how to choose ADC REFERENCE voltage

[satishreddy44]

Hi
most of them choose adc refrence voltage as 2.5v instead some other value.even i have seen people choosing 5v vref.
But many of them going for 2.5v please explain me.

 

[hemnath]

If your ADC input signal is below +2.5V. you can go for the Vref+ to +2.5V. Because to increase the resolution.

i think you are aware of 10Bit adc value. Mostly Internal ADC is about 10 bits. so 2^10 => 1024 steps. ADC counts from 0 to 1023.

If your input signal is 0V. ADC value is 0. If your input signal is maximum say about 2.5V, ADC count will be 1023.

Smallest change in input signal it can detect is about, 2.5V/1024 => 0.00244140625V. Each step increase at every 2.4mV.

If your reference voltage is about 5V, then +5V/1024 => 0.0048828125V .Each step increase at every 4.8mV.

- - - Updated - - -

If your ADC input signal is below +2.5V. you can go for the Vref+ to +2.5V. Because to increase the resolution.

i think you are aware of 10Bit adc value. Mostly Internal ADC is about 10 bits. so 2^10 => 1024 steps. ADC counts from 0 to 1023.

If your input signal is 0V. ADC value is 0. If your input signal is maximum say about 2.5V, ADC count will be 1023.

Smallest change in input signal it can detect is about, 2.5V/1024 => 0.00244140625V. Each step increase at every 2.4mV.

If your reference voltage is about 5V, then +5V/1024 => 0.0048828125V .Each step increase at every 4.8mV.

 

[satishreddy44]

If your ADC input signal is below +2.5V. you can go for the Vref+ to +2.5V. Because to increase the resolution.

i think you are aware of 10Bit adc value. Mostly Internal ADC is about 10 bits. so 2^10 => 1024 steps. ADC counts from 0 to 1023.

If your input signal is 0V. ADC value is 0. If your input signal is maximum say about 2.5V, ADC count will be 1023.

Smallest change in input signal it can detect is about, 2.5V/1024 => 0.00244140625V. Each step increase at every 2.4mV.

If your reference voltage is about 5V, then +5V/1024 => 0.0048828125V .Each step increase at every 4.8mV.

- - - Updated - - -

If your ADC input signal is below +2.5V. you can go for the Vref+ to +2.5V. Because to increase the resolution.

i think you are aware of 10Bit adc value. Mostly Internal ADC is about 10 bits. so 2^10 => 1024 steps. ADC counts from 0 to 1023.

If your input signal is 0V. ADC value is 0. If your input signal is maximum say about 2.5V, ADC count will be 1023.

Smallest change in input signal it can detect is about, 2.5V/1024 => 0.00244140625V. Each step increase at every 2.4mV.

If your reference voltage is about 5V, then +5V/1024 => 0.0048828125V .Each step increase at every 4.8mV.

thank u.as per ur explanation we should choose minimum vref rite?

 

[jayanth.devarayanadurga]

If the voltage range you are inputting to adc pin is 1v then make your Vref+ = 1v and vref- = 0v, then you will get 0 to 1023 adc value for 0v to 1v. If you are using 12 bit adc then you will get 0 - 4095 adc value for 0v to 1v.

 

[hemnath]

thank u.as per ur explanation we should choose minimum vref rite?

It depends on your input signal. Vref+ should be high than your input signal on ADC pin