CE2004 - Lecture-Part II-1 · CE 2004 –Circuits & Signal Analysis 6 more explanations in the next...

CE 2004 – Circuits & Signal Analysis 1

CE2004 II‐1 :Operational Amplifiers

(Op-Amps)Weisi Lin

Email: [email protected] of Computer Science and Engineering

Nanyang Technological UniversitySingapore

Major Topics of Part 2

• Principles of Op‐Amps• General Signals & Systems • Laplace Transform & frequency‐based system analysis


Important Concepts, Logic & Methodology

(Principles of Op-Amps) Introduction to Op-Amps (operational amplifiers)

Concept of Negative Feedback Two Commonly-used Circuits with Negative

Feedback oNon-Inverting AmplifiersoInverting Amplifiers

Recap/Summary


We are starting to talk about amplifiers.

What is an ideal amplifier?

Considerations: Voltage gain Input impedance Output impedance


First op amps built in 1934 Invented by a Bell engineer named Harry Black* Didn’t gain the name operational amplifier until

the computer age began a decade or so later Vacuum Tube Op-Amps

Vacuum Tube Op-Amps Used in WWII to help to strike military targets

- Buffers, summers, differentiators, inverters Took ±300V to ± 100V to power

Transistors replaced vacuum tubes in 1950’s Integrated circuits (ICs) invented in the 1960’s, op

amps were among the first chips to be designed.

*In 1934，Harry Black commuted from his home in NewYork City to work at Bell Labs in New Jersey by way of a railroad/ferry. The ferry ride relaxed Harry enabling him to do some conceptual thinking…


more explanations in the next pages.

Introduction to Operational Amplifiers (Op-Amps) Op-amp is a high gain amplifier having (nearly ideal):

- Very high voltage gain (104 to 106)

- Very high input impedance (typically a few

megohms)

- Low output impedance (less than 100 ohm). An op-amp consists of multi-stage transistor amplifier

fabricated in an integrated circuit (IC) form.


Internal Circuitry of the Standard 741 Op-Amp

Positive DC power supply

An IC consisting of transistors resistors capacitors

Negative DC power supply


With the same Rs,(similar to Vs)

(very different from Vs)

if

i.e,

A voltage source followed by an amplifier

Loading effect minimization: input impedance of an amplifier should be much greater (infinite for an ideal case) than the source impedance.


For amplifier #1:

if Zi2 >>Zo1

An amplifier followed by another amplifier

similar analysis as in the previous page

Loading effect minimization: input impedance of Amplif#2 should be much greater (infinite for an ideal case) than output impedance of Amplif#1.


VEE

VCC

V+

Vout

V-

ZOUTZIN

~

Zout

AdVdZinVin Vout

Op-Amp Equivalent Circuit(see next page)

Vo=AoVin

As represented in the previous pages

Typical Uses of Op-Amps: Voltage amplifiers, Oscillators, Filter circuits, Instrumentation circuits, and so on.


Input port: Modeled as a resistance of value (Very large, typically ).

The input is known as non-inverting input and is the inverting input.

Output port: Modeled as a dependent voltage source, in series with a resistance ( is very small, typically ).

Two DC power supplies: (+ve) and (‐ve). The use of positive and negative voltage allows the output of the op‐amp to swing in both directions.

CCV

.INZ610

v v

ov

EEV

.OUTZ OUTZ100


CCoEE VvvAV )(

Input-Output Relationship:‐ is the internal voltage gain (open loop voltage gain) of the op‐amp.

‐ Typically, will have values (very high)

Output Saturation Limitscan never exceed the op‐amp’s power supply voltages, and (saturation limits of .

) ( vvAV ooutoA

oA 64 10 to10

ovCCV EEV )ov


Voltage transfer characteristics of an op-amp

Vout

Vin=(V+) – (V-)

Positive saturation

Negative saturation

Linear, i.e., Vo=Ao(Vin)

VCC

VEE

VCC= +15V

VEE= -15V

Linear region

The slope of linear region is big (Why?)


Op-Amps

1.The saturation limits of are equal to the op-amp supplyvoltages and2.The gain is so high that a very small nonzerovalue of differential input drives to saturation(non linear operation).3. positive will saturate at its positivesaturation limit4. negative will saturate at its negativesaturation limit

ovCCV .EEV

oA )( oA)( vv ov

)( vv

)( vv ov.CCV

ov.EEV

VEE

VCC

V+

Vout

V-

ZOUTZIN


5.For linear operation, the differential input voltage must be close to zero (very small).6.The internal input resistance Hence the inputcurrents (in +ve and –ve terminals) are assumed to bezero.7.The internal output resistance Hence

Q: What is the main concern in using op-amp as alinear device?A: Saturates for a small Vin.Q: How do we get rid of this problem, and also obtainthe desired voltage gain?A: Negative Feedback

)( vv

.INZ

.0OUTZ .oout VV

VEE

VCC

V+

Vout

V-

ZOUTZIN


Harold Stephen Black (April 14, 1898 – December 11, 1983) invented thenegative feedback amplifier while he was a passenger on theLackawanna Ferry (from Hoboken Terminal to Manhattan) on his way towork at Bell Laboratories (located in Manhattan instead of New Jersey in1927) on August 2, 1927 (US Patent 2,102,671, issued in 1937)—obviously he was thinking while taking ferry.

His invention is considered as an important breakthrough of the 20thcentury in electronics, since it has a wide area of applications. On August8, 1928, Black submitted his invention to the U. S. Patent Office, whichtook more than 9 years to issue the patent. The patent office wasinundated with fraudulent “perpetual motion” (motion of bodies thatcontinues indefinitely) claims, and dismissed Black’s invention at firstsight. Black later wrote: "One reason for the delay was that the conceptwas so contrary to established beliefs that the Patent Office initially didnot believe it would work.”


Concept of Negative Feedback

In negative feedback, the output signal (or part of it) isconnected back (feedback) to the negative input terminal ofan op-amp.

(to “‐” terminal)

-+

SummingNode

Amplifier

1/K

Input Output

output/K

MV

RV

oRMo AVVV )(

ooM AKVV )/(

OVoA


oMo

o AVKA

V

1 oM

oo AV

KAKV

KAKAVV

ooM

o

ooM

ooMo

AKAV

AKKAVV

(As ,KAo )oo AKA

Vo=K VM


Negative feedback reduces the closed loop gain to K（this is under the condition that Ao is very high;notice that K is not related to Ao.）

Using resistors (as to be illustrated with non-inverting &inverting Op-Amps next) to obtain appropriate value of Ksuch that the output will not reach the saturation limits.

Thus, negative feedback ensures linear operation of op-amps, and also yield the required voltage gain.


A question for you to think while we move on…

Why are op-amps designed with very high Ao, while therequired voltage gain of amplifiers is much lower inpractice (so we use negative feedback to reduce the gainand ensure linear operation)? Why not to design op-ampswith a lower Ao?

Hint: thinking along the flexibility provided to users andbenefits to manufacturers.


The Op-amp Golden Rules – with Nagative Feedback Circuits

1.Voltage Rule: Through negative feedback, the outputattempts to do whatever is necessary to make the voltagedifference between the inputs zero.

In a negative feedback op-amp, V+ - V- ≈ 0 (very small).

2. Current Rule: The inputs draw no current, due to the very high input impedance of the op-amp.

The input current is so low (0.08 microamps for thestandard 741 op-amp).

A way to materialize the negative feedback concept presented previously.


Linear (not saturated) Op-Amp Circuits

1. Non-Inverting AmplifierVIN is applied to the +ve inputterminal of the op-amp.

A part of VO is fed back to the–ve terminal of the op-amp.

How does negative feedbackhelp?

If a +ve voltage* is applied to :INV

will increase will increase. )( VV OUTV

VIN +

_VOUT

R1

R2

V-

V+

*Similar analysis for a -ve voltage case.


•A part of is fed back to the terminal.(Determined by the voltage divider resistors R1 and R2).

(1)

• increases decreases decreases.

i.e., due to negative feedback, is decreased,and then is decreased.

OUTV V

212RR

RVV OUT

V )( VV OUTV

)( VV

OUTV


•Equilibrium condition forces to a very small value will not reach the saturation limits.• Op-Amp golden voltage rule: Input difference voltage is very small Approximated as zero (practically, small non-zero).

OUTV )( VV OUTV

VV (2)

Applying (2) in (1):

INOUT VRR

RVV

21

2 (3)

Closed loop-gain, AV:

21

221 1)(

RR

RRR

VVA

INOUT

v

(4)


Significant Parameters of Non-Inverting Op-Amps

AV will be always ≥ 1.

VOUT is always in phase with VIN (i.e., with same polarity, or non-inverting).


Non-Inverting Op-Amps

(to be presented a bit later)

Output Resistance: VOUT (=( )VIN) is independent of RL (no matter what RL is used), so ROUT 0.

2

11RR

RL

since V+=V-Input impedance: ZINis very high. Ideal case = .


Further questions to think…

We know that an op-amp’s voltage gain is very high; doyou think if a different voltage gain value (say, from200,000 to 250,000) matters to a non-inverting circuit’soutput?

Discuss if there is any benefit in op-amp manufacturingprocesses for your conclusion above.


Example: Design a non-inverting amplifier with a gain of +10:

101)(21

221

RR

RRR

VVA

INOUT

v

921

RR

21 9RR ; we choose kR 91 and kR 12

Example: A sensor signal Vin in the range of1mV–50mV has to be amplified with a non-inverting op-amp. If the op-amp power supply is±15V, find the maximum allowable gain withoutcausing distortion at output signal Vout.Hint: To avoid output signal distortion,Vout = (1+R1/R2) x (Vin) ≤ 15Volts(where Vin = 50 mV and (1+R1/R2) = gain)

(Answer: 300)


A variation of the non-inverting Op-Amp circuit as a summer:

V1

Vout

10R

R

VCC

VEE

V2

10R

R

Superposition Theorem can be used to find the relationship of Vout and inputs (you can practise it and then use it in the tutorials).

Additional Example

Calculate all voltages and currents in the circuit, complete with arrows for current direction and polarity markings for voltage polarity.


Additional Example 1 (cont’d)

AV = 1+22k/47k = 1.468Vout=3.2Vx1.468=4.698VIR2= IR1 =Vout/(22k+47k)=68.09AVR2= IR2 x22K=1.498VVR1= IR1 x47K=3.2V (it is verified that V+=V‐)


No current

Determine both the input and output voltage in this circuit:

Vout = 2mA(5k+18k)=46 VAv = 1+18k/5k=4.6Vin = 46V/4.6=10 V


Predict how the operation of this operational amplifier circuit will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

• Resistor R1 is open;• S/C across resistor R2.


(Solution)

• Resistor R1 is open: Output saturates positively

• S/C across resistor R2: Output saturates positively



2. Unity Gain Amplifier

For the Non-Inverting

amplifier, if (short

circuit) and very large (

, i.e. open circuit):

01 R

2R

11/21

RRVVA INOUTv

Hence, INOUT VV Unity gain amplifier or Voltage buffer.

VIN +

_VOUT

V-

V+

VIN +

_VOUT

R1

R2

V-

V+


Why we need an amplifier that seems to do nothing?A unity gain amplifier serves as a buffer circuit, whichprovides a means of: Isolating an input signal from a load by using a stage

having unity voltage gain, with very high input impedanceand low output impedance (previously shown), andwithout phase or polarity inversion, as to be explainedagain in the next page.


Acting as an ideal circuit thus minimizing the loadingeffect (for circuits before & after it)

very high input impedance and low output impedance.

(for the next stage)


VIN-

+VOUT

R1

Rf

V+

V-

2. Inverting Amplifier

which connectsto terminal,

provides the negativefeedback connection.

,fROUTV V

Applying op-amp voltage rule:(V+) – (V-) 0 V+ = V-

Thus, V- will be at the same voltage of V+, i.e. ground

potential (0V)!

V- is at Virtual Ground (i.e. not a real ground) – to be continued in the next page.


At the amplifier input there exists a virtual short circuit orvirtual ground! (not a real ground). Remember, it is only a “virtual” ground, i.e., there is no“real” short circuit between V- and ground (see the InternalCircuitry of the Standard 741 Op-Amp given earlier) There is no current through the amplifier input to ground.

Virtual s/c

Current only goesthrough and . The virtual groundconcept permits a simplesolution to determine theoverall gain (otherwisedifficult to solve due to thefeedback signal mixed withinput signal), as shown asthe next pages.

1R fR


Zin

op-amp


Inverting op‐amp equivalent circuit


I = VIN / R1

VIN / R1 = - Vo/Rf

Vo / VIN = - Rf / R1

N

fo

RVI


Significant Parameters of Inverting Op-Amps

VOUT is 180o out of phase to VIN (inverting amplifier):

VIN-

+VOUT

R1

Rf

0

-+

VIN-

+VOUT

R1

Rf

0-+


Significant Parameters of Inverting Op-Amps (cont’d)

The gain can be made less than unity (i.e., an amplifier that attenuateinput)—different from a non-inverting one. Input impedance: unlike the non-inverting amplifier, the input current ofthe inverting amplifier is non-zero (although the input current of an op-amp itself is always close to zero):I=(VIN-0)/R1=VIN/R1, so the input impedance is finite: 1/ RIVZ ININ

(you will use this in the lab).

Output impedance: very close to zero (like the non-inverting amplifier), because 1/0 RRVRIV fNIfOUT ; i.e., OUTV is independent of RL (i.e.,

does not change when RL changes).OUTV

VIN-

+VOUT

R1

Rf

i=0

I: not zero

RL


Example 1:

Fig. 1

What input voltage results in an output of 2V in the circuit of Fig. 1?

VVk

MVRR

V IIf

o 2 201

1

VI = ‐ 2V/50= ‐ 40 mV


VIN-

+VOUT

6.8kΩ

Rf

V+

V-

Another Example: The input signal Vin of the op-amp circuit in thefigure varies from 0.1V to 0.5V. If the op-amp power supply is±15V, find the maximum allowable RF without causing distortionat output signal Vout.

Hint: To prevent output signal distortion,| Vout | = (RF/R1) x (Vin) ≤ 15Volts

where Vin = 0.5 V and R1 = 6.8kΩ (Answer: RF ≤ 204kΩ)


A further question to think…

Do the 2 circuits below have different voltage gain?

VinVout

R1

R2

Rf

VinVout

R1

Rf

• What happens if two op‐amp circuits are used together: – An inverting circuit followed by a non-

inverting one– A non-inverting circuit followed by an inverting

one– a summer followed by a non- inverting one– An inverting circuit followed by a summer– …



Recap/Summary

Op-amp Golden Rules with Nagative Feedback

1.Voltage Rule: the voltage difference between two inputsis zero.

2. Current Rule: two inputs draw no current.


Recap/Summary (cont’d)Two commonly-used circuits

VIN +

_ VOUT

R1

R2

V-

V+

VIN-

+VOUT

R1

Rf

V+

V-

InvertingNon-Inverting


Recap/Summary (cont’d)

Non-Inverting InvertingVoltage gain 1+R1/R2>1 - Rf/R1<0

Input impedance Very high R1Output impedance Very small Very smallAdditional remarks can form a unity-

gain amplifier; withminimum loadingeffect; a goodvoltage supplier

can form aninverter; a goodvoltage supplier

Both are with negative feedback (to V-terminal).


With op-amps, how to build 1. An amplifier of a positive voltage gain

2. An isolation amplifier (to provide isolation of one part of a circuit

from another)

3.A voltage inverter

4.A voltage summer

5.An Amplifier to minimize loading effect of the previous stage

6.A good voltage supplier

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CE2004 - Lecture-Part II-1 · CE 2004 –Circuits & Signal Analysis 6 more explanations in the next...

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