Op Amps II

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Announcements
• Tuesday’s Lecture next week is cancelled
– October 18th
• Assignment 4 is active, due in my mailbox
by 5pm Friday (October 14th)
• Mid-term Thursday, October 27th
Project Dates
• Project Description - one paragraph, plus circuit diagrams and
web links – email to me by Tuesday Nov 1st
• Guidelines and a few useful links are here:
http://www.physics.udel.edu/~jholder/Phys645/Lab/Projects.ht
m . There are many more sites if you search the web (e.g.
google “hobby electronics”)
• Kits are allowed, but circuits must be assembled on the
prototyping board. $50 maximum budget.
• Once approved, I will need a complete parts list, with digikey
or radioshack catalog part numbers. I will place the orders,
but FINDING THE CORRECT COMPONENTS AND
GETTING THEM IN TIME IS YOUR RESPONSIBILITY!!!
Note that most basic components (resistors, capacitors etc.)
are available in the lab.
• The project should (ideally) contain both analog and digital
• You should work with your lab partner, but individual reports
are required for the projects.
• The project counts for 20% of your final grade (I grade them)
• I will leave copies of some good past projects in the lab
(please don’t take them away)
Lecture 12 Overview
• Op amp circuits
– amplifiers
– Adding/ Subtracting
– Integrating Circuit
– Differentiating Circuit
– Active Filters
Recap: Opamps
• DC coupled, very high gain, differential amplifier.
• Feed part of the output back into the inverting
input to get stable operation in the linear
amplification region
• Golden rules under negative feedback:
• The voltage at the inputs is the same (v+=v-)
• No current flows into the opamp (i+=i-=0)
Op amp circuit 2: Inverting Amplifier
• Signal and feedback resistor,
connected to inverting (-) input.
• v+=v- connected to ground
iS  iF  iin  0
iS  iF
vout  v 
vS  v 

v+ grounded, so:
RF
RS
v  v  0
v 0
vS  0
  out
RF
RS
vout
RF
vS

RS
Gain 
vout
R
 F
vS
RS
Op amp circuit 3: Summing Amplifier
• Same as previous, but add more
voltage sources
i1  i2  ..... iN  iF
vS 1 vS 2
vSN
vout

 .....

RS1 RS 2
RSN
RF
vout
 RF

RF
RF
 
vS 1 
vS 2  .....
vSN 
RS 2
RSN
 RS1

If RS1  RS 2  ...  RSN  RS
vout
RF

(vS1  vS 2  ...  vSN )
RS
Summing Amplifier Applications
• Adds signals from a number of waveforms
• Applications - audio mixer
• http://wiredworld.tripod.com/tronics/mixer.html
• Can use unequal resistors to get a weighted sum
• For example - could make a 4 bit binary - decimal converter
• 4 inputs, each of which is +1V or zero
• Using input resistors of 10k (ones), 5k (twos), 2.5k (fours) and 1.25k (eights)
Op amp circuit 4: A non-inverting amplifier
• Feedback resistor still to inverting input,
but no voltage source on inverting input
(note change of current flow)
• Input voltage to non-inverting input
iS  iF
v  v
since iin = 0
v   0 vout  v 

RS
RF
and v - = v + = v S (since no current flows through R)
vout
vout
 RF
 1 
 RS
 RF
 1 
 RS
 
v


vS

vout
RF
Gain 
 1
vS
RS
Op amp circuit 5: Differential Amplifier (subtractor)
i1  i2  0
vout  v 
v1  v 

R1
R2
v  v
v 
R2
v2  v 
R1  R2
vout 
R2
(v2  v1 )
R1
Useful terms:
• if both inputs change together, this is a common-mode input change
• if they change independently, this is a normal-mode change
• A good differential amp has a high common-mode rejection ratio (CMMR)
Differential Amplifier applications
• Very useful if you have two inputs corrupted with the same noise
• Subtract one from the other to remove noise, remainder is signal
• Many Applications : e.g. an electrocardiagram measures the
potential difference between two points on the body
http://www.picotech.com/applications/ecg.html
The AD624AD is an instrumentation amplifier - this is a high gain, dc
coupled differential amplifier with a high input impedance and high CMRR
(the chip actually contains a few opamps)
Op-amp integrator
Op-amp integrator
iF = -iS
CF
d(v out - 0)
v
=- S
dt
RS
1
v out (t) = RS CF
t
òv
-¥
S
(t)dt
Integrator Application: Ramp Generator
V
Integrator response to a constant voltage:
VIN
This is a ramp generator - very
useful in timing circuits...
t
VOUT
V
• What's the integrator response to a square wave?
Integrator Application: Ramp Generator
V
Integrator response to a constant voltage:
VIN
This is a ramp generator - very
useful in timing circuits...
t
VOUT
V
• What's the integrator response to a square wave?
VIN
t
• ...Useful for waveform generators:
VOUT
What does this do?
Op-amp differentiator
iF  iS
vout
dvS
 CS
RF
dt
dvS
vout (t )   RF CS
dt
Differentiator
• Differentiator response to a square
wave?:
Differentiator Application: edge detection
VIN
• Differentiator response to a square
wave?:
• The differentiator is good at
detecting edges, or transitions - very
useful in digital circuits.
t
VOUT
N.B. Don't confuse the differentiator with a differential amplifier (subtractor)!
Other "mathematical operator"
circuits can be easily constructed:
•
•
•
•
•
Logarithm
Antilogarithm
Multiplier
Divider
Square root function
What about complex impedances?
Vout
Z
 F
VS
ZS
Vout
ZF
 1
VS
ZS
Active low-pass filter
A( j ) 
Vout
Z
 F
VS
ZS
1
1
1


1
Z F RF
j C F
ZF 
RF
1  j  RF C F
A( j )  
Max Amplification: RF/RS
Low pass factor: 1/(1+ jωRFCF)
Cut-off frequency (-3dB = 1/√2)
when ωRFCF=1, ie ω0=1/RFCF
RF / RS
1  j  RF C F
e.g. RF/RS=10; 1/RFCF=1
Active high-pass filter
A( j ) 
Vout
Z
 F
VS
ZS
1
Z S  RS 
jC S
RF
A( j )  
RS 

RF
RS 1 
Max Amplification: RF/RS
High pass factor: 1/(1+ 1/jωRSCS)
Cut-off frequency: ωRFCF=1
1
jC S
1
1
jRS C S
e.g. RF/RS=10; 1/RFCF=1
Active band-pass filter
Combine the two:
jRF CS
A( j )  
(1  jRF CF )(1  jRS CS )
Advantages of active filters:
1)no inductors (large, pick-up)
2)buffered (high input impedance, low output impedance)
– so filter performance independent of source and load; can cascade filters
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