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I read in the LM7805 datasheet that I can change the output voltage to not only 5 V, but to another value by changing the circuit resistance. How do I do that?

From what I understand, to change the output voltage I saw this picture:

However, I can't find the values of R1 and R2 in the datasheet. How do I calculate them?

I want to use the LM7805. However, I read in the datasheet that I can change the output voltage not only 5, but by changing the circuit resistance.

Principally, the LM7805 is a fixed voltage regulator.

The 78xx series does have adjustable versions, but the LM7805 is not one of these: the 05 stands for a fixed output voltage of 5V.

What the circuit does is build a virtual ground. That does work, but it's very undesirable from a regulation quality point of view. (The LM7805 is not symmetrical in what it can quickly adjust if the load changes. This together with a load-dependent virtual ground is a bad situation.)

If you need a different output voltage than 5V, you definitely should use a different regulator. Get a better one – the 7805 is ancient (the datasheet you linked to is literally from 1976), and there's way better-regulating chips out there.

However, I can't find the values of R1 and R2 in the datasheet. How do I calculate them?

You just pick an R2 and then calculate a R1 that fulfills the formula like you'd like it to. Again, this is not a good idea, even if the datasheet suggests it is. It's both not a great regulation that you'll get, and it wastes a lot of power if you want temperature-dependency to not take too much of an effect due to the $I_Q$ in the formula.

In the mid-1970s, there weren't many better ways of regulating an output voltage. 43 years later, there are.

However, I can't find the values of R1 and R2 in the datasheet. How do
I calculate them?

When faced with two unknowns in an equation, picking a value and then using that value to find the other unknown is the best approach.

So lets look at that equation:

$ V_O = V_xx + (fracV_xxR1+I_Q)*R2 $

$I_Q = 500mA$

$ V_xx =$ the fixed value of the regulator

So if I had a 7805 and I wanted 7V -- I would pick r1 to be something that I have, like a 1K resistor :

$ 7V = 5V + (frac5V1000Ω+0.5)*R2 $

Solving for R2 yields 3.9604Ω

R2 is very 'sensitive to large changes in R1, so it might be best to keep R1 relatively low like in the 500Ω to 5kΩ range (that's probably why they used a potentiometer). This also means that R2 will burn up a lot of heat, and the tempco of R2 will affect the regulation so its probably best to go with a real adjustable regulator.

As Marcus already pointed out, it is a terrible idea to depend on the quiescent current. It will vary with the load and give you a bad load regulation.

Just to give a better alternative, if you don't have an LM317 (or better) around but have a zener that just happens to be what you want to add to the output voltage, you can connect it to the common pin, for better results that the ones you get with the resistor (but worse than a proper solution with user defined feedback).

The following plots show the behavior for the solution with the resistor feedback and for the zener. The "sine wave" component is due to the variation of the input voltage from 16 to 20 V. The two curves in each plot show the load regulation (RL = 10k and 10 Ohm).