Maximum Source Resistance for Optimal ADC Performance

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The discussion focuses on calculating the maximum source resistance (Rn) for an 18-bit ADC with a 10 V input range and a bandwidth of 1 MHz, ensuring thermal noise does not affect resolution. The participants clarify the equation for thermal noise voltage, leading to the correct formula: Rn = (Vn^2 / 4kTB). After determining the voltage resolution of the ADC as 38.15 µV, they convert temperature to Kelvin and apply the formula, resulting in Rn being approximately 88.4 kOhms. The final value is deemed reasonable, and the participants express confidence in submitting their findings.
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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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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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