in the input filter.