shanmei
Advanced Member level 1
1:
From this link: https://www.ti.com/lit/wp/snoa829/snoa829.pdf?ts=1593181609815&ref_url=https%3A%2F%2Fwww.google.com%2F
For a 12-bit accuracy, the input signal frequency f should be less than 0.0156fu. fu is the -3dB cut off frequency of the close loop.
f=0.0156fu . if \[ \tau \] is the time constant, so \[ \tau \] =1/fu.
1/f=1/(0.0156fu), we have 1/f=t1, 1/(0.0156fu)=64\[ \tau \], so t1=64\[ \tau \] .
2. From the post link : https://www.edaboard.com/threads/bandwidth-requirement-of-op-amp-for-adc.382864/
For the N-bit resolution accuracy settling, the unit gain bandwidth should be: Wc= 2(N+1)*fs*ln2/beta
the closed loop gain -3dB bandwidth is Wc, and Wc*beta=2(N+1)*fs*ln2, assume beta=1, so Wc=2(N+1)*fs*ln2,
fc=(N+1)*fs*ln2/pi.
For 12bit ADC, 0.5LSB accuracy, t2/\[ \tau \]=13ln(2)/(pi*beta), assume beta=1, t2=13ln(2)/pi\[ \tau \]=2.8\[ \tau \]
Question:
Why t1 is not equal to t2?
Thanks.
From this link: https://www.ti.com/lit/wp/snoa829/snoa829.pdf?ts=1593181609815&ref_url=https%3A%2F%2Fwww.google.com%2F
For a 12-bit accuracy, the input signal frequency f should be less than 0.0156fu. fu is the -3dB cut off frequency of the close loop.
f=0.0156fu . if \[ \tau \] is the time constant, so \[ \tau \] =1/fu.
1/f=1/(0.0156fu), we have 1/f=t1, 1/(0.0156fu)=64\[ \tau \], so t1=64\[ \tau \] .
2. From the post link : https://www.edaboard.com/threads/bandwidth-requirement-of-op-amp-for-adc.382864/
For the N-bit resolution accuracy settling, the unit gain bandwidth should be: Wc= 2(N+1)*fs*ln2/beta
the closed loop gain -3dB bandwidth is Wc, and Wc*beta=2(N+1)*fs*ln2, assume beta=1, so Wc=2(N+1)*fs*ln2,
fc=(N+1)*fs*ln2/pi.
For 12bit ADC, 0.5LSB accuracy, t2/\[ \tau \]=13ln(2)/(pi*beta), assume beta=1, t2=13ln(2)/pi\[ \tau \]=2.8\[ \tau \]
Question:
Why t1 is not equal to t2?
Thanks.
Last edited:
