# Why does this not short circuit?

I recently built this circuit, but didn't understand why these resistors connect voltage-out to ground. My logic would say the path would either go like...

or

• You're assuming that the output terminals are a short-circuit, which isn't true, and that 'electricity follows the path of least resistance' only, which also isn't true, and that the wire to the wiper of the 5k pot doesn't carry any current, which also isn't true. Commented May 14, 2015 at 1:30

I REALLY, REALLY hate that "electricity follows the path of least resistance" expression, since so many beginners seem to see an "only" in there somewhere. In fact, electricity follows all possible paths, with the current in each path determined by that path's resistance. The highest current will flow in the path having least resistance, but current will flow in any other available path as well.

Current follows both paths you show.

R1 and R2 form a voltage divider that determines the output voltage of the LM317. The LM317 attempts to maintain about 1.25 volts between its output and adjust terminals (pin 2 and pin 1).

• I blame my science teachers in elementary school. Did not know this until now, everything makes sense now. Commented May 14, 2015 at 0:55
• It follows both paths if there is a load. Otherwise I don't see how this answer is any different from the one I already provided. Commented May 14, 2015 at 1:31

Assuming you've added a load between the Vout terminals, it follows both paths.

If there is no load, as pictured, it only flows through the resistors. The resistors, R1 and R2, create a voltage divider which serves as an input to the regulator. The regulator uses that input to, well, regulate the output voltage.

It doesn't short circuit because there is no short between the supply and ground. The Vout terminals are not connected together and no current will flow until a load is connected there, that load can be viewed as another resistor.

First all, your formula is wrong, it should be as stated below. Usually the resistor from the output to terminal 1 is labeled R1, and the variable one is R2, which would match your equation. I had to turn it around to match the figure.

Let's take a practical example.

The LM317 provides an internal reference voltage of 1.25 V between the output and adjustments terminals. This is used to set a constant current flow across an external resistor divider (see Figure 6), giving an output voltage V$_{0}$ of:

$$V_0 = V_{REF} \times (1 + R1/R2) + I_{ADJ} \times R1$$

In your case., let's assume the pot is sent right in the middle. According to the datasheet, I$_{ADJ}$ is approximately 100 µA.

so we have:

$$V_{0} = V_{REF} * (1 + 2500 / 240) + 0.0001 * 2500 = 1.25v \times (1 + 10.417) + 0.25 = 14.52v$$

This is the output voltage coming out of the LM317. Let's say the load is 100 Ω. By Ohms Law, we have:

$$I = V / R = 14.52 / 100 = 0.1452 A$$

or 145.2 mA going through the load.

The voltage across the voltage divider is 14.52 and the current is:

$$I = V / R = 14.52 / (240 + 2500) = 14.52 / 2740 = 0.00530 A$$

or 5.30 mA. The voltage across R1 is:

$$V = I \times R = 0.00530 \times 2500 = 13.24v$$

This is the voltage at terminal 1, which is the feedback voltage that controls the output voltage.

The total current coming out of the output of the LM317 is:

$$145.2 mA + 5.3 mA = 150.5 mA$$