Soooo...we had a quiz last night in Electronics II and a rendition of the first problem is below:
We were asked to solve for the voltages out of the four opamps. My thinking is: U1 and U2 are voltage followers, so I just have 8V out of U1 and 5V out of U2. Since these are ideal opamps (as specified in the problem), no current flows into the inverting input of U3. So R's 4 and 5 form a simple voltage divider and I have 4V at both inputs of U3. Moving to R1, with 5V on the left and 4V on the right, I have .1 mA flowing through it. Since that cannot flow into U3, it must all flow through R6, giving me 3V at the output of U4. Since both inputs of U4 are at ground potential, I have 3V across R2 and .3 mA through it. Again, this cannot flow into U4, so it must all go through R3, giving me a voltage of -9V at the output of U3. That was how I answered the problem. I was curious, so I simulated it and got nothing anywhere near those values. Is this just because the simulation doesn't involve ideal opamps, or is it that my solution is wrong?
We were asked to solve for the voltages out of the four opamps. My thinking is: U1 and U2 are voltage followers, so I just have 8V out of U1 and 5V out of U2. Since these are ideal opamps (as specified in the problem), no current flows into the inverting input of U3. So R's 4 and 5 form a simple voltage divider and I have 4V at both inputs of U3. Moving to R1, with 5V on the left and 4V on the right, I have .1 mA flowing through it. Since that cannot flow into U3, it must all flow through R6, giving me 3V at the output of U4. Since both inputs of U4 are at ground potential, I have 3V across R2 and .3 mA through it. Again, this cannot flow into U4, so it must all go through R3, giving me a voltage of -9V at the output of U3. That was how I answered the problem. I was curious, so I simulated it and got nothing anywhere near those values. Is this just because the simulation doesn't involve ideal opamps, or is it that my solution is wrong?