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 nominal value, what is the largest and smallest value that IL may have?

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 nominal value, what is the largest and smallest value that IL may have?

anonymous · Answer

Please help!

kenljw · Answer

Iin RS/(RS + RL) = IL   Current Divider Rule
RS (1 - 2/5) = RL 2/5  : RS = RL 2/3 

60K < RS||RL = (2/3) RL||RL = RL [(2/3)||1] = RL 2/5 < 80K   Effective Load

You should be able to solve from here