Question

1) The resistances of commercially-available discrete resistors are restricted to particular sets. For example, the available values of resistors with 10% tolerance are selections from the E12 set multiplied by a power of ten from 100 through 10s. The E12 set is: E12-10,12, 15, 18,22,27,33,39,47,56,68, 82) Thus, you can buy 10% resistors with a nominal resistance of 3300 or 33M. but not 350n. Furthermore, the tolerance a 10% 390 resistor you can be sure that its resistance is between 351 and 429 means that if you buy In this problem we need to choose 10% resistors to make a current divider that meets a given specification. We are given an input current 1-100μΑ and a linear resistive load represented by the resistor RL-Our resistive load is not regulated for the high current provided by our current source, so we add a resistor RS in parallel to divide the current such that! 0μΑ An additonal requirement is that the Thevenin resistance as seen from the load terminals (with both Rs and R1 connected) is between 60M2 and 80M2. Assume first that the resistors have their nominal resistance. Come up with resistors RS and t such that the divider ratio h n ..s it in 10% of the requirement, where l4arn is the load current using the nominal resistance values. Of course, the resistances you chose are just nominal. Given that they are only guaranteed to have resistances within 10% of the nominal value. what is the largest and smallest value that It may have?

Rs (in Ohms):

RL (in Ohms):

ILmax (in Amps):

ILmin (in Amps):

Answer 1

I_m = 100 mu A, I_c tildeequal 40 mu A R_th = should be between 60 k ohm and 80 k ohm By current division Rule I_c = I_n times R_S/R_S + R_L 40 = 100 times R_S/R_S + R_L R_S + R_L = 2.5 R_S 1.5 R_S = R_L - (1) resistance R_th = R_S times R_L/R_S + R_L Given R_th should be between 80k to 80 k ohm let R_th = 70 k ohm 70 = R_S times R_L/R_S + R_L 70 R_S + 70 R_L = R_s R_L - (2) Put value of R_L from equation (1) to (2) 70 R_S + 70(1.5) R_S = R_S times 1.5 R_S 70 RS + 105 R_S = 1.5 R_S^2 1.5 R_S^2 = 175 R_S R_S = 175/1.5 = 116 ohm \r\n So, R_L = 1.5 R_S = 1.5 times 116.6 R_L almostequal 175 ohm So, from set if we choose R_S = 120 ohm then R_L = 180 ohm Checking R_th
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