Notes on Operational Amplifiers (Op Amps).
The name Op Amp comes from “operational amplifier.”
Op Amp Golden Rules (memorize these rules)
1) The op amp has infinite open-loop gain.
2) The input impedance of the +/− inputs is infinite. (The inputs are ideal
voltmeters). The output impedance is zero. (The output is an ideal voltage
source.)
3) No current flows into the +/− inputs of the op amp. This is really a restatement of
golden rule 2.
4) In a circuit with negative feedback, the output of the op amp will try to adjust its
output so that the voltage difference between the + and − inputs is zero (V+ = V−).
IDEAL OP AMP BEHAVOIR.
The relationship between the input ant the output of an ideal op amp (assumptions:
infinite open loop gain, unlimited voltage).
for V+ − V− 0 : Vout → +
for V+ − V− 0 : Vout → −
for V+ − V− = 0 : Vout = 0
Op Amp Schematic Symbol (The upper input is usually
the inverting input. Occasionally it is drawn with the non-
inverting input on top when it makes the schematic easier
to read. The position of the inputs may vary within the
same schematic, so always look closely at the
schematic!)
Negative Feedback. Most of the basic op amp building blocks rely on negative
feedback. You can easily identify the type of feedback used by the op amp circuit. For
negative feedback, the output is connected to the inverting input (− input). For positive
feedback, the output is connected to the non-inverting input (+ input).
The Input Impedance of the Circuit is defined as the rate of change of Vin with respect
dV
to a change of Iin. This is simply the derivative in The input impedance of the circuit
d Iin
is not in general the same as the impedance of the op amps inputs.
The Output Impedance of the Circuit, for the examples shown here, is the output
impedance of the op amp. Output impedance is defined as the rate of change of Vout with
dVout
respect to a change of Iout. This is simply the derivative . For the ideal op amp, the
d I out
output impedance is zero.
Basic Op Amp Building Blocks
Inverting Amplifier
Rf
Vout = − V in
Rin
Vout R
AV = =− f
Vin Rin
dVin
=R in
d I in
Analysis of the inverting amplifier starts with our op amp golden rules. From rule #4 we
know that V− = V+ and that V− = 0 because V+ is connected to ground. From rule #3 we
know that Iin = I f because no current flows into the inverting input.
V− = V+ V+ = 0 Iin = I f = I
Then we can find the relationship between Vin and Vout using Ohm’s law (OL) and
Kirchhoff’s voltage law (KVL).
Vin
Vin − V− = Iin Rin Vin − 0 = IRin Vin = IRin I =
Rin
V− − Vout = I f Rf 0 − Vout = IR f Vout = −IR f
Rf
Vout = − Vin
Rin
The voltage gain AV is the derivative of Vout with respect to Vin. When the amplifier has
only one input and Vout = 0 when Vin = 0, we will make the assumption that AV = Vout/Vin.
V − IR f = − R f
Av = out =
Vin IRin Rin
Alternatively, we could have started our analysis from the voltage divider formed by Rf
and Rin. The voltage divider will relate the voltage at V− with Vout and Vin. In this case the
total voltage across the divider is Vout − Vin. Because the bottom end of the divider is not
connected to ground, we must add the extra Vin term to offset V−. We arrive at the same
result.
The input impedance of the inverting amplifier is determined by Rin. Note that V− is held
at the same voltage as V+ by the op amp feedback. Because V+ is connected to ground, the
input impedance is just Rin.
Non-inverting Amplifier
Rf
Vout = 1 + Vin
Rin
Vout Rf
AV= =1+
Vin Rin
dVin
→
d Iin
Analysis of the non-inverting amplifier starts with our op amp golden rules. From rule #4
we know that V− = V+ and that V− = Vin because V+ is connected to Vin. From rule #3 we
know that Iin = I f because no current flows into the inverting input.
V− = V+ V+ = Vin Iin = I f = I
Then we can find the relationship between Vin and Vout using Ohm’s law (OL) and
Kirchhoff’s voltage law (KVL).
−
The voltage gain AV is the derivative of Vout with respect to Vin. When the amplifier has
only one input and Vout = 0 when Vin = 0, we will make the assumption that AV = Vout/Vin.
Alternatively, we could have started out analysis from the voltage divider formed by Rf
and Rin. The voltage divider will relate the voltage at V− with Vout and Vin. In this case the
total voltage across the divider is Vout and the we know that V− = Vin. We arrive at the
same result.
The input impedance of the follower is the input impedance of the op amps input. For an
ideal op-amp, the input impedance is infinite.
Voltage Follower Vout = Vin
Vout
AV = =1
Vin
dVin
→
d Iin
This is a special case of the non-inverting amplifier with Rin → ∞ and Rf = 0. The
follower has a very high input impedance. Voltage follower has application when the
source voltage cannot supply very much current, a pH meter for example.
Current-to-Voltage Converter (AKA, I-V Converter, Transimpedance Amplifier).
This circuit takes an input current and converts it to an output voltage. The input
impedance of the ideal current to voltage converter is zero (the ideal current meter).
Vout = −Rf Iin
Vout
AZ= = −R f
I in
dVin
=0
d Iin
Analysis of the current-to-voltage converter starts with our op amp golden rules. From
rule #4 we know that V− = V+ and that V− = 0 because V+ is connected to ground. From
rule #3 we know that Iin = I f because no current flows into the inverting input.
V− = V+ V+ = 0 Iin = I f = I
Then we can find the relationship between Vin and Vout using Ohm’s law (OL) and
Kirchhoff’s voltage law (KVL).
Iin = I f
V− − Vout = I f Rf 0 − Vout = I f Rf Vout = −I f Rf
Vout = −I f Rf = −Iin Rf
The current-to-voltage converter has transimpedance gain. Transimpedance gain is not
unitless, it has units of impedance (Ohms). The transimpedance gain AZ is the derivative
of Vout with respect to Iin. When the amplifier has only one input and Vout = 0 when
Iin = 0, we will make the assumption that AV = Vout/Iin.
Vout − Iin Rf
A
Z
= = = −Rf
Iin Iin
Summing Amplifier. This circuit will add (and subtract) the input voltages.
Subtraction is accomplished by inverting the voltages before adding them. Note that
summing can only occur for inputs to the inverting side of the op amp. This is because of
the V− node is a current summing junction where the input currents sum to the feedback
current.
Vout = AV 1Vin1 + AV 2Vin 2 + AV 3Vin3 +L
Th
d Vout R
the gain of the nth input: A Vn = =− f
dV inn Rinn
d Vinn
the impedance of the nth input: Z inn = = Rinn
d Iinn
This is another look at the summing amplifier that emphases the summing junction.
Analysis of the summing amplifier starts with our op amp golden rules. From rule #4 we
know that V− = V+ and that V− = 0 because V+ is connected to ground. From rule #3 we
know that Iinn = I f because no current flows into the inverting input. (Iinn is the
current of the nth input.)
V− = V+ V+ = 0 I inn = Iin1 + Iin 2 + Iin3 = I f
Then we can find the relationship between Vin and Vout using Ohm’s law (OL) and
Kirchhoff’s voltage law (KVL).
The voltage gain AV is the derivative of Vout with respect to Vin.
Vout = AV 1Vin1 + AV 2Vin 2 + AV 3Vin3 +……..
d Vout = − R f
the gain of the nth input : AVn =
dVinn Rinn
Differential Amplifier. The term differential is used in the sense of difference. Do
not confuse the differential amplifier with the differentiator. One important application of
the differential amplifier over comes the problem of grounding that you encountered in
lab when using the oscilloscope to make measurements. The typical oscilloscope always
performs voltage measurements with respect to its own ground. A differential amplifier
used before the scope input could measure the V+in with respect to V−in. The ground of
the differential amplifier would be connected to the ground of the scope for this
application, so the Vout will be measured correctly.
R2
Vout = (V +in − V −in )
R1
d Vout R
the gain for V+in : AV +in = =+ 2
dV +in R1
d Vout R
the gain for V−in : AV −in = =− 2
dV −in R1
Analysis of the differential amplifier starts with our op amp golden rules. From rule #4
we know that V− = V+ . From rule #3 we know that I1 = I 2 and that I 3 = I 4 because no
current flows into the inputs.
V− = V+ I1 = I 2 I 3= I 4
Then we can find the relationship between V+in, V−in, and Vout using the voltage divider
equations. We recognize that V− = V+ and that V+ will be the output of the voltage
divider formed by the two resistors connected to the non-inverting input. The voltage at
V− is the output of the voltage divider formed by the two resistors connected to the
inverting input.
The voltage gain AV is the derivative of Vout with respect to each input Vin.
d Vout = + R2
the gain for V+in : AV +in =
dV+in R1
d Vout = − R2
the gain for V−in : AV −in =
dV−in R1
The inverting amplifier with generalized impedances. The results derived
above can be extended to general impedances. Note that Zf and Zin can be the impedance
of any network. The following are examples of the inverting amplifier, but ANY of the
previous examples can be generalized in this way.
Zf
Vout = − Vin
Z in
Vout Z
AV = =− f
Vin Z in
dVin
=Z in
d Iin
Integrator. A capacitor as the feedback impedance. j
Vout () = V ( )
Rin C f in
1
Vout () = V ()
RinC f in
1
Rin C f
Vout (t) = − Vin(t) d t
Vout j
AV( ) = =
Vin Rin C f
Vout 1
AV( ) = =
Vin Rin C f
dVin
=R in
d Iin
Analysis of the integrator in the frequency domain is a simple extension of our
generalized result for the inverting amplifier.
Time Domain Analysis of the Integrator starts with our op amp golden rules. From
rule #4 we know that V− = V+ and that V− = 0 because V+ is connected to ground. From
rule #3 we know that Iin = I f because no current flows into the inverting input.
V− = V+ V+ = 0 Iin = I f = I
Then we can find the relationship between Vin and Vout using Ohm’s law (OL) and
Kirchhoff’s voltage law (KVL).
Differentiator. A capacitor as the input impedance. Vout () = − j R f CinVin ()
Vout () = − Rf CinVin ()
dVin (t)
Vout (t) = −R f Cin
dt
Vout
AV () = = − jR fC in
Vin
Vout
AV () = = −R fC in
Vin
Analysis of the differentiator in the frequency domain is a dVin 1
simple extension of our generalized result for the inverting =
d Iin jCin
amplifier.
Time Domain Analysis of the Differentiator starts with our op amp golden rules. From
rule #4 we know that V− = V+ and that V− = 0 because V+ is connected to ground. From
rule #3 we know that Iin = I f because no current flows into the inverting input.
V− = V+ V+ = 0 Iin = I f = I
Then we can find the relationship between Vin and Vout using Ohm’s law (OL) and
Kirchhoff’s voltage law (KVL).
The General Op Amp Circuit
Example: an op amp circuit with 3 inverting and 3 non-inverting inputs
What can we say about such a complicated looking amplifier? Don’t panic, we can use
what we have learned from the above analyses to painlessly arrive at the solution.
Without doing any analysis, what can we say? First, we know that
Vout = AV 1V1 + AV 2V2 + AV 3V3 + AV 4V4 + AV 5V5 + AV 6V6
dV
the gain of the nth input: AVn = dVout
n
Second, we know that the voltage gains for V1, V2, and V3 will be inverting (negative) and
that the voltage gains for V4, V5, and V6 will be non-inverting (positive). We could
simply blaze away at the problem by applying the op amp golden rules just like we did
for the derivations for the basic op amp building blocks, but there is a better way.
The Strategy.
We will make use of the results for the basic op amp building blocks and the principle of
superposition (e.g. the inverting amplifier and the non-inverting amplifier). To apply the
principle of super position, we analyze the gain for one input at a time and turn off all the
other inputs (set them to zero). We treat each input as an ideal voltage source, so that an
input that is turned off is equivalent to a connecting the input terminal directly to
ground. The inverting inputs and the non-inverting inputs will behave differently.
Analysis of the Inverting Inputs by Superposition
Let’s first use analyze the voltage gain AV1 of the input V1. We proceed by connecting all
the other inputs to ground. (Analysis of the voltage gain for the other inverting inputs
(Av2 and Av3) is analogous.)
This simplifies to:
Notice that the non-inverting input, V+ is connected to ground though a resistor. Because
no current flows into the op amp’s inputs, V+ = 0, equivalent to it being connected
directly to ground (see below).
This looks like a summing amplifier with V2 = V3 = 0. The result for the summing
R
amplifier is A = − 7 . We can do the same analysis for each inverting input. (Why
V1
R1
don’t the two resistors R2 and R3 enter this analysis for AV1? Hint: Consider the voltage
across these two resistors.)
Analysis of the Non-Inverting Inputs by Superposition
Now let’s use analyze the voltage gain AV4 of the non-inverting input V4. We proceed by
connecting all the other inputs to ground. (Analysis of the voltage gain for the other non-
inverting inputs (Av5 and Av6) is analogous.)
This simplifies to:
This is a non-inverting amplifier. The resistors on the non-inverting side, R4, R5, R6, and
R8, form a voltage divider that reduces the voltage seen by V+. It is the voltage at V+ that
is seen by the op amp. The voltage gain of V+ is determined by the resistors on the
inverting side, R1, R2, R3, and R7. Hence the voltage gain AV4 of the V4 input has a term
from the voltage divider, relating V4 to V+, and a term for the voltage gain of V+, relating
V+ to Vout. The voltage gain AV4 is the product of the two term and relates V4 to Vout.
The voltage gains for the other non-inverting inputs can be found in this way.
An important observation is that multiple inputs into the non-inverting side of the op
amp do not sum in the simple way that they do for inverting inputs. Thus, the summing
amplifier that we listed as a basic building block does not have a non-inverting analog!
(If we need a non-inverted sum, we just follow the summing amplifier with a unity gain
inverting amplifier.)
