Lecture14 OP Amps
description
Transcript of Lecture14 OP Amps
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Chapter 14: OperationalAmplifiers
1. Ideal op-amps2. Negative feedback3. Applications4. Useful designs5. Integrators, differentiators and filters
Operational Amplifiers
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Introduction: Ideal Operational Amplifier
Operational amplifier (Op-amp) is made of many transistors, diodes, resistors and capacitors in integrated circuit technology.Ideal op-amp is characterized by:Infinite input impedanceInfinite gain for differential inputZero output impedanceInfinite frequency bandwidth
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Equivalent circuit of the ideal op-amp can be modeled by:Voltage controlled source with very large gain AOL
known as open loop gainFeedback reduces the gain of op-ampIdeal op-amp has no nonlinear distortions
Ideal Operational Amplifier
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Ideal Operational Amplifier
A real op-amp must have a DC supply voltage which is often not shown on the schematics
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Op-amp are almost always used with a negative feedback:Part of the output signal is returned to the input with negative signFeedback reduces the gain of op-ampSince op-amp has large gain even small input produces large output, thus for the limited output voltage (lest than VCC) the input voltage vx must be very small. Practically we set vx to zero when analyzing the op-amp circuits.
Inverting Amplifier
with vx =0 i1 = vin /R1
i2 = i1 and
vo = -i2 R2 = -vin R2 /R1
soAv=vo /vin =-R2 /R1
i1
i2
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Inverting Amplifier
Since vo = -i2 R2 = -vin R2 /R1
Then we see that the output voltage does not depend on the load resistance and behaves as voltage source.Thus the output impedance of the inverting amplifier is zero.The input impedance is R1 as Zin=vin/i1=R1
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Inverting Amplifier with higher gain
Inverting amplifier gain vo = -i2 R2 = -vin R2 /R1
Is limited due to fact that it is hard to obtain large resistance ratio.
Higher gains can be obtained in the circuit below where we have:i1=vin/R1=i2 from KCL at N2 we have: i2 + i3= i4
N2
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Inverting Amplifier with higher gain
Higher gains can be obtained in the circuit below where we have:i1=vin/R1=i2 from KCL at N2 we have: i2 + i3= i4
Also from KVL1: –vo=i2R2+i4R4 => i4=(-vo-i2R2)/R4
and from KVL2: i2R2=i3R3 => i3=i2R2/R3
KVL1
KVL2
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Inverting Amplifier with higher gain
Finally using: i2 + i3= i4 andi4=(-vo-i2R2)/R4
i3=i2R2/R3 we havei2+i2R2/R3 =(-vo-i2R2)/R4 => i2(1+R2/R3 +R2/R4)= -vo/R4
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Inverting Amplifier with higher gain
i2(1+R2/R3 +R2/R4)= -vo/R4
Substitutei2=vin/R1 => vin/R1 *(1+R2/R3 +R2/R4)= -vo/R4
to get the voltage gain
vo/vin=-R4/R1 *(1+R2/R3 +R2/R4)
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Inverting Amplifier with higher gain
So if we chose R1=R3=1kandR2=R4=10 kthen the voltage gain is
Av=vo/vin==-R4/R1 *(1+R2/R3 +R2/R4)==-10*(1+10+1)==-120
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Summing Amplifier
The output voltage in summing amplifier is
vo=-if*Rf since vi=0
+vi
-
iA
iB
if
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Summing Amplifier
The output voltage in summing amplifier is
vo=-if*Rf since vi=0
+vi
-
iA
iB
if
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Summing Amplifier
The output voltage in summing amplifier is
vo=-if*Rf since vi=0
if=iA+iB=vA/RA+vB/RB => vo=-(vA/RA+vB/RB)*Rf
+vi
-
iA
iB
if
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Summing Amplifier
The output voltage in summing amplifier is
vo=-if*Rf since vi=0
if=iA+iB=vA/RA+vB/RB => vo=-(vA/RA+vB/RB)*Rf
For n inputs we will havevo=- Rf i(vi/Ri)
+vi
-
iA
iB
if
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Exercise 14.2
Find the currents and voltages in these two circuits:
a) i1=vin/R1=1V/1k=1mA
i2=i1=1mA from KCLvo=-i2*R2=-10V from KVLio=vo/RL=-10mA from Ohms law
ix=io-i2=-10mA-1mA=-11mA
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Exercise 14.2
Find the currents and voltages in these two circuits:
b) i1=vin/R1=5mA
i2=i1=5mAi2*1k= i3*1k=> i3=5mAi4=i2+i3=10mA
vo=- i2*1k- i4*1k=-10 V
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Exercise 14.2
Find expression for the output voltage in the amplifier circuit:
i1
i2
i4
i3
i5
iL
i1=v1/R1=v1/10k
i2=i1=v1/10mA
v3 =- i2*R2=- v1/10k *20k =-2v1
+V3
-
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Exercise 14.2
Find expression for the output voltage in the amplifier circuit:
i1
i2
i4
i3
i5
iL
v3 =- i2*R2=- v1/10k *20k =-2v1
i5=i3+i4=v3/10k +v2/10k
vo =- i5*R5=-(v3/10k +v2/10k )*20k =-2v3 -2v2 =4v1 -2v2
+V3
-
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Positive Feedback
When we flip the polarization of the op-amp as shown on the figure we will get a positive feedback that saturates the amplifier output.This is not a good idea.
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Noninverting amplifier
v1=vin
i1=v1/R1
i2=-i1
vo = v1 - i2*R2= v1 + i1*R2 =
= v1 + R2 *v1/R1 =v1(1+ R2 /R1)
Thus the voltage gain of noninverting amplifier is:
Av= vo / vin = 1+ R2 /R1
iL
i2-i1
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Voltage Follower
=>
vo = v1(1+ R2 /R1)
vo = v1
Special case of noninverting amplifier is a voltage follower
so when R2=0
Since in the noninverting amplifier
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Exercise 14.4
Find voltage gain Av=vo/vin and input impedancea.With the switch openb.With the switch closeda.From KVL: vin=i1*R+i1*R+vo
i2=0 and i1*R=i2*R => i1=0so vin=vo and Av=vo/vin =1
i1
i1
i2
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Exercise 14.4
Find voltage gain Av=vo/vin and input impedancea.With the switch openb.With the switch closeda.Input impedance: Zin=vin/iin =vin/0 =inf
i1
i1
i2
iin
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Exercise 14.4
Find voltage gain Av=vo/vin and input impedancea.With the switch openb.With the switch closed
b. for closed switch: i2=vin/Rand i1*R=i2*R => i1=i2 => vin=i1*R+i1*R+vo so vin=vin/R*R+vin/R*R+vo => -vin=vo
and Av=vo/vin =-1
i1
i1
i2
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Exercise 14.4
Find voltage gain Av=vo/vin and input impedancea.With the switch openb.With the switch closed
b. i2=vin/R Input impedance: Zin=vin/iin =vin/(i1+i2)and i1=i2 => Zin=vin/iin =vin/(2*vin/R)=R/2
i1
i1
i2
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Voltage to Current Converter
Find the output current io as a function of vin
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Voltage to Current Converter
Find the output current io as a function of vin
vin =io*Rf
soio=vin/Rf
io
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Exercise 14.6
a) Find the voltage gain vo/vin
b) Calculate the voltage gain vo/vin for R1=10 k, R2 = 100 k
c) Find the input resistance
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Exercise 14.6
a) Find the voltage gain vo/vin
b) Calculate the voltage gain vo/vin for R1=10 k, R2 = 100 k
c) Find the input resistanceFrom KCL1: vin/R1 =(v2-vin)/R2 => v2/R2= vin(1/R2+1/R1)From KCL2: (v2-vin)/R2+v2/R1+(v2-v0)/R2 =0 =>
v2
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Exercise 14.6
a) Find the voltage gain vo/vin
b) Calculate the voltage gain vo/vin for R1=10 k, R2 = 100 kc) Find the input resistanceFrom KCL1: vin/R1=(v2-vin)/R2 => v2/R2= vin(1/R2+1/R1) (*)From KCL2: (v2-vin)/R2+v2/R1+(v2-v0)/R2 =0 => v2(2/R2+1/R1)= (vin+v0)/R2
(*) v2= vin(1+R2/R1) => vin(1+R2/R1)(2/R2+1/R1)= (vin+v0)/R2
vin (R2 (1+R2/R1)(2/R2+1/R1)-1)=v0
v2v0 / vin =131
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Exercise 14.6
a) Find the voltage gain vo/vin
b) Calculate the voltage gain vo/vin for R1=10 k, R2 = 100 kc) Find the input resistanceFrom KCL1: vin/R1=(v2-vin)/R2 => v2/R2= vin(1/R2+1/R1) (*)From KCL2: (v2-vin)/R2+v2/R1+(v2-v0)/R2 =0 => v2(2/R2+1/R1)= (vin+v0)/R2
(*) v2= vin(1+R2/R1) => vin(1+R2/R1)(2/R2+1/R1)= (vin+v0)/R2
vin (R2 (1+R2/R1)(2/R2+1/R1)-1)=v0 => v0 / vin =100(1+10)0.12-1
v2v0 / vin =131
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Example
Matrix equations for op-amp circuits
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EE 616
Example
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Design of Simple Amplifiers
Practical amplifiers can be designed using op-amp with feedback.
We know that for noninverting amplifierAv= vo / vin = 1+ R2 /R1
so to obtain Av=10 we could use R1=1and R2=9But such low output resistance will draw too much current from the power supply
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Design of Simple Amplifiers
The same gain can be obtained with large resistance values.
But for high output resistance are sensitive to bias current and we must use a filtering output capacitor to remove the noise.
Av= vo / vin = 1+ R2 /R1
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We consider the following op-amp imperfections:1)Nonideal linear operation,2)Nonlinear characteristics3)Dc offset values.
Op-Amp Imperfections in a Linear Mode
Input and output impedances:1)Ideal opamp have Rin=0,
2)Real op-amp has
0; outin RR
1001
;101 12
out
in
R
MR
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We consider the following op-amp imperfections:1)Nonideal linear operation,2)Nonlinear characteristics3)Dc offset values.
Op-Amp Imperfections in a Linear Mode
Voltage gain:1)Ideal op-amp has infinite gain and bandwidth, 2)Real op-amp has the gain that changes with frequency.3)Open loop gain:
BOL
OLOL
ff
j
AfA
1
0
Open loop bandwidth
Terminal frequency ft – gain drops to 1
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Op-Amp Imperfections in a Linear Mode
Negative feedback is used to lower the gain and extend the bandwidth.
Open loop gain
From KVL
id
oOL V
VfA
oidin VVV oOL
oin V
A
VV
OL
OL
OL
in
oCL A
A
AV
VA
111
21
1
RR
R
where
So the closed loop gain
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Op-Amp Imperfections in a Linear Mode
Using open loop gain
We get
OLBOL
OL
OL
OLBOL
OL
OL
OL
in
oCL
Aff
j
AA
Af
fj
A
A
A
V
VA
0
0
0
0
0
11
1
11
So we will get closed loop dc gain
BOL
OL
id
oCL
ff
j
A
V
VfA
1
0
OL
OL
in
oCL A
A
V
VA
0
00 1
closed loop bandwidth
OLBOLBCL Aff 01
closed loop voltage gain
BCL
CLCL
ff
j
AfA
1
0
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Op-Amp Imperfections in a Linear Mode
Comparing to open loop, the closed loop gain is reducedAnd closed loop bandwidth is larger
The gain*bandwidth product stays the same
BOLOL
OLBOLOL
OL
BCLCL
fA
AfA
A
fA
0
00
0
0
11
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Nonlinear Limitations
Nonlinear limitations:1)Output voltage swing is limited and depend on power supply voltage
for
2)Maximum output current is limited
VVtvVVV oDD 12,12,15,15
mAmAtiamplifierAfor o 40,40741
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Nonlinear Limitations
When voltage or current limits are exceeded, clipping of the output signal occurs causing large nonlinear distortions
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Nonlinear Limitations
Another nonlinear limitation is limited rate of change of the output signal known as the slew-rate limit SR
SRdt
dvo Using slew rate we can find maximum frequency known as full-power bandwidth.Assuming
tVtv omo sin
SRVf
tVdt
dv
om
omo
2
cos
omFP V
SRf
2
So the full-power bandwidth
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There are three dc offset values related to op-amp design:1)Bias currents IB+, IB- – related to differential inputs2)Offset current – ideally zero value3)Offset voltage – results in nonzero output for zero inputThey can be represented as additional dc sources in the op-amp model
Dc offset values
2
BBB
III
BBoff III
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Industrial op-amp
741 Amplifier is the most popular amplifier it hasAOL=100000
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741 Amplifier BJT transistor level schematic
Industrial op-amp
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