Maximum Source Resistance for Optimal ADC Performance

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Homework Statement



A simplified model of ADC noise refers the noise to a noisy source resistance Rn while assuming the rest of the signal path to be noiseless. Figure 3 represents a particular 18-bit ADC that has a 10 V input voltage range. The ADC has a bandwidth of 1 MHz.

Calculate the maximum value of Rn if the resolution of the ADC is not to be adversely affected by thermal noise. Assume the ADC operates at 25 degrees C.

[N.b. The voltage resolution of an ADC is equal to its overall voltage measurement range divided by the number of discrete values possible on its output.]
ADC.png
FIG. 3

Homework Equations

The Attempt at a Solution



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The issue of resolution has totally thrown me off but here is what I'm thinking.

Rn = (Vn \ 4kTRB), where k is the Boltzmann constant.

Vn = ? I'm not sure whether Vn = resolution or full scale voltage.

It's all I got for now, please help!
 
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MrBondx said:
Rn = (Vn \ 4kTRB)

That equation does not look correct. What is the equation for the thermal noise voltage in terms of of k and T and R and B?
 
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berkeman said:
That equation does not look correct. What is the equation for the thermal noise voltage in terms of of k and T and R and B?

Thanks, Yea my mistake

Vn = sqrt(4kTBR)

Re-arranging

R = (Vn^2 / 4kTB)
 
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MrBondx said:
Thanks, Yea my mistake

Vn = sqrt(4kTBR)

Re-arranging

R = (Vn^2 / 4kTB)

Ah, much better. I was getting vertigo trying to decipher what you wrote. :smile:

So now you're pretty much done. Take the input voltage range and divide by the total resolution (18 bits is how many steps?). Then apply the formula...
 
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:smile:
Input voltage resolution

18bits = 2^18 steps = 262144

Voltage range / steps = (10 / 262144)

Vn = 38.14697 x 10^-6 V

Is that correct?
 
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MrBondx said:
:smile:
Input voltage resolution

18bits = 2^18 steps = 262144

Voltage range / steps = (10 / 262144)

Vn = 38.14697 x 10^-6 V

Is that correct?

Yes, I get 38.15uV as well. So what is the equivalent resistance to make that noise voltage at (absolute) room temperature?
 
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Converting Celsius to Kelvin

25degrees = 298.15 K

Plugging numbers into equation

Rn = (38.15 x 10^-6)^2 / (4 x (1.38 x 10^-23) x 298.15 x 1000000)

= 88433.17

Rn = 88.4 kOhms
 
MrBondx said:
Converting Celsius to Kelvin

25degrees = 298.15 K

Plugging numbers into equation

Rn = (38.15 x 10^-6)^2 / (4 x (1.38 x 10^-23) x 298.15 x 1000000)

= 88433.17

Rn = 88.4 kOhms

Looks like a reasonable value. Do you know if it's correct?
 
I hope it is correct, will send it for marking. Thanks for your help, much appreciated.