Binary Weighted Resistor DAC MCQ Quiz - Objective Question with Answer for Binary Weighted Resistor DAC - Download Free PDF

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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 b<sub>7</sub> to b<sub>0</sub> are represented, where b<sub>7</sub> is the Most Significant Bit (MSB) and b<sub>0</sub> is the Least Significant Bit (LSB). To achieve the correct analog output voltage (V<sub>0</sub>) 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 (b<sub>7</sub> to b<sub>4</sub>) and the lower 4 bits (b<sub>3</sub> to b<sub>0</sub>).
• 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 V<sub>0</sub> that corresponds to the 8-bit binary input.

Analysis:

The higher nibble (b<sub>7</sub> to b<sub>4</sub>) represents the most significant part of the binary input. Its contribution to the output voltage should dominate over the lower nibble (b<sub>3</sub> to b<sub>0</sub>). To achieve this, the output of the DAC handling the higher nibble is scaled by a factor of 16 (since 2<sup>4</sup> = 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<sub>H</sub> (higher nibble) and V<sub>L</sub> (lower nibble), respectively. The final output voltage V<sub>0</sub> is given by:

V<sub>0</sub> = 16 × V<sub>H</sub> + V<sub>L</sub>

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 (b<sub>3</sub> to b<sub>0</sub> or b<sub>7</sub> to b<sub>4</sub>) 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.

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• 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

MSB			LSB
1	1	0	0

 

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

Concept:

R – 2R ladder DAC:

3 bit R – 2R DAC  by using Inverting op-amp is shown below:

Output voltage is defined as:

V0 = - I1(RF) ⋯ (i)

Current is calculated as:

     ⋯ (ii)

Current drawn from the source is

I = V / R ⋯ (iii)

Current is calculated as:

The final output voltage is

For n bit DAC output voltage is

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

V0 = -3.125 V

DAC ( Digital to Analog converter) is an electronic circuit that is used to convert any digital signal into an analog signal.

The digital binary data exists in the form of bits. Each bit is either 1 or 0 & they represent its weight corresponding to its position.

The weight is 2n where the n is the position of the bit from the right-hand side & it starts from 0.

The output voltage range is calculated by using the formula,

Output voltage range = Maximum DAC output voltage - Minimum DAC output voltage

Output voltage range = 9.5 V - 3.5 V

Output voltage range = 6 V

Conclusion: Option 4 is correct.

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:

The resolution of an n-bit DAC with a range of output voltage from 0 to V is given by:

Calculation:

Number of bits (n) = 8

Resolution 

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:

The resolution of an n-bit DAC with a range of output voltage from 0 to V is given by:

Calculation:

Number of bits (n) = 8

Resolution 

Concept:

R – 2R ladder DAC:

3 bit R – 2R DAC  by using Inverting op-amp is shown below:

Output voltage is defined as:

V0 = - I₁(RF) ⋯ (i)

Current is calculated as:

     ⋯ (ii)

Current drawn from the source is

I = V / R ⋯ (iii)

Current is calculated as:

The final output voltage is

For n bit DAC output voltage is

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

V₀ = -3.125 V

• 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

MSB			LSB
1	1	0	0

 

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

Concept:

R – 2R ladder DAC:

3 bit R – 2R DAC  by using Inverting op-amp is shown below:

Output voltage is defined as:

V0 = - I1(RF) ⋯ (i)

Current is calculated as:

     ⋯ (ii)

Current drawn from the source is

I = V / R ⋯ (iii)

Current is calculated as:

The final output voltage is

For n bit DAC output voltage is

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

V0 = -3.125 V

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_p-p = 10 V

Resolution of the D/A will be:

Input to the D/A converter in (13 A)₁₆ i.e.

(13 A)₁₆ = (314)₁₀

Now, the output of the converter will be:

Output = Resolution × Input

Output 

Output = 3.069

DAC ( Digital to Analog converter) is an electronic circuit that is used to convert any digital signal into an analog signal.

The digital binary data exists in the form of bits. Each bit is either 1 or 0 & they represent its weight corresponding to its position.

The weight is 2n where the n is the position of the bit from the right-hand side & it starts from 0.

The output voltage range is calculated by using the formula,

Output voltage range = Maximum DAC output voltage - Minimum DAC output voltage

Output voltage range = 9.5 V - 3.5 V

Output voltage range = 6 V

Conclusion: Option 4 is correct.

Given,

The increment in the output voltage on changing just one bit is 0.625 V.

Hence, the resolution is 0.625 V.

Now the decimal equivalent of 1111 is 15.

Thus, the output voltage is 15 × 0.625 = 9.375 V.

Digital to Analogue Converters (Weighted resistor):

DAC’s convert binary or non-binary numbers and codes into analog ones with its output voltage (or current) being proportional to the value of its digital input number.

In an n-bit weighted resistor DAC, there are n-resistors

The resistances are: 20R, 21R, 22R ……….., 2n-1R

Where,

R is the resistance used

20R is the resistor value of the corresponding MSB 

2n-1R is the resistor value of the corresponding LSB 

n is the number of bits

Calculation:

Given,

The resistor value corresponding to LSB is 40 kΩ

2n-1 R = LSB 

For a 5 bit comparator, n = 5

2^5-1 R = 40 kΩ 

R = 2.5 kΩ (where R is the MSB of the weighted resistor D/A converter)

Hence, the resistor value corresponding to MSB will be 2.5 kΩ

In a n-bit weighted resistor DAC, there are n-resistors

The resistances are: 2⁰R, 2¹R, 2²R ……….., 2^n-1R

In a 8-bit DAC,

Smallest resistance = R = 500 Ω

Largest resistance = 2^n-1 R = 2⁷ (500) = 64 kΩ

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:

The resolution of an n-bit DAC with a range of output voltage from 0 to V is given by:

Calculation:

Number of bits (n) = 8

Resolution