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Introduction & Overview
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Quick Overview
Standard
In this section, students learn about the significance of the R-2R Ladder DAC in digital-to-analog conversion, focusing on key calculations such as Least Significant Bit (LSB) voltage, expected analog output voltages for various digital inputs, and practical examples that illustrate the concept clearly.
Detailed
R-2R Ladder DAC Calculations
The R-2R Ladder DAC is critically important in digital-to-analog conversion. This section delves into the essential calculations necessary for understanding and implementing this type of DAC, a widely used architecture in mixed-signal systems.
Key Calculations
- Resolution: The resolution of a DAC refers to the smallest change in output voltage that corresponds to a one-bit change at the input. It is determined by the formula:
$$V_{LSB} = \frac{V_{FS}}{2^N}$$
where:
- \(V_{FS}\) is the full-scale output voltage (maximum analog output)
- \(N\) is the number of bits.
For example, if \(V_{FS} = 5V\) and the R-2R DAC is 3-bits, the calculation would result in:
$$V_{LSB} = \frac{5V}{2^3} = 0.625V$$
- Expected Analog Output Voltage: The output voltage for given digital input combinations can be computed using:
$$V_{out} = V_{REF} \times \left( \frac{D_{N-1}}{2} + \frac{D_{N-2}}{4} + \cdots + \frac{D_0}{2^N} \right)$$
Let’s consider a 3-bit R-2R DAC with the binary input represented as \(D_2, D_1, D_0\) corresponding to MSB to LSB.
*## Example Calculation - Digital Input
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LSB Voltage (Resolution)
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V_LSB=V_REF/2N = [Your Calculation] V
Detailed Explanation
The term LSB stands for Least Significant Bit, which refers to the smallest change in output that can be represented by the DAC. The resolution of a DAC defines how finely we can change its output voltage with a change in the digital input. It can be calculated using the formula: LSB Voltage = V_REF (Reference Voltage) divided by 2 raised to the power of N (number of bits). For instance, if V_REF is 5V and N is 3, the LSB would be calculated as 5V / 2^3 = 5V / 8 = 0.625V. This means that the smallest output voltage change corresponds to 0.625V. Understanding this concept helps in realizing how precise the DAC can be in terms of output voltage level.
Examples & Analogies
Imagine you are using a dimmer switch to control the brightness of a light bulb. Each step you turn the dimmer knob can be thought of as an LSB. If the dimmer has a fine resolution, you can turn the knob slightly to create a small change in brightness. If it has a coarse resolution, you can only make big jumps between brightness levels, leading to a less smooth adjustment.
Expected Analog Output Voltage Calculation
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V_out=V_REFtimesleft(fracD_N−12+fracD_N−24+cdots+fracD_02Nright) (Show one example calculation for a specific digital input, e.g., '101')
Detailed Explanation
To calculate the expected analog output voltage of the R-2R ladder DAC based on a given digital input, we use the formula: V_out = V_REF multiplied by the sum of the fractions that correspond to the binary digits of the input. Each bit D_i contributes to the output based on its position in the binary number. For example, if we consider a 3-bit input '101', which equals 5 in decimal, we can plug in the values: V_out = V_REF times (D_2/2 + D_1/4 + D_0/8). This results in V_out = V_REF times (1/2 + 0/4 + 1/8) = V_REF times (0.5 + 0 + 0.125) = V_REF times 0.625. If we take V_REF as 5V, then V_out = 5V * 0.625 = 3.125V. This process illustrates how digital input values convert to corresponding analog voltages through the R-2R ladder network.
Examples & Analogies
Think of it like a pizza being sliced into different portions. Each slice represents a binary bit in the 3-bit input. If we 'order' the pizza in a way that the biggest slice gets the most cheese (the most significant bit), the second slice gets slightly less, and the tiniest slice has just a bit of cheese. The total amount of cheese you get on your plate corresponds to the voltage output based on how many 'slices' you have ordered.
Quiz Questions
Choose the best answer for each multiple-choice question or indicate True/False.
-
If a 6-bit R-2R Ladder DAC has a $V_{REF}$ of 3.3V, what is its approximate resolution ($V_{LSB}$)?
a) 0.051V
b) 0.103V
c) 0.206V
d) 0.025V -
For a 4-bit R-2R Ladder DAC with $V_{REF} = 8V$, which digital input code would produce an output voltage of 6V?
a) "1010"
b) "1100"
c) "0110"
d) "1000" - True or False: In an R-2R Ladder DAC, the current contribution from the Most Significant Bit (MSB, $D_{N-1}$) is always the smallest.
-
A 3-bit R-2R DAC with $V_{REF} = 5V$ produces an output of 1.25V. What was the digital input code?
a) "001"
b) "010"
c) "100"
d) "011"
Solutions (Do not look until you've completed the practice questions!)
Exercise Solutions
Easy:
- Resolution refers to the smallest change in analog output voltage that corresponds to a 1-bit change in the digital input. It signifies the smallest step size a DAC can produce.
- The resolution of a DAC is determined by the full-scale output voltage ($V_{FS}$) and the number of input bits (N).
Medium:
-
Given: N = 4 bits, $V_{FS} = 10V$.
a) Calculate LSB Voltage:
$V_{LSB} = \frac{V_{FS}}{2^N} = \frac{10V}{2^4} = \frac{10V}{16} = \textbf{0.625V}$ b) This LSB voltage signifies that the DAC can change its analog output voltage in increments of 0.625V. For example, changing the digital input from "0000" to "0001" will increase the output by 0.625V. -
Given: N = 3 bits, $V_{REF} = 5V$. Digital Input = "110" ($D_2=1, D_1=1, D_0=0$).
Calculate Expected Analog Output Voltage:
$V_{out} = V_{REF} \times \left( \frac{D_2}{2} + \frac{D_1}{4} + \frac{D_0}{8} \right)$
$V_{out} = 5V \times \left( \frac{1}{2} + \frac{1}{4} + \frac{0}{8} \right)$
$V_{out} = 5V \times \left( 0.5 + 0.25 + 0 \right)$
$V_{out} = 5V \times (0.75)$
$V_{out} = \textbf{3.75V}$
Hard:
-
Given: Desired Resolution $\le 0.1V$, $V_{FS} = 8V$.
a) Minimum Number of Bits (N):
We need $V_{LSB} \le 0.1V$.
$\frac{V_{FS}}{2^N} \le 0.1V$
$\frac{8V}{2^N} \le 0.1V$
$2^N \ge \frac{8}{0.1}$
$2^N \ge 80$
To find N, we can test powers of 2:
$2^6 = 64$
$2^7 = 128$
Since $2^N$ must be at least 80, the minimum number of bits required is $\textbf{N = 7 bits}$. b) Exact LSB Voltage for N=7:
$V_{LSB} = \frac{V_{FS}}{2^N} = \frac{8V}{2^7} = \frac{8V}{128} = \textbf{0.0625V}$ c) Analog Output Voltage for "1010" (assuming 4-bit context and $V_{REF}=V_{FS}=8V$):
Note: The question asks to use "1010" as a 4-bit representation, even though 7 bits were calculated for the minimum resolution. This implies calculating as if it were a 4-bit DAC with the given $V_{REF}$ for this part.
Digital Input = "1010" ($D_3=1, D_2=0, D_1=1, D_0=0$)
$V_{out} = V_{REF} \times \left( \frac{D_3}{2} + \frac{D_2}{4} + \frac{D_1}{8} + \frac{D_0}{16} \right)$
$V_{out} = 8V \times \left( \frac{1}{2} + \frac{0}{4} + \frac{1}{8} + \frac{0}{16} \right)$
$V_{out} = 8V \times \left( 0.5 + 0 + 0.125 + 0 \right)$
$V_{out} = 8V \times (0.625)$
$V_{out} = \textbf{5V}$
Quiz Answers
-
a) 0.051V
- $V_{LSB} = \frac{3.3V}{2^6} = \frac{3.3V}{64} \approx 0.05156V$
-
b) "1100"
- Calculate LSB: $V_{LSB} = \frac{8V}{2^4} = \frac{8V}{16} = 0.5V$.
- For 6V, the number of LSB steps is $6V / 0.5V = 12$ steps.
- Convert 12 to binary (4-bit): "1100" ($1 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 0 \times 2^0 = 8 + 4 + 0 + 0 = 12$).
-
False.
- The MSB ($D_{N-1}$) contributes the largest current (or voltage weight) as it corresponds to $V_{REF}/2$, whereas the LSB ($D_0$) contributes the smallest, $V_{REF}/2^N$.
-
b) "010"
- For 3-bit, $V_{LSB} = \frac{5V}{2^3} = \frac{5V}{8} = 0.625V$.
- Output 1.25V corresponds to $1.25V / 0.625V = 2$ LSB steps.
- Decimal 2 in 3-bit binary is "010".