Title: MAE 170 Experimental Techniques Lecture 4 OpAmps and Wheatstone Bridge Prof' McKittrick
1MAE 170Experimental TechniquesLecture 4OpAmps
and Wheatstone BridgeProf. McKittrick
2Announcements
 Midterm exam next week, Oct. 23 at 5 pm
 Consists of 2025 multiple choice and/or T/F
questions (includes today's lecture)  No LabView questions
 8 of your grade
 Lecture will follow immediately after the exam
 Lab 5, Thermocouples and heat transfer
3Topics
 Operational amplifiers
 OPerational AMPlifiers (opamps)
 Basically a differential amplifier having a large
voltage gain, high input impedance and low output
impendence  Used for many applications e.g. filters,
voltage and current regulators, A/D converters,
waveform generators.  Wheatstone bridge
 How to measure an unknown resistance
 Based on Ohms Law (VIR) and on Kirchoffs Law
(current in current out)
4Objectives of this weeks lab experiments
 Construct OpAmp amplifier circuit
 Inverting mode
 Measure frequency response
 Construct another opamp circuit
 e.g. noninverting opamp
 Construct a Wheatstone bridge circuit
 Measure an unknown resistance
Implement above using LabView VIs
5General characteristics of signal amplification
 Many transducers produce signals with low
voltage (mV)  Difficult to transmit low signals
 Many processing systems require voltages in the
range 110 V  Amplitude can be increased with an amplifier
6Signal amplification
 Simple amplifier
 Vi is input differential voltage
 Vo is output voltage, usually higher
 Degree of amplification is the gain, G
 G Vo/Vi
7Gain
 Gain can be lt1 (attenuation) or gt1
 Commonly expressed in a log scale in decibels
 GdB 20 log G 20 log (Vout/Vin)
 Amplifier can distort signal
 E.g. frequency distortion
 V(t) GVmisin(2 pft f)
 Ami in the amplitude of the input sine wave
 And f is the phase angle (either 0 or 180)
8OpAmps
 Practical signal amplifiers are constructed with
opamps. An opamp is a large integrated
circuit, which consists of gt 50 transistors  An opamp has
 noninverting input ()
 an inverting input ()
 one output
9OpAmps
V
Gain g Vout/Vin
Vout
V
Vin V  V
Opamps are written as a triangle
Bode Plot
g
Voltage gain of up to 106 (1 mV ? 1 Volt)
10Types of opamps
11OpAmpsideal open loop
 Two input terminals (,)
 Two power supply voltages (Vs ,)
 One output voltage
Vout g(V  V)
12Ideal opamp
 Infinite voltage gain
 Infinite input impedance
 Zero output impedance
 Infinite bandwidth
 Zero input offset voltage (i.e., exactly zero
out if zero in).
13741 opamp
 Input resistance rd 2 MW
 Ouput resistance ro 75 W
 Gain g 200,000
 CMRR 70 dB
14Real and ideal OpAmps
Parameter Ideal Typical
OpAmp OpAmp Differential
voltage gain, g ? 105109 Common mode voltage
gain 0 105 Gain bandwidth product, GBP ? 120
MHz Input resistance ? 1061012 Output
resistance 0 1001000 ?
15Frequency responseopen loop
 Low frequencies gain constant
 Above 6 Hz, frequency drops at  6 dB/octave
or  20 dB/decade  An octave is a doubling, a decade is a 10 fold
increase in frequency  The cutoff frequency is fc 6Hz
fc
Slope 6 dB/octave
Bode Plot
16Characteristics of opamps
 Gain bandwidth product, GBP
 Product of openloop gain and frequency is a
constant  GBP gopenloopfc
 For the 741, GBP is 1 MHz
 Commonmode rejection ratio, CMRR
17Common mode gain
 The opamp amplifies the difference between two
input signals V and V Vout G(VV)  If both signals are the same, Vout 0
 For real opamps, any signal common to both
inputs will be amplified by a common mode gain  Vo GcmVcm
18Common mode rejection ratio CMRR
 Ideal opamps reject voltages present to both
inputs  However in real opamps small output voltage
results from a change to input commonmode
voltage  Due to mismatching of (transistors, heating of IC
circuits, noise etc.) at the input commonmode
voltage  produces a differential error voltage at the
input error gets amplified along with any other
signal present at the input
19CMRR
 CMRR Gdiff/Gcm 20 log (Vdiff/Vcm) dB
 Gdiff Vo/(VV)
 Gcm Vo/Vcm, Vcm V V
 Then VV Vcm/CMRR
 CMRR
 Gives how much differential error voltage is
produced at the input given Vcm and CMRR  Thus want a large CMRR
20Review principals of feedback
Xo A Xi Input is amplified Xf b
Xo Output is fedback to the input Now, Xi
(Xs Xf), positive feedback, 
negative feedback
21Why feedback?
 Coupling the output back to reinforce/cancel
some of the input  Better control
 Reduce the effect of noise
 Reduces output distortion
 Gain is independent of signal level
22Examples of positive and negative feedback
 Positive feedback
 oscillators
 Negative feedback
 opamps
23An opamp schematic of the 747 opamp
24Golden rules for opamps
 Voltage rule
 The output attempts to do whatever is necessary
to make the voltage difference between the inputs
zero.  DV 0 around any loop
 Current rule
 The inputs draw no current.
 Si 0 at any junction
25Inverting and noninverting opamps
 Inverting opamps
 Output voltage is the opposite sign as input
voltage  Noninverting opamps
 Output voltage is the same sign as input voltage
26Noninverting opamp
Iin
A
Vin

Vout
B
R2
R1
Current flow from B?opamp negligible (high input
impedance from opamp) Current flowing through R1
IR1 Vout/(R1R2) Vn VB IR1R1
27Noninverting opamp
28Noninverting opamp
 The circuit shows a feedback loop
 Output connected to one of the input terminals
 Signal input to the terminal
 Results in a closed loop configuration
29Inverting opamp
For an inverting amplifier, the current rule
tries to drive the current to zero at point A.
This requires This gives the voltage
amplification
30Wheatstone bridge
 Used to determine an unknown resistance in a
circuit
31Original Wheatstone bridge
32Wheatstone bridge
Due to their outstanding sensitivity, Wheatstone
bridge circuits are very advantageous for the
measurement of resistance, inductance, and
capacitance.
Wheatstone bridges are widely used for strain
measurements. (This will be done for Lab 7)
33Kirchoffs laws
 Kirchoffs laws
 The sum of the potential drops around any circuit
loop must equal the sum of the potential
increases  DV 0 around any loop
 At a junction point in a circuit where the
current can divide, the sum of the currents into
the junction must equal the sum of the currents
out of the junction.  Si 0 at any junction
34Wheatstone bridge
The potential drop is zero if I1R3I3R3 and
I2R2I4R4 and as the potential drop 0 I1I2
and I3I4 Unknown resistance is
35What does some of this look like in the lab?
 Protoboard already configured to make inverting
opamp measurements
36In more detail
 Opamp is 8pin black DIP (Dual Inline Package)
 Red 15 V supply
 Green 15 V supply
 Red Vin 5V
 Black ground
 White output