ES250: Electrical Science - Clarkson University...• Sli fSolving for v1, we h ave: • The node eqn. at the inverting node of theat the inverting node of the op amp is:op amp is:

ES250:Electrical Science

HW6: The Operational Amplifier

• This chapter introduces the operational amplifier or op ampIntroduction

This chapter introduces the operational amplifier or op amp• We will learn how to analyze and design circuits that contain

op amps, including models of varying accuracy and complexity, the simplest of which is the ideal op amp:– circuits that contain ideal op amps can be solved using

d tinode equations– circuits that perform mathematical operations such as

gain summing or integration can be made using op ampsgain, summing, or integration can be made using op amps• Real op amps have properties not captured by the ideal op

amp model, including: input offset voltage, bias current, dc gain, input resistance, and output resistance– more complicated models are needed, however, to

f haccount for these properties

• The op amp is designed to be used with other circuit The Operational Amplifier

elements to perform signal‐processing operations

– A μA741 operational amplifier is shown to the left, while its pin connections are shown to the right:

• An op amp circuit is shown below including power suppliesThe Operational Amplifier

An op amp circuit is shown below including power supplies v+ and v− used to bias the device, i.e., create the conditions required for the op amp to function properly:

• Op amp circuits are often analyzed using KCL, without theThe Operational Amplifier

Op amp circuits are often analyzed using KCL, without the power supplies shown in the prior circuit; as such, it is easy to overlook the circuit’s power supply currents

• Applying KCL to the ground node of the prior circuit yields:

– this also implies that the sum of all currents into the op amp are zero

– if the op amp power supplies are not shown (as is often the case for simplicity), it is best not to apply KCL at the ground node to avoid “overlooking” the supply currents

• To be linear, the op amp output voltage, vo, and current, io,The Ideal Operational Amplifier

To be linear, the op amp output voltage, vo, and current, io, must satisfy three conditions:

– where the saturation voltage v the saturation currentwhere the saturation voltage, vsat, the saturation current, isat, and the slew rate limit, SR, are all op amp parameters, e.g., a μA741 op amp biased with +/‐15 V supplies has:

– these restrictions reflect that op amps cannot producethese restrictions reflect that op amps cannot produce arbitrarily large voltages or currents, or change output voltage arbitrarily quickly

• The ideal op amp is a linear model characterized byThe Ideal Operational Amplifier

The ideal op amp is a linear model characterized by restrictions on its input currents and voltages, as shown:

h d l d d h– note, the output current and voltage depends on the circuit in which the op amp is used

• Find the output voltage, vo, in terms of the input voltage, vs :Example 6.3‐1: Ideal Op Amp

Find the output voltage, vo, in terms of the input voltage, vs :Solution:Apply KVL around the outer loop to obtain:

To verify conditions0V+

−

0A→

0A→ To verify conditions for linearity are satisfied, apply KCL at 0V

+ , pp ythe output node to obtain:

0V−

Since i1 = 0,Since i1 0,

⇒

• To satisfy the linearity conditions for a μA741 op ampExample 6.3‐1: Ideal Op Amp

To satisfy the linearity conditions for a μA741 op amp requires:

• Thus if vs = vo = 10 V and RL = 20 kΩ, then:

which is consistent with the use of the ideal op amp model

• To satisfy the linearity conditions for a μA741 op ampExample 6.3‐1: Ideal Op Amp

To satisfy the linearity conditions for a μA741 op amp requires:

• On the other hand, when vs = vo = 10 V and RL = 2 kΩ, then:

so it is not appropriate in this case to model the μA741 asso it is not appropriate in this case to model the μA741 as an ideal op amp, i.e., when vs = 10 V, we require RL > 5 kΩin order to satisfy the current limit for linearityy y

• Find the ratio vo/vs for the ideal op amp (see red references)Exercise 6.3‐1

Find the ratio vo/vs for the ideal op amp (see red references) circuit below:

i0V+ 0A→

0Ai0V− 0A→

• Solution by KVL: ( ) ( )0v R R i v R R i− − + = ⇒ = − +• Solution by KVL: ( ) ( )

( )

1 2 1 2

1 1

00

o o

s s

v R R i v R R iv R i v R i+ = ⇒ = +

− − = ⇒ = −

( )1 2 2

1 1

1o

s

R Rv Rv R R

+∴ = = +

• Find the ratio vo/vs for the ideal op amp (see red references)Exercise 6.3‐2

Find the ratio vo/vs for the ideal op amp (see red references) circuit below:

i

0V+ 0A→

0A→

1i

− 0A→

i2i

• Solution by KVL:Exercise 6.3‐2

Solution by KVL:

( ) ( )1 2 2 21 2

0 ss

vv R R i iR R

− + + = ⇒ =+( )

( )

1 2

2 23 1 2 2 1 20 0 s

R RvR RR i R i i i

R R R R

+

+ + = ⇒ = − = − ⋅+( )

( ) ( ) ( )3 3 1 2

3 423 4 1 0 0 3 4 10

R R R R

R RRR R i v v R R i v

+

++ + = ⇒ = − + = ⋅( ) ( ) ( )

( )

3 4 1 0 0 3 4 13 1 2

3 40 2 2 4

0 sR R i v v R R i vR R R

R Rv R R R

+ + ⇒ ++

+

( )( ) ( )

3 40 2 2 4

3 1 2 1 2 3

1s

v R R Rv R R R R R R

∴ = ⋅ = + + +

• 0 4

2 13

Note when , then 1s

v RR Rv R

≈ +

Questions?Questions?

• It is convenient to analyze op amp circuits using nodeNodal Analysis of Ideal Op Amp Circuits

It is convenient to analyze op amp circuits using node equations; three things to remember about ideal op amps:

– node voltages across the op amp inputs are equal; thus, g p p p q ; ,one node voltage can be eliminated from the node eqns.

– currents flowing into an ideal op amp are zero

– the output current of an ideal op amp is not zero; therefore, a KCL node eqn. should be included at the op amp output node only if the output current is to be found (skip otherwise as it adds an unknown variable to solve)

• Assume an ideal op amp and use node equations to analyzeExample 6.4‐1: Difference Amplifier

Assume an ideal op amp and use node equations to analyze this circuit below to determine vo in terms of va and vb:

• The node eqn. at the noninverting node of the op amp is:

• Since v2 = v1 and i2 = 0, this equation becomes:Example 6.4‐1: Difference Amplifier

Since v2 v1 and i2 0, this equation becomes:

S l i f h• Solving for v1, we have:

• The node eqn at the inverting node of the op amp is:The node eqn. at the inverting node of the op amp is:

h b• Since v1 = 0.75vb and i1 = 0, this equation becomes:

• Thus, solving for vo in terms of va and vb we obtain:

• This circuit is called a difference amplifier since the output voltage is a scaled version of the input voltage difference

• For the bridge amplifier below, determine the output Example 6.4‐2: Analysis of a Bridge Amplifier

voltage, vo, in terms of the source voltage, vs:

Note: the “bridge”Note: the bridge portion of the circuit is shown in black, while the ideal op amp and resistors R5 and R6 are

d l f hused to amplify the bridge output voltage, v are shown in redvab, are shown in red

• Here is an opportunity to use Thévenin's theorem to replace Example 6.4‐2: Analysis of a Bridge Amplifier

the bridge circuit with its Thévenin equivalent circuit:

Note: to determine R look into theNote: to determine Rt look into the circuit with the voltage source, vs, shorted, i.e., vs = 0 V; therefore:s

( ) ( )1 2 3 4tR R R R RR RR R

= +

3 41 2

1 2 3 4

R RR RR R R R

= ++ + 0V⇒ =

• We can solve for vab = voc using KVL mesh currents, as Example 6.4‐2: Analysis of a Bridge Amplifier

shown:( )

1

1 2 1

Mesh around :0s

iv R R i− + + =

( )11 2

sviR R

⇒ =+

( )2

3 4 2

Mesh around :0s

iv R R i+ + =

1i

( )23 4

sviR R

⇒ = −+

2 1 4 2

Loop around :0

ab oc

ab

v vv R i R i

=− + + =

2i

( ) ( )2 4

1 2 3 4ab s

R Rv vR R R R

⇒ = −

+ +

• Therefore, the Thévenin equivalent circuit for the bridge Example 6.4‐2: Analysis of a Bridge Amplifier

circuit is given by:

• Inserting the Thévenin equivalent into the op amp circuit:Example 6.4‐2: Analysis of a Bridge Amplifier

Using KVL, node voltage va is given by:

Since v1 = 0 and i1 = 0,

The node equation at node a yields:

Since v = v and i = 0:Since va = voc and i1 = 0:

⇒

• Find the value of the voltage measured by the voltmeter:Example 6.4‐3: Op Amp Analysis Using Node Eqns.

• To begin, rewrite the circuit assuming an ideal op amp:Example 6.4‐3: Op Amp Analysis Using Node Eqns.

By KVL:By KVL:

⇒

By KVL:

⇐

KCL @ nodes 2 and 1 yields:

⇒⇒ ∴

• Find the value of the voltage measured by the voltmeter:Example 6.4‐4: Analysis of an Op Amp Circuit

• To begin, rewrite the circuit assuming an ideal op amp:Example 6.4‐4: Analysis of an Op Amp Circuit

By KVL:

⇒⇒

0V

KCL @ nodes 2 and 3 yields:

0VBy KVL:

⇐KCL @ nodes 2 and 3 yields:

⇒∴

and

∴


• One of the early applications of op amps was to build Design Using Operational Amplifiers

circuits that performed mathematical operations; the following figures show several standard op amp circuits:

• The following figures show several standard op amp circuits:Design Using Operational Amplifiers








• Using a voltage follower to prevent circuit loading:Example 6.5‐1: Using a Voltage Follower

Using a voltage follower to prevent circuit loading:0≠

Circuit 2 loads Circuit 1 changing its desired operating point0≠

Circuit 2 loads Circuit 1 changing its desired operating point

• A voltage follower can be used to prevent circuit loading:Example 6.5‐1: Using a Voltage Follower

• A common application of op amps is to scale, i.e., multiply, a Example 6.5‐2: Amplifier Design

voltage by a constant, K, called the gain of the amplifier:

• The choice of amplifier circuit determines the value of gain;Example 6.5‐2: Amplifier Design

four cases are shown: K = −5 (i.e., an inverting amplifier):

5 inv= −

• The choice of amplifier circuit depends on the value of gain;Example 6.5‐2: Amplifier Design

four cases are shown: K = 5 (i.e., an noninverting amplifier):

5 inv=


four cases are shown: K = 1 (i.e., follower/buffer amplifier):

↓

0V+0A↓

inv=

−0A→

in

By KVL around the op amp:y p po inv v=


four cases are shown: K = 0.8 (i.e., less than unity, use a front end voltage divider with a follower/buffer amplifier):

v +0A↓

av 0V+

−0A→

0.8 inv=

By V80k

÷ By KVL:80k 0.820k 80ka in inv v v

= =+

o av v=


• This section describes a procedure for designing op ampOp Amp Circuits and Linear Algebraic Eqns.

This section describes a procedure for designing op amp circuits to implement linear algebraic eqns.:

– a voltage or current used to represent something is called g p ga signal; here some of the op amp node voltages will be used to represent the variables in the algebraic eqn.

– for example, the eqn.:

– will be represented by an op amp circuit that has node voltages vx, vy, and vz that are related by the eqn.:

– in this case, vx, vy, and vz are signals representing the variables x, y, and z; but signals can also be used to

h l hrepresent physical measurements such as temperature, pressure, speed, etc.

Op Amp Circuits and Linear Algebraic Eqns.• Block diagrams are often used to represent signals andBlock diagrams are often used to represent signals and

systems, including algebraic eqns. such as:

• Consider designing an op amp circuit to implement the block diagram above:block diagram above:

– blocks representing a constant gain can be implemented using either inverting or noninverting amplifier (or a g g g p (buffered voltage divider) depending on the sign and magnitude of the desired gain, as illustrated previously

Op Amp Circuits and Linear Algebraic Eqns.• Op amp circuits that implement the individual blocks thatOp amp circuits that implement the individual blocks that

make up the block diagram are shown below, starting with the noninverting amplifier to implement the positive gain:

Op Amp Circuits and Linear Algebraic Eqns.• An inverting amplifier is used to implement the negativeAn inverting amplifier is used to implement the negative

gain block, as shown below:

Op Amp Circuits and Linear Algebraic Eqns.• Note, the preferred way to make a constant voltage sourceNote, the preferred way to make a constant voltage source

is to use of the op amp power supplies whenever possible, i.e., using a voltage divider with buffer cirucit, as shown:

130K = Ω0V+0A↓

By V ÷

0V−0A→

= 20KΩ20k= 15V130k 20k

ΩΩ+ Ω

• Let K1 = K2 = K3 = 1/(n + 1)=1/4, K 4 R R 20kΩ d

Op Amp Circuits and Linear Algebraic Eqns.

K4 = 4, Rb = R = 20kΩ, and Ra = R/(n + 1) = R/4 = 20k/4 = 5kΩ

• A noninverting summingsumming amplifier is used to combine the various signals

Op Amp Circuits and Linear Algebraic Eqns.

• This circuit show how to combine the various op amp circuits to achieve the overall desired algebraic operation


• The ideal op amp is the simplest model obtained by ignoringCharacteristics of Practical Op Amps

The ideal op amp is the simplest model obtained by ignoring some imperfections of practical op amps

• This section considers some of these imperfections and pprovides alternate op amp models to account for these imperfections , namely:

– nonzero bias currents, ib1 and ib2

– nonzero input offset voltage, vos

– finite input resistance, Ri

– nonzero output resistance , Roo

– finite voltage gain, A

• The “offsets and finite gain model” of an operationalCharacteristics of Practical Op Amps

The offsets and finite gain model of an operational amplifier is shown below:

• The above model can be simplified to the “offsets model” or the “finite gain model” depending on the assumed parameter values

• The “offsets model” of an operational amplifier assumes Characteristics of Practical Op Amps

that Ri = infinite, Ro = 0 Ω, and A = infinite, to produce ideal op amp characteristics, as shown below:

↓

0V+

0A↓

−0A↑

• Note, the values of ib1, ib2, vos and , b1, b2, osthe elements connected to this op amp model determine vo

• The “finite gain model” of an op amp assumes that ib1 and Characteristics of Practical Op Amps

b1ib2 = 0A, and that vos = 0V, as shown below:

• Note, the values of Ri, Ro, A and the elements connected to this op amp model determine the value of vo

• In this example, the “offsets model” of an op amp will be Example 6.7‐1 Offset Voltage and Bias Currents

used to analyze an inverting amplifier circuit, as shown:

• This circuit can be analyzed using superposition, i.e., by Example 6.7‐1 Offset Voltage and Bias Currents

considering one independent source operating at a time, e.g., vin, vos, ib1 and ib2, and disabling all others:

0A→

0V+−0A→

• This circuit can be analyzed using superposition, i.e., by Example 6.7‐1 Offset Voltage and Bias Currents

considering one independent source operating at a time, e.g., vin, vos, ib1 and ib2, and disabling all others

• Analysis with only vin active, i.e., ideal inverting amp result:

0A↓

0V+−

↑0A↑

• Analysis with only vos active:Example 6.7‐1 Offset Voltage and Bias Currents

os

0A↓

0V+−

↑

0A

0A

• Analysis with only ib1 active:Example 6.7‐1 Offset Voltage and Bias Currents

b1

0V+−

0A→

0A↑

• Analysis with only ib2 active:Example 6.7‐1 Offset Voltage and Bias Currents

b2

0V+−→

0A→

0A→

• Summing the individual outputs yields the output solution Example 6.7‐1 Offset Voltage and Bias Currents

for op amps with “offset voltage and bias currents” model:

• The input offset voltage and the bias current for a μA741 op ill b 5 V d 500 Aamp will be at most 5 mV and 500 nA, so:

• As shown previously, the voltage follower circuit has a unity Example 6.7‐2: Finite Gain Model

gain when analyzed using an ideal op amp model:

0V+− 0A→

0A→

sv=

• What effect does the input resistance, output resistance and Example 6.7‐2: Finite Gain

finite voltage gain associated with the “finite gain” op amp model have on the voltage follower circuit below:

• Assume R1 = 1 kΩ; RL = 10 kΩ; and the op amp parameters Solution

1 LRi = 100 kΩ, Ro = 100 Ω and A = 105V/V:

• If vo = 10 V, output current,

• Apply KCL at the top node of RL to obtain:Solution

L

• Given the large input resistance, Ri, it is reasonable to assume that i1 will be much smaller than both io and iL; therefore, we will assume that i1 = 0

and check the validity of this assumption later⇒

• KVL around the mesh containing the VCVS, Ro, and RL yields:

• Solving for (v2 − v1) yields:

• Current i1 can now be calculated using Ohm's law as: 1 g

0 ≅

• Applying KVL to the outside loop yields:Solution

• Using algebra to determine vs yields:g g s y

• Solving for the circuit gain yields:

• For the specified A, Ro, and Ri, we have :Solution

o i

• Thus the input resistance output resistance and voltageThus, the input resistance, output resistance, and voltage gain of the practical operational amplifier have only a small, essentially negligible, combined effect on the performance y g g pof the buffer amplifier

• In the finite gain model, the voltage of the dependent Common Mode Rejection Ratio

source is given by:

• In practice, the dependent source voltage is more accurately expressed as:

wherewhere

• The common mode rejection ratio (CMRR) is defined to be the ratio of A to Acm:

• CMRR can be added to the finite gain model as shown:g

• In most cases, negligible error is caused by ignoring the Common Mode Rejection Ratio

CMRR of the operational amplifier

• The CMRR does not need to be considered unless accurate measurements of very small differential voltages must be made in the presence of very large common mode voltages

• The finite gain model implies that the gain, A, of the Gain Bandwidth Product

operational amplifier is a constant; suppose that the differential input voltage is given by:

• The voltage of the dependent source in the finite gain model is given by:g y

• The amplitude, AM, of this sinusoidal voltage does not depend on the frequency, ω; practical op amps do not work p q y, ; p p pthis way, i.e., the gain is a fcn. of frequency, say A(ω)

• For many practical amplifiers, A(ω) can be adequately represented as:

• No need to know how this fcn. behaves (see Ch 13), justNo need to know how this fcn. behaves (see Ch 13), just realize the parameter B, called the op amp gain bandwidth product, describes the gains dependence on frequency

• Table below summarizes various parameters associated with Characteristics of Practical Op Amps

several types of commonly available op amps:


ES250: Electrical Science - Clarkson University...• Sli fSolving for v1, we h ave: • The node eqn. at the inverting node of theat the inverting node of the op amp is:op amp is:

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