Minimum load current of LM317

glorimda said:

What am I following wrong?

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The UPPER resistor value determines the minimum load current. The minimum load current is 5mA for the more expensive LM117 (the resistor should be 240 ohms or less) and is 10mA for the less expensive LM317 (the resistor should be 120 ohms or less). The upper resistor always has 1.25V across it.

The LOWER resistor will have the same current as the upper resistor and Ohm's Law is used to simply calculate its value for setting the output voltage. The output voltage is the voltage across the lower resistor plus 1.25V.

@audioguru,
I fear whether we are thinking properly as, the 240 or 120 ohm resistor is only reaching the ADJ pin and not ground return. There would be another resistor whose value depends on the output voltage. The actual current is to be calculated after adding that resistor also.

Perhaps in this specific case, where the output needed is 1.3V, the bottom resistor may have a low value.

Since this is so simple, why not just try it with real components, with a breadboard, the IC LM317 or 117 if you have it, two resistors, another resistor a dummy load and a dmm to prove these, how hard could this be ?

C'mon, even without breadboard, a dead bug style soldering is not that difficult either.


Just love the mral's siggy -> "One test is worth a thousand expert opinions"

the concept is not around building one alone. The O P and many others gain by understanding how to design around.
As analysis is possible by electrical method, so we discuss. It is question of approach and calculation. After all full fledged datasheet and internal diagram also available. There are willing seniors like Audioguru and others, let us learn the designing concept and approaches.

Remember: on the 117/317, the internal circuitry forces 1.23V across the TOP resistor (connected from the output pin to adjust pin) which creates the current flowing through the bottom resistor. hence, the top resistor is the one providing the minimum load which is 5mA or 10mA depending on which version.

BravoV said:

Since this is so simple, why not just try it with real components, with a breadboard, the IC LM317 or 117 if you have it, two resistors, another resistor a dummy load and a dmm to prove these, how hard could this be ?

Click to expand...

Because the minimum load current of each IC is different so the spec's say for the upper resistor to always use 120 ohms or less for an LM317 and use 240 ohms or less for an LM117 then all ICs will work perfectly.

And it varies over temperature range. You could find plenty that work OK at room temp at 5mA that flake out at temp extremes.

I recently grabbed a couple of LM317 from the junk box and tested minimum load, they regulated 1.25 fine with a 1k resistor, so were able to be used as 1.25mA constant current sources. Another brand needed a 820 ohm resistor to regulate 1.25v.

The 5mA (or old spec of 10mA) is only needed to gurantee regulation for the full range of input voltages, AND full range of currents, AND full range of temperature spec. But if you just need to drive a LED at 2mA or something where the variables are relatively fixed then you can violate the spec and just use the right value of resistor and see if it works. Which it probably will.

Not to be used in life support applications!

I usually use a 120 ohm resistor from out to adjust and a pot from adj to ground or often use just a 1K pot with the wiper on the adj and one side to out and the other side to ground. My math says if you use a 120 ohm from out to adj and a 5 ohm from adj to ground, you'll have your 1.3 volts. Most of the time, I get a measurement of 1.25 V between out and adj. Also, using 120 ohms satisfies the 10ma minimum current.

Hi again,

I mostly use the LM317L version, which can go lower on the min current. I still used 220 ohms though for the upper resistor.

MrAl said:

Hi again,

I mostly use the LM317L version, which can go lower on the min current. I still used 220 ohms though for the upper resistor.

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You are correct!
The LM317L is not shown on the datasheet for the LM317, it has its own datasheet.
It is shown with a 240 ohm upper resistor because it has a 5mA max minimum load current.

MrAl,

So let me clear it.

1. Load current is the one coming out from Vout pin.

2. The component determining current is R1. No matter of what the value of R2 I use, the current is only determined by R1. And to see the reason of this, I guess need to interpret the internal schematic.

3. And Vout is determined by the value of R2.

4. On your first reply with example of using 250, 750R and 5V output voltage, it seemed that the current is determined by whole impedance being seen from the output terminal. Cuz by calculation 5V/1K = 5mA.
' 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 if R2 doesn't do anything on determining the current, then this is wrong. we would rather do 1.25V/250R=5mA first and find the R2 to get 5V.

5. The current flowing through output terminal is always 5mA unless changing the 250R.

I hope everything above is right then I don't need to ask again.
Thanks again!

The minimum load current is 5mA for an LM117 and a little LM317L, or 10mA for an LM317.

yep, you're right. was my mistake. But what I'm focusing on is not whether it is LM117 or 317 or 5mA or 10mA. Just I'm wondering about the relations between resistor, current.
Thanks for reply though.

glorimda said:

But what I'm focusing on is not whether it is LM117 or 317 or 5mA or 10mA. Just I'm wondering about the relations between resistor, current.

Click to expand...

The relation between the current and the upper resistor is calculated with simple arithmatic using Ohm's Law:
1.25V/5mA= 250 ohms. 240 ohms is the nearest standard value.
1.25V/10mA= 125 ohms. 120 ohms is the nearest standard value.

The value for the lower resistor is also calculated with simple arithmatic using Ohm's Law.

glorimda said:

MrAl,

So let me clear it.

1. Load current is the one coming out from Vout pin.

2. The component determining current is R1. No matter of what the value of R2 I use, the current is only determined by R1. And to see the reason of this, I guess need to interpret the internal schematic.

3. And Vout is determined by the value of R2.

4. On your first reply with example of using 250, 750R and 5V output voltage, it seemed that the current is determined by whole impedance being seen from the output terminal. Cuz by calculation 5V/1K = 5mA.
' 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 if R2 doesn't do anything on determining the current, then this is wrong. we would rather do 1.25V/250R=5mA first and find the R2 to get 5V.

5. The current flowing through output terminal is always 5mA unless changing the 250R.

I hope everything above is right then I don't need to ask again.
Thanks again!

Click to expand...

1. Yes.
2. R1 is usually calculated from 5ma=1.25/R1, but we have to remember that R2 is also required.
Without R2 no current will flow from the output pin, so the voltage will rise. Thus if R2
becomes disconnected the regulator doesnt work, because the load current flows through R1 and R2
at the same time, with very little current into the regulator ADJ pin. But to approximate this
working, you can use R1=1.25/0.005 and just make sure R2 can never becomes disconnected.
(See also below).
3. Yes, that's about right, once you determine R1.
4. Yes, the current flows through the whole network, not just R1. But as an approximation
many people use R1=1.25/0.005 (or R1=1.25/0.010) and go from there.
5. The current flowing through the output terminal is determined by the total resistance from
the output terminal to ground, not just R1 alone. The current *set point* is usually determined by
i=1.25/R1.

You can look at a behavioral model rather than the complete internal schematic of the LM317 in order
to much better understand how this little regulator works. You'll get a lot more from that. If you'd like
to see a behavioral model i'll post one here. This kind of model is much more simple and so you can get
the exact idea how all this works by examining that.

Be happy, you'll know everything there is to know about this regulator before we're done here

Ok. I guess I'm getting to know clearly.

At first sight, I guessed simply if I increase the value of R2, Iout would be decreasing. But I'm seeing that even if I increase the R2 value to get higher Vout, minimum current is almost fixed by calculation or even increased. So R2 affects on Iout, but mostly determined by R1.

I'd appreciate if you can give me what you called 'behavioral model'. Really thank you for your lots of help.

Hi again,


The model is shown in the attached diagram.
The parts are as follows:
X2 is an amplifier with a gain of +100000
X1 is a subtractor (summing junction with one inverted input)
V1 is a 1.25v battery (reference)
The "50ua" is a 50ua current source
The terminals of the LM317 are labeled:
VCC, OUT, and ADJ as per the device.

The currents shown are:
iOut: This is the total current out of the device, this must be at least 5ma (or 10ma depending on real world device)
iR1: Current through R1
iR2: Current through R2
iRL: Current through the load resistor RL.

From this model we can see that iR2 is equal to iR1 plus 50ua.
We also see that iOut is the sum of iR1 plus iRL.

This model can be used to better understand how the LM317 works inside without knowing all the details, however
it can not be used to analyze the current limiting function nor how the changes in Vcc affect the output. Since
our main concern in this thread is the minimum output current, this model is appropriate. It can also be used to
understand how the bias current (50ua) can change the output with temperature, but that's not important right
now.

audioguru said:

The minimum load current is 5mA for an LM117 and a little LM317L, or 10mA for an LM317.

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The LM317 is made by different manufacturers and each has it's own datasheet. Most manufacturers now use improved LM317 die in their process.

From ON semiconductor LM317 datasheet;
Minimum Load Current to Maintain Regulation (VI-VO = 40 V) ILmin (typ =3.5mA) (max =10 mA)
so 3.5mA is considered a typical load value to achieve 40v regulation and one I tested regulated fine at 12v even with a load current of 1.25mA.

I understand it's professional to allow 10mA to be within spec for all possible range of input voltages and all possible temperatures and all possible manufacturing variations etc, but for many simple hobby tasks (like the LED constant current source from a 12v battery etc) the LM317 WILL still work fine at much less load current, especially if you hook one up with a resistor and test it.

MrAl // Thanks for attachment. I don't have time right now, but I'll look into it through this weekend. Really appreciate all you gave me.
Mr RB// Thanks for ur comment. Since this time, I realized that I need to look at datasheet detailedly before doing designing circuit. And this time helped me a lot for that.