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

220 + 680 ohm gives you 5.11 V
270 + 820 = 5.046 V

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

Check the first page of the datasheet.

See also my power supply for my headphone amps.

For those of you who wants to calculate min-max values (including tolerances) for LM317 send me an email. I have made an Excel file for this calcualtion.
nice seeing you here Peranders. :)

Well after I posted this thread I came across LM2940T-5; a fixed +5V voltage regulator. Do you think this is a better solution than using a variable regulator like the LM317T?

I'm thinking of using a 10000uF 10V electrolytic. What's the difference if I use 100uF or 1000uf instead of 10000uF? Would it be better if I used 10000uF instead?




2002-04-12 4:41 am
The LM2940T-5 is a switching regulator and it doesn't dissipate a lot of heat.

When selecting what capacitor to use for DC filtering, look for what the IC recommends on DC filtering. I think either 100uf or 1000uf is good. The power supply is not too great so 10000uf will not help much.
I have a look at the first page of the LM317 datasheet. There's a sample circuit there to built a circuit based on the voltage needed.

The only problem is Vin says >=+28V. The LM317 is going to be used as a power supply for the Kwak-clock. I don't think I can get +28V inside a CD player.

Is there any way to input <=+12V and get +5V using LM317.

Thank you. :)



2002-01-31 5:48 pm
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.



2001-06-12 10:15 am

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.

low value of the feedback resistor

"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."

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.