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:
32 comments Page 1 of 4.
Sethu said:
8 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.
Jr.R said:
9 years ago
Resolution: V/2^n -1.
In the problem resolution is given r=5mV(5x10^-3), and we're to find the n-bit.
By equating the formula of resolution:
We Let x=n as a variable of unknown.
5*10^-3=5/2^x -1 ---> input in your calculator then shift solve for X then press = you can get 9.967 or approximately equal to 10.
In the problem resolution is given r=5mV(5x10^-3), and we're to find the n-bit.
By equating the formula of resolution:
We Let x=n as a variable of unknown.
5*10^-3=5/2^x -1 ---> input in your calculator then shift solve for X then press = you can get 9.967 or approximately equal to 10.
Rakshatha priya P said:
2 years ago
Resolution = V/2^N-1.
5mV = 5/2^N-1
2^N-1 = 5/5*10^-3V
2^N-1 = 1/10^-3V
2^N-1 = 10^3V
2^N-1 = 1000
2^N = 1000+1
2^N = 1001.
The nearest power to this value for 2^N is 2^10.
i.e,2^9=512, 2^10=1024, then the value 1024 is nearest to 1001,
then 2^N=2^10,
=> N=10.
5mV = 5/2^N-1
2^N-1 = 5/5*10^-3V
2^N-1 = 1/10^-3V
2^N-1 = 10^3V
2^N-1 = 1000
2^N = 1000+1
2^N = 1001.
The nearest power to this value for 2^N is 2^10.
i.e,2^9=512, 2^10=1024, then the value 1024 is nearest to 1001,
then 2^N=2^10,
=> N=10.
(18)
Smarak tripathy said:
1 decade ago
Resolution = Analog Input/Number of Steps.
Number of Steps = (2^N)-1.
Resolution = 5 m.
Analog Input = 5.
N = 9.96 something.
In terms of resolution always round off to higher number. So 10 the simplified explanation.
Number of Steps = (2^N)-1.
Resolution = 5 m.
Analog Input = 5.
N = 9.96 something.
In terms of resolution always round off to higher number. So 10 the simplified explanation.
Krishna said:
1 decade ago
Resolution = Analog Input/Number of Steps.
Number of Steps = (2^N)-1.
Resolution = 5m.
Analog Input = 5.
N = 9.96 something.
In terms of resolution always round off to higher number. So 10.
Number of Steps = (2^N)-1.
Resolution = 5m.
Analog Input = 5.
N = 9.96 something.
In terms of resolution always round off to higher number. So 10.
Palaniappan said:
5 years ago
Resolution =(vmax-vmin)/(2^N).
((2^N)-1) = (vmax-vmin/Resolution).
(2^N) = (5-0/5m)+1.
(2^N) = 1001.
Nx(log2) = (log (1001)).
N = (log (1001))/(log2).
N = 9.96.
((2^N)-1) = (vmax-vmin/Resolution).
(2^N) = (5-0/5m)+1.
(2^N) = 1001.
Nx(log2) = (log (1001)).
N = (log (1001))/(log2).
N = 9.96.
(9)
Shamly said:
8 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.
Sudheer said:
9 years ago
Resolution = Vr / 2^n -1.
5m = 5/2^n -1.
2^n -1 = 5/5m.
2^n-1 = 1000.
2^n = 1001.
So, n should be 10.
5m = 5/2^n -1.
2^n -1 = 5/5m.
2^n-1 = 1000.
2^n = 1001.
So, n should be 10.
Karthickraja said:
7 years ago
5v is 5mv which means the 5v hv n number of 5mv so;
5v=n*5mv,
5v=10*5mv.
5v=n*5mv,
5v=10*5mv.
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