We are given an input current Iin=100μA 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 IL≈40μA. An additional requirement is that the Thevenin resistance as seen from the load terminals is between 60kΩ and 80kΩ. Assume first that the resistors have their nominal resistance. Come up with resistors RS and RL such that the divider ratio IL/Iin is within 10% of the requirement. Of course, the resistances you chose are just nomin

Question

We are given an input current Iin=100μA 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 IL≈40μA. An additional requirement is that the Thevenin resistance as seen from the load terminals is between 60kΩ and 80kΩ. Assume first that the resistors have their nominal resistance. Come up with resistors RS and RL such that the divider ratio IL/Iin is within 10% of the requirement.  Of course, the resistances you chose are just nomin

anonymous · Answer

I = E/R
  = (E/Rl) + (E/Rs)
  = E*[(1/Rl)+(1/Rs)]
  = 100 mA
 
(E/Rl) = 40 mA
(E/Rs) = 60 mA

E = 40Rl
  = 60Rs

Rl = (60/40)Rs
   = 1.5*Rs

Take middle of 60 Ohms and 80 Ohms => 70 Ohms

70 = (Rl*Rs)/(Rl+Rs)
   = (1.5*Rs)*(Rs) / (1.5Rs + Rs)
   = (1.5*Rs) / 2.5

Rs = 70 (2.5/1.5)
   = 116.7 Ohms

Rl = 1.5*Rs
   = 175 Ohms

Resistor values have denominations of 10  12  15  18  22  27  33  39  47  56  68  and 82 Ohms

Choose Rl = 182 Ohms
       Rs = 120 Ohms

Check (Rl*Rs)/(Rl+Rs) = (180*120) / (180+120)
                      = 72.3 Ohms