How do I use this voltage regulator with a digital pinout?

Question

I'm trying to learn about electronics, so please bear with me and keep in mind that my idea may be completely nonsensical...

I have this voltage regulator: T317CM

I want to regulate the voltage of a circuit using the digital pinout of a computer board, which is on a separate circuit. If the digital input is high, the voltage of the circuit should be reduced, and if the digital input is low the voltage should be unaffected. Or vice-versa, it doesn't matter. I want a setup something like this:

I can't make any sense of the datasheet, but with a few packs of cells and a multimeter I can reduce 7V to 2V, and keep at 7V if I short two of the pins. If I just make the diagram above the voltmeter always reads 7V, regardless of the pinout voltage. The pinout circuit only connects to the circuit being regulated with that wire - does the ground wire need to be connected somewhere too?

Here's a table of expected inputs/outputs:

V of Digital Pin  |  V in  |  V at voltmeter
       0V             7V            7V
       2V             7V            5V
       5V             7V            2V

These are just rough values and it can be the other way around if that makes the circuit simpler (eg, V at voltmeter = 7V when digital pin is 5V)

Answer 1

It's probably too late for OP, but I'll give the answer to the actual problem: How can a 317 regulator be switched to a low voltage under control of a digital signal. The answer is actually shown in the linked datasheet:

When the "TTL control" signal is high, the lower resistor in the adjustment divider is effectively shorted, and the output voltage will drop to about 1.2 V (maybe 1.4 V, accounting for the Vce-sat of the transistor).

For a 3.3 V logic signal, the 1k resistor in the base of the transistor may have to be reduced.

If the "reduced" output level needs to be higher than 1.2 V, an extra resistor could be added at the collector of the transistor.

Answer 2

The voltage output of adjustable voltage regulators from the LM317 family is controlled by the current input to the ADJ pin. Typically, the ADJ current is programmed by two resistors (see R1 and R2 in the datasheet) forming a voltage divider.

Please see the schematic for "Digitally Selected Outputs" under "Typical Applications" on page 22 of the LM317 datasheet that can be found here: http://www.ti.com/lit/ds/symlink/lm317-n.pdf

Answer 3

You say that you cannot understand the datasheet.
The DS has very clear an very specific examples of how to use the part as shown below.
If you cannot understand simple circuits such as this one then you need to learn to do so as it is fundamental to learning the art.

The diagram that you supplied does not make full sense. I and others can guess at what youare trying to show but it is not certain. You would be betteroff taking the correct diagram as shown below and modifying it to convey your point.

To answer your incomplete and implied question:

Rather than using R1 and R2 to set the voltage at Vadj,
if you instead applied a voltage of Vmatt to pin Vadj
and if Vij was connected to a voltage at least 3 volts greater than Vmatt,
then
Vout = Vmatt + 1.25V.

LM317

When

• Vin < Vinmax (about 40V often) and

• Vin > ~ 4.5 V and

• Vin >V_out_desired + 1.25

• The minimum load requirement from the datasheet is met (usually done by having R1 is some maximum value)

Then Vout = Vadj + 1.25V.

Many of the comments people are making about currents are relevant BUT the above is the LM317s fundamental defining equation.