Calculating required resistor for LM338 in parallel

Thread Starter

tinker123

Joined Mar 6, 2020
15
I have two questions about the sample circuit I found, pictured below. What are the requirements for R1\R2? Is the value of 0.1 Ohm somewhat arbitrary? How should the "correct" value be determined? The same question goes for R3, how did they pick 2K?

I am building a circuit with a required VO of 8.5 V, any reason why R4 = 470 Ohms and R5 = 2.7K Ohms wouldn't work fine? If R5 is increased does R3 need to match it or was it just a coincidence they picked the same value for the sample?

SampleVRegCircuit.jpg
 

ElectricSpidey

Joined Dec 2, 2017
1,067
The .1 ohm resistors are there to help the regulators share the load, and provide bias to the op-amp, they can be changed depending on the current needed, the voltage loss that is acceptable, and the wattage. (but should remain matched)

R4 is that value to provide the minimum load current required by the 338. (1.25 divided by 120= 10mA) It can be changed if the regulator is always under load.

Have you determined how you are going to get rid of that 21plus watts, in that small enclosed space? (I think it was 21.something)

Is there a good reason you cannot start with a lower voltage supply, or get rid of the excess power at the supply, rather than in that confined space?
 
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Thread Starter

tinker123

Joined Mar 6, 2020
15
The .1 ohm resistors are there to help the regulators share the load, and provide bias to the op-amp, they can be changed depending on the current needed, the voltage loss that is acceptable, and the wattage. (but should remain matched)
Thanks for getting back to me. I have a lot of experience with software but not hardware. In software design reviews we always question "magic" numbers that are hard coded. Why is 2 seconds the right time to wait, why is 3 the right number of retries, etc.

Why is 0.1 the right resistance? Not denying it, just want to understand the choice. How does one decide what voltage loss is both necessary and acceptable? What math should be done to evaluate 0.1 vs 0.05 Ohms?

I understand R4 is basically there to prevent the pot from being turned down to the point of short circuit and the overall values selected so the current through them is 10 to 1000 times greater than Iadj in order to get proper regulation (read that somewhere after my post above).

How do you think the value for R3 was selected?

Have you determined how you are going to get rid of that 21plus watts, in that small enclosed space? (I think it was 21.something)

Is there a good reason you cannot start with a lower voltage supply, or get rid of the excess power at the supply, rather than in that confined space?
Thanks for remembering my use case :)

You can buy 12V 8A laptop supplies but they don't sell high current 8.5V supplies, at least not affordably. A secondary external regulator is possible but not user friendly, so that is a last resort. The duty cycle I mentioned was worst case, I can likely get away with a small heat sink and depend on the LM338s to self limit. If they die in testing I can add a thermoregulation circuit and enforce a lower duty cycle.

I'm learning a lot here and having fun with it... Supply list is almost ready to order.
 

Dodgydave

Joined Jun 22, 2012
9,080
Lm338 is a Variable regulator, the resistor R4 across the Output and Adjust pins sets the Constant Current through the preset resistor, ideally it should be 10mA, which makes R4 120 ohms, ( Constant Voltage of 1.25V across Vout and Vadj), this makes it easier to calculate the desired output voltage.

Any laptop charger can be modified to go lower, it's just a matter of altering the feedback voltage, which is usually a TL431 variable zener.
 

ElectricSpidey

Joined Dec 2, 2017
1,067
No tinker, R4 is there to provide a voltage divider to the adjust pin, not to limit the current on the pot.

The value for VR1 and R3 is chosen to allow voltage adjustment to 20 volts, a higher value would allow for a higher output range.

As far as R1 and R2 hard to say exactly why that value was chosen, probably low enough to still do the job without losing too much voltage across them, and needing a bigger size. (wattage)

And by “preset resistor” he means VR1.:)
 
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Thread Starter

tinker123

Joined Mar 6, 2020
15
Lm338 is a Variable regulator, the resistor R4 across the Output and Adjust pins sets the Constant Current through the preset resistor, ideally it should be 10mA, which makes R4 120 ohms, ( Constant Voltage of 1.25V across Vout and Vadj), this makes it easier to calculate the desired output voltage.
Does that value of 10 mA come off the design spec somewhere? Thanks for explaining that aspect, I was missing the relationship to the 1.25V offset.


Any laptop charger can be modified to go lower, it's just a matter of altering the feedback voltage, which is usually a TL431 variable zener.
I salute your willingness to think outside the box. From what I've seen those bricks don't open easily or cleanly and there isn't much room for changing the circuitry. The higher current ones are $20 bucks so there is a limit to how much experimentation I can afford. I'll put that on the third tier option list.
 

Thread Starter

tinker123

Joined Mar 6, 2020
15
No tinker, R4 is there to provide a voltage divider to the adjust pin, not to limit the current on the pot.

The value for VR1 and R3 is chosen to allow voltage adjustment to 20 volts, a higher value would allow for a higher output range.

As far as R1 and R2 hard to say exactly why that value was chosen, probably low enough to still do the job without losing too much voltage across them, and needing a bigger size. (wattage)

And by “preset resistor” he means VR1.:)
Understood about R4, I was thinking in terms of why not use all three pins of the pot as opposed to the separate R4 resistor. I had to look that up to understand (trivial and kind of irrelevant, sorry)

The part of the circuit I don't entirely understand is how current flows between the op amp and the regulators' Iadj over R3. I understand the op amp is trying to keep it's inputs equal but I don't know how to calculate the exact voltage drop over R3.
 

MrAl

Joined Jun 17, 2014
7,491
Hi,

Why the complicated circuit? Do you need ultra good regulation?

The more typical way of doing this is to add a small series resistance to each output before placing them in parallel. The small resistance acts to help balance the current so they share the load better.
The other point is that if two regulators just barely meet the current requirement, then use three regulators with the small output resistors so you have three sharing the load current.

Most applications can put up with a small variation in voltage, and because you are using chips with onboard voltage reference this implies that your application can put up with a somewhat significant voltage deviation from turn on to several minutes run time that stems from the voltage reference heating up over time.
 

ElectricSpidey

Joined Dec 2, 2017
1,067
That circuit is straight out of the LM138 data sheet. (edited model #)

But what's missing is the note that states the minimum 100 mA load.

As far as calculating the voltage drop across R3...Why? Are you worried about dissipation?
 
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Thread Starter

tinker123

Joined Mar 6, 2020
15
Hi,

Why the complicated circuit? Do you need ultra good regulation?

The more typical way of doing this is to add a small series resistance to each output before placing them in parallel. The small resistance acts to help balance the current so they share the load better.
The other point is that if two regulators just barely meet the current requirement, then use three regulators with the small output resistors so you have three sharing the load current.

Most applications can put up with a small variation in voltage, and because you are using chips with onboard voltage reference this implies that your application can put up with a somewhat significant voltage deviation from turn on to several minutes run time that stems from the voltage reference heating up over time.
No, my load is a device that is normally powered by two 18650's in series so is very forgiving of input voltage variations (on the downside at least) and probably doesn't care much about ripple either. I saw a suggested circuit that used only output resistors but all discussion on the topic said that approach is ineffective. It also used 0.3 Ohm on the output which would burn a lot of power. When you say this approach is typical can you provide any pointers to reference material?

My load normally draws slightly over 5 amps (but says 8 A on the label) so I might actually get away with just a single LM338 (I'll test that). I'm just being thorough. I also want to be sure I order all the parts I'll need in one go so this doesn't take forever, already seems like it is :)
 

rtgear

Joined Sep 8, 2013
1
Some other things to think about. Heat 2.5A through each 0.1Ω Resistor is 625mW each plus the heat through the regulators.
TI has a great tool, web bench power designer.
Looking at your circuit I found a different power design for myself.
I like the Buck converters, no heat loss. Almost a wash on parts. LM338 at >$2, $3.78 for the 3150
I use a 2N7002 (pull a separate resistor on feedback pin to ground) to change feedback so voltage at my charger is either 13.1V or 14.7V
 

Attachments

BobaMosfet

Joined Jul 1, 2009
1,063
In the case of the circuit shown by the thread poster, R1 & R2 have a resistor value of 0.1 Ohms, and a corresponding high wattage value, because their purpose is to NOT limit the flow of current very much. Raising the resistor value will reduce the amount of current being provided to the input on both voltage regulators (LM338). so R1 & R2 must be left alone.

The values of R3, R4, and R1 are therefore defined by calculation in the datasheet for the LM338.
 

Thread Starter

tinker123

Joined Mar 6, 2020
15
Thank you for that link MrAI, it's great to see source documentation for this pattern. It also provided me with good key words for further research. This approach isn't recommended when operating close to the current limits as needed in my scenario.

This link https://www.analog.com/en/technical-articles/paralleling-linear-regulators-made-easy.html has a good analysis with the following warning.
"However, the voltage drop across the current balancing resistors is much too large at full load (1.1A* 2Ω = 2.2V drop ) "
 

Thread Starter

tinker123

Joined Mar 6, 2020
15
Some other things to think about. Heat 2.5A through each 0.1Ω Resistor is 625mW each plus the heat through the regulators.
TI has a great tool, web bench power designer.
Looking at your circuit I found a different power design for myself.
I like the Buck converters, no heat loss. Almost a wash on parts. LM338 at >$2, $3.78 for the 3150
I use a 2N7002 (pull a separate resistor on feedback pin to ground) to change feedback so voltage at my charger is either 13.1V or 14.7V
Thanks for the suggestions rtgear. This thread was follow up specifically about linear regulators in parallel. I had an earlier post where I discussed my challenge of trying to build the entire circuit inside the 18x65mm tube that normally houses an 18650 cell. None of the high current, >5A, buck converters are "complete" regulators on a chip, they all require physically large inductors and caps. Sometimes there is no easy answer...
 

WBahn

Joined Mar 31, 2012
25,556
I'm coming in lately, but it seems like there is still some key information that hasn't been spelled out.

From looking at the TS's posts, it appears that he is looking for an 8.5 V power supply that can deliver 5 A (though the label on the load argues for an 8 A capacity). Currently the power is provided by two 18650 cells in series, so that's a nominal voltage of 7.2 V.

Before trying to modify a circuit to do what you want, it's a good idea to get a good idea of just what it is you NEED the circuit to do.

Why 8.5 V when the terminal voltage of the source it is replacing is 7.2 V?

How much tolerance can your output voltage have? A Lithium ion cell has a "full charge" voltage of about 4.2 V to a "dead cell" voltage of 3.0 V, so over it's charge life the two-cell pack would go from about 8.4 V down to about 6.0 V. If your normal load can be successfully powered by the batteries, then it would seem likely that you can get by with any voltage output from your power supply within that range and get by, which also implies that good voltage regulation probably isn't that important.

It appears you want to power this with a 12 V input source. Why? Is it just because you think they are cheap and easy to come by? May so, maybe not. But don't force yourself into a bad choice just because you don't think there are others. If your output is 8 V and you are delivering 5 A, then your linear regulator circuit is going to have to dump 40 W of heat somewhere. If that is coming out of a tube meant for the two cells, that is going to get really hot really fast.

Why do you want to put everything in the tubes where the batteries go? It's a cute idea, but what does it really accomplish? You still have to have an external source to provide the 12 V to the tubes, so what is really gained?

Why not just use a commercial DC-DC converter? What are your constraints in terms of size and cost?
 

MrAl

Joined Jun 17, 2014
7,491
Thank you for that link MrAI, it's great to see source documentation for this pattern. It also provided me with good key words for further research. This approach isn't recommended when operating close to the current limits as needed in my scenario.

This link https://www.analog.com/en/technical-articles/paralleling-linear-regulators-made-easy.html has a good analysis with the following warning.
"However, the voltage drop across the current balancing resistors is much too large at full load (1.1A* 2Ω = 2.2V drop ) "
Hello,

If somebody designs a circuit to balance loads and it drops 2.2v out of say 10v, then they dont understand the concepts behind doing it this way. The resistors only have to balance less than 0.1v at full load so that should tell you something. I also mentioned that in an application that takes 2 regulators you may have to go to 3 so that you can balance the currents properly without loosing too much voltage.
 

ElectricSpidey

Joined Dec 2, 2017
1,067
“It appears you want to power this with a 12 V input source. Why? Is it just because you think they are cheap and easy to come by? May so, maybe not. But don't force yourself into a bad choice just because you don't think there are others. If your output is 8 V and you are delivering 5 A, then your linear regulator circuit is going to have to dump 40 W of heat somewhere. If that is coming out of a tube meant for the two cells, that is going to get really hot really fast.”

This is wrong…

Supply is 12 volts
Output is 8 volts
That leaves 4 volts across the regulator
4 times 5 is 20

The load is 40 watts…not the regulator.
 

Thread Starter

tinker123

Joined Mar 6, 2020
15
I'm coming in lately, but it seems like there is still some key information that hasn't been spelled out.
Thanks for trying to help WBahn. This post was really just to help me understand the parallel regulator circuit. The overall design problem was covered in another post and the topic beaten to death there.

There are, as always, lots of different options. I was checking how far down the road it was possible to get with this one. I have a much better understanding of the trade offs now than when I started planning this project (which is just for a hobby hack).
 
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