Ideal 6-bit DAC: Understanding its Function

A Digital-to-Analog Converter (DAC) is an electronic device responsible for converting a digital binary code into an analog signal, which is typically a voltage or current. When we refer to an ideal DAC with zero offset, it means that its analog output voltage is directly proportional to the numerical value of the digital input. There is no initial voltage present when the digital input is zero.

The smallest change in the output voltage corresponds to a change of one Least Significant Bit (LSB) in the digital input. This smallest voltage change is often called the step size or resolution of the DAC. For an ideal DAC, the output voltage \(V_{out}\) can be calculated using the formula:

\[V_{out} = V_{LSB} \times D\]

Where:

• \(V_{LSB}\) is the voltage corresponding to one LSB (the step size).
• \(D\) is the decimal equivalent of the digital input code.

DAC Output Voltage Calculation Method

We are given an ideal 6-bit DAC. The problem provides that for an input of ‘000010’, the output voltage is 0.1 V. Our goal is to determine the output for a different input, ‘001010’.

Input '000010': Converting to Decimal

First, we need to convert the given 6-bit binary input ‘000010’ into its decimal equivalent to understand its numerical value that the DAC processes:

• \(000010_2 = (0 \times 2^5) + (0 \times 2^4) + (0 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (0 \times 2^0)\)
• \( = 0 + 0 + 0 + 0 + 2 + 0 \)
• \( = 2_{10}\)

So, a decimal input value of 2 produces an output voltage of 0.1 V.

Resolution: Calculating DAC Step Size (\(V_{LSB}\))

Since the output voltage is directly proportional to the decimal value of the input, we can use the given information to find the DAC's step size, \(V_{LSB}\). This value represents how much voltage each decimal unit (or LSB) contributes to the output.

Using the formula \(V_{out} = V_{LSB} \times D\):

• \(0.1 \text{ V} = V_{LSB} \times 2\)
• To find \(V_{LSB}\), we rearrange the equation: \(V_{LSB} = \frac{0.1 \text{ V}}{2}\)
• \(V_{LSB} = 0.05 \text{ V/bit}\)

This means that for every increment of 1 in the decimal equivalent of the digital input, the output voltage increases by 0.05 V.

Input '001010': Converting to Decimal

Next, we convert the second digital input, ‘001010’, to its decimal equivalent:

• \(001010_2 = (0 \times 2^5) + (0 \times 2^4) + (1 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (0 \times 2^0)\)
• \( = 0 + 0 + 8 + 0 + 2 + 0 \)
• \( = 10_{10}\)

Now we know that the second input corresponds to a decimal value of 10.

Output Voltage: Final Calculation

Finally, we use the calculated step size (\(V_{LSB} = 0.05 \text{ V/bit}\)) and the decimal equivalent of the new input (\(D_{new} = 10\)) to find the corresponding output voltage:

• \(V_{out} = V_{LSB} \times D_{new}\)
• \(V_{out} = 0.05 \text{ V/bit} \times 10\)
• \(V_{out} = 0.5 \text{ V}\)

Therefore, for the input ‘001010’, the output voltage of the ideal 6-bit DAC will be 0.5 V.

Calculation Summary Table

The output for the input ‘001010’ is 0.5 V.