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, 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.
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.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.
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.....I wish manufacturers would destroy failing parts and sell only the good ones. Then the the range of spec's will be narrower.
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.sell only the good ones.
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?
Lets see if we can clear this mystery up once and for all
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/IHi 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.
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?
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