This document discusses analyzing an op-amp circuit to determine the output voltage. It presents the circuit and walks through applying Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL) along with device equations to set up 12 equations with 12 unknowns. It solves the equations step-by-step to arrive at the "right equation" of vout = 14/3vin - 2vin. Alternatively, the document shows the circuit can be analyzed using superposition to obtain the same output voltage equation.

1. 2/21/2011 Example An op amp circuit analysis lecture 1/23
Jim Stiles The Univ. of Kansas Dept. of EECS
Example: An op-amp
circuit analysis
Let’s determine the output voltage vout (t) of the circuit below:
R2 =3KR1 = 1K
+
-
ideal
R3 =1K
vout (t)
vin (t)
I=2 mA

2. 2/21/2011 Example An op amp circuit analysis lecture 2/23
Jim Stiles The Univ. of Kansas Dept. of EECS
Without this step, your answer (and
thus your grade) mean nothing
The first step in EVERY circuit analysis problem is to label all currents and voltages:
1
i
3
i
2
i
i−
i+
1
v+ − 2
v+ −
3
v
+
−
R2 =3KR1 = 1K
+
-
ideal
R3 =1K
vout
vin
I=2 mA
v+
v−

3. 2/21/2011 Example An op amp circuit analysis lecture 3/23
Jim Stiles The Univ. of Kansas Dept. of EECS
The search for a template…
Q: I looked and looked at the notes, and I even looked at the book, but I can’t
seem to find the right equation for this configuration!
A: That’s because the “right equation” for this circuit does not exist—at least
yet.
It’s up to you to use your knowledge and your skills to determine the “right
equation” for the output voltage out
v !

4. 2/21/2011 Example An op amp circuit analysis lecture 4/23
Jim Stiles The Univ. of Kansas Dept. of EECS
You have the tools to determine this
yourself—no need to find a template!
Q: OK, let’s see; the output voltage is:
out
v = ????
I’m stuck. Just how do I determine the output voltage?
A: Open up your circuit analysis tool box. Note it consists of three tools and
three tools only:
Tool 1: KCL
Tool 2: KVL
Tool 3: Device equations (e.g., Ohm’s Law and the virtual short).
Let’s use these tools to determine the “right” equation!
First, let’s apply KCL (I’m quite partial to KCL).

5. 2/21/2011 Example An op amp circuit analysis lecture 5/23
Jim Stiles The Univ. of Kansas Dept. of EECS
The first KCL
Note there are two nodes in this circuit. The KCL for the first node is:
1 2
i i i I−
= + +
Note the potential of this node (with respect to ground) is that of the inverting
op-amp terminal (i.e., v−
).
v−
v+
1
i
3
i
2
i
i−
i+
1
v+ − 2
v+ −
3
v
+
−
R2 =3KR1 = 1K
+
-
ideal
R3 =1K
vout
vin
I=2 mA

6. 2/21/2011 Example An op amp circuit analysis lecture 6/23
Jim Stiles The Univ. of Kansas Dept. of EECS
The second KCL
The KCL of the second node is:
3
I i i+
= +
Note the potential of this node (with respect to ground) is that of the non-
inverting op-amp terminal (i.e., v+
).
v−
v+
1
i
3
i
2
i
i−
i+
1
v+ − 2
v+ −
3
v
+
−
R2 =3KR1 = 1K
+
-
ideal
R3 =1K
vout
vin
I=2 mA

7. 2/21/2011 Example An op amp circuit analysis lecture 7/23
Jim Stiles The Univ. of Kansas Dept. of EECS
The first KVL
Now for our second tool—KVL.
We can conclude:
1 1in in
v v v v v v− −
− = ⇒ = −
1
i
3
i
2
i
i−
i+
1
v+ − 2
v+ −
3
v
+
−
R2 =3KR1 = 1K
+
-
ideal
R3 =1K
vout
vin
I=2 mA
v+
v−

8. 2/21/2011 Example An op amp circuit analysis lecture 8/23
Jim Stiles The Univ. of Kansas Dept. of EECS
The second KVL
And also:
2 2out out
v v v v v v− −
− = ⇒ = −
1
i
3
i
2
i
i−
i+
1
v+ − 2
v+ −
3
v
+
−
R2 =3KR1 = 1K
+
-
ideal
R3 =1K
vout
vin
I=2 mA
v+
v−

9. 2/21/2011 Example An op amp circuit analysis lecture 9/23
Jim Stiles The Univ. of Kansas Dept. of EECS
The third KVL
And likewise:
3 3
0v v v v+ +
− = ⇒ =
1
i
3
i
2
i
i−
i+
1
v+ − 2
v+ −
3
v
+
−
R2 =3KR1 = 1K
+
-
ideal
R3 =1K
vout
vin
I=2 mA
v+
v−

10. 2/21/2011 Example An op amp circuit analysis lecture 10/23
Jim Stiles The Univ. of Kansas Dept. of EECS
There are seven device equations
Finally, we add in the device equations.
Note in this circuit there are three resistors, a current source, and an op-amp
From Ohm’s Law we know:
1
1
1
v
i
R
= 2
2
2
v
i
R
= 3
3
3
v
i
R
=
And from the current source:
2I =
And from the op-amp, three equations!
0i−
= 0i+
= v v− +
=

11. 2/21/2011 Example An op amp circuit analysis lecture 11/23
Jim Stiles The Univ. of Kansas Dept. of EECS
12 equations and 12 unknowns!
Q: Yikes! Two KCL equations, three KVL equations, and seven device
equations—together we have twelve equations. Do we really need all these?
A: Absolutely! These 12 equations completely describe the circuit. There are
each independent; without any one of them, we could not determine out
v !
To prove this, just count up the number of variables in these equations:
We have six currents:
1 2 3
, , , , ,i i i i i I+ −
And six voltages:
1 2 3
, , , , , out
v v v v v v+ −
Together we have 12 unknowns—which works out well, since we have 12
equations!
Thus, the only task remaining is to solve this algebra problem!

12. 2/21/2011 Example An op amp circuit analysis lecture 12/23
Jim Stiles The Univ. of Kansas Dept. of EECS
Don’t ask the calculator
to figure this out!
Q: OK, here’s where I take out my trusty programmable calculator,
type in the equations, and let it tell me the answer!
A: Nope. I will not be at all impressed with such results (and your grade will
reflect this!).
Instead, put together the equations in a way that makes complete physical
sense—just one step at a time.

13. 2/21/2011 Example An op amp circuit analysis lecture 13/23
Jim Stiles The Univ. of Kansas Dept. of EECS
i3 = 2 mA
First, we take the two device equations:
0i+
= and 2I =
And from the second KCL equation:
3 3 3
2 0 2I i i i i+
= + ⇒ = + ⇒ =
1
i
2
2
i
i−
0
1
v+ − 2
v+ −
3
v
+
−
R2 =3KR1 = 1K
+
-
ideal
R3 =1K
vout
vin
I=2 mA
v+
v−

14. 2/21/2011 Example An op amp circuit analysis lecture 14/23
Jim Stiles The Univ. of Kansas Dept. of EECS
So v3 = 2.0 V
Now that we know the current through 3
R , we can determine the voltage across
it (um, using Ohm’s law…).
( )3 3 3
2 1 2v i R= = =
1
i
2
2
i
i−
0
1
v+ − 2
v+ −
2
+
−
v+
v−
R2 =3KR1 = 1K
+
-
ideal
R3 =1K
vout
vin
I=2 mA

15. 2/21/2011 Example An op amp circuit analysis lecture 15/23
Jim Stiles The Univ. of Kansas Dept. of EECS
Thus v- = 2.0 V
Thus, we can now determine both v+
(from a KVL equation) and v−
(from a device
equation):
3
2v v+
= = and 2v v− +
= =
1
i
2
2
i
i−
0
1
v+ − 2
v+ −
2
+
−
2v+
=
2v−
=
R2 =3KR1 = 1K
+
-
ideal
R3 =1K
vout
vin
I=2 mA

16. 2/21/2011 Example An op amp circuit analysis lecture 16/23
Jim Stiles The Univ. of Kansas Dept. of EECS
And now i1 =i2 +2
Now, inserting another device equation:
0i−
=
into the first KCL equation:
1 2 1 2 1 2
0 2 2i i i I i i i i−
= + + ⇒ = + + ⇒ = +
2
2i +
2
2
i
0
0
1
v+ − 2
v+ −
2
+
−
2v+
=
2v−
=
R2 =3KR1 = 1K
+
-
ideal
R3 =1K
vout
vin
I=2 mA

17. 2/21/2011 Example An op amp circuit analysis lecture 17/23
Jim Stiles The Univ. of Kansas Dept. of EECS
So that v2 = 2- vout
From KVL we find:
1 1 1
0in in in
v v v v v v v−
= − ⇒ = − ⇒ =
and:
2 2
2out out
v v v v v−
= − ⇒ = −
2
2i +
2
2
i
0
0
( )2in
v+ − − ( )2 out
v+ − −
2
+
−
2v+
=
2v−
=
R2 =3KR1 = 1K
+
-
ideal
R3 =1K
vout
vin
I=2 mA

18. 2/21/2011 Example An op amp circuit analysis lecture 18/23
Jim Stiles The Univ. of Kansas Dept. of EECS
Now we can find vout
So, from Ohm’s law (one of those device equations!), we find:
1
1 2 2
1
2
2 4
1
in
in
vv
i i i v
R
−
= ⇒ + = ⇒ = −
and:
2
2 2
2
2
3
out
vv
i i
R
−
= ⇒ =
Equating these last two results:
2
4 14 3
3
out
in out in
v
v v v
−
− = ⇒ = −

19. 2/21/2011 Example An op amp circuit analysis lecture 19/23
Jim Stiles The Univ. of Kansas Dept. of EECS
The “right equation” !
Thus, we have at last arrived at the result:
14 3out in
v v= −
2in
v −
2
4in
v −
0
0
( )2in
v+ − − ( )3 12in
v+ − −
2
+
−
2v+
=
2v−
=
R2 =3KR1 = 1K
+
-
ideal
R3 =1K
14 3out in
v v= −
vin
I=2 mA

20. 2/21/2011 Example An op amp circuit analysis lecture 20/23
Jim Stiles The Univ. of Kansas Dept. of EECS
An alternative: superposition
Note an alternative method for determining this result is the application of
superposition.
First we turn off the current source (e.g., 0I = )—note that this is an open
circuit!!!!!!!!
R2 =3KR1 = 1K
+
-
ideal
R3 =1K
out
v
vin
I=0

21. 2/21/2011 Example An op amp circuit analysis lecture 21/23
Jim Stiles The Univ. of Kansas Dept. of EECS
It’s just an inverting amp
Note this is the same configuration as that of an inverting amplifier!
Thus, we can quickly determine (since we already know!) that:
2
1
3
3
1out in in in
R
v v v v
R
= − = − = −
R2 =3KR1 = 1K
+
-
ideal
R3 =1K
3out in
v v= −
vin

22. 2/21/2011 Example An op amp circuit analysis lecture 22/23
Jim Stiles The Univ. of Kansas Dept. of EECS
See if you can prove this result!
Likewise, if we instead set the input source to zero ( 0in
v = —ground potential!),
we will find that the output voltage is 14 volts (with respect to ground):
R2 =3KR1 = 1K
+
-
ideal
R3 =1K
14out
v =I=2 mA

23. 2/21/2011 Example An op amp circuit analysis lecture 23/23
Jim Stiles The Univ. of Kansas Dept. of EECS
Look; the answer is the same!
From superposition, we conclude that the output voltage is the sum of these two
results:
14 3out in
v v= −
The same result as before!
2in
v −
2
4in
v −
0
0
( )2in
v+ − − ( )3 12in
v+ − −
2
+
−
2v+
=
2v−
=
R2 =3KR1 = 1K
+
-
ideal
R3 =1K
14 3out in
v v= −
vin
I=2 mA