Question

a) What is the resolution (minimum distinguishable input voltage) of a 8-bit ADC with VRH 5V and Suppose that there is a 12-bit AD converter with VRL = 1V and VRH-5V. Find the corresponding voltage values for the A/D conversion results of 12, 180, and 1200 b) Assume that we have a 20-bit Successive Approximation ADC, driven by a 2 KHz clock. What is the minimum conversion time of one sample with this ADC? Express your answer in seconds. c)

Answer 1

(a) Given 8-bit ADC with V_RH = 5v and V_RL = 1V Resolution = V_RH - V_RL/2^n where n is no.of bits Resolution = 5 - 1/2^8 = 15.6 mv (b) 12 - bit A/D Converts with V_RL = 1v & V_RH = 5v Count = 12 Implies input voltage = V_RL + Count times (V_RH - V_RL)/2^12 = 1 + r 2 times (4)/2^12 [input voltage = 1.011 volts] [Resolution = 5 - 1/2^12 = 0.976 mv \r\n for count = 180 input voltage = V_RL + Count times V_RH - V_RL/2^12 = 1 + 180(4)/2^12 [input voltage = 1.17 v] [Resolution = V_RH - V_RL/2^n] [Resolution = 5 - 1/2612 = 0.976 mv] for Count = 1200 input voltage = R_RL + Count (V_RH - V_RL)/2^12 = 1 + 1200(4)/2^12 [input voltage = 2.17 v] [Resolution = 0.976 mv] \r\n (c) minimum Conversion time for successive Approximation ADC = n times T_clk = 20 times 1/2 kHZ [minimum conversion time = 0.01 sec)