Estimating input noise on comparator circuit

[melkord]

This is exercise 1 from the Analog IC Design book by Carusone, Johns, Martin, Ch. 10 Comparator.
I do not understand how we can estimate rms input noise from the plot and where 7.0344 mV comes from.

Appreciate any tips.

 

[FvM]

You are asking for the 1-10^-12 confidence interval of the given normal distribution. Review statistics or signal theory text book how to calaculate it. Hint, the curve plotted above is a gaussian integral.

 

[D.A.(Tony)Stewart]

Assume it is a "Normal Distributed Distribution Function" https://en.m.wikipedia.org/wiki/Normal_distribution

If you were given the log P version of this graph or an assumption of the spectral density of the noise it would make more sense. Perhaps in the book but excluded here. It could be Gaussian only or with shot 1/f noise. The exponential curve becomes a straight line on log scale so 1e-12 can be extrapolated but no it from this linear curve. The level must be this log probability plus the offset of 3 mV.

The Pk/rms is known for sine waves, but pulses like Shot noise depends on the crest factor not mentioned in this text so assumed ignored. Thus the log slope of error probability becomes a constant (x mV/20dB) plus offset.

But here assuming Gaussian Noise the log error probability from 50 to 90% was given to extrapolate to 99.9999999999 % Or 1e-12 error rate

Looking at the answer here 7.03 * Vrms means for Gaussian noise each sine wave is \$\sqrt{2}\$ or 7.03 * 1.414= 9.94 ~ 10 mV per 12 decades of log error probability or 0.833 mV per decade based on Gaussian SNR of the signal then adding offset of 3mV you can extrapolate to any probability. Removing the offset of 50% makes the 90% threshold now 80% of the peak. To me this just reflects the linear gain of the comparator. This ratio the the linear asymptote of of ~ 5.5 mV above the 90% P = 4.28 means a ratio of (5.5-4.2) mV / 3 mV = 43.3% appears to be the short term ratio. What SNR is this and does that represent a Gaussian shape above 3mV?

When I was a HDD Test Engineer measuring soft and hard error rates in the old days (80's) , we had to compute this. Rather than using 50 to 90% we used 90% 100% measuring a shrinking window to measure margin for a BER of 1 bit per track or per sector of bits which was also pattern dependent with offset. Then with different patterns symmetric and asymmetry remove offset error then were able to extrapolate the exponential curve accurately taking multiple tracks to compare with 1 track to estimate the slope of the curve. Bit errors in the medium then are represented by offsets at a log level between the mean spacing of bit errors. So it got more complicated than this curve, but proved to be more accurate than the method demonstrated here. FWIW.

 

[melkord]

Assume it is a "Normal Distributed Distribution Function" https://en.m.wikipedia.org/wiki/Normal_distribution

If you were given the log P version of this graph or an assumption of the spectral density of the noise it would make more sense. Perhaps in the book but excluded here. It could be Gaussian only or with shot 1/f noise. The exponential curve becomes a straight line on log scale so 1e-12 can be extrapolated but no it from this linear curve. The level must be this log probability plus the offset of 3 mV.

The Pk/rms is known for sine waves, but pulses like Shot noise depends on the crest factor not mentioned in this text so assumed ignored. Thus the log slope of error probability becomes a constant (x mV/20dB) plus offset.

But here assuming Gaussian Noise the log error probability from 50 to 90% was given to extrapolate to 99.9999999999 % Or 1e-12 error rate

Looking at the answer here 7.03 * Vrms means for Gaussian noise each sine wave is \$\sqrt{2}\$ or 7.03 * 1.414= 9.94 ~ 10 mV per 12 decades of log error probability or 0.833 mV per decade based on Gaussian SNR of the signal then adding offset of 3mV you can extrapolate to any probability. Removing the offset of 50% makes the 90% threshold now 80% of the peak. To me this just reflects the linear gain of the comparator. This ratio the the linear asymptote of of ~ 5.5 mV above the 90% P = 4.28 means a ratio of (5.5-4.2) mV / 3 mV = 43.3% appears to be the short term ratio. What SNR is this and does that represent a Gaussian shape above 3mV?

When I was a HDD Test Engineer measuring soft and hard error rates in the old days (80's) , we had to compute this. Rather than using 50 to 90% we used 90% 100% measuring a shrinking window to measure margin for a BER of 1 bit per track or per sector of bits which was also pattern dependent with offset. Then with different patterns symmetric and asymmetry remove offset error then were able to extrapolate the exponential curve accurately taking multiple tracks to compare with 1 track to estimate the slope of the curve. Bit errors in the medium then are represented by offsets at a log level between the mean spacing of bit errors. So it got more complicated than this curve, but proved to be more accurate than the method demonstrated here. FWIW.

Hi,
thanks for your reply. I will go back to understand your answer later.
here is my calculation.
I assume 10^-12 is a typo error and error rate should be 10^-2 = 0.01.

From the graph, I can extract:
input offset voltage = 3 mV
1.2816σ= 4.28 mV --> from Normal Distribution

for error rate = 0.01, it means CDF=0.99 --> erf(1.644976357) --> x = 10.7691mV
So for error rate = 0.01 or 99% confidence level, the input must be
x<-4.7691 mV or x>10.7691 mV.

I still do not understand how to get rms input noise = 1 mV as writen in the book.

 

[D.A.(Tony)Stewart]

-

Hi,
thanks for your reply. I will go back to understand your answer later.
here is my calculation.
I assume 10^-12 is a typo error and error rate should be 10^-2 = 0.01.

From the graph, I can extract:
input offset voltage = 3 mV
1.2816σ= 4.28 mV --> from Normal Distribution

for error rate = 0.01, it means CDF=0.99 --> erf(1.644976357) --> x = 10.7691mV
So for error rate = 0.01 or 99% confidence level, the input must be
x<-4.7691 mV or x>10.7691 mV.

I still do not understand how to get rms input noise = 1 mV as writen in the book.

1e-12 is not a mistake, rather a common value for hard error rates at one time before FEC came along.

- consider what the BER is for 74HCxx logic with linear Gaussian Noise of xx mV and a rise time of x to xx ns clipped at Vdd with a threshold of Vdd/2 +/- 25% what SNR would you extrapolate to and at 1dB per decade what BER in years.

--- Updated Sep 10, 2021 ---

I'm not sure either and I cannot assume the deviation is "standard" with what was shown. Except rms = 0.707 of peak of any sinusoid and these may or not be random, except it looks symmetrical so it assumes equal push pull RdsOn or Rce and load C and thus BW or SNR, with a certain linear gain of around Vo/ (1.28x2) and a fixed threshold of 3mV that might not be fixed over all variables. So some assumptions are not shown.