1. Introduction to ADC Calculation:

An Analog-to-Digital Converter (ADC) translates an analog input voltage ($V_{in}$) into a digital output ($D_{out}$). The formula for calculating the digital output for an $N$-bit ADC is:

$D_{out} = \left(\frac{V_{in}}{V_{ref}}\right) \times (2^N - 1)$

Where:

• $N$ is the number of bits (in this case, $N = 10$).
• $V_{ref}$ is the reference voltage (here, $V_{ref} = 2.56\,V$).
• $V_{in}$ is the input voltage.

2. Calculation of $D_{out}$ for $V_{in} = 0.2\,V$:

$D_{out} = \left(\frac{0.2}{2.56}\right) \times (2^{10} - 1)$

$D_{out} = \left(\frac{0.2}{2.56}\right) \times 1023 \approx 79.92 \approx 80 \text{ (rounded to the nearest integer)}$

3. Calculation of $D_{out}$ for $V_{in} = 0\,V$:

$D_{out} = \left(\frac{0}{2.56}\right) \times (2^{10} - 1) = 0$

4. Variation Between $V_{in} = 0.2\,V$ and $V_{in} = 0\,V$:

The variation is simply the difference in the digital output between the two scenarios:

$\text{Variation} = 80 - 0 = 80$

Summary:

• $D_{out}$ for $V_{in} = 0.2\,V$ is approximately 80.
• $D_{out}$ for $V_{in} = 0\,V$ is 0.
• The variation between these two digital outputs is 80.

Would you like more details or have any questions?

Here are 5 related questions to consider:

1. How does changing the reference voltage ($V_{ref}$) affect the digital output ($D_{out}$)?
2. What is the impact of increasing the bit resolution of the ADC on the accuracy?
3. How does the ADC handle input voltages exceeding the reference voltage ($V_{ref}$)?
4. What would be the digital output for $V_{in} = 1.28\,V$ with the same ADC settings?
5. What is the smallest voltage increment that can be detected by this 10-bit ADC?

Tip: When working with ADCs, it's crucial to ensure that $V_{in}$ does not exceed $V_{ref}$, as it could lead to inaccurate conversions or even damage to the ADC.