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Two 10-bit ADCs, one of successive approximation type and other of single slope integrating type, take Ta and Tb time respectively to convert 3V analog input signal to digital output. If the input analog signal is increased to 6V, the approximate time taken by the two ADCs will respectively be
  • a)
    2Ta, 2Tb
  • b)
    Ta, Tb
  • c)
    Ta, 2Tb
  • d)
    2Ta, Tb
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Two 10-bit ADCs, one of successive approximation type and other of sin...
  • For the successive approximation ADC, the conversion time is proportional to the input voltage range, so for a 6V input signal, the conversion time will be 2 times Ta. Therefore, the approximate time taken by the successive approximation ADC for a 6V input signal will be 2Ta.
  • For the single slope integrating ADC, the conversion time is also proportional to the input voltage range, so for a 6V input signal, the conversion time will be 2 times Tb. Therefore, the approximate time taken by the single slope integrating ADC for a 6V input signal will be Tb.
  • Therefore, the approximate time taken by the two ADCs for a 6V input signal will be 2Ta for the successive approximation ADC and Tb for the single slope integrating ADC.
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Community Answer
Two 10-bit ADCs, one of successive approximation type and other of sin...
Explanation:
To understand why the approximate time taken by the two ADCs will be 2Ta and Tb respectively when the input analog signal is increased to 6V, let's first understand how each ADC works.

Successive Approximation ADC:
1. The successive approximation ADC works by comparing the input analog signal with a reference voltage and iteratively approximating the digital output.
2. It starts by assuming the most significant bit (MSB) of the digital output as 1 and compares the result with the input signal.
3. If the input signal is greater than the result, the MSB is kept as 1. Otherwise, it is set to 0.
4. The process is repeated for the remaining bits, considering each bit's weight and adjusting the approximation.
5. The conversion time of a successive approximation ADC is determined by the number of bits and the clock frequency used for comparison.

Single Slope Integrating ADC:
1. The single slope integrating ADC works by integrating the input analog signal for a fixed period of time and then comparing the integrated voltage with a reference voltage.
2. The input signal is integrated using a capacitor during the integration period.
3. After the integration period, the voltage across the capacitor is compared with the reference voltage using a comparator.
4. If the integrated voltage is greater than the reference voltage, the digital output is set to 1. Otherwise, it is set to 0.
5. The conversion time of a single slope integrating ADC is determined by the integration period and the clock frequency.

Conversion Time Calculation:
1. Let's assume Ta is the conversion time for the successive approximation ADC and Tb is the conversion time for the single slope integrating ADC when the input analog signal is 3V.
2. When the input analog signal is increased to 6V, the successive approximation ADC will take the same amount of time to convert the signal as it works on a bit-by-bit approximation.
3. However, the single slope integrating ADC will take twice the time to convert the signal because the integration period needs to be doubled to accommodate the increased voltage range.
4. Therefore, the approximate time taken by the two ADCs will be 2Ta and Tb respectively.

Conclusion:
When the input analog signal is increased to 6V, the successive approximation ADC will take 2 times the original conversion time (2Ta), while the single slope integrating ADC will take the same conversion time as before (Tb).
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Two 10-bit ADCs, one of successive approximation type and other of single slope integrating type, take Ta and Tb time respectively to convert 3V analog input signal to digital output. If the input analog signal is increased to 6V, the approximate time taken by the two ADCs will respectively bea)2Ta, 2Tbb)Ta, Tbc)Ta, 2Tbd)2Ta, TbCorrect answer is option 'D'. Can you explain this answer?
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