Question about voltage divider

[mannok]

I am seeking for help regarding to question about voltage divider.

Someone in the Internet said that if I want to get 1.5V from a 9V source I can connect the 9V source to 2 resistors (500ohms and 100ohms) in series just like this picture.

By KVL, I know that Vout takes same voltage of Z2. However, my problem is why Vout will take 1.5V in this case. In my view, Vout may connected to some loads and finally connected to the GND, we can view it as a resistor too! In this situation, we should know the resistance of Vout and Z2 in order to calculate the total resistance of them as in series and compare with Z1 to find out the voltage consuming ratio.

So my question is why we can know Vout without knowing the resistance beyond Vout?

 

[KrisBlueNZ]

Yes, you're right. A voltage divider will only produce the designed output voltage if there is no load on it. The more it is loaded, the more the voltage will drop. If the load is resistive, you can calculate the effect by putting the load resistance in parallel with Z₂.

So we can't know V_OUT unless the load resistance is infinite.

There are various ways of buffering the output of a voltage divider so the circuit can supply current without the voltage changing (much). If you have a specific application in mind, tell us what it is and we can advise how best to do that.

 

[mannok]

Thank you KrisBlueNZ!!
Actually, I am playing with Raspberry-PI. I have a module which will output 5V when it sense obstacle. The problem is R-PI input pin only accept 3.3V at most, so I need to convert 5V to 3.3V,

 

[KrisBlueNZ]

OK. What is the input impedance of the Raspberry Pi's digital input? Is it just a direct connection to the CPU or is there a pulldown resistor, for example? Anyway, you understand how to do the calculations.