Electronics and Communication Engineering - Digital Electronics - Discussion
Discussion Forum : Digital Electronics - Section 1 (Q.No. 1)
1.
The resolution of an n bit DAC with a maximum input of 5 V is 5 mV. The value of n is
Answer: Option
Explanation:
1000 = 5 or N = 10.
Discussion:
33 comments Page 2 of 4.
Shamly said:
9 years ago
Resolution of DAC =[full scale range]/[(2^n)-1],
Given full scale range 5v.
Resolution of DAC =5mv.
So,
5 * 10^-3 = 5/(2^n-1).
You will get the answer.
Given full scale range 5v.
Resolution of DAC =5mv.
So,
5 * 10^-3 = 5/(2^n-1).
You will get the answer.
(1)
Sneha .khot said:
8 years ago
((2^N)-1) = (Input/Resolution ).
(2^N) = (5/5m)+1.
(2^N) = 1001.
Nx(log2) = (log (1001)).
N = (log (1001))/(log2).
N = 9.96.
(2^N) = (5/5m)+1.
(2^N) = 1001.
Nx(log2) = (log (1001)).
N = (log (1001))/(log2).
N = 9.96.
Sethu said:
9 years ago
Resolution = V / 2^n -1. (Given resolution=5*10^-3, V=5, n=?) substitute values to the equation.
5*10^-3= 5/2^n -1.
2^n -1 = 5/5*10^-3
2^n-1 = 5*10^3 /5
=10^3
=1000
2^n =1000+1
= 1001
2^n = 1001. (Take log on both sides)
nlog2 = log(1001)
n = log(1001)/ log2
= 9.96722 (round to 10) =10.
5*10^-3= 5/2^n -1.
2^n -1 = 5/5*10^-3
2^n-1 = 5*10^3 /5
=10^3
=1000
2^n =1000+1
= 1001
2^n = 1001. (Take log on both sides)
nlog2 = log(1001)
n = log(1001)/ log2
= 9.96722 (round to 10) =10.
Dhiru said:
8 years ago
Nice solution, Thanks @ Sneha.
Chaithanya said:
8 years ago
Anyone know a shortcut to find the log values?
Aparna said:
8 years ago
Thanks @Sneha.
Kunar said:
1 day ago
Nice explanation. Thanks a lot.
Manoj said:
9 years ago
iIf n^2 = 1001,
So n = 31.46.
So n = 31.46.
Sairam said:
1 decade ago
Will you please give me a complete formula you have used?
Venky said:
9 years ago
Please clarify to understand.
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