hi guys i have a doubt in this question. the input signal to ADC is varying between 0-15V and the output of the ADC is giving in a 4 bit word. the input is scanned 100 times n 1s. what will be the sampling rate and how to calculate the resolution? and if the input voltage given is 10.5 V, what would be the output signal ?? i have doubt in 3rd bit and 4 bit word output, how are they different to each other and how we distinguish and calculate the problems for these types of questions ?
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The sampling rate is 100Hz.The resolution is 2^4=16. If the ADC input measures between 0V and 10.5V then each step is 10.5/16=0.65625. Here is a list of input voltage to output codes.
Input...............digital output
0 to 0.65625 0000
0.65625 to 1.3125 0001
1.3125 to 1.96875 0010
1.96875 to 2.625 0011
2.625 to 3.28125 0100
3.28125 to 3.9375 0101
3.9375 to 4.59375 0110
4.59375 to 5.25 0111
5.25 to 5.90625 1000
5.90625 to 6.5625 1001
6.5625 to 7.21875 1010
7.21875 to 7.875 1011
7.875 to 8.53125 1100
8.53125 to 9.1875 1101
9.1875 to 9.84375 1110
9.84375 to 10.5 1111
This is what you will get if it is perfectly accurate. It will be close but not exact.
If you have a voltmeter and a potentiometer then you can connect the potentiometer as a voltage divider. One of the pot to 10.5 V or more and the other end to 0 V. The wiper is to be connected to the ADC input. The pot can be turned up and down and the code from the ADC can be read with whatever you have, maybe LEDs. Make a table of input values to output codes and compare with the above list.
Clarify what I might have misunderstood in your question and I will try again.
Input...............digital output
0 to 0.65625 0000
0.65625 to 1.3125 0001
1.3125 to 1.96875 0010
1.96875 to 2.625 0011
2.625 to 3.28125 0100
3.28125 to 3.9375 0101
3.9375 to 4.59375 0110
4.59375 to 5.25 0111
5.25 to 5.90625 1000
5.90625 to 6.5625 1001
6.5625 to 7.21875 1010
7.21875 to 7.875 1011
7.875 to 8.53125 1100
8.53125 to 9.1875 1101
9.1875 to 9.84375 1110
9.84375 to 10.5 1111
This is what you will get if it is perfectly accurate. It will be close but not exact.
If you have a voltmeter and a potentiometer then you can connect the potentiometer as a voltage divider. One of the pot to 10.5 V or more and the other end to 0 V. The wiper is to be connected to the ADC input. The pot can be turned up and down and the code from the ADC can be read with whatever you have, maybe LEDs. Make a table of input values to output codes and compare with the above list.
Clarify what I might have misunderstood in your question and I will try again.