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DavidG25
DavidG25
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I agree with Voltage Spike's answer: the comparator and DAC in a SAR ADC is not very similar to a high gain forward amplifier in a traditional feedback system. It makes more sense to me as a binary search algorithm that is directed by the loop. Still, there are some uncanny similarities—for example, dividing the DAC output by 2 would make the "closed loop gain" of the SAR ADC equal to 2 or the inverse of the "feedback factor".

One analogy I can attempt to make would be based on the error signal at the summing node which is minimized in both the traditional and SAR case. In a traditional feedback system with unity feedback the error signal is $$|x_{e}| = x_{in}/(1 + A)$$ In a SAR ADC the residual signal is $$|x_{e,SAR}| \le LSB/2 = V_{supply}/2^{N+1}$$ where \$N\$ is the bit depth of the DAC.

Crudely equating these and solving for \$A\$ gives $$x_{in}/(1 + A) = V_{supply}/2^{N+1}$$ $$A = 2^{N+1} \frac{x_{in}}{V_{supply}} - 1$$

Another attempt could take the approach of linearizing the system by representing the quantization error as an additive noise source. I might work through this later.

I agree with Voltage Spike's answer: the comparator and DAC in a SAR ADC is not very similar to a high gain forward amplifier in a traditional feedback system. It makes more sense as a binary search algorithm that is directed by the loop. Still, there are some uncanny similarities—for example, dividing the DAC output by 2 would make the "closed loop gain" of the SAR ADC equal to 2 or the inverse of the "feedback factor".

One analogy I can attempt to make would be based on the error signal at the summing node which is minimized in both the traditional and SAR case. In a traditional feedback system with unity feedback the error signal is $$|x_{e}| = x_{in}/(1 + A)$$ In a SAR ADC the residual signal is $$|x_{e,SAR}| \le LSB/2 = V_{supply}/2^{N+1}$$ where \$N\$ is the bit depth of the DAC.

Crudely equating these and solving for \$A\$ gives $$x_{in}/(1 + A) = V_{supply}/2^{N+1}$$ $$A = 2^{N+1} \frac{x_{in}}{V_{supply}} - 1$$

Another attempt could take the approach of linearizing the system by representing the quantization error as an additive noise source. I might work through this later.

I agree with Voltage Spike's answer: the comparator and DAC in a SAR ADC is not very similar to a high gain forward amplifier in a traditional feedback system. It makes more sense to me as a binary search algorithm that is directed by the loop. Still, there are some uncanny similarities—for example, dividing the DAC output by 2 would make the "closed loop gain" of the SAR ADC equal to 2 or the inverse of the "feedback factor".

One analogy I can attempt to make would be based on the error signal at the summing node which is minimized in both the traditional and SAR case. In a traditional feedback system with unity feedback the error signal is $$|x_{e}| = x_{in}/(1 + A)$$ In a SAR ADC the residual signal is $$|x_{e,SAR}| \le LSB/2 = V_{supply}/2^{N+1}$$ where \$N\$ is the bit depth of the DAC.

Crudely equating these and solving for \$A\$ gives $$x_{in}/(1 + A) = V_{supply}/2^{N+1}$$ $$A = 2^{N+1} \frac{x_{in}}{V_{supply}} - 1$$

Another attempt could take the approach of linearizing the system by representing the quantization error as an additive noise source. I might work through this later.

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DavidG25
DavidG25
  • 1.4k
  • 9
  • 23

I agree with Voltage Spike's answer: the comparator and DAC in a SAR ADC is not very similar to a high gain forward amplifier in a traditional feedback system. It makes more sense as a binary search algorithm that is directed by the loop. Still, there are some uncanny similarities—for example, dividing the DAC output by 2 would make the "closed loop gain" of the SAR ADC equal to 2 or the inverse of the "feedback factor".

One analogy I can attempt to make would be based on the error signal at the summing node which is minimized in both the traditional and SAR case. In a traditional feedback system with unity feedback the error signal is $$|x_{e}| = x_{in}/(1 + A)$$ In a SAR ADC the residual signal is $$|x_{e,SAR}| \le LSB/2 = V_{supply}/2^{N+1}$$ where \$N\$ is the bit depth of the DAC.

Crudely equating these and solving for \$A\$ gives $$x_{in}/(1 + A) = V_{supply}/2^{N+1}$$ $$A = 2^{N+1} \frac{x_{in}}{V_{supply}} - 1$$

Another attempt could take the approach of linearizing the system by representing the quantization error as an additive noise source. I might work through this later.