I'm a beginner with designing circuits and I'm attempting this because I'm trying to design my first PCB.
The basic idea is that I have a 12V battery as a power source and I want an LED to light up when the voltage drops below 3.2V (as an example). I got the idea from a retired SparkFun product called the "Uh-Oh" Battery Level Indicator.
However, I'm struggling to understand two things:
- How the TL431 works as an adjustable shunt
- How the input voltage at which it will allow the LED to light up is calculated
How It Works
Based on this answer from another question posted on EESE, I am understanding that at a voltage above the internal reference voltage of the TL431, it will conduct current due to low impedance but at a voltage below it's internal reference voltage, it will not conduct current due to high impedance. If that thinking is correct, is the following diagram true?
According to the SparkFun schematic I already linked earlier, the output voltage of the TL431 is calculated as:
V_out = V_in * R_2/(R_1+R_2) and since the schematic uses a potentiometer, I am assuming that the two resistors are replaced by a single potentiometer, which is why R_1+R_2 is replaced by the maximum resistive value of the potentiometer, which is 10kOhms. This just leaves R_2 to be calculated.
However, another video I was watching on the TL431, shows a different calculation which can be seen at 4:38, the equation shown is different. There, it is V_out = V_in * (1 + (R_1/R_2)).
I am assuming that there is something fundamental I am misunderstanding here about how to calculate the resistive values for which I want the LED to light up at an input voltage of 3.2V. So, what am I missing? Why are these two equations different?
Side Note: I don't know if this is required, but I am designing the circuit to handle a maximum current of 1A.