Successive Approximation ADC Question

[cjm181]

Homework Statement

.[/B]

The full range input of a12-bit, successive-approximation type ADC is 1 volt. Determine:

1. 
   
   1. a) the maximum input change required to give a one bit change in output of the ADC
   2. b) The number of approximations made to complete the conversion of an input signal of 0.8125 V and the final binary result of the conversion process.
      

Homework Equations

The Attempt at a Solution

a) Vres = Vref / ((2^n)-1) = 1 / ((2^12)-1) = 2.442x10^-4

b) Going to type this backwards!

0 to 1v input range, output range 0 to 4095 hex for a 12 bit system. So, to find the hex value for an input of 0.8125v:

4095 * 0.8125 = 3327.188 or 3327. So convert this to binary gives an output of 11 00 11 11 11 11.

So the number of approximations made! the bit i am struggling with!

As 3327 was not reached until all 12 bits had run thru the system, does this mean 12 approximations were made?

Say the hex input was 2048 (10 00 00 00 00 00), would the number of approximations be 1?

Thanks
Craig

 

[BvU]

bit strange they ask for the maximum change. what if v went from 0 to 0.5 volt; would that count as a one bit change in output ?

 

[cnh1995]

what if v went from 0 to 0.5 volt; would that count as a one bit change in output ?

It means the digital output is 011111111111 (or 2047 decimal). That's a lot of change in output!

bit strange they ask for the maximum change

I think it should be minimum change..

 

[cjm181]

sorry, i went thru part a, it is the minimum change. had that confirmed from uni. its poorly written part a.

Any thoughts on b?

kr
craig

 

[rude man]

Homework Statement

.[/B]

The full range input of a12-bit, successive-approximation type ADC is 1 volt. Determine:

1. 1. a) the maximum input change required to give a one bit change in output of the ADC
   2. b) The number of approximations made to complete the conversion of an input signal of 0.8125 V and the final binary result of the conversion process.

0 to 1v input range, output range 0 to 4095 hex
for a 12 bit system.

not hex!

So, to find the hex value for an input of 0.8125v:

4095 * 0.8125 = 3327.188 or 3327. So convert this to binary gives an output of 11 00 11 11 11 11.

about right (+/-1, I don't know)

So the number of approximations made! the bit i am struggling with!

Me too! The converter makes 12 comparisons no matter what the voltage being converted.

 

[cjm181]

Thats what i thought! unless once it hits the figure does it consider all other numbers to the left 0?

 

[rude man]

Thats what i thought! unless once it hits the figure does it consider all other numbers to the left 0?

Even if all the LSB's below 0.8125V are zeros the converter still has to go thru the motion to determine that fact. It always has to do 12 comparisons.

 

[Tom.G]

Thats what i thought! unless once it hits the figure does it consider all other numbers to the left 0?

It starts at the left (most significant) bit, and works downward in value (to least significant).

Begin with high-order bit.
When considering a bit:
SET the bit.
IF the approximation is less than the input THEN NEXT bit
ELSE bit is reset.
NEXT bit

 

[cjm181]

I don't know what i am writing about hex for? brain fade!

replace my use of the word hex with decimals!

So is 11 00 11 11 11 11 correct for an input of 0.815V?

As said above, i am saying 12bit successive approximation ADC uses 12 approximations

 

[Tom.G]

So is 11 00 11 11 11 11 correct for an input of 0.815V?

My calculator agrees with that answer.

 

[cjm181]

nice! thanks everyone!