I think this is a simple question for those of you that know what
resistors are readily available. I want to end up with 5v, but not sure
what I need for R1 & R2.
Vo = 1.25 * (1 + (R2 / R1)) + Iadj(R2)
I'm not really sure what Iadj(R2) is.. but a couple examples I googled
seemed to just ignore it. This example also used 100 for R1, but the
only way to get 5v is to use 300 as R2. Any reason not to do this?
ok, these might be really stupid questions.. but I've never used a
regulator before.. so cut me some slack.. :)
PS, my EasyProg just arrived.. YEAH!!
If you only need 5V, just run to radio shack and get a LM7805.... you input
a voltage between 5.5 and 15 (??) and get out 5V. Simpler than an adjustable
about the 317, I think Iadj is the current going into adj pin and R2 * Iadj
would determine an offset voltage. You could make R2 variable and have a
range of output voltages.
On 11/18/05, Shay <highstyleweb.com> wrote: shay
David Van Horn
> Vo = 1.25 * (1 + (R2 / R1)) + Iadj(R2)
> I'm not really sure what Iadj(R2) is..
Get the Iadj current and multiply by the value of R2.
It's an error term. You can also put a pot in the circuit and just
adjust it out.
but a couple examples I googled
> seemed to just ignore it. This example also used 100 for R1, but the
> only way to get 5v is to use 300 as R2. Any reason not to do this?
If the values work.. Check it into a resistor load before firing up a
real circuit. Watch getting into silly values like 1 ohm and 3 ohms, or
1M and 3M. That error term will get really interesting in the latter
Iadj is the small current that goes into the adjust pin. It's ideally
zero. A typical value is 50uA with a rated maximum of 100uA. To ignore
Iadj, the current through the voltage output voltage divider needs to be
substantially more than Iadj. The 100 ohms that was suggested gives a
current of 1.25V/100 = 12.5mA, substantially more than 50uA or 100uA.
I don't like calculating voltage dividers backwards, so I go back to Ohm's
Law in using LM317s.
The output of an LM317 is 1.25V above the adjust pin. That puts 1.25V
across the top resistor. If we use the 100 ohms suggested, the current
through the top resistor is 1.25V/100ohms or 12.5mA. We assume the Iadj is
neglegible or about zero, so, that 12.5mA also goes through the bottom
resistor (Kirchoff's Current Law). Further, if we want 5V on the output
(top of the top resistor), and we have 1.25V across the top resistor
(negative side of the voltage across the resistor at the ADJ pin), we have
5V-1.25V at the top of the bottom resistor. Since the bottom of the bottom
resistor is at ground (0V), we have 3.75V across the bottom resistor.
Therefore, since R=V/I, the bottom resistor must be 3.75V/12.5mA = 300
It takes longer to describe than it takes to do (at least on an RPN
And, I agree, the LM7805 (or LM340-T5) is a simpler solution.
(The ideal design has zero parts)
If LM317T ends with LM317T5 then you can ground it and it
will output 5 volts
James Humes wrote:
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