Counting type ADC

ADC tutorial (counting type)

ADC Math

0-5V Analog I/P (This is the maximum voltage we can give to PIC)

0-1023 Digital Count (The equailent digital count 2ⁿ n stands for number of bits example 2⁸ = 2x2x2x2x2x2x2x2x2x2 = 1023)

Resolution = (5-0)/(1023-0) = 0.00489 V/Count  (or)  5/1020(or) 2^{n -1}

To measure the 5 volt using counting type ADC

I/P voltage = Va = 0.00489 (V/Count)  * Digital Count (we know it takes 1023 count to convert 5 volt to equalent binary value from 0000000000 to 1111111111)

Consider the below example

                       

In the above circuit we want to measure 20 volt but our PIC will handle maximum voltage of 5Volt. So we want to do some voltage divider network

            Consider the above equation and find the resistance value here the roll of zener is it will ensure only the voltage is around 5.1 volt.

All the things are same except input voltage its 20V instead of 5V

so we want to do some mathematical equation in Va (Input voltage)

So Va (I/P voltage) = 4(to multiply to convert become 20)*Va = 4* Digital Count * 0.00489(resolution) = 0.01956 * Digital Count

10 bit ADC Binary and equailent analog count

0000000000  (binary minimum count)      0.00489 V/Count 

0000000001  (binary minimum count)      0.00489 V/Count

0000000010  (binary minimum count)      0.00978 V/Count (0.00489+0.00489)

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1111111111 (binary minimum count)       1023

Make the thing clear 0.00489 (V/Count) x 1023 = 5.0024 (app 5V)

To avoid floating point, use I/P voltage = 196*Digital Count. This number is a long integer.
Example, suppose
Vin = 13.6V. Then,
Va = 0.25*Vin = 3.4V
=> Digital Count = 3.4/0.00489 = 695
=> Calculated I/P Voltage = 196*695 = 136220 = 13.6V  (First 3 digits of 6 digit product)

Sources of error
The above math looks pretty easy but when you implement it, you will not errors in output measurements  because the above calculations are based on following ideal conditions:

• Vcc supply voltage to PIC12F683 is exactly 5V,
• R1 and R2 are exact 1.3K and 3.9K respectively

So, let's revise the math above with real figures. I measured the supply voltage to PIC and it is 5.02V. R1 and R2 are measured to be 1267 and 3890 Ohms. So this gives,

 0 - 5.02 V Analog I/P ---> 0-1023 Digital Count

=> Resolution = (5.02-0)/(1023-0) = 0.004907 V/Count

Va = 1267*Vin/(1267+3890) = 0.2457*Vin

=> I/P voltage = 4.07*Va = 4.07* Digital Count * 0.004907 

                       = 0.01997 * Digital Count 

                       = 0.02*Digital Count (Approx.)

To avoid floating point, use I/P voltage = 2*Digital Count. No need for Long integer.

Example, suppose 

Vin = 4.6V. Then, 

Va = 0.2457*Vin = 1.13V

=> Digital Count = 1.13/0.004907 = 230

=> Calculated I/P Voltage = 2*230 = 0460 = 04.6V  (First 3 digits of 4 digit product)

Supose u want to make a volt meter it's maximum range will be 5 Volt(as u say)
Ur ADC converter has 10 Bit resulotion it mean in it has maximum value in decimal 1023.
It means as u apply maximum voltage(as u say 5 volt) to ADC it will give u 1023 in decimal.
So the formula is

Voltage = Applied voltage * maximum range / ADC resulotion

Voltage > measured voltage (we don't know it's value)

Applied Voltage > This is our applied voltage which we apply on ADC leg it rage will be 0 ~ 1023.

Maximum range > for which range u want to scale it( as u say 5 volt)

ADC resulotion > 10 Bit it means it maximum value in binary = (1111111111) or in Decimal 1023.

Now we start,
u want to scale it for 5 Volt( it mean it's maximum reading will 5 volt)
The formula will be

Voltage = Applied voltage * 5 / 1023

for example u apply 5 volt to ADC.
then ADC will convert this value into binary so it will be (1111111111) or if we see it in decimal then it will be 1023
It means

Applied Voltage = 1023

So we put this value in our formula

Voltage = 1023 * 5 / 1023

So the answer will be

Voltage = 5

Now an other example
Supose u apply 2.5 volt to ADC
So it will conver it into binary (0111111111) or it will eqaul to 511 in decimal.

Applied Voltage = 511

So we put this value in our formula

Voltage = 511 * 5 / 1023

So the answer will be

Voltage = 2.497