Vo is needed, but I have problem with this:
1) What is the voltage in Node 1?

2) And to check this: Voltages in other nodes are: (V2=V; V4=V5=2V; V6=Vo). Right?

 

Vo is needed, but I have problem with this:
1) What is the voltage in Node 1?

What does the ground symbol tied to that node mean?

2) And to check this: Voltages in other nodes are: (V2=V; V4=V5=2V; V6=Vo). Right?

If you don't know what the voltage is on Node 1, then how can you determine that the voltage on Node V2 is V or that the voltage on Nodes V4 and V5 are 2 V?

 

What does the ground symbol tied to that node mean?

Omg... What had I asked... :O
Sorry.
Yes, yes, I got it. See you when I'm done with the problem.

 

Omg... What had I asked... :O
Sorry.
Yes, yes, I got it. See you when I'm done with the problem.

Don't worry, we all need a good head slap from time to time.

 

Don't worry, we all need a good head slap from time to time.

That's true, haha.

By the way, I've done this problem and here is my solution. I hope you can read it.



And here is how I chose the currents:

 

I only looked at your work enough to see that you claim that the output if four times the input (taking the voltage at the right to be the input).

So the next step is to check the correctness of the answer from the answer itself. To do this you assume the answer is correct and then see if it results in all of the voltages and currents being self-consistent. The best way is to use the generic form of 4·V at the output, but almost as useful is to just pick one or two specific voltages and see what happens.

One value that is often of interest, and usually fairly easy to check, is set Vin to 0 V (I'm going to use Vin instead of V so as to avoid confusion with the unit for volts).

If Vin = 0 V, then V2 = 0 V and I1 = 0 A.

Since V5 = 2 V, I2 = 2 V / 4 R = 0.5 V/R.

Since I1 = 0 A, I2 = I3, also I5 = I3; furthermore, since I2 = I3, no current crosses the bridge between node 4 and node 5, making I4 = I5.

The end result is that Vo = - I2·8R = -4 V

But this doesn't agree with your result that, for Vin = 0 V, that Vo = 4·Vin = 0 V

 

I only looked at your work enough to see that you claim that the output if four times the input (taking the voltage at the right to be the input).

So the next step is to check the correctness of the answer from the answer itself. To do this you assume the answer is correct and then see if it results in all of the voltages and currents being self-consistent. The best way is to use the generic form of 4·V at the output, but almost as useful is to just pick one or two specific voltages and see what happens.

One value that is often of interest, and usually fairly easy to check, is set Vin to 0 V (I'm going to use Vin instead of V so as to avoid confusion with the unit for volts).

If Vin = 0 V, then V2 = 0 V and I1 = 0 A.

Since V5 = 2 V, I2 = 2 V / 4 R = 0.5 V/R.

Since I1 = 0 A, I2 = I3, also I5 = I3; furthermore, since I2 = I3, no current crosses the bridge between node 4 and node 5, making I4 = I5.

The end result is that Vo = - I2·8R = -4 V

But this doesn't agree with your result that, for Vin = 0 V, that Vo = 4·Vin = 0 V

Ahhh... Yes... You are right. I'll try to find the mistake. :/

 

Ahhh... Yes... You are right. I'll try to find the mistake. :/

Even more valuable will be to understand how I went about finding that there WAS a mistake.

 

Even more valuable will be to understand how I went about finding that there WAS a mistake.

Yes, of course. I remember you had already told me that way of checking the result, but I forgot to try.

 

Yes, of course. I remember you had already told me that way of checking the result, but I forgot to try.

It's easy to forget stuff that you are new at, so the reminders will bring you back until it starts becoming habit.

If you keep at it, eventually, you'll feel very uncomfortable when you don't check an answer.

 

Woah, look, I wrote it badly in the picture and made a big big mistake.

Right voltage IS Vin, yes. And the left voltage is 2Vin, so the left voltage depends on the right voltage.

If we make Vin=0 - also the other (left) voltage will be zero => V4 and V5 will be zero and V0 will be zero which is OK. Right?

 

Woah, look, I wrote it badly in the picture and made a big big mistake.

Valuable lessons come from our mistakes. You can now see how critically important it is to prepare your starting point carefully. If you don't, you end up spending a bunch of time chasing an answer that's guaranteed to be wrong.

The way you annotated your latest diagram with the current definitions (label, location, and direction) bodes very well going forward.

Right voltage IS Vin, yes. And the left voltage is 2Vin, so the left voltage depends on the right voltage.

I thought the left voltage was just a fixed two volts. It's a bit unusual to have two related voltages in a problem, but it's certainly not impossible to create a circuit that behaves that way.

If we make Vin=0 - also the other (left) voltage will be zero => V4 and V5 will be zero and V0 will be zero which is OK. Right?

Yes, if both source voltages are zero, then the output will be zero.

Now that you've corrected the problem, take another shot at getting Vo in terms of Vin.

 

Since I1 = 0 A, I2 = I3, also I5 = I3; furthermore, since I2 = I3, no current crosses the bridge between node 4 and node 5, making I4 = I5.

The part in red is only true if the lower opamp output sources or sinks no current. Why do you believe that to be the case?

 

The part in red is only true if the lower opamp output sources or sinks no current. Why do you believe that to be the case?

You are correct. In bouncing back and forth between browser tabs as I wrote I thought Node 3 was going to an opamp input.

 

I made another mistake. That's what I get for trying to get something out while tending the stove.

 

I made another mistake. That's what I get for trying to get something out while tending the stove.

No problem. Thanks for helping me always.
I'm sorry for taking you away so much time but my exam is tomorrow so I hope I won't be so tiresome anymore.

 

Have you been able to solve for the gain?

 

Have you been able to solve for the gain?

Yes, everything is fine.

 

What is your result for the gain?

 

What is your result for the gain?

I just wanted to say that I've passed the exam (80/100) and to thank you.
Among the others, I got a problem with an op amp and it was completely correct.

Thanks!