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I am having some difficulty figuring out V_out. I know that it is going to be inverted because V+ = 0 and I know that V- ≠ 0. I don't really know how to go about finding the currents in the system to calculate V-.

enter image description here

Doing KCL I get stuck with 2 equations and 3 unknowns (the third is Vout)

Circuit Simulation

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    \$\begingroup\$ I'm voting to close this question as off-topic because it appears to be a homework question with no attempt at a solution included herein. \$\endgroup\$
    – Fizz
    Oct 7, 2015 at 6:32
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    \$\begingroup\$ Tip: you only have negative feedback here... so are your assumptions about voltage(s) really correct? \$\endgroup\$
    – Fizz
    Oct 7, 2015 at 6:37
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    \$\begingroup\$ You know how much current will be going through the 10ohm resistor. Now figure out how much current will be going through each of the other resistors. \$\endgroup\$ Oct 7, 2015 at 6:37
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    \$\begingroup\$ @LvW, don't make it hard. Clearly this is in like Chapter 3 of the book. \$\endgroup\$
    – Daniel
    Oct 7, 2015 at 7:41
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    \$\begingroup\$ @Daniel I don't see how either of my assumptions could be wrong. I've modeled this circuit and both my statements are true. What book? \$\endgroup\$ Oct 7, 2015 at 7:48

2 Answers 2

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I know that V- ≠ 0

Bad assumption because negative feedback is being used and negative feedback attempts to maintain the two input voltages at the the same potential. Given that the non-inverting input is 0V then it is safe (for a theoretical assessment) to assume the voltage at the inverting input terminal is also 0V.

You then have to ask your self what voltage does the output have to produce to give 0V (equilibrium) at the inverting input. Clearly the voltage at the junction of the 10, 20 and 40 ohm resistors has to be -4V (by visual inspection and simple brain math) but you can work that out by ignoring the 10 and 40 ohm resistors and imagining what voltage needs to be on the 20 ohm resistor terminal that makes 0V at the inverting input.

I'll leave the rest for you to solve because you've had more than enough hints.

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Since you've added a simulation in that (fairly terrible) web sim... If you change the min and max values for the opamp output to some sensible values (right click on the opamp to do that), you'll see that subsequently V- gets very close to 0 (some microvolts). You've set the rails of the opamp to +2 and -2V, so it's saturating, that's why the inputs are fairly far apart. Set the opamp rails/limits to something like +15 and -15V.

I have no idea if this link will show (or not) the edits I've made reflecting the above, but hopefully it does.

Here's a blow-by-blow how to solve it by hand (no system of equations needed):

  1. Calculate the current through the leftmost 10-ohm resistor since you know the potentials of both its ends (V- really is 0 because the [ideal] opamp is configured only with negative feedback).
  2. The current through the 20-ohm resistor is going to be same as the one though the leftmost 10-ohm resistor (since there's no current flowing into the [ideal] opamp pins)
  3. Calculate the voltage at the T-join of the 3 resistors using Ohm's law for the 20-ohm resistor.
  4. Calculate the current through the 40-ohm resistor since you now know the potentials of both its ends.
  5. Add the currents for the 20- and 40-ohm resistors to obtain the current through the right-most 10-ohm resistor
  6. Calculate the voltage at Vout by using Ohm's law for the right-most 10-ohm resistor; you've calculated the current through it in the previous step and you know the potential at the T-joint from the 3rd step.
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  • \$\begingroup\$ Serve it on a plate dude LOL \$\endgroup\$
    – Andy aka
    Oct 7, 2015 at 20:13
  • \$\begingroup\$ Given how confused he was about everything (did you see number of back and forth comments under the question?) and how he managed to confuse himself even more with the sim... it seemed necessary. Anyway, my experience from answering on M.SE (where homework is the bulk of traffic) is that people who post homework questions post them pretty close to the deadline... and I let enough time pass. I'd have expected him to have marked your answer as accepted if he'd gotten sufficiently enlighten by that alone... \$\endgroup\$
    – Fizz
    Oct 7, 2015 at 20:21
  • \$\begingroup\$ He'll only accept it if it comes up at his exam!!! \$\endgroup\$
    – Andy aka
    Oct 7, 2015 at 20:46
  • \$\begingroup\$ At least I didn't type a whole bunch of LaTeX in, as done in a question solving basically the same circuit symbolically: electronics.stackexchange.com/a/194480/54580 \$\endgroup\$
    – Fizz
    Oct 9, 2015 at 21:14

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