Help me calculate Vout for these op amps

[samphiebel]

How do I calculate Vout for the following circuits? I'm lost and I don't know how to start, I would like to know how to find the equation for Vout

 

[safwatonline]

op amp analysis

superposition

 

[samphiebel]

Re: op amp analysis

could you help me get started

 

[safwatonline]

op amp analysis

Ok it seems long so i will tell steps that i think will solve it but i wont solve it (too lazy for that job )
1-let negative pin of the OPAMP Vn and the positive Vp and the twwo terminals of the resistor Rg are Vrgu "uppper" and Vrgl "lower".
2- apply superposition , disable first v2 (short circuit)
3- Vp=Vn
4-find Vn in terms of Vrgu and V1 (potential divider and superposition)
5-Ix is the current passing through R1 from Vp is Vp/R1
6- Iy is the current passing through R2 from Vrgl to grnd is Vrgl/R2
7-Vrgl = Ix(R1+R2)
8- Vrgu= Vrgl+(Ix+Iy)Ry
9- now u can get Vrgu and Vp in terms of R1 and R2 and V1
10- having all voltages and currents calculate Vout
11- redo all steps again for V2 is active and V1 is disabled
12 add the two calculated Vout

 

[Euler's Identity]

Re: op amp analysis

No time right now, but use nodal analysis. W/in 10 minutes I had four equations and four unknowns.

Nodal is where you assign currents and then sub each current with an E/R.

Realize that V+=V- when the op amp isn't saturated (is in linear operation, where you need it). Doing this will reduce two unknowns to one.

I'm sorry I don't have to time to lay it all out right now, but holler if you need more assistance and when I get back I'll help you, or another here can help.

However, a quick review...

Say you have this circuit:

Vin ---- R1 ----- Vo ------ R2 ------ GND

The nodal analysis would look like:

I1 =

Vin - Vo
----------
R1

I2 =

Vo - GND
----------- =
R2

Vo
----
R2

Since it's a series circuit,

I1 = I2.

Now substitute.

Vin - Vo
---------- =
R1

Vo
----
R2

Now solve for the unknown Vo.

R2(Vin - Vo) = R1Vo

R2Vin - R2Vo = R1Vo

R2Vin = Vo(R1 + R2)

Vo =

VinR2
----------
R1 + R2

...the familiar voltage divider equation...

That's how nodal analysis works. The only thing I didn't show were the current directions (the arrows) I assigned at the start -- both were to the right. However, it matters not how you assign the curents, for as long as you assume that the voltage at the start of the arrow is greater than the voltage at the end of the arrow, then you'll get the correct solution.

Like I said, using nodal I was able to get four equations and four unknowns; therefore, given the time, I can solve for Vo, as my number of equations matches my number of unknowns.

Give a shout if you need more.

 

[Euler's Identity]

Re: op amp analysis

Oops! I left out one more piece you'll need.

Assume that the V+ and V- terminals of the op amp draw no current. What this does is it allows you to set the current through R1 (V1 side) equal to the current through R2. The same will also be true on the V2 side even though you're talking about different R1 and R2.

Note: When I drew the circuit I made the voltage at the upper node (of Rg) Vb and the one at the lower Vd. Likewise, I made V- Va and V+ Vc. Therefore, the current through the upper R2 is

Va - Vb
----------
R2

and the lower is

Vc - Vd
----------
R2

Bear in mind something else though. Since we're assumng linear operation (with all those R's, we obviously want the op amp running linear), we're saying that no current is flowing into the + and - terminals. With this the case, R1 and R2 could be lumped into single components R1 + R2. This would mean the current from V1 is

V1 - Vb
-----------
R1 + R2

and the current from V2 is

V2 - Vd
----------
R1 + R2


Finally, to clarify, although superposition does handle many op amp circuits, with this circuit I'd use nodal. Why? Because there is current through Rg. Without Rg, the circuit would be a (plain old) diff amp, and superposition would've been best. However, with Rg, the leg voltages will be effected. Therefore, nodal analysis makes more sense, at least to me that is.

 

[Euler's Identity]

Re: op amp analysis

Well I had a little time to spend on it, so here it is.

Vo =

2

(R2/R1)

(1 + R2/Rg)

(V2 - V1)

In other words, as Rg decreases, the gain goes up, towards infinity, and as Rg increases, the gain approaches twice the ratio of R2/R1. One could easily follow the circuit with a 2-1 voltage divider to cancel the 2 and have a very simple diff amp whose minimum gain is set by R2/R1 but which can be easily boosted by varying Rg.

Incidentally, when using nodal on this, I found that I needed to sub the quantity Vd-Vc (the upper and lower nodes at the R2 junctions for Rg) into a new variable, which I called, Vn. This caused two eq's and three unknowns to reduce to two and two which gave the solution.

Holler if you want more.

 

[Euler's Identity]

Re: op amp analysis

Well, it's been some time since I've been here. Yet this circuit (the first one that is) came to mind just the other day. I knew I'd derived it here back when, but I'd lost the derivation, so I had to do it again. Oddly enough I kept running into a divide by zero. Hence, I decided to come back here for a reminder.

Since I've done it again, this time I'm going to explain some more for the next researcher who happens by.

Using the circuit diagram above, I described the system this way:

Set V(inv)=V(noninv)=V, which reflects an ideal op amp in linear operation.

Define the nodes above and below Rg as Vb and Vd. (Incidently, I called Rg Rf this time around.)

Define/describe seven currents:

I1 =

V1-V
-----
R1


I2 =

V-Vb
-----
R2


I3 =

Vb-Vo
-------
R2


I4 =

V2-V
------
R1


I5 =

V-Vd
------
R2


I6 =

Vd
----
R2


If =

Vb-Vd
-------
Rf


Now use KVL to relate those currents:

I1 = I2 (eq1)
I2 = I3 + If (eq2)
I4 = I5 (eq3)
I6 = I5 + If (eq4)

Create a linear equation of three voltages from eq1:

V(R1+R2) = R2V1 + R1Vb

Create a linear equation of three voltages from eq3:

V(R1+R2) = R2V2 + R1Vd

Notice that these two equations can be combined to lose V(R1+R2), and then express the result as differential voltages:

R1(Vb-Vd) = R2(V2-V1) (eq I)

Next create a linear equation from eq2, but leave in terms of differential voltages:

Rf(V-Vb) = Rf(Vb-Vo) + R2(Vb-Vd) (eq II)

Then, to complete this thought, create a linear equation from eq4, again leaving it in terms of differential voltages:

RfVd = Rf(V-Vd) + R2(Vb-Vd) (eq III)


Had we set these equations up for a matrix, it wouldn't have been easily noticed that the term (Vb-Vd) occurs in all three equations. That's the key to solving this system.

Once the common term is noticed, then the common term can be assigned a new variable, Vn:

Vn = (Vb-Vd)

After this, equations I, II, and III can be rewritten in terms of the new variable:

eq I: R1Vn = R2(V2-V1)
eq II: Rf(V-Vb) = Rf(Vb - Vo) + R2Vn
eq III: RfVd = Rf(V-Vd) + R2Vn

Now expand II and III to get

eq II: RfV = 2RfVb - RfVo +R2Vn
eq III: RfV = 2RfVd - R2Vn

Then, lose the common RfV terms by subtracting III from II, which also creates the magic differential voltage term (Vb-Vd), which we'll then substitute, again, with Vn.

This gives

RfVo = 2Vn(Rf+R2) (Eq A)

Now, instead of all those unknowns, we just have two unknowns: Vn and Vo.

To get Vo, all we need to do is solve equation I for Vn, equation A for Vo, and then substitute Vn's solution into Vo's solution.

Vn =

R2(V2-V1)
------------
R1


Vo =

2(1 + R2/Rf)Vn


So we get

Vo =

2

R2
---
R1

(1 + R2/Rf)

(V2-V1)


Just remember to note that I changed Rg to Rf (I was thinking about a fed back current), and you'll have the solution from last time around.

Note: Three guesses why I decided to go through this again, here. ;-)


P.S. If anyone out there gets something from this, how about making my day and letting me know. Yes?

P.P.S. This is a really nifty circuit, because it gives you a differential amplifier and also a gain adjust to set it just so. That is difficult with the standard op amp differential amplifier, as you'd have to change two resistors, equally, not just one, or you'd lose V1-V2.

Enjoy!

Added after 16 minutes:


I was just thinking. If I had to give this circuit a name, based on Vo,

Vo =

2

R2
---
R1

(1 + R2/Rf)

(V2-V1)

I'd call it a "combo differential amp."

Notice how it includes the inverting amp gain factor (factor out neg sign from (V2-V1)) and the non-inverting amp gain factor. Hence, it's a combination of an inverting amp (A=-R2/R1), a non-inverting amp (A=1 + R2/Rf), and a differential amp with A=2.

It's what I'd call cool, nifty, neato, and right-on.