I have a doubt, if anyone here can help me...

How can I find the minimum value of 'io' that can be achieved by only modifying 'R'?

Thanks in advance

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  • \$\begingroup\$ This questions is unclear, almost as if you are misunderstanding something fundamental. \$\endgroup\$
    – DKNguyen
    Nov 28 '20 at 22:35
  • 1
    \$\begingroup\$ Perhaps your prof means "if you size R to minimize \$i_o\$, what is the resulting value of \$i_o\$?" (That's awkwardly put, but I'm standing by it!) \$\endgroup\$
    – TimWescott
    Nov 28 '20 at 22:57
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    \$\begingroup\$ I would assign 1.25V to the inverting terminal, strip away everything but the 3 resistors and Io, and find the minimum through derivatives. \$\endgroup\$
    – DKNguyen
    Nov 28 '20 at 23:00
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    \$\begingroup\$ The inverting terminal must always have certain fixed voltage on it, so the resistor from inverting terminal to ground consumes certain fixed current. Then the output voltage defines the load current, so you need to minimize the current to the 10k load, and that is dependent on output voltage, so how do you minimize output voltage with R? \$\endgroup\$
    – Justme
    Nov 28 '20 at 23:08
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    \$\begingroup\$ Mathematically, the minimum magnitude of current is 0uA at R = -30k\$\Omega\$ \$\endgroup\$ Nov 28 '20 at 23:30

Voltage at non-inverting input is 3.75V and therefore voltage at inverting input is also 3.75V.

Current through top 30k resistor is 3.75/30k = 125uA and this will be the current through R irrespective of its value.

With R = 0 Ohms, current through 10k load is 3.75/10k = 375uA

Therefore with R = 0 Ohms, Io = 375uA + 125uA = 500uA which is the minimum value of Io.

Any increase in R above 0 Ohms increases the voltage across 10k load and therefore increases Io above its minimum value of 500uA.

  • \$\begingroup\$ thankss now I get it \$\endgroup\$
    – Ognum
    Nov 28 '20 at 23:19
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    \$\begingroup\$ @James, while the calculation is correct, I wish to point out that the original question was simply a request for help in finding the answer for homework. Now you simply gave directly a full solution that can be copy-pasted without further thought. \$\endgroup\$
    – Justme
    Nov 29 '20 at 0:52
  • \$\begingroup\$ @Justme Thanks for not down voting. \$\endgroup\$
    – James
    Nov 29 '20 at 10:36

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