Transcript Lecture 7 Overview - University of Delaware

Announcements
• Assignment 3 due now, or by tomorrow
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• Assignment 4 posted, due next week
– Thursday in class, or Friday 5pm in my
mailbox
• mid-term: Thursday, October 27th
Lecture 11 Overview
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Amplifier impedance
The operational amplifier
Ideal op-amp
Negative feedback
Applications
– Amplifiers
– Summing/ subtracting circuits
Impedances
• Why do we care about the input and output impedance?
• Simplest "black box" amplifier model:
ROUT
VIN
RIN
AVIN
VOUT
• The amplifier measures voltage across RIN, then generates a voltage
which is larger by a factor A
• This voltage generator, in series with the output resistance ROUT, is
connected to the output port.
• A should be a constant (i.e. gain is linear)
Impedances
• Attach an input - a source voltage VS plus source impedance RS
RS
VS
ROUT
VIN
RIN
AVIN
• Note the voltage divider RS + RIN.
• VIN=VS(RIN/(RIN+RS)
• We want VIN = VS regardless of source impedance
• So want RIN to be large.
• The ideal amplifier has an infinite input impedance
VOUT
Impedances
• Attach a load - an output circuit with a resistance RL
RS
VS
ROUT
VIN
RIN
AVIN
• Note the voltage divider ROUT + RL.
• VOUT=AVIN(RL/(RL+ROUT))
• Want VOUT=AVIN regardless of load
• We want ROUT to be small.
• The ideal amplifier has zero output impedance
VOUT
RL
Operational Amplifier
• Integrated circuit containing ~20 transistors, multiple amplifier stages
Operational Amplifier
• An op amp is a high voltage gain, DC amplifier with high input
impedance, low output impedance, and differential inputs.
• Positive input at the non-inverting input produces positive output,
positive input at the inverting input produces negative output.
Operational Amplifier
• An op amp is a high voltage gain, DC amplifier with high input
impedance, low output impedance, and differential inputs.
• Positive input at the non-inverting input produces positive output,
positive input at the inverting input produces negative output.
• Can model any amplifier as a "black-box" with a parallel input
impedance Rin, and a voltage source with gain Av in series with an
output impedance Rout.
Ideal op-amp
• Place a source and a load on the model
RS
+
vout
RL
-
• Infinite internal resistance Rin (so vin=vs).
• Zero output resistance Rout (so vout=Avvin).
• "A" very large
• iin=0; no current flow into op-amp
So the equivalent circuit of an
ideal op-amp looks like this:
Many Applications e.g.
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Amplifiers
Adders and subtractors
Integrators and differentiators
Clock generators
Active Filters
Digital-to-analog converters
Applications
Originally developed for use in analog computers:
http://www.youtube.com/watch?v=PBILL8UypHA
Applications
Originally developed for use in analog computers:
http://www.youtube.com/watch?v=PBILL8UypHA
Using op-amps
• Power the op-amp and apply a voltage
• Works as an amplifier, but:
• No flexibility (A~105-6)
• Exact gain is unreliable (depends on chip, frequency and temp)
• Saturates at very low input voltages (Max vout=power supply voltage)
• To operate as an amp, v+-v-<VS/A=12/105 so v+≈v• In the ideal case, when an op-amp is functioning properly in the
active region, the voltage difference between the inverting and noninverting inputs≈0
Noninverting Amplifier
vO  A(v   v  )

R2 

vO  A vIN  vO
R1  R2 


AR 2 

  Av IN
vO 1 
 R1  R2 
AvIN
vO 
AR2
1
R1  R2
When A is very large:
Take A=106, R1=9R, R2=R
AvIN
vO 
1  AR2
R1  R2
AvIN
vO 
R2
A
R1  R2
vO  vIN
R1  R2
R2
>>1
106 vIN
vO 
6
10
R
1
9R  R
106 vIN
vO 
6 1
1  10 
10
vO  vIN 10
• Gain now determined only by resistance ratio
• Doesn’t depend on A, (or temperature,
frequency, variations in fabrication)
Negative feedback:
• How did we get to stable operation in the linear
amplification region???
• Feed a portion of the output signal back into the input
(feeding it back into the inverting input = negative feedback)
• This cancels most of the input
• Maintains (very) small differential signal at input
• Reduces the gain, but if the open loop gain is ~, who
cares?
• Good discussion of negative feedback here:
http://www.allaboutcircuits.com/vol_3/chpt_8/4.html
Why use Negative feedback?:
• Helps to overcome distortion and non-linearity
• Improves the frequency response
• Makes properties predictable - independent of
temperature, manufacturing differences or other
properties of the opamp
• Circuit properties only depend upon the
external feedback network and so can be easily
controlled
• Simplifies circuit design - can concentrate on
circuit function (as opposed to details of
operating points, biasing etc.)
More insight
• Under negative feedback:
 R1  R2 

vIN
vO  R1 


v v 

0
A
A
v  v
• We also know
• i+ ≈ 0
• i- ≈ 0
• Helpful for analysis (under negative feedback)
• Two "Golden Rules"
1) No current flows into the op-amp
2) v+ ≈ v-
More insight
• Allows us to label almost every point in circuit terms of vIN!
1) No current flows into the op-amp
2) v+ ≈ v-
Op amp circuit 1: Voltage follower
• So vO=vIN
•or, using equations
vO  vIN
R1  0
R2  
• What's the gain of this circuit?
R1  R2
R2
Op amp circuit 1: Voltage follower
• So vO=vIN
•or, using equations
vO  vIN
R1  R2
R2
R1  0
R2  
• What's the application of this circuit?
•Buffer
Useful interface between different circuits:
voltage gain = 1
Has minimum effect on previous and next
input impedance=∞
circuit in signal chain
output impedance=0
RS
VS
ROUT
VIN
RIN
AVIN
VOUT
RL
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
• Applications - audio mixer
• Adds signals from a number of waveforms
• 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: Another 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
and v   v   vS
vout
vout
v   0 vout  v 

RS
RF
 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)