ch 6 lecture slides

8/12/2019 Ch 6 Lecture Slides

1/25

Copyright 2013 The McGraw-Hill Companies, Inc. Permission required for

reproduction or display.1

Chapter 6

The Operational

Amplifier


2/25


reproduction or display. 2

The operational amplifier oropamp for short,

finds daily usage in a large variety of

electronic applications.


3/25



op amps have three principal terminals:

the outputthe non-inverting input

theinverting

input


4/25

Ideal Op Amp Rules

No current ever flows into either

input terminal. There is no voltage difference

between the two input terminals.

The op amp actsto make this

happen!




5/25

Apply KVL, Ohms

law, and the ideal op

amp rules to find

Copyright 2013 The McGraw-Hill Companies, Inc. Permission required forreproduction or display. 5

vout Rf

R1

v in


6/25

Example: vin(t)=5sin3tmV,Rf=47 k,R1=4.7 k


vout

(t) = -50sin3tmV


7/25


vout 1Rf

R1

v in

To solve, use KVL, KCL,and op amp rules.

Suggested circuit variables to

perform the circuit analysis


8/25

Example: vin(t)=5sin3tmV,Rf=47 k,R1=4.7 k


vout

(t) = 55sin3tmV


9/25

vout(t) =vin(t)

this design allows connection of apractical voltage source to a load without

experiencing voltage droop!Copyright 2013 The McGraw-Hill Companies, Inc. Permission required forreproduction or display. 9


10/25


vout Rf

R(v1 v2 v3)

This amplifier performs the operationof adding.

It also introduces a gain ofRf/R


11/25


vout

Rf

R

R2

R1

(v1 v

2) This voltage is notaffected by thecircuit on the right.

Op amps can be combined in stages to create the

desired relationship between the outputs and the inputs.


12/25

Zener diode: i=0 if v


13/25

With a reference voltage source Vref, we candrive a constant currentIs=Vref/ Rrefthrough

any loadRL.



14/25

The op amp can be modeled as a dependent

voltage source, with the following components

as shown: input resistanceRi output resistanceRo open loop gainA



15/25

Example:vin(t)=5sin3tmV,

Rf=47 k,

R1=4.7 k


For a 741op amp (A=200,000, Ri=2M, Ro=75

vout(t) = -49.997 sin 3t mV.

An ideal op amp produces vout(t) = -50 sin 3t mV.[Analyze the detailed op amp model using nodal analysis.]


16/25

When A=,Ro=0 , and Ri= , the op amp

behaves according to the ideal op amp rules.

(vd=0 and iin=0)



17/25

Whenv1= v2= vCM, the output should be zero,

but real op amps produce a small common

mode voltage voCM. ACM=| voCM / vCM |



18/25

The enormous but unpredictable gain of the op amp is made

usable through negative feedback.

When vingoes up, vdgoes down, and the op amp reacts by

lowering voutuntil the unwanted non-zero vdis pushed backto zero.


this feedback

resistor allows the

output to affect the

input terminal.


19/25

An op amp requires power supplies.

Usually, equal and opposite

voltages are connect to the V+

andV-terminals.

Typical values are 5 to 24 volts.

The power supply ground must be

the same as the signal ground.


in this example +18V is connected to V+

and -18 V is connected to V-


20/25

vout=10vin,but only up to the18 V supplies



21/25

Non-zero output offsets can be removed:



22/25

Slew rate is the maximum V/s for output.

examples: input (green) and output (red)



23/25

Op amps in open loop can be used to make

decisions. In this case, is vin>2.5 V?



24/25

Design a circuit that provides a logic 1 5 V output if a

certain voltage signal drops below 3 V, and zero volts

otherwise.

Answer:



25/25

This device allows precise amplification of

small voltage differences:

vout=K(v+-v-)

Copyright 2013 The McGraw-Hill Companies, Inc. Permission required forreproduction or display 25

ch 6 lecture slides

Documents