Tony Richardson
Introduction
Ideal op amp analysis by the “two rules” method is fast and easy, but can be confusing. The difficulty lies in knowing where to start and when to apply the rules. A method for ideal op amp analysis is presented here which many find to be more straightforward than the “two rules” method.
The disadvantage to this method is that it requires a bit more algebra than the “two rules” approach.
The method is based on the “two rules” and nodal analysis (Kirchhoff's current law).
The Nodal Analysis Method
The notation that will be used to represent the op amp terminal voltages and currents is shown in
Figure 1.
i
1 v
1 v
2
–
+ i o v o i
2
Figure 1: Definitions of Voltages and Currents
There are two steps in the nodal analysis method:
1) Use nodal analysis (Kirchhoff's current law) to sum currents at both the inverting and noninverting op amp input terminals. Express the currents in terms of the voltages v
1
, v
(and other node voltages as necessary). Remember that in an ideal op amp, currents zero. (That is one of the rules in the “two rules” method.) Solve each of the equations for v
2 i
1
2
, and
and i v
2 v o
are
1
and
respectively. In many cases, one of the input terminals is directly connected to a voltage source or ground, the corresponding terminal voltage is then known and it is not necessary to apply Kirchhoff's current law.
2) Set the expressions for v1 and v2 equal to one another. (Remember that in an ideal op amp they are equal – that is the other rule in the “two rules” method.) Solve the resulting equation for the output voltage v o
.
Inverting Amplifier Example
Consider the op amp circuit shown in Figure 2. (This circuit is called an inverting amplifier.)
Since the noninverting terminal is connected to ground we know that v
2
is equal to zero. We sum currents at the inverting node (I treat currents that leave the terminal as positive) and obtain the second of the following two equations (remember that the current entering the op amp is equal to zero): v
1
− v a
R a v
2
= 0
 v
1
− v o
R f
= 0
10/09/2007 1 of 4

Solving the second equation for v
1
yields: v
1
= v a v
2
= 0
/ R a
 v o
/ R f
1 / R a
 1 / R f v a
R a v
1 v
2
–
+
R f
R o
+ v o
–
Figure 2: An Inverting Amplifier
Setting the expressions on the right side of each equation equal to one another gives: v a
/ R a
 v o
/ R f
1 / R a
 1 / R f
= 0
Finally, solving for v o
produces the desired result: v o
=−
R f
R a v a
Summing Amplifier Example
Now consider the summing amplifier shown in Figure 3. Recognizing that the voltage at the noninverting terminal is also equal to zero and summing currents at the inverting terminal yields the following set of equations: v
1
− v a
R a
 v
2
= 0 v
1
− v
R b b
 v
1
− v o
R f
= 0
Solving the second equation for v
1
gives: v
1
= v a v
2
= 0
/ R a
 v b
/ R b
 v o
/ R f
1 / R a
 1 / R b
 1 / R f
After setting the right sides of these two equations equal to each other and solving for v o
we obtain: v o
=−
 R f
R a v a

R f
R b v b

10/09/2007 2 of 4

v b v a
R b
R a v
1 v
2
–
+
R f
R o
+ v o
–
Figure 3: A Summing Amplifier
Noninverting Amplifier Example
A noninverting amplifier is shown in Figure 4. Since i
2
is equal to zero, there is no voltage drop across R a
and v
2
is equal to equation in the set below: v a
. Summing currents at the inverting terminal yields the second v
1
R
1
 v
2 v
1
= v a
−
R f v o
= 0 v a
R a v
2 v
1
+
– v o
R
1
R f
Figure 4: A Noninverting Amplifier
Solving the second equation for v
1
, v
2
= v a v
1
= v o
/ R f
1 / R
1
 1 / R f
Setting the right sides equal to each other and solving for v o
gives: v o
=

1 
R f
R
1
 v a
=
R
1
 R f
R f v a
10/09/2007 3 of 4

A Final Example
Consider the amplifier shown in Figure 5. We obtain the following set of equations after summing currents at both nodes: v
1
− v
R a
R
1 a
 v
2
− v s v

1
− v
R v
2
R
2 b o
= 0
= 0 v v a s
R a
R
1 v
1 v
2
–
+
R b
R
2 v o
Figure 5: A Final Example
Solving for v
1
and v
2
yields: v
1
= v a
1
/
/
R
R a a
 v o
/ R
 1 / R b b
= v a v
2
= v s R
1
R
2
 R
2
R b
 v o
R a
 R b
R a
Setting the right hand sides of both equations equal to each other gives: v a
R b
 v o
R a
 R b
R a
= v s
R
2
R
1
 R
2
Solving for the output voltages produces: v o
=
R a
 R b
R a
 v s
R
2
R
1
 R
2
− v a
R b
R a
 R b

In the special case where R a
= R b
and R
1
= R
2
this reduces to v o
= v s
− v a and the output voltage is equal to the difference of the two input voltages.
10/09/2007 4 of 4