In an assignment, I need to find the settling Time of the ADC Input Circuit time to reach 1/4 of the LSB.

The ADC has 12 bits. I also know that C=12 pF and R=10 kohm. I found a formula that i used which worked:

t=-ln(1/(2^N*4))*τ where the 4 symbolizes the 4 in 1/4. N is the wordlength N=12 τ=RC=10*10^3*12*10^-12=1.2*10^-7. => t=1164.448 ns

Now the task is to use the 6 most significant bits, still with 1/4 of the LSB, C and R is the same. My original idea was to use the same formula but now with N=6. t=-ln(1/(2^6*4))*1.2*10^-7=665.421 ns

According to a correcting system this was wrong, so i don't have access to any correct answers.

What am i doing wrong here? How can i do right to find the settling time to reach 1/4 of the LSB when only using 6 most significant bits of a 12 bit ADC?

I see that the ADC is still a 12 bit system, so maybe that is where i am going wrong when i do N=6, changing the ADC to a 6 bit system. But still, i don't know how to specify the time calculation to just find the 6 most significant bits out of the 12.

  • \$\begingroup\$ Are you changing any settings that might change the capacitive value presented by the ADC circuitry? How did your number "turn out wrong"? Where did you find the formula? \$\endgroup\$ – Asmyldof Jul 14 '15 at 9:53
  • \$\begingroup\$ This is an assignment so that is how i know that it turned out wrong, so therefor no settings is changed its just fictional. The formula was found derived here:silabs.com/Support%20Documents/TechnicalDocs/an119.pdf \$\endgroup\$ – elmaggan Jul 14 '15 at 9:56
  • \$\begingroup\$ In which case, you should show some of your working and the result using the formula and what result your assignment says it should be, else we cannot (literally and ethically) help you. If we don't know what you typed in and why and what result you got, we can't see, nor explain what went wrong and just throwing out a number at you will not help you understand. \$\endgroup\$ – Asmyldof Jul 14 '15 at 10:20
  • \$\begingroup\$ I tried to specify a bit now on my way through the assingment. \$\endgroup\$ – elmaggan Jul 14 '15 at 10:33
  • \$\begingroup\$ I'm assuming you mean a fully automated digital correcting system? Did the same correcting system approve the 1164.448ns and how does it work with number precision? As a point or certain possible relevance. \$\endgroup\$ – Asmyldof Jul 14 '15 at 10:39

The problem is that each bit, from most significant to least significant, must have the same analog resolution, so the input settling time requirement is unchanged regardless of how many bits are being acquired.

As an example, let's say you are using a 12-bit ADC with a 10-volt input. 1 lsb is 10v/4095, or 2.422 mV, and the transition of the msb occurs at a nominal 5.00122 volts. The transition of the MSB at a count of 1FFH to 200H must be located to the same resolution as the lsb. If it is not, the entire accuracy of the 12 bit word will suffer.

So acquiring a signal by 1-bit ADC (and such things do exist and are extremely useful - look up "sigma-delta adc") with 12-bit precision requires exactly the same input accuracy as acquiring all 12 bits. It's quicker, but needs to be just as precise.

  • \$\begingroup\$ So are you saying that it is the same settling time for all 12 bits (that i have got right) as for the 6 MSBs of the 12-bit ADC? \$\endgroup\$ – elmaggan Jul 14 '15 at 15:06
  • \$\begingroup\$ Yes, as long as you realize that this is the analog input settling time. There are other issues, such as internal interactions with the actual digitization, but they are not addressed in your question. Each bit requires the exact same analog accuracy as every other bit. Try calculating the accuracy of an ADC if the accuracy of each bit is some percentage (like 1/2 bit) of the voltage range implied in the bit position - like if the msb of a 12-bit, 10 volt ADC can vary by +/- 2.5 volts. \$\endgroup\$ – WhatRoughBeast Jul 14 '15 at 15:13

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