Digital To Analog Converters MCQ Quiz - Objective Question with Answer for Digital To Analog Converters - Download Free PDF

Explanation:

8-bit Digital-to-Analog Converter (DAC) using Two 4-bit DACs

Problem Understanding: In the given question, an 8-bit DAC is implemented using two identical 4-bit DACs with equal reference voltage. The binary inputs b₇ to b₀ are represented, where b₇ is the Most Significant Bit (MSB) and b₀ is the Least Significant Bit (LSB). To achieve the correct analog output voltage (V₀) corresponding to an 8-bit DAC, we need to determine the value of the resistor R in the circuit. The operational amplifier (op-amp) is assumed to be ideal.

Working Principle:

To understand the operation of this circuit, note the following:

• The 8-bit input is divided into two 4-bit groups: the higher 4 bits (b₇ to b₄) and the lower 4 bits (b₃ to b₀).
• Each 4-bit group is fed into one of the two identical 4-bit DACs.
• The higher 4 bits produce an analog output proportional to their binary value, scaled to the range of the higher nibble.
• The lower 4 bits produce an analog output proportional to their binary value, scaled to the range of the lower nibble.
• The outputs of the two DACs are combined using a resistor network and an operational amplifier to produce a final output voltage V₀ that corresponds to the 8-bit binary input.

Analysis:

The higher nibble (b₇ to b₄) represents the most significant part of the binary input. Its contribution to the output voltage should dominate over the lower nibble (b₃ to b₀). To achieve this, the output of the DAC handling the higher nibble is scaled by a factor of 16 (since 2⁴ = 16) relative to the output of the DAC handling the lower nibble.

The resistor R in the circuit is responsible for ensuring this scaling. Specifically, the resistor network and the ideal op-amp configure the circuit such that:

• The output voltage of the higher nibble DAC is multiplied by 16.
• The output voltage of the lower nibble DAC is added directly.

Mathematical Derivation:

Let the outputs of the two 4-bit DACs be V_H (higher nibble) and V_L (lower nibble), respectively. The final output voltage V₀ is given by:

V₀ = 16 × V_H + V_L

Both DACs have the same reference voltage and identical resistor networks. Therefore, the scaling factor of 16 is achieved by appropriately choosing the value of resistor R. For a standard 4-bit DAC, the output voltage is proportional to the binary input (b₃ to b₀ or b₇ to b₄) scaled by the reference voltage and the DAC’s resolution.

The resistor R is chosen such that the higher nibble’s output voltage is effectively scaled by 16 relative to the lower nibble’s output. This is achieved by using a resistor value of 1 kΩ.

Correct Answer: The value of R should be 1 kΩ.

Additional Information

To further understand the reasoning, let’s analyze why the other options are incorrect:

Option 1: 8 kΩ

This value is too high. If R were 8 kΩ, the scaling factor would not properly match the requirement of multiplying the higher nibble’s output by 16. This would result in incorrect analog values at the output.

Option 2: 0.25 kΩ

This value is too low. A resistor value of 0.25 kΩ would result in an incorrect scaling factor, leading to a mismatch between the contributions of the higher and lower nibbles.

Option 4: 0.5 kΩ

Similar to Option 2, this value is also too low. The scaling factor would not correctly match the requirement, resulting in incorrect output voltages.

Option 5: Not provided in the question

As the correct scaling requires R = 1 kΩ, any other value not listed in the options would also fail to produce the correct output.

Conclusion:

The resistor R plays a crucial role in ensuring the correct scaling of the higher nibble’s output relative to the lower nibble’s output in the 8-bit DAC circuit. By choosing R = 1 kΩ, the circuit achieves the desired scaling, resulting in accurate analog values corresponding to the 8-bit binary inputs. This understanding is essential in designing such circuits to ensure proper functionality and accuracy.

Explanation:

Analysis of the 4-bit Unipolar DAC and Comparator System:

The given problem involves a 4-bit unipolar DAC (Digital-to-Analog Converter) with a 10V reference voltage and a comparator whose threshold voltage is set at 7.5V. The DAC counts from 0 to 15 (since it is a 4-bit DAC, it can represent 2⁴ = 16 levels). The comparator monitors the output of the DAC and determines when it crosses the threshold voltage.

It is observed that when the DAC count reaches 8, the comparator erroneously indicates threshold crossing. Upon analysis, it is found that one of the input bits of the DAC is stuck at '1'. We need to determine which bit is stuck at '1'.

Step-by-Step Solution:

1. Understanding the DAC Operation:

A 4-bit unipolar DAC converts a digital value (binary) into an analog voltage. The output voltage is given by:

V_out = (Digital Value / Maximum Digital Value) × Reference Voltage

Since the DAC is 4-bit, the maximum digital value is 15 (binary 1111). The reference voltage is 10V. Therefore:

V_out = (Digital Value / 15) × 10

For each binary value, the corresponding analog voltage is calculated as follows:

2. Comparator Threshold Analysis:

The comparator is set to a threshold voltage of 7.5V. It should ideally indicate threshold crossing when the DAC output voltage exceeds 7.5V. From the table above, this happens when the binary value exceeds 1011 (decimal 11), as for binary 1100 (decimal 12), V_out becomes 8V.

However, the problem states that the comparator erroneously indicates threshold crossing at binary 1000 (decimal 8). This suggests that the DAC is producing an incorrect output voltage due to one of its input bits being stuck at '1'.

3. Identifying the Stuck-at-'1' Bit:

To determine which bit is stuck at '1', we analyze the effect of each bit being stuck at '1' on the output voltage:

• B₃ (MSB): If the most significant bit (B₃) is stuck at '1', the output voltage would be significantly higher than expected for lower binary values, as B₃ contributes the largest weight (8 × Reference Voltage / 15). For binary 1000 (decimal 8), the output voltage would be correct, so B₃ is not the stuck bit.
• B₂: If B₂ is stuck at '1', it adds an additional weight (4 × Reference Voltage / 15) to the output voltage. For binary 1000 (decimal 8), the output voltage would incorrectly include this additional weight, resulting in an erroneous threshold crossing at the comparator. This matches the problem description, so B₂ is the stuck bit.
• B₁: If B₁ is stuck at '1', it adds a smaller weight (2 × Reference Voltage / 15). For binary 1000 (decimal 8), the output voltage would not exceed the threshold, so B₁ is not the stuck bit.
• B₀ (LSB): If the least significant bit (B₀) is stuck at '1', it adds the smallest weight (Reference Voltage / 15). This would not cause the erroneous threshold crossing at binary 1000 (decimal 8), so B₀ is not the stuck bit.

Correct Answer: The stuck-at-'1' bit is B₂.

Important Information

To further understand the analysis, let’s evaluate the other options:

• B₀ (LSB): The least significant bit contributes the smallest weight to the output voltage. If it is stuck at '1', the error in output voltage would be minimal and insufficient to cause the threshold crossing at binary 1000 (decimal 8).
• B₁: The second least significant bit contributes a slightly larger weight than B₀. If it is stuck at '1', the output voltage would still not exceed the threshold at binary 1000 (decimal 8).
• B₃ (MSB): The most significant bit contributes the largest weight to the output voltage. If it is stuck at '1', the output voltage would be significantly higher than expected for lower binary values, but it would not match the observed erroneous behavior at binary 1000 (decimal 8).

Conclusion:

By analyzing the effect of each bit being stuck at '1', we identify that B₂ is the stuck bit, as its incorrect contribution to the output voltage causes the comparator to erroneously indicate threshold crossing at binary 1000 (decimal 8). This analysis highlights the importance of understanding DAC operation and bit weighting in digital systems.

• MSB stands for most significant bit is the bit position in a binary number having the greatest value.
• The MSB is sometimes referred to as the high-order bit or left-most bit due to the convention in positional notation of writing more significant digits further to the left.
• The MSB can also correspond to the sign bit of a signed binary number. In one's and two's complement notation, "1" signifies a negative number and "0" signifies a positive number.

Example:

(12)10 = (1100)2

According to the question, the most significant bit of '100010' binary data is '1'. 

The correct option is 2

Concept:

D/A converter:

D/A converter converts digital or binary data into its equivalent analog data. This analog data is required to drive motors and other analog devices. The converted analog value is either in voltage or current form.

There are two types of D/A converters:

• Weighted Resistor or Resistive Divider type
• R-2R Ladder-type

Weighted resistor DAC:

It uses a summing amplifier with a binary-weighted network as shown below.

For an n-bit ADC,

• If the weight of MSB resistor is 2R, then the weight of the LSB resistor is 2nR.
• We require n number of resistors.
• We require n different values of resistors i.e. 2R, 22R, 23R, …, 2nR.

The accuracy and stability depend on the accuracy of resistors.

The requirement of a wide range of resistor values restricts the use up to 8-bit.

R-2R ladder resistor DAC:

It uses a summing amplifier with an R-2R ladder network as shown below.

For n-bit DAC, it requires only 2 different values of resistors i.e. R and 2R.

• The smallest change in the input signal that can be detected by an instrument is referred to as its "resolution."
• The term resolution describes the finest detail that a device or system can detect or measure.
• It is a key parameter for systems that deal with digital signals, as it is directly linked to the quality or level of detail of the output.

Concept:

For a ladder-type 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 full-scale 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 input.

The percentage resolution (%R) of an n-bit DAC is:

\(\%R = \frac{1}{{{2^n} - 1}} \times 100\)

The resolution of an n-bit 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 \( = \frac{{1}}{{{2^8} - 1}} = \frac{{1}}{{255}}\)

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 n-bit DAC is:

\(\%R = \frac{1}{{{2^n} - 1}} \times 100\)

The resolution of an n-bit 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:

• The accuracy of the A/D converter determines how close the actual digital output is to the theoretically expected digital output for given analog input.
• In other words, the accuracy of the converter determines how many bits in the digital output code represent useful information about the input signal.

% Accuracy of a n bit ADC = (1 / 2n ) × 100

Concept:

For a ladder-type 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

V₀ ≈ 6.5 V

Note:  If the reference voltage is given, then:

\(Resolution=\frac{{{V}_{ref}}}{{{2}^{n}}}\)

Concept:

The output for an 'n' bit DAC is given by:

V_o = Resolution × (Binary equivalent of digital input)

The resolution is obtained by:

\(R = {V_{ref}\over 2^n}\)

Calculation:

Given, resolution = 20 mV/LSB

The binary equivalent of the digital input is:

(10000000)₂ = 1× 2⁷ = 128 V

V_o = 20 × 10^-3 × 128

Vo = 2.56 V

Concept:

Maximum error (E_r) of D/A is given as,

E_r = (V_FS x Accuracy)

Where,

V_FS: full-scale 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

\The correct answer is option 2):(Flash type)

Concept:

Flash ADC (Fastest)

• The flash ADC is the fastest type available. A flash ADC uses comparators, one per voltage step, and a string of resistors.
• Flash-type ADC requires no counter For an n-bit ADC, flash-type ADC requires (2ⁿ – 1) comparators
• A 4-bit ADC will have 15 comparators, and an 8-bit ADC will have 255 comparators.
• The following figure shows a 3-bit flash ADC circuit.

• It is formed of a series of comparators, each one comparing the input signal to a unique reference voltage.
• The comparator outputs connect to the inputs of a priority encoder circuit, which then produces a binary output.
• Vref is a stable reference voltage provided by a precision voltage regulator as part of the converter circuit.
• As the analog input voltage exceeds the reference voltage at each comparator, the comparator outputs will sequentially saturate to a high state.
• The priority encoder generates a binary number based on the highest-order active input, ignoring all other active inputs.

Additional Information

• The successive approximation A/D converter has a shorter conversion time compared to the counter ramp A/D converter.
• Counter type ADC and successive approximate ADC use DAC Counter type ADC uses linear search and successive approximation type ADC uses binary search
• A ring counter is used in successive approximation-type ADC
•  Dual slope ADC is the most accurate

Concept:

• Analog-to-Digital Converters (ADCs) transform an analog voltage to a binary number (a series of 1’s and 0’s).
• Then eventually to a digital number (base 10) for reading on a meter, monitor, or chart.
• The ADC resolution depends upon the number of bits used to represent the digit number. 
• As the number of bits increases the resolution of an Analog to Digital Converter improves and the quantization error decreases.

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' Full-scale voltage

n = number of bits

2ⁱbⁱ 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' Full-scale voltage

Resolution:

Resolution of ADC is a change in analog voltage corresponding to a 1-bit 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