allright. I pulled up the LM317 spec sheet and found the following

so....here is the work I have based on this equation

I am wayyyyy off base here. what am I doing wrong?

 

The adjustment pin current is typically a max of ~5uA for the LM317. You appear to have Iadj = 0.01A - which is way off.

 

In R2 of your schematic (your diagram and the application notes have reversed designations) you have 100Ω. Since the method the LM317 uses is to maintain 1.25V between the output pin and adjust you should be able to figure current. The current through R2 is the same as through R1. From there it is gravy.

 

Iadj is 50uA typical, and 100uA=0.0001A max.

But this circuit can be treat as a voltage divider.
Since R2 = 100Ω and voltage across R2 is equal 1.25V we have
IR1 = 1.25V/100Ω = 12.5mA
And the current that is flow through R1 is equal
IR2 = Iadj + IR1 = 12.6mA
And we want Vo = 15V----> VR1 = 15V - 1.25V = 13.75V

R1 = 13.75V/12.6mA = 1.09126984KΩ

\(R1=R2*\frac{Vout-1.25V }{1.25V+I_{ADJ}*R2}\)

Or

\(R1=R2*(\frac{Vout }{1.25V}-1)\)

 

The position of R2 and R1 are different in your problem and in the example you posted. Go back a redo the math and notice carefully how the fraction made by the resistor values is arranged in the example and how the Resistors in the problem are the reverse.

 

Iadj is 50uA typical, and 100uA=0.0001A max.

Agreed - I misread the data sheet. Looked at the adjustment pin current change row rather than the actual value. I can only put it down to deteriorating vision.

Apologies for the misinformation.

 

Iadj is 50uA typical, and 100uA=0.0001A max.

But this circuit can be treat as a voltage divider.
Since R2 = 100Ω and voltage across R2 is equal 1.25V we have
IR1 = 1.25V/100Ω = 12.5mA
And the current that is flow through R1 is equal
IR2 = Iadj + IR1 = 12.6mA
And we want Vo = 15V----> VR1 = 15V - 1.25V = 13.75V

R1 = 13.75V/12.6mA = 1.09126984KΩ

\(R1=R2*\frac{Uwy-1.25V }{1.25V+I_{ADJ}*R2}\)

Or

\(R1=R2*(\frac{Uwy }{1.25V}-1)\)

what is the variable "Uwy" referring to?

 

Well in my country "Uwy" means exactly the same as Vout in english.
And sorry for confusion

 

Well in my country "Uwy" means exactly the same as Vout in english.
And sorry for confusion

no worries... what country is that?

 

I'm from Poland
http://en.wikipedia.org/wiki/Poland

 

In R2 of your schematic (your diagram and the application notes have reversed designations) you have 100Ω. Since the method the LM317 uses is to maintain 1.25V between the output pin and adjust you should be able to figure current. The current through R2 is the same as through R1. From there it is gravy.

thats a great explanation.

I = 0.0125A

R = 13.75/0.0125 = 1100

which is basically the same answer that was derived by Jony130

however, this answer was listed as wrong

what are we doing wrong here?

 

what is the dropout voltage? could that have anything to do with it?

 

however, this answer was listed as wrong

what are we doing wrong here?

Maybe answer is wrong

As for the dropout voltage some times called headroom voltage read this
http://en.wikipedia.org/wiki/Dropout_voltage

 

per wikipedia

"In electronics, the dropout voltage of a voltage regulator is the smallest possible difference between the input voltage and output voltage to remain inside the regulator's intended operating range."

so if I understand correctly the dropout voltage has no effect on what the output voltage will be.

 

I am to the point now where I am mindlessly staring at this problem... ugh

 

Hello,

The PDF in the following link will give you some more insight in the voltage regulator:
http://www.techlearner.com/Apps/basregulators.pdf

The LM317 you will find at page 22.

Bertus

 

Hello,

The PDF in the following link will give you some more insight in the voltage regulator:
http://www.techlearner.com/Apps/basregulators.pdf

The LM317 you will find at page 22.

Bertus

but my schematic looks different from the one that is in that PDF

for example

what values should I use for Rwp and Rwn?

 

Hello,

Rwn and Rpn are the resistances of the wires from the regulator to the load.
These will be dependend on the used wire, but can be dismissed when using small loads, as the values are mosttimes very small.

Bertus

 

allright so in the equation they are subtracting (IL*Rwp), which in reality is the voltage drop of the wire. since this loss is negligible we will disregard it.

R2 should be -88

also incorrect

 

Check your math.

15 = 1.25 * ( 100 + R2)/100
1500 = 1.25*(100 + R2)
1500 = 125 + 1.25R2
R2 = 1375/1.25 = 1100