Question

An 8-bit counter type ADC has worst case conversion time equal to 10 times of that of an 8-bit successive approximation type ADC, if AND gate of counter type ADC is clocked with a $5MHz$ clock signal. This conversation time is used for an analog input of $2V$ of SAC type ADC. Then conversation time for a $4V$ input will be

• A. $17.2\mu sec$
• B. $3.6\mu sec$
• C. $1.8\mu sec$
• D. $5.1\mu sec$

Answer 1

The correct option is D $5.1\mu sec$

$Giventhat10\times \left(t_{c}ofSACtypeADC\right)=t_{c}ofcountertypeADCsince,{}^{\prime }{t}_c^{\prime }ofcountertypeADC=(2^n-1)T_{CLK}\Rightarrow t_{c}ofcountertypeADC=255\times \frac{1}{5}\times {10}^{-6}=51\mu secso,{}^{\prime }{t}_c^{\prime }ofSACtypeADC=5.1\mu sec$