ENGR 1121
Real World Measurements
Franklin Olin College of Engineering, Spring 2011
The Op Amp and Differential
Amplifier – the workhorses of
Analog Sensor Interfaces
Hazards of not understanding electrical interfaces
From the Arkansas Democrat Gazette, July 25, 1996:
“Two Local Men Injured in Freak Truck Accident Cotton Patch, Ark.”
“Two local men were seriously injured when their pick-up truck left the road and struck a tree near
Cotton Patch on State Highway 38 early Monday morning.
“Thurston Poole, 33, of Des Arc and Billy Ray Wallis, 38, of Little Rock are listed in serious condition at
Baptist Medical Center.
“The accident occurred as the two men were returning to Des Arc after a frog gigging trip. On an
overcast Sunday night, Poole's pick-up truck headlights malfunctioned. The two men concluded
that the headlight fuse on the older model truck had burned out. As a replacement fuse was not
available, Wallis noticed that the .22 caliber bullet from his pistol fit perfectly into the fuse box next
to the steering wheel column. Upon inserting the bullet, the headlights again began to operate
properly and the two men proceeded on east-bound toward the White River bridge.
“After traveling approximately twenty miles and just before crossing the river, the bullet apparently
overheated, discharged and struck Poole in the right testicle. The vehicle swerved sharply to the
right exiting the pavement and striking a tree. Poole suffered only minor cuts and abrasion from
the accident, but will require surgery to repair the other wound. Wallis sustained a broken clavicle
and was treated and released.
"Thank God we weren't on that bridge when Thurston shot his nuts off or we might both be dead stated
Wallis. "I've been a trooper for ten years in this part of the world, but this is a first for me. I can't
believe that those two would admit how this accident happened", said Snyder.
“Upon being notified of the wreck, Lavinia, Poole's wife, asked how many frogs the boys had caught
and did anyone get them from the truck.”
Factoid I :
Almost all modern sensors are intended to ultimately
generate an electrical output.
Factoid 2 :
The sensor generates one of the following, which is a
function of the phenomenon being measured:
• a very small voltage or current
• a variable resistance
• a variable capacitance or inductance
• a variable periodic pulse width or frequency
Basic types of sensor interface circuits
• Differential Amplifier- Best for sensors with neither
output wire internally grounded. Often produces very
small voltages or currents susceptible to noise.
• Balanced Resistive (Wheatstone) Bridge - For
variable resistance sensors, or measuring a voltage
against a reference.
• Balanced Resonant Bridge - For variable inductance
or capacitance sensors.
• Optical Interface - For measurement of variable light
transmission
• Counter/timer Circuits - For totaling, timing or digitally
integrating outputs of sensors that produce pulses.
• Direct Digital Interface - When the native phenomena
is already in binary format
Sensors, detectors and transducers that produce
small voltages or currents
Examples: Thermocouple, photovoltaic (solar) cell, automotive
exhaust oxygen sensor, microphone, guitar pickup, phono cartridge,
piezo-electric (crystal) strain sensors, magnetic tape or computer hard
drive head, pH probe, ECG or EEG measurements from skin contacts,
capacitive switch, Geiger–Müller counter, magnetic position sensor.
Sensor impedance
+
Sensor usually
produces a weak
voltage.
Thevenin equivalent circuit
_
Induced noise
appears on
both input
wires if they
are run close
together
+
_
Example: Thermocouple
Two different metals welded together to form a junction
(e.g., Iron-Constantan, Chromium-Alumel)
Some commonly used thermocouples:
Type
Material
Wire Color
Temperature Range (deg C, deg F)
J
Iron Constantan*
Red/White
0 to 750C
32 to 1382F
K
Chrome Alumel**
Red/Yellow
200 to 1250C
328 to 2282F
T
Copper Constantan Red/Blue
200 to 350C
328 to 662F
E
Chrome Constantan Red/Purple
200 to 900C
328 to 1652F
* 55% copper and 45% nickel alloy
** 95% nickel, 2% manganese, 2% aluminium, 1% silicon alloy
http://www.omega.com/techref/themointro.html
Measurement of small voltages: single-ended
connection
Single-ended (grounded) sensor connection.
Noise reduction method: passive EMI shielding
shield
Single
signal wire,
often run in
a shielded
cable
_
+
A voltage-producing
sensor or transducer
grounded sensor
Op amp circuit
amplifies signal and
induced noise equally
Improved measurement of small voltages: differential
connection
Differential (floating) sensor connections.
Noise reduction method: Common Mode Rejection (CMR).
Example: Thermocouple
Measurement
junction
Floating inputs –
neither is
grounded so
induced noise
voltage appears
equally on both.
_
+
Ungrounded
sensor
Both signal wires
run close to each
other as a twisted
pair or in a
shielded cable
Reference
junction
Op amp circuit
amplifies only the
difference between
the two floating
inputs
Sensors, detectors and transducers that produce
variable resistances
Examples: Thermistor, piezoresistor or semiconductor strain gauge or
pressure gauge, carbon grain microphone (telephone), CDS photocell,
photodetector or phototransistor, skin resistance, thoracic impedance
(for adaptive pacemaker, start of combustion (ionization) sensor, Hall
effect sensor, simple switch.
Usually less susceptible to EMI noise due to lower impedance.
Vref
Rpull-up
Sensor
resistance
varies with
phenomenon
to be
measured
Voltage
varies with
phenomena to
be measured
+
_
Noise voltage
added to input
voltage.
Example:
Strain gauge using a balanced bridge circuit.
VCC
R1
Strain
gauge
Differential
voltage
measurement
R3
Adjustable pot to
“balance the
bridge”
The Operational Amplifier –
a fundamental Analog IC
Almost every analog sensor requires one or more
op-amps for its interface.
They are typically used as differential amplifiers,
with very high input impedance, low output
impedance, and high gain. With negative
feedback, they are highly linear.
Non-Inverting Input v+
Inverting Input v-
+
_
Output y
The Operational Amplifier
(a quick review of analysis and applications)
The Ideal Op-Amp Assumptions:
1. Zero output impedance
2. Infinite input impedance
3. Infinite* differential gain, GD or just G
* implies that v+  v- for VOUT in linear range
4. Zero common mode gain, GCM
These assumptions are only valid at low-medium frequencies (bandwidth limited),
and if we don’t care too much about the maximum rate that the output voltage can
change (slew rate). Every op amp is characterized by a Gain-Bandwidth (GBW)
Product that may limit the use of the device in a high-frequency application.
The Operational Amplifier
In an inverting configuration...
V  V  0
Rf
KCL at V node yields :
VOUT  
Ri
VIN
v- _
v+
+
VOUT
Rf
VIN
Ri
The Operational Amplifier
Ideal Op-Amp analysis of inverting configuration...
IRi  IRf  0
Rf
Ri
VIN
IRi
v-
v+
IRf
( 1)
IRi 
VIN  V
Ri
IRf 
VOUT  V
Rf
Substitute into (1). Solve for V- .
VOUT  G(V  V )  G( 0  V )
VOUT =
G x (V+ - V-)
_+
( 2)
Combine (1) and (2) to eventually get :
G
VOUT 
V
Ri IN
1  ( G  1)
RF
as G  ,
VOUT  
RF
VIN
Ri
The Operational Amplifier
In a non-inverting configuration...
v+
VIN
+
v- _
VOUT
IRf
I Ri  I Rf  0
(1)
I Ri 
0  V
Ri
I Rf 
VOUT  V
Rf
Subst it ut e int o(1). Solve for V- .
VOUT  G (V  V )  G (0  V )
( 2)
Combine(1) and (2) t o event uallyget :
IRi
Ri
Rf
VOUT 
G
Ri
1 G
RF  Ri
as G  ,
VIN
 R 
VOUT  1  F VIN
Ri 

The Operational Amplifier
A special case of the non-inverting
configuration: Unity gain follower...
v+
VOUT  G (VIN  VOUT )
+
VIN
v-
_
VOUT
G
VOUT 
VIN
1 G
as G   VOUT  VIN
No gain, so what good
is this circuit?
The Operational Amplifier
An analog weighted summing amplifier...
Rf
R1
V1
VOUT
Rf 
 Rf
   V1 
V2 
R2 
 R1
v- _
V2
R2
v+
+
VOUT
The Differential Amplifier
Ideally, only amplifies the difference between the two inputs.
Used when neither side of a sensor connection is grounded.
REJECTS COMMON-MODE NOISE
R2
R1
V1
_
V2
+
VOUT
VOUT
 R2 

G 
 R1  R2  V  V 

2
1
 R1 

1  G 
 R1  R2 
For G   VOUT
R1
R2
R2
V2  V1 

R1
Differential Mode v.s. Common Mode
Differential Mode: amplifies V1 - V2
Common Mode: amplifies V1 and V2
R2
Differential Mode Gain : VOUT
R2
V2  V1 

R1
CommonMode Gain : VOUT  GCommon (V2  V1 )
R1
V1
_
V2
+
R1
R2
where G Common is very verysmall
VOUT
Ratio of Common
Mode Gain to
Differential Mode
Gain is the CMRR
Demonstration: Single-ended v.s. differential mode
sensor interface
Demonstration Circuit 1. Single-ended (grounded)
sensor connection, 1kHz signal plus 10kHz noise.
Rf
VOUT
Rf
 Rf

   VSignal  VNoise 
R2
 R1

R1
Vsignal
v- _
VNoise
R2
v+
+
VOUT
Demonstration Circuit 2. Differential amplifier (neither
signal input grounded), 1kHz signal plus 10kHz noise.
VOUT
R1
 Rf
  
 R1

(VSignal   VSignal  )

Rf
Vsignal +
R2
VNoise
Vsignal -
R2
R1
Note: R2
doesn’t
matter
v- _
v+
+
VOUT
What if single-supply OP amp?
(output can’t go below ground)
Use VRef instead of ground for non-inverting input.
Note that new pseudo-ground will be Vref .
R2
R1
V1
_
V2
+
VOUT
VOUT
 R2 

G 
 R1  R2  V  V 

2
1
 R1 

1  G 
 R1  R2 
VOUT  VRe f
R1
R2
VRef
R2
 V2  V1 
R1
Q: V1 = 1 and V2 = 5, what is VOUT?
Note that the input resistances are not equal.
R3=3K
R1=1K
V1
_
V2
+
R2=2K
R3=3K
VO
Q: V1 = 1 and V2 = 5, what is VOUT?
Note that the input resistances are not equal.
A: Vout = 9 V
R3=3K
R1=1K
V1
_
V2
+
R2=2K
R3=3K
KCL at V-
V1  V V  Vo

, V 
R1
R3
KCL at V
V2  V V  0

R2
R3
set V-  V  0 , some algebra...
R
VO  3
R1
VO
 R1  R3

4

V

V

3
5

1
2
1

5
  9 Volts
R

R


3
 2

Improving the input impedance and
gain of the differential amplifier
3-Op Amp Instrumentation Amplifier. Commonly used for
sensor interfacing, especially biomedical applications.
V1
R4
+
_
R3
R2
R1
R2
_
V2
_
+
+
R3
R4
VOUT
Why is it so awesome?
High CMRR, high gain, and high input impedance.
V1
R4
+
_
R3
R2
_
R1
R2
A sensor that
produces a tiny
voltage, like a
strain gauge or
thermocouple
_
+
V2
+
R3
R4
Q: R2=R3=R4=100K. Specify R1 such that the
instrumentation amplifier below produces a differential
gain of 40 dB.
V1
R4
+
_
R3
R2
_
R1
R2
_
V2
+
+
R3
R4
VOUT
Q: R2=R3=R4=100K. Specify R1 such that the
instrumentation amplifier below produces a differential
gain of 40 dB.
V1'  V2'
V  V2
First stage :
 1
R1  2R2
R1
A: R1= 2020 ohms
R1  2R2
V1  V2
R1

V1'  V2' 
V1
R4
+
Second stage : Vo  
_
R3
R2
_
R1
R2
_
V2
+
+
R3
R4
R
Vo   4
R3
 R1  2R2

 R1



R4
V1'  V2'
R3

 V1  V2





VO
1  R  20K
Want 20 log10   1
R1
1 
1  R  20K
or   1
R1
1 

  40dB



  100  R1  202 ohms


Q: What if R2a not the same as R2b ? How does this
affect the differential gain?
V1
R4
+
_
R3
R2a
_
R1
R2b
_
V2
+
+
R3
R4
VOUT
Q: What if R2a not the same as R2b ? How does this
affect the differential gain?
A: Doesn’t matter!
V1'  V2'
V  V2
First stage :
 1
R 2a  R1  R2b
R1
V1'  V2' 
R 2a  R 2b  R1
V1  V2
R1

Second stage : Vo  
V1
+
_
R2a
R1
V’1
i
R4
R3
+
R3
V’2
R
4
VOUT

 R 2a  R 2b  R1 

 V1  V2
R1



R3
R2b
+
Vo  
R4
V1'  V2'
R3
_
_
V2
R4


So why do we
set R2a = R2b ?

Q: Why do we set R2a = R2b ?
A: Maximum internal
headroom.
Consider V’1 and V’2
individually...
i
V1  V2
R1




V1'  V1  iR 2a  V1 
R 2a
V1  V2
R1
V2'  V2  iR 2b  V2 
R 2b
V1  V2
R1
Suppose V1  1, V2  1
V1
+
_
R2a
R1
V’1
i
_
R2b
+


 center at 0 volts
V2'  1  2  3

Unequal : R 2a  20K , R 2b  0, R1  10K
V1'  1  2  3
R3
+
V’2
VO
V1'  1  2  2  5 

 center at  2 volts
'
V2  1  0  1

R  30K 
In both cases,Vo   4 
 V1  V2
R3  10K 

R3
_
V2
Balanced case : R 2a  R 2b  R1  10K
R4
R
4
