Minimum load current of LM317

[glorimda]

I'm using the LM317(adjustable voltage regulator). I had some problem getting the right output I want and finally I figured it out.
At first, to get the output of certain value I want, I referred the specification document and followed the application schematic that I attached. I put the R1 with 3K and R2 with 300 to get 1.3V output. But it didn't work. And I've struggled to figure out the reason. Finally, my colleague told me I need to use R1 with 240ohm and find out the value of R2 with respect to R1.He also said it is related to minimum load current.
And I did as he said and it's working now. And now I'm trying to figure out about what the minimum load current is and why I need to put 240ohm in R2.
I'm now seeing the inside schematic of LM317(attached), but I can't understand.
Can anybody explain about this?

 

[ericgibbs]

I'm using the LM317(adjustable voltage regulator). I had some problem getting the right output I want and finally I figured it out.
At first, to get the output of certain value I want, I referred the specification document and followed the application schematic that I attached. I put the R1 with 3K and R2 with 300 to get 1.3V output. But it didn't work. And I've struggled to figure out the reason. Finally, my colleague told me I need to use R1 with 240ohm and find out the value of R2 with respect to R1.He also said it is related to minimum load current.
And I did as he said and it's working now. And now I'm trying to figure out about what the minimum load current is and why I need to put 240ohm in R2.
I'm now seeing the inside schematic of LM317(attached), but I can't understand.
Can anybody explain about this?

hi,
R1 is fixed at 240R, R2 is chosen using the formula Vout= Vref[1+(R2/R1)]

So if you have R1=R2 =240R then the Vout will be set a 1.25*2 = 2.5V

The Minimum current into an external load is 3.5mA to 10mA,,, so just add a load resistor of 2.5/.01 = 250R.

What Vout do you require.??

 

[alec_t]

The IC has no direct current path to ground, so its operating current has to be drawn via the load. That is why there is a minimum load current.

 

[glorimda]

hi, ericgibbs. I aim to get 1.3V from the output. And I now solved how to get it. But what I asked was, why I have to use 240R for R1 instead of any other value, then I can just induce the output voltage by using that fomula with proper R2 . But actually, with R1 other than 240R, it outputs wrong value.
And my colleague told me the reason I have to use 240R is relevant to minimum load current. This term is what I'm wondering about.
What this term means and what is the relation between that 240R and minimum load current.
And to figure out this, I looked at the inside schematic of LM317. But I can't seem to find the reason myself.

 

[ericgibbs]

hi, ericgibbs. I aim to get 1.3V from the output. And I now solved how to get it. But what I asked was, why I have to use 240R for R1 instead of any other value, then I can just induce the output voltage by using that fomula with proper R2 . But actually, with R1 other than 240R, it outputs wrong value.
And I guess the reason I have to use 240R is relevant to minimum load current. This is what I'm wondering about.
And to figure out this, I looked at the inside schematic of LM317. But I can't seem to find the reason myself.

The LM317 requires that the internal 1.25V ref has a 100uA enabling current.
So a 240R from Voutput pin to Vadj pin is required
The R2 is for setting the Vout value.

With Vadj at 0V for example the Vout will be approx 1.25V, so to get 1.3Vout add a low value resistor from Vadj to 0V.

With reference to the Minimum load current the LM317 requires a current load of at least 10mA for the internal circuits of the LM317 to work correctly.

The R1 at 240R is a Basic requirement of the LM317 when used as a voltage regulator.

 

[glorimda]

alec_t//Thanks for reply, but I can't get it exactly. Is the current you're saying the current flowing from ADJ pin to Vout pin? But why should it be there? Is it just being there?
Yes, there's not path to ground so I understand that operating current should be through R2. And then? Honestly, even I can't ask exactly what I want cuz I don't understand the term 'minimum load current' ..... sorry for my bad.

 

[glorimda]

ericgibbs //So If I understood right, 240R is to get 100uA Adjustment pin current and Minimum load current is the minimum current required to operate the LM317.
Then as a matter of fact, there's no relation between 240R and minimum load current?

 

[mvs sarma]

are you interested in 1.3V or a constant current for an LED?

 

[glorimda]

mvs sarma //1.3V. This is what I need. And already got 1.3V, but wanna know the relation between 240R and minimum load current.

 

[mvs sarma]

as the240 0hms is outside the device, it adds to load current as you see the current at output pin.
this has been mentioned n the maths for calculation of output voltage, but negligible as the current is too small.

 

[audioguru]

Look at the datasheet!
R1 is 240 ohms in all schematics because they show the more expensive LM117, not the less expensive LM317 that has a higher minimum load current.
If R1 is 120 ohms then it draws enough current for the LM317 then a load is not needed to hold down the output voltage.

 

[bountyhunter]

Look at the datasheet!
R1 is 240 ohms in all schematics because they show the more expensive LM117, not the less expensive LM317 that has a higher minimum load current.
If R1 is 120 ohms then it draws enough current for the LM317 then a load is not needed to hold down the output voltage.

Correct, the vast majority of parts (both 117 and 317) will be OK with 5 mA (the 240 Ohm resistor) but there would be small yield loss to that 5mA limit at final test so they opened it to 10 mA on the junk grade part to be able to sell the few that don't make it at 5mA.

 

[audioguru]

I wish manufacturers would destroy failing parts and sell only the good ones. Then the the range of spec's will be narrower.

 

[bountyhunter]

I wish manufacturers would destroy failing parts and sell only the good ones. Then the the range of spec's will be narrower.

They areen't failing parts, they are "outlying" parts on the bell curve distribution. If you tested 10,000 117 and 10,000 317 parts, the distribution would be centered at the same value (the dies are identical) but the 117 parts would have the far edge of the bell curve snipped off. It would not be a good business decision to lose significant yield to such an irrelevant spec because that forces price up. In 20 years, I never heard anybody complain about quiescent current on it. But they sure complain about PRICE.....

sell only the good ones.

Interesting philisophical question... at NS in the last years when the lunatics were completely running the asylum.... the new EDICT FROM GOD was that no new product could release unless the "A" grade yield was over 90%. As I pointed out, that means there is NO MORE "A" GRADE. In the old days, A grades were the "cherry parts" with tight specs that were reaped out of the center of the distributions. If 90% are A grade..... then they are all really C grade. It just forced them to open limits wider.

But sometimes when real A grade yield was high, it generated scrap because you had to build so many parts to get enough of the A's, some "C" grades were left unwanted (scrap). And scrap is one of the metrics every manager hated. I was at a meeting where the manager howled about how the LP2951A generated so much scrap... and I asked him:

"So.... you don't like a part that generates $200 MILLION a year in profit because it also creates a million in scarp?"

Affirmative. Sometimes people are so stupid they can't comprehend how stupid they are.

And they handed the LP2951A over to a green designer with the orders to get the A grade yield up to 90%... and he screwed it up entirely.

Oh, well....

 

[MrAl]

I'm using the LM317(adjustable voltage regulator). I had some problem getting the right output I want and finally I figured it out.
At first, to get the output of certain value I want, I referred the specification document and followed the application schematic that I attached. I put the R1 with 3K and R2 with 300 to get 1.3V output. But it didn't work. And I've struggled to figure out the reason. Finally, my colleague told me I need to use R1 with 240ohm and find out the value of R2 with respect to R1.He also said it is related to minimum load current.
And I did as he said and it's working now. And now I'm trying to figure out about what the minimum load current is and why I need to put 240ohm in R2.
I'm now seeing the inside schematic of LM317(attached), but I can't understand.
Can anybody explain about this?

Hi there,


Lets see if we can clear this mystery up once and for all

To start with, just forget about the 240 ohm resistor, forget about it completely for now. For one thing, designs can be made without using a 240 ohm resistor. Let me try to explain a little better here.

The output of the regulator MUST be LOADED to at least 5ma (10ma for some versions, but lets assume 5ma for this discussion). This is the minimum load talked about on the data sheet. It has almost nothing to do with the 240 ohm resistor. It's a specification independent of everything else. The output must see a load current of 5ma out of the output terminal, period. Note we dont have to mention any resistor values to understand that spec.

Then there is a second 'spec' that we have to ensure gets satisfied. The second spec is the voltage at the "ADJUST" terminal has to be equal to the desired output voltage minus 1.25 volts (approximate). This means if we want 5.00 volts output then we have to make sure we have 3.75 volts at the ADJ pin.

So we have to make sure of two different things at the same time...
1. Output is loaded to at least 5ma (for any example)
2. ADJ pin is 3.75 volts (for this example where we want 5v output)

Now lets say we have access to any parts we want and we can use anything we want to build this regulator.
We can easily get the ADJ pin to be 3.75 volts by simply connecting a battery or accurate power supply to that terminal with the positive connected to ADJ and the negative connected to ground. So we satisfy spec #2.
Now since we have 5v output, and 5ma is 5/1000, all we have to do is connect a 1k load resistor to the output to ground. That means we'll meet the 5ma minimum load condition. If the load we use already draws more than 5ma with no possibility of going lower, we would not need any resistor on the output, but we'll assume we want to make sure the regulator stays within regulation even without a load.
Since we met both spec's we're done, that's it.

Ok, so we met the spec's, and that regulator would run forever, but in doing so we used a voltage reference or battery to set the ADJ pin voltage, and normally we dont want to have to do that.

To fix that situation, we know that we have 5v coming out of the regulator and it's regulated, so we can connect a resistor divider from the output to ground and use the divider center tap to feed the adjust pin.

We know we need 3.75v at the ADJ pin, so lets start with a lower divider resistance of 3.75k. Because we have 5v output, that means the upper resistor of the divider will have to be 1.25k so that we get the required 3.75v at the ADJ terminal. Connecting those two resistors, we are again done, but this time we didnt need a separate voltage supply as we replaced it with a resistive voltage divider.

But note now we have three resistors, two for the divider, and one for the minimum load (5ma) requirement. Wouldnt it be nice if we could just use two resistors instead of three? What's the difference? Note that the resistive divider provides a DC path to ground for the output, and the 1k load resistor also provides a DC path to ground for the output. We dont need two DC paths to ground, so we try to get by with only one.

The only difference is that the three resistors provide BOTH a proper voltage reference AND a minimum load for the regulator. But to get by with only two resistors all we have to do is LOWER the total resistance of the resistive divider until it's total resistance is equal to the load resistor of 1k, and then that will provide the min current requirement. But we have to keep the voltage at the ADJ pin (divider center tap) at 3.75 volts to keep the output at 5v.
One way to accomplish this is to transform the impedance of the resistive divider to 1k without changing the RATIO of the two resistors. This is easy to do. Since 1.25K+3.75k is equal to 5k, if we want 1k we have to divide BOTH resistors by 5 (pretty simple).
Doing this with the 3.75k resistor we get 750 ohms, and doing this with the 1.25k resistor we get 250 ohms. So for the final design we use a 750 ohm resistor for the lower resistor and more notably, a 250 ohm resistor for the upper resistor.

Note that once we transformed the network we ended up with a resistor value close to 240 ohms for the top resistor. That's why the recommended value is usually 240 ohms.

Note also that the lower resistor changes when we want to change the output voltage to something other than 5 volts, but the upper resistor (250 ohms) does not change. Because that resistor does not change it is recommended for all designs, because using a 250 ohm resistor (or 240 ohms) ALWAYS FORCES us to choose a lower resistor who's value when combined with the upper resistor will always draw at least 5ma from the output.

So that's the reason behind using a 240 ohm resistor for the upper resistor. As noted, sometimes 120 ohms is required for the devices that insist on a 10ma min load current.

There's a small addition to this reasoning, and that is we also want to ensure that the divider resistance current swamps the bias current of the regulator so that we maintain decent output voltage temperature independence. 5ma does this pretty well, so in making sure we have 5ma flowing through the divider we also ensure getting decent temperature characteristics.

 

[audioguru]

Lets see if we can clear this mystery up once and for all

Now you are confusing things because you selected only 5mA for the minimum load current which is used for the expensive LM117. It can use 240 ohms for the upper resistor.

But nearly all of us use the less expensive LM317 that needs 10mA for its minimum load current and uses 120 ohms for the upper resistor. The internet is full of wrong circuits that use 240 ohms with the LM317.

Note that not all LM317 ICs need 120 ohms, only some of them and only when there is 40V from input to output.

 

[bountyhunter]

Hi there,


Lets see if we can clear this mystery up once and for all

Note that once we transformed the network we ended up with a resistor value close to 240 ohms for the top resistor. That's why the recommended value is usually 240 ohms.

NO. The reason it's specified as 120 Ohms is even more simple: the LM117/317 always forces 1.25V across the resistor, and getting 10mA means 120 Ohms (nearest standard size). R = V/I

It is true the "divider resistors" can be much higher Ohmage if there is an output load present at all times providing the minimum load. The 120 Ohm resistor means no other output load is required.

 

[glorimda]

MrAl, thank so much. It did help me a lot understanding it. I followed your explanation until this.

'because using a 250 ohm resistor (or 240 ohms) ALWAYS FORCES us to choose a lower resistor who's value when combined with the upper resistor will always draw at least 5ma from the output.'

If I understood) right, you meant the lower part resistor should be lower than 750 then the whole resistance can be lower than 1K and it should be( to meet the minimum load current of 5ma. But I'm seeing that both higher or lower than 750 are used to get the output needed(for instance, to get 10V output, we need higher lower part resistance). If lower part resister is higher than 750, impedance being seen from the output is higher than 1K and that cause lower load current than 5ma.

What am I following wrong?

 

[ericgibbs]

hi
Do as BountyHunter suggests a 120R for R1, leaves you free to have R2 or not, also covers the minimum load current requirement, neat and simple [ and cheap]

 

[MrAl]

MrAl, thank so much. It did help me a lot understanding it. I followed your explanation until this.

'because using a 250 ohm resistor (or 240 ohms) ALWAYS FORCES us to choose a lower resistor who's value when combined with the upper resistor will always draw at least 5ma from the output.'

If I understood) right, you meant the lower part resistor should be lower than 750 then the whole resistance can be lower than 1K and it should be( to meet the minimum load current of 5ma. But I'm seeing that both higher or lower than 750 are used to get the output needed(for instance, to get 10V output, we need higher lower part resistance). If lower part resister is higher than 750, impedance being seen from the output is higher than 1K and that cause lower load current than 5ma.

What am I following wrong?

Hi again,

glorimda:

By stating that
"the 250 (or 240) ohm resistor always forces us to choose a lower resistor who's value when combined with the upper resistor will always draw at least 5ma from the output"

what that means is that although we dont think about it, we end up ALWAYS using a lower resistor that causes a 5ma draw from the output. In other words, after we choose the lower resistor (like 750 ohms) to get 5v output we have (possibly without realizing it) chosen a resistor that draws 5ma. That's all that means.
So once you use a 240 ohm resistor then you select your lower resistor based on the little formula they give you, and you are accomplishing two things:
1. Drawing the required 5ma
2. Getting your required output voltage.

See how that works now? Once the output is set, you have your 5ma and you're ready to go.
Also as i pointed out, some versions require 10ma min load current.

Others:
I made it perfectly clear that some versions will required 10ma not 5ma. Still confused?

bounty:
It's not that internal 1.25v that keeps 5ma through the resistor, it's the combination of the lower resistor and the upper resistor that actually draws the 5ma. You can go ahead and make the mistake of calculating 1.25v/R1 if you like, but you'll find it doesnt always work, that's why i dont present it that way. That's why transforming the network worked so well, because then it draws the required 5ma.
Dont believe it? Do some circuit simulations to prove or try to disprove it.