Maximum Source Resistance for Optimal ADC Performance

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Discussion Overview

The discussion revolves around calculating the maximum source resistance (Rn) for an 18-bit ADC to ensure optimal performance without adversely affecting its resolution due to thermal noise. Participants explore the relationship between thermal noise, resistance, and temperature, while addressing specific calculations related to the ADC's specifications.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant presents the formula Rn = (Vn / 4kTRB) but expresses uncertainty about the definition of Vn.
  • Another participant corrects the formula for thermal noise voltage to Vn = sqrt(4kTBR) and rearranges it to R = (Vn^2 / 4kTB).
  • Participants calculate the input voltage resolution based on the ADC's bit depth, determining it to be 262144 steps for 18 bits.
  • One participant computes the voltage noise (Vn) as approximately 38.15 µV based on the voltage range and steps, seeking confirmation of this value.
  • Another participant confirms the calculated Vn and proceeds to calculate the equivalent resistance (Rn) using the thermal noise formula, arriving at a value of approximately 88.4 kOhms.
  • Participants express uncertainty about the correctness of the final resistance value and discuss its reasonableness.

Areas of Agreement / Disagreement

Participants generally agree on the calculations and formulas used, but there remains uncertainty about the correctness of the final value for Rn. No consensus is reached on whether the calculated resistance is definitively correct.

Contextual Notes

Participants rely on specific assumptions regarding temperature and the definitions of voltage noise and resistance. The discussion does not resolve whether the calculations are free from errors or if additional factors should be considered.

MrBondx
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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



[/B]
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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Oh, and remember that T is absolute temperature...
 
: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?
 
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I hope it is correct, will send it for marking. Thanks for your help, much appreciated.
 

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