How to improve regulation of a power supply

[IanP]

Guys,
I am not questioning the series/series solution.
Just the sequence was wrong, and I can see that you have come to this conclusion.
However, you will still have problems at low-voltage range..and if the aim was to build something for wide voltage and current ranges why not to do it properly???

 

[ME]

yes, the last circuit should work, and I agree it's one of the easiest method.
But I think the output can't be regulated down to 2V, can it?
It's good work from u, ME..

Regulating down to 2V should be no problem, it can be regulated down to Vref, which is 1.24V for LM338 or 1.25V for LM317.

If you set R2 to 0 Ω in the formular then you get Vout = Vref

 

[lastdance]

ME, u need 0.7V for current limit

 

[ME]

ME, u need 0.7V for current limit

Yes, but the 600mV current drop is "on the other side" of GND, so this has no influence on the output voltage. The voltage at the negative input of the rectifier is 6-700mV more negative than GND; GND is not the most negative point in this design because it's not connected directly to the rectifier, but the current resistor is in between GND and the rectifiers negative return parth.
The current has no influence on the output voltage between Vout and GND.
Vout varies compared to the negative pin of the rectifier, but GND is "floating".
R1 and R2 makes a voltage divisor that refernces to Vout and GND. So using the forumlar with only R1 and R2 is fine, it doean't matter which voltage and resistors there's "on the other (negative) side" of GND. When the GND voltage rises the voltage over R1 + R2 will fall, this will make LM338 turn up for the output voltage so Vout rises as much as GND rises.
Remeber to connect the rest of you circuit to GND and don't connect anything else to the negative side of the rectifier as this volatage varies comapered to GND.

GND is not nessasarly the most negative votlage in a supply, but a point you select as reference for all your other voltages.The regulator regulates the voltage referenced to GND and not referenced to the negative side of the rectifier which is more or less negative than GND, depeding on the current.
This trick with GND on the positive side of the current sensing resistor is often used in such circuits.
In his original design he also had GND at the positive side of the current sensing resistor. The only problem was that he used the negative side of the current sensing resistor as reference for the voltage regulator. This makes the voltage variying with various loads and hence provides a very bad load regulation.



Actually there's a very similar example shown in both the LM317 and LM338 datasheets. R3 is placed slightly different in this example:

 

[Borber]

Actually, the emitter of 2N2222 should be connected to -1.25V respect to ground and you will have 0V output when current limit is activated at short circuit on output.

 

[ME]

Actually, the emitter of 2N2222 should be connected to -1.25V respect to ground and you will have 0V output when current limit is activated at short circuit on output.

Yes, if you want to be able to go all the way down to 0V on the output when the current limiter activates. That's why they have used both R3 and R4 in the circuit instead of only R3.
When there's a 0.6V voltage drop across R3 (Basis-Emtitter), there should be about 1.25V - 0.6V = 0.65V across R4. This way you will have -1.25V a the emitter when the transistor activates at 0.6V B-E. This way there will be 0V on the output when the transistor activates at about 0.6V B-E.
R4 also has another function: it prevents a heavey current to flow from basis to emitter. If R4 was not inserted, then there would flow a lot of current throug the B-E junction and the transistor would bourn.

 

[Borber]

R4 is only 10Ω and is not meant to limit base current of transistor but to acheive total -1.25V drop on emmiter. Instead of R4 diode can be used. Transistor is protected if you put 100Ω resistor in series with base as it is in last circuit. Nevertheless when current limit is activated the voltage drop on sense resistor remains practically constant. New problem can appear if unregulated voltage is not high enough to compensate additional voltage drop of 1.25V. In this case it is neccessary to use small resistor for current sensing and OPAMP. Of cause resistor should be connected in input circuit of regulator. Connecting anything in series to regulator output circuit will degrade regulation, increasing internal resistance of the regulator.

 

[ME]

Btw. R1 in the origianl circuit in the first post is probably too high (2.2 kΩ). This can be too high because too little current will flow into R2 from R1 compared to the current flowing from ADJ pin into R2. Then the current from ADJ to GND will cause a too high voltage rise over R2 if R2 is too big and this causes an error in the output voltgage. There's always a constant value of 1.25V from the Vout pin to the ADJ pin.
240 ohm is the usual value for R1.

1.25V / 240 Ω = 5.2mA
This current from R1 is sufficient to make the current thorugh R2 to be substanially larger than the current flowing from the ADJ pin into R2. This is very important to keep in mind.
This is why the error voltage Iadj * R2 is added to the formular. Iadj is almost constant (45µA typical), so the higher R2 value, the higher error voltage you will get.
If you choose a lower value for R1 then you will get a lower value for R2 too.

So first choose R1 to a value about 240 Ω and then calculate R2 to get the desired value.

 

[Borber]

I propose this circuit. Current limit is set to 1A. R3 needs to be changed for different current limit. R2 is to be changed for desired output voltage.

 

[smartsarath2003]

Thanks for all the sujjestions

First of all I haven't amended my actual ckt diagram. The only changes from my original ckt is R1=220ohm and a 2K2 pull up at the comparator output. I got it the above ckt working with these changes. Also regarding ON/Off oscillations, I can't see any and I thought bcos of the continuous feedback

Coming to regulation problem I don't want to use 2 regulators in series.
The circuits posted by ME and Borber seems interesting.
But I would like to avoid seeing any voltage at the output, when its short circuited.
Thats why I added series transister. As I mentioned I got most of my drop across Vce of Q1. The drop across 0.25R is in millivolts.
But It looks like I have to use the series transister, if I want to isolate power supply completely in case of short. Are there any other power transisters that has very small drop across Vce@3A. Or any alternative designs/modifications to completely isolate power supply in case of short.
How do I produce -1.25V, by using another regulator at ground???
To minimise the drop across the 0.25R, I think I can move it before the regulator and the actual ground being before 0.25R. Any comments about this change?

Thanks

 

[Borber]

The above circuit works as you wanted. When the output is shortcircuited the output voltage is 0V. If load is excessive the output voltage will be according to ohms law Vout=Ilimit*Rload. Circuit does not need -1.25V, it operates from -5V.

 

[ME]

You don't have to produce -1.25V with any regulators.
Take the example I posted:
GND is more positive than the negative side of the rectifier.When the current through R3 and R4 are high enough, then you will have -1.25V at the negative side of the rectifier (refenrced to GND).

If you used the negative pin of the rectifier as reference for your measuremet, then you would have +1.25V at GND.
GND is not the most negative volage in the circuit.
The most negative point in the circuit will always be the negative pin of the rectifier. GND is more or less positive compared to the negative side of the rectifier depending on the current flowing in R3 and R4. This does not influence the load regulation, because the LM338 senses the output voltage reference to GND and not referenced to the negative side in the rectifier.
This is what make this circuit much better than yours. In your design, the output voltage will drop proportinal to the current flowing through your current sense resistor.
This will not hapen in my example.

If you design my example with the right values for R3 and R4 then you will have 0V out when the current limiter turns on, so this is not a problem.

It seems you don't understand how this works, but be careful not to be fooled by the fact that GND isn't placed directly at the negative side of the rectifier and hence GND is not the most negative point in the circuit. That's why you can have -1.25V in the circuit. Remeber -1.25 is measured referenced to the output GND point, not the negative side of the rectifier.
Try to read my posts againg and studdy the circuit until you understand how it works and why you can have negaive voltages in this circuit.
And try to build the circuit yourself and experiment with it. It's very simple to build.
Choose R1 to 240 ohm. Then calculate R2 for the desired output voltage in the circuit using the formular.
Then calculate R3 to have a 0.6V (V b-e) voltage drop acrooss it at the desired current limit. Then calculate R4 to have a voltage drop of Vref - (V b-e).
Vref =1.25V and (V b-e) = 0.6 V.
It's actually quite simple to find the right values. But you might need to adjust R3 and R4 a little depending on how big (V b-e) is needed before the transistor turns on. But usually it's 0.6V - 0.7V.
R3 determines the current limit.
R4 makes sure you have 0V at the output when the current lmit activates if you use the right value for R4.

If you tell which output voltage and current limit you want, then I can calculate the values of R1 to R4. But try to do it yourself first and then ask if it is correctly.

I don't see a good reason to make borbers circuit which requires an extra -5V.
The circuit shown in my post should work properly and it's simpler and cheaper than borbers circuit.
It's not a circuit I came up with, but it's taken from the LM317 datasheet.

 

[VVV]

See my last post to this thread:
That circuit should help you get true current limit.

The -1.25V can be obtained simply with two diodes in series, instead of a zener. So you just build a Zener regulator, where you replace the Zener diode with two FORWARD-biased diodes in series.

 

[smartsarath2003]

• If you design my example with the right values for R3 and R4 then you will have 0V out when the current limiter turns on, so this is not a problem.

This circuit works very fine, but to have adjustable current limt from 1mA-3A, I have to use a pot from 0.2R-600R. And I think its bit hard to get a pot with that precise and large values. Thats what is the only problem, i found with that design.
But it is pretty good design with fixed current limit

you tried to connect -1.25V at adjust pin, using resisters which is depend upon the value of the current flowing in the circuit.
But if i use a regulator like LM317with its input conncted to gnd,adjust pin joined to your 2N2222 transister, while the its base is driven by LM339 in my original ckt and the Rsense moved before the transister. I can take off the power transister Q1. and have a precise current adjustment and no series elements after regulator
I think its basically joining your ckt(modified) and borbers ckt
Will that works with my requirement
I'll try to draw the schematic and post it tommorow(attaching a rough ckt)
Thanks
I should have a resistance between Adj and out pins of LM317 as well

 

[smartsarath2003]

Here I am attching what is intended to be my idea
But I didn't had any impression that this ckt works firsthand and looks bit strange to myself. 8O
But the basic idea is connect -1.25V at adjust pin of the regulator, which brings the output down to 0V automatically

Two querries I have are
1.Does the LM317 maintain -1.25V at the adjust pin as it is floating(may blow even)
2.Will there be any voltage drop between Vin,Vout(1.25V) in the const current configuration.

Or I am going completely wrong way :!:

 

[ME]

Your circuit seems very strange to me.
I don't think you have designed it right.
First of all the LM317 can't handle as much current as LM338, so why do you use LM317???
All the current that runs through the LM338 also runs through the LM317, so I don't understand why you got this strange idea of using LM317.
And for what purspose do you want to use LM317?
IS it because you want to be able to adjust the output voltgae all the way down to 0V or what?
I didn't thought you wanted an adjustable output voltage.
What are your requirements for the circuit? Which output voltage and current range do you want? What is the input voltage?

It's almost impossible to see how you have connected the LM317, but it looks like you have conneted the adjust pin to the emitter of the transistor.
If this is the case ythen you have connected it wrong. The adj pin should not be conneted to a higer voltage than the output.

But it seems like the LM317 is completely misplaced.
There's absolutely no pursope at all to create a voltage drop where LM317 is placed. Ther should be no voltage drop where LM317 is placed.
I can't see any purpose of the the resistor placed at the emitter and LM317 either.
I think you have misunderstood the whole concept of where the 1.25V voltage drop could be desireble and why a 1.25V voltage drop could be a good idea. The 1.25V voltage drp should be where your current sense resitor is. The current sense resistor itself will create the voltage drop itself. Set the resistor divider at your comparator to 1.25V. This could be made accurate by replacing the lower resistor with a 1.25V reference voltsage diode.

If you want to be able to adjust the output votage all the way down to 0V and not only 1.25V, then you need to connect R2 to a stable -1.25V voltage instead of GND. but this will make the regulation worse. Because the output will be depandable of how satble the -1.25V voltage is, when R2 is not connected to the output GND.

The comparator is also conneted wrong.
You have switched IN+ and IN- in the schematic.
At low currents or no current at all, the comparator output will be high, which turns on the transistor on and turns down the output voltage of LM338, so there's only 1.25V at the output. So your desing will never work.

I don't know if you still use the LM339 comparator, but if you do, then you have connected the output wrong, because it has open collector output, which has already been mentioned several times in this topic.
If you use LM339, then there's no need

The way you have designed the circuit, it will not work at all.

I don't know which kind of experience you have in electronics, but it seems you are not that used to working with electronics.
It looks like you miss some of the basics of how different circuit works, but that's ok.
I think you should try to fully understand how the example from the LM317 dataheets works, before you try to change it.
To me it seems like you don't understand how it works, after looking at your schematic.

Next time, try to upload a schematic in a higher resolution and try to save it as a *.PNG file instead of a *.JPG file. PNG files has no quality loss like JPG, but the PNG filesixe is still smaller than JPG files for images like screenshots etc.

 

[smartsarath2003]

Hai,
I myself mentioned above ckt doesn't work. i want to convey my idea, sothat I can get some comments/feedback from others, to get it working.

The only point I am trying to make there is connecting a -1.25V at adjust pin of LM338, sothat the output voltage is 0V, when current limit is over.
As the output of LM317 is grounded,it adj pin will be at -1.25V.
I definitely know LM317 blows,because the adjust pin is floating normally, and I am expecting a better ideas...

coming to LM339 comparator, yes I got it wrong: inputs interchanged and no pull-up at output. I just forgot/mistaken.

 

[lastdance]

smartsarah, first of all, can you help answer the following questions?
1. Are you trying to build a power supply that has adjustable output voltage and
current limit?
2. Why do you want the output voltage to be zero when there's current limit? If
you look at a normal power supply, the output voltage during current limit is
Ilimit X Roverload
If it's a simple power supply that you wanted to build, most of the ingredients were already here, and you do not have to be noble to come up with all the weird ideas. You can be creative, but electronics fundamental is something u can't change.

 

[ME]

As the output of LM317 is grounded,it adj pin will be at -1.25V.

None of the LM317 pins are grounded. Ground in the circuit is at the -output on the right side of the current sense resistor.
As I said earlier, LM317 has absolutely no purpose where it is placed and it doesn't create any 1.25V voltage drop. LM317 is a positive voltage regulator but it seems lieke you are trying to use it like a negative voltage regulator allthough it is still connected wrong.
But you still need to answer a lot questions asked in the previous post.

What are the requirements for the circuit in and outputs. And try to upload a schemaitic where it is actually possible to read the text.

As the previous poster mentioned, you can't change circuit fundementals, but it seems like that is why you are trying to do. You seem to lack some basic circuit understandings.

Why don't you want to use the cicruit I suggested first? You still haven't answered that I think. This circuite is much simpler than u
yours and it actually works as opposed to yours.
Your idea about the 1.25V voltage drop is wrong, and there's no reason at all to try to create a 1.25V voltage drop where you have placed LM317. You misss some ciccuit fundementals here. If you want a 1.25V voltage drop, then it should not be wheree you have placed LM317. I'm not even sure between which two pins of the LM317 you want the voltage drop to be. But LM317 is misplaced no matter which pin is which in the circuit.


GND is not where you have shown t in this cirtuit. GND is on the other side of the 0.25V resistor. Your circuit should be referenced to the right side or the 0.25 V resistor, not the left side. Otherwise the current senes resistor serves no purpose.
But it seems you miss some circuit fundementals here.
GND howe ever is not the most negative voltage in this circuit.
The most negative voltage will alway be at the engative side of your rectifier. No matter if you have a LM317 or not connected to it.
LM317 is misplaced even in this basic circuit.

 

[smartsarath2003]

Why don't you want to use the cicruit I suggested first? You still haven't answered that I think. This circuite is much simpler than u
yours and it actually works as opposed to yours.

This circuit works very fine, but to have adjustable current limt from 1mA-3A, I have to use a pot from 0.2R-600R. And I think its bit hard to get a pot with that precise and large values. Thats what is the only problem, i found with that design.
But it is pretty simple & good design with fixed current limit

Any way I am leaving my previous idea after much criticism and also as it involves too many components. I think it'll work by using centre-tap transformer and using centertap as ground. This is some what similar to VVV's post.

Here is my latest version,but didn't tested this design(will test next week). Basically I removed BD 677 series transister and gone for IRF540 Mosfet,which has very samll Rds-ON(44mohms). So at maximum current ,this will introduce a maximum of 0.132V series drop only, which is less than my spec. Also I moved the Rsence before the regulator as per the previous sujjestions. Also removed the unnessacary transister driven by LM339 and conncted the output directly to Mosfet gate.
The only question I had is its given that Vgs(threshold)=2-4V and Vgs(max)=20V and i am applying 5V. Will I get Rds(min) at this gate voltage

The abasic spec's for this design are 5-30V output, upto 3A adjustable output current, 1% load regulation at max output ,short circuit protection,Digital Voltmeter and ammeter for voltage and current display

Any other comments are welcome 8O
Thanks