Analog Signal Processing

Using Operational Amplifiers

Ping Guo
5/10/2015

Objectives
Understand the input/output characteristics of a linear
amplifier
Understand how to use the model of an ideal operational
amplifier (op amp) in circuit analysis

Know how to design op amp circuits

Understand difference between ideal and real op amps

Analog Signals: Introduction

Acquisition and processing of electric signals from
transducers
Eventually we want to turn them in to digital input for a
micro-controller but
Signal may be too small (mV)

May be noisy due to electromagnetic interference

Have a DC offset, usually due to the transducer and
instrumentation design
Most of these problems can be remedied -> Signal
processing
The simplest and most common form of signal processing
is amplification
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Amplifiers
Ideal amplifier increases the amplitude of a signal without
affecting the phase relationships of different components of
the signal
An amplifier is modeled as a two-port device
Gain:

I/O Impedances:
Zin = Vin / Iin
Zout = Vout / Iout

Operational Amplifier (op amp)

Op amp: A low-cost and versatile integrated circuit
consisting of many internal transistors, resistors, and
capacitors manufactured into a single chip of silicon
Basic building block for:
Amplifiers
Integrators
Summers
Differentiators
Comparators
A/D and D/A converters
Active filters
Sample and hold amplifiers

Integrated circuit based amplifier

Ideal Model for Op Amp

Two inputs are called the inverting input (-) and the
noninverting input (+)
Very high gain
Active device, requires external power (usually 15V)

Ideal Model for Op Amp

Op amp equivalent circuit:

Assumptions:
1.

It has infinite impedance at both inputs. No current is drawn from in

put circuits: I+=I-=0

2.

It has infinite gain. The difference between the input voltages must
be 0:
V+=VIt has zero output impedance. The output voltage dose not depend
on the output current.

3.

Inverting Amplifier
iF

i+=0,i-=0
i1

V+=0
V+=V-=0
i1=iF+i-iF
Vin=Ri1+VVout=V--RFiF

Vout
RF
Au

Vin
R

Non-Inverting Amplifier
iF

i+=i-=0
V+= Vin
i1=iF-i-iF

i1

+
V+= V

=
= 1 +

Vout
RF
Au
1
Vin
R
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Voltage Follower
High input impedance and low output impedance
Isolate the source from the rest of the circuit

RF=0 and R=

Vout
RF
Au
1
1
Vin
R

Vout Vin
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Inverting Summer

Vout

RF
RF
( V1
V2 )
R1
R2

R1 R2 RF
Vout (V1 V2 )

12

Noninverting Summer
R1

= 1+
1

RF

1 + 2 + 3
3

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Difference Amplifier

Superposition: The sum of the individual responses is equivalent

to the overall response to the multiple inputs.
When the inputs are ideal voltage sources, the other sources are
shorted.
When the inputs are ideal current sources, the other sources are
replaced with open circuits
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1.

Replace V2 with a short circuit, effectively grounding R2.

Inverting amplifier

RF
Vout1 V1
R1

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2.

Replace V1 with a short circuit, effectively grounding R1.

V+ is derived form a voltage divider

RF
V V3
V2
R2 RF
Noninverting amplifier

Vout 2

RF
RF
RF
(1
)V3 (1
)(
)V2
R1
R1 R2 RF

Superposition

Vout

RF
RF
RF
RF
V1 (1
)(
)V2
(V2 V1 ) ( R1 R2 R )
R1
R1 R2 RF
R
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Difference Amplifier
Analysis with ideal op amp model

i-=0
V

Vout R1 V1RF
R1 RF

i+=0
RF
V
V2
R2 RF

V-=V+
RF
Vout R1 V1RF
V2
R2 RF
R1 RF
R
Vout F (V2 V1 )
R
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Example
R1=10k , R2=20k , V1=-1V , V2=1V , Vout?

V1
R1
V2

Vo1 V1 1V

R1
R1

R2

+ V (1 R2 )V 3V
o2
2

R2

R2
Vout
Vout

R2

(Vo 2 Vo1 )
R1
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(3 1) 8V
10

R1

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Integrator

V-= V+ =0
iin=Vin /R
iout=CdVout/dt
iin=-iout
Vin
dVout
C
R
dt
t

Vout

Vin ( )d

RC 0
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Discussions:
If a DC voltage is applied
as an input to an ideal
integrator, how does the
output change over time?
What is the output given
a sinusoidal input?

t
t

0
90

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Differentiator
iout
iin

V-= V+ =0
iout=Vout/R
iin=CdVin/dt
iin=-iout
Vout

dVin
RC
dt

Integrators and Differentiators

Differentiators accentuate the effect of noise
Integrators tend to smooth signals over time

21

Discussions:
How is the output signal
look like if (a) a square
wave, (b) a triangular
wave, and (c) a sine wave
input signal is applied?

22

Real Op Amp
The input impedance of a real op amp is not infinite
Does not have infinite gain
The maximum output voltage can be obtained from the amplifier is
about 1.4 V less then the supply voltage
Rail-to-rail input (and/or output) op-amps can work with input
(and/or output) signals very close to the power supply rails.

Delay in response
Slew rate/Rise time

Has a finite bandwidth, which is a function of the gain

established by external components
Slew rate/Rise time

Common-mode rejection ratio (CMRR)

Ad
1
)dB
Vout Ad (V V ) Acm (V V ) CMRR 20 log10 (
Acm
2
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Time Response of Real Op Amp

Slew rate: the maximum time rate of change possible for the
output voltage
Rise time: the time required for the output voltage to go from
10% to 90% of its final value
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Bandwidth
Gain bandwidth product (GBP): the product of the open loop
gain and the bandwidth at the at gain
GBP is a constant along the open-loop gain curve

Closed loop gain

Fall-off frequency
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Sample Datasheet

26

27

Instrumentation Amplifier
Very high input impedance
Large CMRR
Capability to amplifier low-level signals in a noisy environment
Consistent bandwidth over a large range of gains

28

i-= i+ =0
V-=V+
V3 V1 I1R2

V2 V4 I1R2
V1 V2 I1R1
V3 I ( R3 R4 ) Vout
i-= i+ =0
R4V3 Vout R3
V-=V+ V V3 IR3

R2
R2
V3 ( 1)V1 V2
R1
R1
R2
R2
V4 V1 ( 1)V2
R1
R1
R5 ( R3 R4 )
R4
Vout
V4 V3
R3 ( R3 R5 )
R3

R3 R4
R5
V
V4 V
R3 R5

Vout

R4
R2
[ (1 2 )](V2 V1 )
R3
R1
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Comparator
No negative feedback, and the circuit exhibits infinite gain

The output of the comparator is given by

where Vsat is the saturation voltage of the comparator,

Vref is the reference voltage
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Comparator
Vref

Vin

Vin

Vref
Vout

Vout
+Vsat

+Vsat

0
-Vsat

Vin
0

Vref

Vin
Vref

-Vsat

What if the reference voltage is zero?

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Schmitt Trigger
A comparator with hysteresis
To convert analog signals to digital signals (signal
conditioning, function generation)

Non inverting

Inverting

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Inverting Schmitt Trigger

Positive feedback, the output is saturated.
When Vout is positively saturated:
R1
V
Vsat VTH
R1 R2
When Vout is negatively saturated:

R1
V
Vsat VTL
R1 R2

33

Initial Condition:
Vout=+Vsat
V+=VTH
When Vin
Vin>VTH
Vout: +Vsat -Vsat
V+=VTL

Vout
+Vsat

VTL

O
-Vsat

VTH

Vin

When Vin
Vin<VTL
Vout: -Vsat +Vsat
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VTH High threshold

VTL Low threshold

Vout
+Vsat

VTL

O
-Vsat

VTH

Vin

(1) Changing to Vsat

when crossing up the
high threshold
(2) Changing to Vsat
when crossing down
the low threshold

R1
VTH
Vsat
R1 R2
R1
VTL
Vsat
R1 R2
35

Example
The input is a sinusoidal
signal. Whats the output?

Vi
VTH

t
VTL

Vout
Vsat

t
-Vsat

36

Non-inverting Schmitt Trigger

When Vout=-Vsat

R2
R1
Vin
Vsat 0
R1 R2
R1 R2
VTH

R1
Vsat
R2

When Vout=+Vsat

R2
R1
Vin
Vsat 0
R1 R2
R1 R2
R1
VTL Vsat
R2
37

VTH High threshold

VTL Low threshold

Vout
Vsat
VTH

VTL

O
-Vsat

Vin

(1) Changing to Vsat when

crossing up the high
threshold
(2) Changing to Vsat
when crossing down
the low threshold

R1
VTH Vsat
R2
R1
VTL Vsat
R2
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Square Wave Generator

Inverting Schmitt trigger
+
RC circuit

R1
VTH
Vsat
R1 R2
R1
VTL
Vsat
R1 R2

39

VC

Vsat
VTH

Vout

Vsat
0

-Vsat
Assuming Vout=+Vsat -> V+=VTH
The capacitor is being charged by Vout (Vc(0)=0)
Vc when Vc<VTH, V-<V+ -> Vout unchanged
when Vc>VTH, V->V+

-> Vout switch to -Vsat

40

VC
VTH
t

VTL
-Vsat

When Vout=-Vsat -> V+=VTL

The capacitor is discharging
Vc when Vc<VTL, V-<V+

-> Vout switch to +Vsat

Another cycle starts

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VC
Frequency and Period V
TH

2R1
T = 2RC ln 1+

R2

VTL
Vout

1
f=
T

Vsat
t

0
- Vsat

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Vout?

R2=3k

V1
R1=1k

Vout

2 mA

R3=1k

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