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If
Vout = 1.25 * (1 + R2/R1) + Iadj * R2
is the formula for the LM317 v regulator,
and i need a 5v circuit, and have a 9v input, does the input voltage not effect the output voltage,
and what is a common resistor to start with for R1,
No, this is the purpose of a voltage regulator - it gives a constant output voltage, regardless of the input voltage (within limits) - input must be a couple of volts higher than output, and not more than 38V (I think... can't quite remember)
I think the datasheet recommends 240ohm for R1. Since this is a non-standard value, try 220 ohm. Also, when calculating, you can ignore Iadj (It is very small)
there is a resistor value for 620ohms or 680ohms, but if you cant find it, just as what Phasor said use a 1k pot. its fun watching the needle or the display change value, as you said its your first. besides it makes your regulator more versatile. :lol: :lol: :lol:
Guy, forget the lm317, just use 7805, feed the 9v at the input and you get a clean 5v output.
but if yu really want to use the lm317 then do this:
have r1 = 220 ohms, and for r2 -> have 470 ohm or 560 ohm resistor in series with 1k pot, this way you'll get a total resistance of r2 between apx 500ohms-1500 ohms, and that'll give you a nice variable voltage between 4 and 9 volts
the reason for the 470 or 560 resistor is that, you don't want to have r2 as 0 ohms if yu turn the pot all the way in one direction
The LM317 is a very versatile IC regulator and much can be used from it specially adjustable ones. Why not use LM7805 coz it really serves its purpose of regulating down to 5V. Its just a suggestion.
as an added safety feature try to put diode on the input and output of the device LM317.
for input short circuit protection:
place in between the input (pin 1) and output (pin 2) diode IN4001 or IN4002 with the cathode at the input side.
for output short circuit protection:
place in between the output (pin 2) and current adjust (pin 3) with diode IN4001 or IN4002 with the cathode at the output side.
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