Maximum difference of voltage in converter

[axcelenator]

I have a successive approximation ADC converter (left) and is connected to a DAC converter.
The Vref of the ADC is 10V and the Vref of the DAC is brought fom an internal leg of the SAR.
This internal leg is the output of the SAR's internal DAC.

The question is: what can be the maximum voltage difference between Vin and Vout?

(I know that the maximum voltage difference between Vin and output of the SAR's internal DAC is LSB but here I have a kind of chain and it' a problem...)

Thanks for looking!

 

[CWatters]

Work through it stage by stage.

Full scale input = vref = 10V
The ADC has 12 bits so each digital value could represent an input voltage +/- x
What output voltage would the DAC produce when fed by that digital value.
Vout could differ from Vin by x

 

[axcelenator]

what do you mean x?

 

[CWatters]

Deleted

 

[rude man]

You don't want to work with digitization (lsb's, thresholds, etc.).

Draw the diagram: an a/d converter with its reference of 10V. Now, the critical question is, what is the voltage output from the SAR "leg", given an input voltage v? That voltage is going to be the reference voltage for the d/a converter. You're going to have to think deeply about this I believe.

Let the a/d output = x (a number between zero and 2^N - 1 where N is the size of the a/d converter output register. Let v_out be the output of the d/a converter. What is its output in terms of its digital input x (assume N same for both the a/d and d/a converters) and the SAR output voltage? After some algebra, find v - v_out in terms of v alone & do the usual calculus to find v for max. difference, and the difference itself.

This problem has nothing to do with resolution other than the assumption of N being the same for both converters. The answer is a nice healthy voltage!

 

[CWatters]

Ah heck I totally missunderstood the circuit. My tired old eyes saw a complete A-D converter followed by a D-A converter. It's not that at all. The bit on the left is just the SAR part of an A-D converter.

 

[axcelenator]

hello I got the Vout but look at it: I have an answer with a part which is taken only after calculation as a number without partions so there is no way to do some algebra on: vin-vout.

But: The question was how to find what will be the maximum difference and this is the problem

 

[axcelenator]

This is what i got

 

[rude man]

This is what i got

What you did was fine.
Now finish up by computing Vin - Vout. Your Vout will be in terms of Vin per you last equation which of course has to be cleaned up a bit.

 

[axcelenator]

it can't be cleaned up because the part in brackets [ ] is integer. i can't take out of that nothing
for example:
vin=6.3 and n=12 →vout=3.9675volt

there is know way to clean it up

 

[rude man]

it can't be cleaned up because the part in brackets [ ] is integer. i can't take out of that nothing
for example:
vin=6.3 and n=12 →vout=3.9675volt

there is know way to clean it up

OK, going to take you one step closer:

Vout = (10/2^2n)[(Vin/10)*2^n]^2 ... your last equation

= (10/2^2n)[(Vin^2)*(2^2n)/100]
= (Vin^2)/10.
So now take Vin - Vin^2/10 and maximize w/r/t Vin.

 

[axcelenator]

sorry but can't understand how do you take the ^2 and make it inside the integer in brackets.
and what is w/r/t?

 

[rude man]

sorry but can't understand how do you take the ^2 and make it inside the integer in brackets.
and what is w/r/t?

w/r/t means 'with respect to', often used in math.

You need to work the other question out by yourself. It's just algebra. How about {2ⁿ}² = 2²ⁿ, does that help? I don't know where your're having trouble, really.

 

[axcelenator]

hey, this is the trouble : look at this part. [(Vin/10)*2^n]^2
lets take Vin=6.3Volt --> [2580.48]^2=2580^2=6656400/
and now: this is what you suggest: [(Vin^2)*(2^2n)/100] --> [6658877.03]=6658877
what we get is that: 6656400 not equals 6658877 )-:
thats what I am trying to explain that i can't hand out algebra with integers...

 

[rude man]

hey, this is the trouble : look at this part. [(Vin/10)*2^n]^2
ers...

When you multiply "this part" with 10/(2^2n) you wind up with (Vin^2)/10. Why are you messing with those large numbers?