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I have the following circuit:

enter image description here

I have to solve for VO. I know that the input voltage to the left op amp is -1 at both terminals and -2 for the right op amp. I also know that the input current to both is 0.

However, I don't know where to start to find VO. Can someone point me in the right direction?

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    \$\begingroup\$ Please label the components and nets in your schematic. That way we can talk about "R1" and "U2" instead of "the 10 kOhm resistor" and "the second op-amp". \$\endgroup\$ – The Photon Jan 27 '14 at 21:36
  • \$\begingroup\$ This question is identical to another asked very recently: electronics.stackexchange.com/questions/97679/… \$\endgroup\$ – Spehro Pefhany Jan 27 '14 at 21:37
  • \$\begingroup\$ @SpehroPefhany, I'd call that a duplicate. \$\endgroup\$ – The Photon Jan 27 '14 at 21:38
  • \$\begingroup\$ Just about all of your assumed premises are wrong. Neither the input voltages nor the currents are as you say. There is too much to unravel when someone comes here with such large misconceptions. Go back and study inverting opamp configurations again, which is probably the point of the exercise anyway. \$\endgroup\$ – Olin Lathrop Jan 27 '14 at 21:39
  • \$\begingroup\$ @ThePhoton Flagged. \$\endgroup\$ – Spehro Pefhany Jan 27 '14 at 21:40
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First, the input voltages to the op-amps positive terminals are not -1 and -2 volts, they're +1 and +2 volts.

In general, the process of working these is pretty straightforward. Start with what you know, solve for what you don't. You know that an op-amp with negative feedback (a path from the output to the negative input) tries to make its inputs equal. (In the real world there would be concerns about the voltage rails of the op amp, but this is obviously a textbook problem so we won't worry about such.) The positive input voltages are given to you, fixed by the voltage sources. From this fact, we also know the negative input voltages.

So that means we know the voltage on both sides of the 10k resistor: 1V and 0V. There's 1V across this resistor, and it's 10k, so the current through it is 100 uA, flowing left. We have current leaving a circuit node. It has to also enter that node somewhere, and there are only two possible paths: through the 20k resistor, or through the input of the op-amp.

Another assumption we make about op-amps is that current can't flow into or out of their input terminals. In the real world, that's not true, but for simple analysis of DC circuits with resistances in this range it's probably close enough.

So we know there's 100 uA going through the 20k as well, flowing left. We know the left side of the 20k is at 1V, and it's got 100 uA x 20k = 2V drop across it, so we get a voltage of 3V on the right side of the 20k.

Continue the process. The left side of the 30k resistor has 3V on it, and the right side has 2V. That means there's one volt across it, giving 33 uA through it, flowing right.

That 33 uA can't flow into the input of the op amp, so it has to go through the 40k. 33 uA through a 40k resistor gives a voltage drop of 1.33 volts. The left side of the 40k resistor is at 2V, and the right side is 1.33V lower, giving .66 volts.

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I also know that the input current to both is 0.

The input current isn't zero (if you mean the current through the 10k resistor) - you must ask yourself what the voltage is at the left op-amp's inverting input. Here's a clue - assume the op-amp has infinite gain and needs virtually zero volts between inverting and non-inverting input - ask yourself - what voltage is on the inverting input given there is 1 volt on the non-inverting input.

When you have that answer, you can work out the current through the 10kohm resistor - one side is X (which by now you should realize what it is because I've given you good clues) and the other side is ground/0V.

Where does this current flow from? That's the next question. Here are some choices: -

  • From the inverting input
  • From the stars above
  • From the op-amps's output
  • It leaks across from the non-inverting input

Only one answer above is correct and reasonable.

When you have the right answer it should then be a simple matter to work out what the output voltage of the left op-amp is.

PS just in case you get stuck - the output from the left op-amp is 3V.

The right hand op-amp is a little trickier because it has 3V feeding it from the left op-amp and 2V on the non-inverting input.

Hope this helps.

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  • \$\begingroup\$ Ok so the voltage on the first inverting input should be -1 as well, correct? So the current through the 10k resistor is -0.0001A? \$\endgroup\$ – David Jan 27 '14 at 21:36
  • \$\begingroup\$ Do you really need me to answer this now? \$\endgroup\$ – Andy aka Jan 27 '14 at 21:51

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