Ideal Operation amplifier

WBahn

Joined Mar 31, 2012
24,964
It is connected in the upper middle ...
Then, in your work, why do you only circle the vertical one since the horizontal one is the same node/

yes theres a voltage source between 2 non reference nodes its a supernode
Whether it is a supernode or not is beside the point. You have indicated that the voltage of the upper left node is -2 V. Why? Just because it is connected to negative side of a -2 V supply?
 

Thread Starter

Asad ahmed1

Joined Feb 10, 2016
68
i wasn't thinking of super node that time you made me think of that point ..your so much helpful I wish to be like you helping others.. thanks for your replies
 

Thread Starter

Asad ahmed1

Joined Feb 10, 2016
68
By making it a super-node would the quation be like First on the lower side current remains same in series so same current is going to the node so would the equation be like this if i take the lower opamp output as V3 and current be like

0-V1 = V1-V3 the output voltage of the amplifier would become 2V1=V3 and the current flowing towards node 2V1/5
10 10

And these two
V2- 0 + V1-0
10 10
and the supernode one V1=-2+V1
 

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WBahn

Joined Mar 31, 2012
24,964
Please check my node equation..
If I check your node equations it will be to see if they are in agreement with your diagram. There's no point in doing that until your diagram is unambiguous.

Which of the following is the intended wiring within the red circle?

Option1.jpg Option2.jpg

Option 1 on the left (or the top) or Option 2 on the right (or the bottom)?
 

WBahn

Joined Mar 31, 2012
24,964
Okay, thanks.

The next thing you need to do is make your node equations clear and tied to the diagram. I see this "v2" and "v3" running around but don't see either one of them defined in your diagram. I'm not going to waste my time guessing what you meant. If you want people to invest their efforts in checking your work, it is up to YOU to make your work easy to follow. The first step in doing that is to clearly annotate each node in your diagram (at least those that are used in any of your work) with the label your work uses for that node. Don't make people guess, engineering is not about guessing.
 

Thread Starter

Asad ahmed1

Joined Feb 10, 2016
68
now on simplifying i got 6V1 - 2 = -V0

i need the equation in V1/Vo so dividing bothsides by Vo

6V1 - 2 = -1
V0

6V1 = -1 + 2/Vo
Vo

how to write it ..
and in the question they asked for the block diagram what is the block diagam
 

WBahn

Joined Mar 31, 2012
24,964
When I said that your equation was good I was wrong (although what you intended was correct). Look at your second term. The divisor is just 10.

Also, you need to start tracking your units -- the divisors are all resistances, so they should have been 10 kΩ, etc.

Your simplified equation in the post above should be

6·V1 - 2 V = - V0

You can't get it in the form of Vo/V1 because you have a DC offset. The best you can do it get in in the form of

Vo = gain·V1 + Voffset

The 'gain' is the "incremental gain" and tells you how much the output changes by given a change in the input, while Voffset tells you what the output is when the input is 0 V.

As for the block diagram -- that depends on what they are looking for in a block diagram. Often what they want are gain blocks and summing junctions, but I have no way of knowing what your instructor wants.

Note that another way of writing the above equation is

Vo = gain·(V1 + Vinoff)

Where Vinoff = Voffset/gain and is called the input offset voltage.
 
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