An Analog-to-Digital Converter (A/D converter or ADC) is an electronic circuit that converts a continuous analog signal into a discrete digital signal. The time it takes for an ADC to perform this conversion is known as the conversion time. This time is a crucial parameter, as it determines how fast the ADC can sample the input signal. The conversion time depends on the ADC architecture and the clock frequency provided.
The question provides the following information:
We need to find the maximum conversion time required.
First, let's calculate the period of the input clock signal. The clock period ($T_{clk}$) is the inverse of the clock frequency ($f_{clk}$):
Given $f_{clk} = 1 \text{ MHz} = 1 \times 10^6 \text{ Hz}$, the clock period is:
The conversion time of an A/D converter varies significantly based on its architecture (e.g., Successive Approximation, Integrating, Flash, Sigma-Delta). The number of clock cycles required for a conversion depends on this architecture and the desired resolution (number of bits, N).
For common ADC types:
Given the options are in milliseconds (2 ms, 4 ms, 6 ms, 8 ms), a SAR ADC with a conversion time of 12-14 µs doesn't fit. The values suggest a type of converter whose conversion time is proportional to a power of 2 related to the number of bits N, such as an integrating or counting type.
Let's examine the options in terms of clock cycles:
Comparing these clock cycle counts to powers of 2 for N=12:
We can see that the options are approximately $2^{11}$, $2^{12}$, and $2^{13}$ clock cycles. Since N=12, the option 2 ms corresponds closely to $2^{11}$ clock cycles (2048 cycles), the option 4 ms corresponds closely to $2^{12}$ clock cycles (4096 cycles), and the option 8 ms corresponds closely to $2^{13}$ clock cycles (8192 cycles).
Based on the options provided and the given correct answer, it appears the intended calculation method for the maximum conversion time is approximately $2^{N-1}$ clock cycles for this specific 12-bit A/D converter.
Using this assumption:
Maximum Conversion Time $\approx 2^{N-1} \times T_{clk}$
Substituting N = 12 and $T_{clk} = 1 \text{ µs}$:
This calculated value of 2.048 ms is very close to the option 2 ms.
If we were to consider other possibilities:
The calculation based on $2^{N-1}$ clock cycles most closely matches one of the given options (2 ms), suggesting this was the intended model for this question.
Therefore, for a 12-bit A/D converter with a 1 MHz input clock, the maximum conversion time is nearly 2 ms, assuming an architecture where this calculation applies.
| Parameter | Value |
|---|---|
| Resolution (N) | 12 bits |
| Clock Frequency ($f_{clk}$) | 1 MHz ($1 \times 10^6$ Hz) |
| Clock Period ($T_{clk}$) | 1 µs ($1 \times 10^{-6}$ s) |
| Assumed Cycles (approx) | $2^{N-1} = 2^{11} = 2048$ cycles |
| Calculated Max Conversion Time | $2048 \times 1 \text{ µs} = 2048 \text{ µs} = 2.048 \text{ ms}$ |
| Closest Option | 2 ms |
| Term | Description |
|---|---|
| A/D Converter (ADC) | Converts analog voltage or current into a digital representation. |
| Resolution (N) | The number of bits in the digital output, determining the number of discrete levels ($2^N$). A 12-bit ADC provides $2^{12} = 4096$ levels. |
| Clock Frequency | The speed at which the internal operations of the ADC are timed. A higher clock frequency generally allows for faster conversions (lower conversion time). |
| Conversion Time | The time taken by the ADC to convert one analog sample into a digital code. Lower conversion time means a higher sampling rate is possible. |
The conversion time of an A/D converter is primarily influenced by its architecture and the clock frequency. Different architectures have vastly different speed characteristics:
In this problem, the relationship between the conversion time (in ms) and the number of bits (N=12) when clocked at 1 MHz strongly suggests an integrating or counting type of ADC where the conversion time is proportional to a power of 2 of N, likely $2^{N-1}$, $2^N$, or $2^{N+1}$. The calculated value using $2^{N-1}$ cycles closely matches the option 2 ms.
When once a pocket of smoke, containing air pollutants, is released into the atmosphere from a source like an automobile or a factory chimney, it gets dispersed into the atmosphere into various directions depending upon the
1. prevailing winds
2. temperature
3. pressure conditions
Select the correct answer.
During the compaction test, the weight of compacted soil specimen along with mould is 38.2 N. The volume and weight of mould are 0.95×10-3 m³ and 20.5 N respectively and the water content is 12%. The dry unit weight of the compacted specimen will be nearly