#### Steve1992

Joined Apr 7, 2006
100
In a DAC practical circuit I read the 'step' change in Vout can be given as:

Vstep = Vref/2^n

where n is the number of bits in the digital word.

Eg.
Vref = 5V
n = 8-bits

5V/256 = 19.5mV

41(decimal) x 19.5mV = 800mV.

Is it OK to use this same procedure for ADC successive approximation circuit calculations?

Eg.
Vsig = 3.6V

5V/256 = 19.5mV

Vsig/19.5mV = 183(decimal)

#### MrChips

Joined Oct 2, 2009
24,179
Yes we can. The problem is: How many steps are there?
Is it 2^n steps or 2^n -1 steps?

#### Steve1992

Joined Apr 7, 2006
100
Is that a rhetorical question?

From what Ive read its Vref/2^n

thanks.

#### Steve1992

Joined Apr 7, 2006
100
Sorry, I had got confused because I didn't know this process is known as quantization. #### Steve1992

Joined Apr 7, 2006
100
Sorry, I got a little lost because I didn't know this process is known as quantization.

#### MrChips

Joined Oct 2, 2009
24,179
Let us take a simple example.
Suppose we have a 2-bit DAC with full-scale range of 0-3V.
Is the step size 3/4 V or 3/3 V?