Comparator calculation

Discussion in 'Homework Help' started by screen1988, May 9, 2013.

Mar 7, 2013
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Could you check if my understanding is right or not?

1LSB = 1.5 /(2^12 -1) = 0.366mV
And I don't know why Av is equal to VDD divided by 0.18V not 0.36V.

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2. t_n_k AAC Fanatic!

Mar 6, 2009
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Yes your LSB value is more accurate.

Presumably the goal is to be able to detect a minimum input value of 1 LSB. If the designer set the comparator gain to cause an output transition at exactly 1 LSB it may not happen in practice, due to small differences between predicted and actual values. Increasing the comparator sensitivity such that it would notionally transition at 1/2 X LSB value would increase the likelihood of catching an actual 1 X LSB change from the ADC. Conversely, increasing the sensitivity by too great a factor [say 1/10 X LSB] might lead to false comparator transitions.

I suspect this is an oversimplification of a design concept. For instance, there is no suggestion that some comparator hysteresis might be advisable in practice. Nor is there any reference to the point that high gain amplifiers [such as op-amps for example] may make inferior comparators.

screen1988 likes this.

Mar 7, 2013
310
3
Thanks!
Can you explain why it isn't good to do that? I mean why it can lead to false transilation. I guess it can be affected by noise but not sure.
And why in the picture Av is minimum? Vout = Vdd =1.8V (maximum)
Vin = 0.18V (minimum)
Then I think that Av has to be maximum?

4. t_n_k AAC Fanatic!

Mar 6, 2009
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Without seeing the bigger picture with respect to the image you posted it's difficult to comment further. If it is copied from a textbook what is the author trying to convey to the reader? You haven't given a lot in terms of the context.

Why is the required gain indicated a minimum? If the goal is to obtain an amplifier output of 1.8V with 180uV input then one needs a gain of at least 10,000.