need a circuit of a single +5V power supply using LM317T

[UrSv]

The output voltage of a 317 is roughly Vref*(1+R2/R1) which means that is independent of the input voltage (under normal conditions) which means that if you use 820 for R2 and 270 for R1 you get 1.25*(1+3.03) which is close to 5 V. Or you could use a trimmer for R2 and adjust to whatever you want. The input voltage is an example and not specified as such. The difference is that when you have 28 V input the voltage drop across the regulator is higher and the power dissipated much higher (P=Vdrop*Iout). I would go for something like 10-12 V in for 5 V out.

/UrSV

[Electro]

jarthel, I suggest you look at Questlink. You have to register to view the electronic component database although registering is free at Questlink.

There are other electronic component database sites but this is a start.

[peranders]

I have posted jarthel my Excel-file. Maybe someboby else find it useful.

The file was personal use, therefore very simple.

The file is for calculation of R1 and R1 and also which errors you will get.

[ALW]

You must have rather low value of the feedback resistor, between "output" and "adjust".

Just curious as to the above - why?

The data sheet doesn't state this anywhere, and since a higher impedance divider network makes the adj. terminal decoupling capacitor much more effective, reducing noise, o/p impedance and improving line rejection I'm puzzled.

Andy.

[HarryHaller]

"Since the 100 µA current from the adjustment terminal represents an error term, the LM117 was designed to minimize IADJ and make it very constant with line and load changes. To do this, all quiescent operating current is returned to the output establishing a minimum load current requirement. If there is insufficient load on the output, the output will rise."

http://www.national.com/ds/LM/LM117.pdf

You want to make sure that the ajustment current is large in comparison to error current of 100uA, Dividing the reference voltage of 1.25 volts by 240 ohms give a program current of about 5 mA through the resistor from the adjustment terminal to ground which sets your output voltage. This current is 50 times as great as the error current from the adjustment terminal swamping its error contribution.

H.H.