Electronic – ADC Calibration – Vdd calculation

In documents and lots of online sources there is a formula as follows,

$$ Vdd = 3.3V * \frac{VREFINT_{CAL}}{VREFINT_{DATA}}$$

But I cannot get why CAL value is in the numerator and RAW data is in the denominator. I have this logic in my mind,

In the factory, producers used a preknown/fixed 3.3 V as Vdd and then they get this reference CAL value from ADC. But in reality, my Vdd may be different because of non-idealities. So I read Vref_int channel via my ADC and it gives me RAW. I get the CAL value by reading the device memory address provided by manufacturer. ADC reading and voltage level has a direct relationship. So, I conclude that;

3.3 V ------> CAL(from device memory)

x V ------> RAW(read from Vref_int)

Then, x * CAL = 3.3 * RAW

So, x = 3.3 * RAW/CAL should be my formula to get my current Vdd value so that then I know that a maximum ADC reading (4095 for 12 bits reading for ex.) corresponds to x volts.

However, in the above formula it states that x = 3.3 * CAL/RAW.

Please help me with that, thanks.

My board is STM32F4-DISC.

——————– edit ————————————

With all my thanks to the ones who answered, I now understood(I hope) what is going on and want to explain here in detail.

First, what is what :

VREFINT_{CAL} or CAL : The ADC reading of the manufacturer from Vref_int when an exact 3.3V Vdda is used (unitless, 12 bits integer)

VREFINT_{DATA} or RAW : The ADC reading of the user from Vref_int when the Vdda is unknown (unitless, 12 bits integer)

V_{DDA} : Kind of the voltage base for our MCU, when you give a digital high output then it should be exactly equal to Vdda (unit = volts)

Now,

1- In the factory, the manifacturer gets the ADC conversion data from
the Vref_int channel and stores it into a memory location as a
12-bit integer. $$ \frac{{VREFINT_{voltage}}}{3.3(V_{DDA})}= \frac{{VREFINT_{CAL}(stored\,value\,,CAL)}}{4095} \quad (1)$$
The logic is simple here, 4095 for 3.3V and CAL for Vref_int.

2- When you want to calibrate your voltage level, since you don't know
your exact voltage base level, you read the Vref_int yourself and
get another ADC conversion value(RAW), let's call it $$VREFINT_{DATA}$$
Then $$ \frac{VREFINT_{voltage}}{V_{DDA}(unknown)}= \frac{{VREFINT_{DATA}(measurement\,of\,user\,,RAW)}}{4095} $$

Since $$VREFINT_{voltage}$$ is fixed, replace it with its equivalent from eqn. (1).

Then you can see that the equation yields to $$V_{DDA}(unknown) = \frac{3.3*CAL}{RAW}$$

Finally, you can use this Vdda value as your voltage reference(base) on your calculations.

This edit may include some mistakes, please let me know.