ADC Calibration - Vdd calculation

Question

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.

Answer 1

ADC reading and voltage level has a direct relationship.

The ADC reading on the V_REFINT channel and V_DDA have inverse relationship.

An ADC reading is the ratio of the measured voltage to V_DDA, scaled up to 4095 (the maximum value that can be expressed on 12 bits). E.g. if you applied V_DDA/3 to an ADC input, the result would be 4095/3=1365.

So, the conversion result, ADCx = 4095*V_ADCINx/V_DDA

You cant't measure V_DDA directly, because if you apply a voltage equal to V_DDA to an ADC channel, the conversion result is always 4095.

V_REFINT is a fixed voltage generated inside the MCU, around 1.2 volts. When you convert the V_REFINT channel of the ADC, you get the ratio of this voltage to V_DDA, i.e. 4095*V_REFINT/V_DDA. Note that V_DDA is now in the denominator. If the supply voltage goes down, the conversion result goes up, and vice versa.

Because V_REFINT is not an exact voltage source, they measure it in the manufacturing process, using a good stabilized 3.3V supply, and write the conversion result (V_REFINTCAL) in the system memory area.

Now if you'd read the V_REFINT channel, and got the exact same value that's stored in the system memory, it would mean that V_DDA=3.3V, like in the factory calibration process. This is also what the formula in the question gives you when V_REFINTCAL = V_REFINTDATA.

Answer 2

VREFINTcal is the only known value. The value of VREFINT, when measured with a known good 3.3V Vdd.
If you divide data with cal you know how many bits Vdd is off, assuming VREFINTcal is still valid.

If VREFINTcal is 1000, and data is 1020, this means you've measured VREFINT to be 20 bits higher than it's supposed to be. Indicating a lower Vdd value.
Hence 1000/1020 = 0.98. A 2% difference. Thus 3.3*0.98 = 3.24 V.

You are not measuring the change in VREFINT, but the change of the ADC reference.

link to drawing

Answer 3

The relationship between ADC input voltage and output value is:

$$IN = \frac{V_{REF} * OUT}{ 4095 }$$

Therefore:

$$VREFINT = \frac{3.3*CAL}{4095} = \frac{V_{DD}*RAW}{4095}$$

Therefore:

$$V_{DD} = \frac{3.3*CAL}{RAW}$$