Concept:
For a laddertype D/A Converter:
Output Voltage (V0) = Resolution × Decimal Equivalent of binary input.
Where Resolution is given by:
\(Resolution=\frac{{{V}_{ref}}}{{{2}^{n}}}\)
Application:
Given n = 5 and the Digital input = 11010
∵ The Resolution will be:
\(R=\frac{{{V}_{ret}}}{{{2}^{n}}}=\frac{10}{{{2}^{5}}}=~0.3125\)
Since the decimal Equivalent of 11010 = 26
So, V0 = 26 × 0.3125
V0 = 8.125 V
Note: If the fullscale voltage is given, then:
Resolution \(=\frac{{{V}_{fs}}}{{{2}^{n}}1}\)
Resolution: It is defined as the smallest change in the analog output voltage corresponding to a change of one bit in the digital output.
The percentage resolution (%R) of an nbit DAC is:
\(\%R = \frac{1}{{{2^n}  1}} \times 100\)
The resolution of an nbit DAC with a range of output voltage from 0 to V is given by:
\(R = \frac{V}{{{2^n}  1}}volts\)
Calculation:
Number of bits (n) = 8
Resolution \( = \frac{{1}}{{{2^8}  1}} = \frac{{1}}{{255}}\)
A 6bit ladder D/A converter has a maximum output of 10 V. The output for input 101001 is approximately
Concept:
For a laddertype D/A Converter:
Output Voltage (V0) = Resolution × Decimal Equivalent of binary input.
Where Resolution is given by:
\(Resolution=\frac{{{V}_{fs}}}{{{2}^{n}}1}\)
Where, V_{fs }= Full scale voltage or maximum voltage
Application:
Given n = 6 and the Digital input = 101001
∵ The Resolution will be:
\(R=\frac{{{V}_{ret}}}{{{2}^{n}1}}=\frac{10}{{{2}^{6}1}}=~0.1587\)
Since the decimal Equivalent of 101001 = 41
So, V0 = 41 × 0.1587
V0 = 6.5067 V
V_{0 }≈ 6.5 V
Note: If the reference voltage is given, then:
\(Resolution=\frac{{{V}_{ref}}}{{{2}^{n}}}\)
Concept of Resolution:
It is defined as the smallest change in the analog output voltage corresponding to a change of one bit in the digital output.
The percentage resolution (%R) of an nbit DAC is:
\(\%R = \frac{1}{{{2^n}  1}} \times 100\)
Calculation:
As we know the formula resolution,
\(\%R = \frac{1}{{{2^n}  1}} \times 100\)
250 = 2^{N} 1
2^{N} = 251 ≈ 255
i.e., 2^{8} = 255
N = 8
Hence the minimum value of N satisfying the condition.
Concept:
Resolution for n – bit A/D converter will be:
\(R= \frac{V_{FS} \ \times \ (2^ib^i)}{{{2^n}  1}}\)
Where
R = Resolution
VFS is reference voltage 'or' Fullscale voltage
n = number of bits
2^{i}b^{i} gives output voltage value.
Calculation:
Given:
V_{FS} = 5.8 V
n = 4
R = 05 V
\( 5= \frac{5.8 \ \times \ 2^ib^i}{{{2^{4}}  1}}\)
\( 2^ib^i= \frac{5 \ \times \ 15}{{5.8}}\)
2ibi = 12.93 V ≈ 12 V
In bits, the output voltage will be 1100
Hence option (1) is the correct answer.
Important Points
The resolution of DAC is a change in analog voltage corresponding to the LSB bit increment at the input.
The resolution (R) is calculated as:
\( R= \frac{{{V_{FS}}}}{{{2^N}  1}}\)
No. of levels = 2N – 1
Vr is reference voltage 'or' Fullscale voltage
Concept:
R – 2R ladder DAC:
3 bit R – 2R DAC by using Inverting opamp is shown below:
Output voltage is defined as:
V0 =  I_{1}(RF) ⋯ (i)
Current is calculated as:
\(I_1 = \left[ {\frac{{{a_2}}}{2} + \frac{{{a_1}}}{4} + \frac{{{a_0}}}{8}} \right]I\) ⋯ (ii)
Current drawn from the source is
I = V / R ⋯ (iii)
Current is calculated as:
\(I_1 = \left[ {\frac{{{a_2}}}{2} + \frac{{{a_1}}}{4} + \frac{{{a_0}}}{8}} \right]\frac{V}{R}\)
The final output voltage is
\({V_0} =  \frac{V}{R}\left[ {\frac{{{a_2}}}{2} + \frac{{{a_1}}}{4} + \frac{{{a_0}}}{8}} \right]{R_F}\)
For n bit DAC output voltage is
\({V_0} =  \frac{V}{R}\left[ {\frac{{{a_{n  1}}}}{{{2^1}}} + \frac{{{a_{n  2}}}}{{{2^2}}} + \cdots + \frac{{{a_0}}}{{{2^n}}}} \right]{R_F}\)
Calculation:
Given digital input is 1010 and the reference voltage is 5 V
a3 = 1, a2 = 0, a1 = 1, a0 = 0
considering the value of the feedback resistor for Op amp is R Ω
Output voltage is
\({V_0} =  \frac{{5R}}{R}\left[ {\frac{1}{2} + \frac{0}{4} + \frac{1}{8} + \frac{0}{{16}}} \right]\)
\({V_0} = 5\times\frac{5}{8}\)
V_{0} = 3.125 V
Resolution:
Resolution of ADC is a change in analog voltage corresponding to a 1bit increment.
Resolution is the number of bits per conversion cycle that the converter is capable of processing.
\(R=\frac{V_{range}}{2^n}\)
n = No. of bits of ADC
V_{range }= V_{max}  V_{min}
Analog output = Reslotion x Decimal equivalent to binary
V_{o} = R x D
Calculation:
V_{range} = V_{max}  V_{min}
V_{max} = 2.56 V
V_{min} = 0 V
⇒ V_{range} = 2.56 V
n = 8
Resolution is
\(R=\frac{V_{range}}{2^n}\)
⇒ \(R=\frac{2.56}{2^8}\)
∴ R = 0.01.
Output voltage V_{o} = 1 V
⇒ 1 = 0.01 x D
∴ D = 100
The binary value of 100 is
The binary equivalent of 100 is (011 001 00)_{2}
The present resolution of an 8bit D/A converter is
Resolution: It is defined as the smallest change in the analog output voltage corresponding to a change of one bit in the digital output.
The percentage resolution (%R) of an nbit DAC is:
\(\%R = \frac{1}{{{2^n}  1}} \times 100\)
The resolution of an nbit DAC with a range of output voltage from 0 to V is given by:
\(R = \frac{V}{{{2^n}  1}}volts\)
Calculation:
Full scale output voltage (V) = 10 V
Number of bits (n) = 8
Resolution \( = \frac{{1}}{{{2^8}  1}} = \frac{{1}}{{255}}\)
Concept:
Maximum error (E_{r}) of D/A is given as,
E_{r }= (V_{FS} x Accuracy)
Where,
V_{FS}: fullscale input voltage
Calculation:
Given: V_{fS }= 5V , Accuracy= 0.2%
\({E_r} = \frac{{5 \times 0.2}}{{100}}\)
\({E_r} = \frac{1}{{100}} \)
E_{r} = 10 mV
Resolution: It is defined as the smallest change in the analog output voltage corresponding to a change of one bit in the digital output.
The percentage resolution (%R) of an nbit DAC is:
\(\%R = \frac{1}{{{2^n}  1}} \times 100\)
The resolution of an nbit DAC with a range of output voltage from 0 to V is given by:
\(R = \frac{V}{{{2^n}  1}}volts\)
Hence the difference between analog voltage represented by two adjacent digital codes of an analog to digital converter is called resolution.
Hence option (2) is the correct answer.
Important Points
Accuracy:
% Accuracy of a n bit ADC = (1 / 2n ) × 100
Concept:
Digital to analog (D/A) conversion is the process of taking a value represented in digital code (such as straight binary or BCD) and converting it to a voltage or current which is proportional to the digital value.
In general,
Analog output = K × digital input
Where K is the resolution and it is a constant value for a given DAC.
Application:
Given D/A is of 10 bits.
Also, V_{pp} = 10 V
Resolution of the D/A will be:
\(Res. = \frac{{10}}{{1023}}\)
Input to the D/A converter in (13 A)_{16} i.e.
\({\left( {13A} \right)_{16}} = {16^2} \times 1 + 3 \times {16^1} + 10 \times {16^0}\)
(13 A)_{16} = (314)_{10}
Now, the output of the converter will be:
Output = Resolution × Input
Output \(= \frac{{10}}{{1023}} \times 314\)
Output = 3.069
Explanation:
Resolution (R):
Resolution of Digital to Analog Converter(DAC) is a change in analog voltage corresponding to 1 LSB bit increment at the input.
\(R = \frac{V_r}{2^n1}\)
Where
V_{r} = Reference voltage
n = number of bits
Resolution should be below 0.1% of the maximum value
\(\frac{V_r}{2^n1}≤\frac{0.1\ \times \ V_r}{100}\)
(2^{n}  1) ≥ 1000
2^{n} ≥ 1001
n = 10 satisfies the above equation.
Hence n = 10.
The following waveform shows the output of a 4 bit DAC with 5 V reference voltage.
The 4bit digital input of DAC is connected to 4 bit up counter, the onebit input of DAC Is stuck at ‘0’, which is this bit?
Concept:
In the case of DAC, the output is defined as = K * Decimal equivalent of Binary number.
K = Resolution (Step size) \( = \frac{{{V_R}\left( {Reference\;voltage} \right)}}{{{2^n}}}\)
n = no. of Bits in DAC
Calculation:
V_{R} = 5 V, n = 4
Step size \( = \frac{{{V_R}}}{{{2^n}}} = \frac{5}{{16}} = .3125\;V\)
Analog Voltage 
No. of steps 
Correct o/p 
DAC o/p 







B_{3} 
B_{2} 
B_{1} 
B_{0} 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
.3125 
1 
0 
0 
0 
1 
0 
0 
0 
1 
.6250 
2 
0 
0 
1 
0 
0 
0 
1 
0 
.9375 
3 
0 
0 
1 
1 
0 
0 
1 
1 
1.250 
4 
0 
1 
0 
0 
0 
0 
0 
0 
1.5625 
5 
0 
1 
0 
1 
0 
0 
0 
1 
1.875 
6 
0 
1 
1 
0 
0 
0 
1 
0 
2.1875 
7 
0 
1 
1 
1 
0 
0 
1 
1 
2.5 
8 
1 
0 
0 
0 
1 
0 
0 
0 
By truth table, we can see Bit 2 (B_{2}) is stuck at 0. It is not changing with respect to the input.
Conclusion:
In the diagram in 4th clock pulse (0100), the right output is 1.250V (0100) but it is showing output 0 volts (0000) similarly in the case of 5th clock pulse correct output voltage is 1.5625 V (0101) but it is showing output .6250 which corresponds to 2 step count(0010) we can observe that bit2 is not changing i.e. it is stuck at 0.
Concept:
For a DAC
Analog output = Resolution × Digital equivalent in decimal
Resolution = \(\frac{{{V_{{\mathop{\rm Re}\nolimits} f}}}}{{{2^n}  1}}\)
Analysis:
Given,
For digital input 10100, analog output is 10 mA
10100_{2} = (20)_{10}
Analog output = Resolution × Digital equivalent in decimal
10 × 10^{3} = Resolution × 20
Resolution = 0.5 × 10^{3}
For digital input 11101
Decimal equivalent = 29
Analog output = Resolution × Digital equivalent in decimal
= 0.5 × 10^{3} × 29
=14.5 × 10^{3}
=14.5 mA
A 10bit DAC provides an analog output which has a maximum value of 10.23 volts. Resolution of the DAC is
Concept:
The resolution of DAC is a change in analog voltage corresponding to LSB bit increment at the input.
The resolution (R) is calculated as:
\( R= \frac{{{V_{FS}}}}{{{2^n}  1}}\)
No. of levels = 2n – 1
Vr is reference voltage 'or' Fullscale voltage
Calculation:
Given n = 10 bits
V_{FS} = 10.23 V
\(R= \frac{10.23}{{{2^{10}}  1}}\)
R = 10mV
Concept:
Output Voltage (V_{o}) = Resolution × Decimal Equivalent of binary input.
Resolution is given by:
\(Resolution=\frac{{{V}_{fs}}}{{{2}^{n}}1}\)
Where V_{fs} = Fullscale voltage or maximum voltage
Application:
Given:
V_{fs }= 10 V
Resolution = 50 mV
\( 50\;mV= \frac{10~V}{{{2^n 1}}} \)
2^{n}  1 = 200
n = log_{2}(201)
n = 7.65
∴ \(n \approx 8\)
Concept:
R – 2R ladder DAC:
3 bit R – 2R DAC by using Inverting opamp is shown below:
Analysis:
Output voltage is defined as:
V_{0} =  I_{1}R_{F} ⋯ (i)
Current is calculated as:
\(I_1 = \left[ {\frac{{{a_2}}}{2} + \frac{{{a_1}}}{4} + \frac{{{a_0}}}{8}} \right]I\) ⋯ (ii)
Current drawn from the source is
I = V / R ⋯ (iii)
Current is calculated as:
\(I_1 = \left[ {\frac{{{a_2}}}{2} + \frac{{{a_1}}}{4} + \frac{{{a_0}}}{8}} \right]\frac{V}{R}\)
The final output voltage is
\({V_0} =  \frac{V}{R}\left[ {\frac{{{a_2}}}{2} + \frac{{{a_1}}}{4} + \frac{{{a_0}}}{8}} \right]{R_F}\)
For n bit DAC output voltage is
\({V_0} =  \frac{V}{R}\left[ {\frac{{{a_{n  1}}}}{{{2^1}}} + \frac{{{a_{n  2}}}}{{{2^2}}} + \cdots + \frac{{{a_0}}}{{{2^n}}}} \right]{R_F}\)
Calculation:
Given digital input is 0110 and the reference voltage is 3 V
a_{3} = 0, a_{2} = 1, a_{1} = 1, a_{0} = 0
considering the value of the feedback resistor for Op amp is R Ω
Output voltage is
\({V_0} =  \frac{{3R}}{R}\left[ {\frac{0}{2} + \frac{1}{4} + \frac{1}{8} + \frac{0}{{16}}} \right]\)
\({V_0} =  \frac{3}{4}\left[ {1 + \frac{1}{2}} \right]\)
\({V_0} =  \frac{3}{4} \times \frac{3}{2} =  \frac{9}{8}\)
V_{0} =  1.125 V
Weighted resistor DAC:
It uses a summing amplifier with a binaryweighted network as shown below.
For an nbit ADC,
The accuracy and stability depend on the accuracy of resistors.
The requirement of a wide range of resistor values restricts the use up to 8bit.
R2R ladder resistor DAC:
It uses a summing amplifier with an R2R ladder network as shown below.
For nbit DAC, it requires only 2 different values of resistors i.e. R and 2R.
Resolution is the number of pieces or parts that the output or displayed reading from a sensor or measuring instrument can be broken down into without any instability in the signal or reading.
Resolutions can be expressed in a number of ways, the two most common being:
1. Decimal Places
Our 10kg kitchen scale has a resolution of kilograms to 3 decimal places or ‘3dp’.
2. Parts/Divisions/Counts
Our 10kg kitchen scale has a resolution of ‘1 part in 10,000’, ‘10,000 divisions’ or ‘10,000 ‘counts’.
There are four types of resolution are as follows:
1. Radiometric Resolution:
2 Spatial Resolution:
3. Spectral Resolution:
4. Temporal Resolution:
Examples – Landsat, GOES, and NOAA
Concept:
FullScale Output
It is the total voltage level given to the DAC or in simple terms, it is calculated as:
FSO (VFS) = Total steps × Step size
\(FSO~ (V_{FS})= \left( {{2^N}  1} \right) × step\;size\)
The resolution of DAC is a change in analog voltage corresponding to the LSB bit increment at the input.
The resolution (R) is calculated as:
\( R= \frac{{{V_{FS}}}}{{{2^N}  1}}\)
No. of levels = 2^{N} – 1
Vr is reference voltage 'or' Fullscale voltage
Calculation:
Given n = 10 bits
\(FSO~ (V_{FS})= \left( {{2^N}  1} \right) × step\;size\)
VFS = (1023) × 10 × 10^{3} = 10.23 V
VFS = 10.23 V
\(R= \frac{10.23}{{{2^{10}}  1}} \times 100\)
R = 1%