ece 342 – jose schutt-aine 1 ece 342 solid-state devices & circuits 18. operational amplifiers...
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ECE 342 – Jose Schutt-Aine 1
ECE 342Solid-State Devices & Circuits
18. Operational Amplifiers
Jose E. Schutt-AineElectrical & Computer Engineering
University of [email protected]
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Operational Amplifiers
General terminal configuration with bias
• Universal importance (e.g. amplification from microphone to loudspeakers)
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Common configuration with bias implied but not shown
Operational Amplifiers
2 1( )outv A v v
Gain=A
• Signaling1. Differential input stage2. Difference between input is amplified
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2. Zero output impedance
Operational Amplifiers
Also, op amps are dc (or direct coupled) amplifiers since they are expected to amplify signals with frequency as low as DC.
1. Infinite input impedance
• Ideal Op Amp
4. Infinite CMRR or zero common-mode gain
3. Infinite open-loop gain Ainf
5. Infinite bandwidth
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Differential & Common-Mode Signals
- Differential input signal vID=v2-v1
- Common-mode input signal vICm=0.5(v1+v2)
1 2ID
ICm
vv v
2 2ID
ICm
vv v
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Ideally, vICM should be zero to achieve high CMRR.
• Amplifier will amplify the difference between the two input signals
Operational Amplifiers
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Practical ConsiderationsThe output voltage swing of an op amp is limited by the DC power supply. Since op amp can exhibit high gain, power supply voltage fluctuations must be minimized use decoupling capacitors from power supply
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Inverting Configuration
We introduce RF (or R2) to reduce gain (from inf)
• When RF is connected to terminal 1, we talk about negative feedback. If RF is tied to terminal 2, we have positive feedback
Terminal 2 is tied to ground
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Inverting Configuration
Need to evaluate vo/vI
Assume ideal Op-Amp
2 1( ) 0ov v vA
Since gain is infinite:
v1 is virtual ground
1 2, 0Thus v v
11
1 1
I Iv v vi
R R
Note: A is open-loop gain
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Since input impedance of OP amp is infinite, current through RF is i1
11
Ivi iR
1
0I o o I
F F
v v v vi
R R R
1 1
o I Fo I
F
v v Rv v
R R R
Inverting Configuration
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1
o F
I
v RG
v R
Closed-Loop gain
Observe that the closed-loop gain is the ratio of external components we can make the closed-loop as accurate as we want. Gain is smaller but more accurate.
Inverting Configuration
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2 1 1 /o ov v A v v v A
We assumed that the OP-amp was ideal. If we assume that the gain A is finite = A
Inverting Configuration
11 1
( / ) /in o in ov v A v v Ai
R R
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1
/i o Fo oo F
v v A Rv vv iR
A A R
Still assume infinite input impedance
Inverting Configuration
1
1 1
/
1 / / 1o F F
I F F
v R R ARG
v R R A R A R
1 1F
F
ARG
R A R
Closed-loop gain
forinverting configuration
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11
1
/
/
oR
oo
v A RvR
vi v AG
The reflected impedance of RF is given by
Inverting Configuration
1
1R
RR
A
G
since 11 (1 )
F
A RA
G R
1F
R
RR
A
small
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Inverting Configuration
Since the reflected impedance is so small, v1 is thus very small and the inverting terminal is said to be a virtual ground in this configuration
1
, FRas A GR
We see that
, 0Ras A R Note: To minimize the closed-loop gain (G) on the value of the open-loop gain (A), make 1+RF/R1 << A
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11 1/I I
iI
v vR R
i v R
Input and Output Impedances
- If high gain is required, input impedance will be low- Output impedance is zero
Inverting Configuration
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Find closed-loop gain for A=103, A=104 and A=105 assuming R1=1 kW and RF=100 kW. Assuming vI=0.1 V, find v1.
Example
A |G| v1
Using formulas
Note: Since output of inverting configuration is at terminal of VCVS, output impedance of closed-loop amp is zero.
103 90.83 -9.08 mV104 99.00 -0.99 mV105 99.90 -0.1 mV
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Non-Inverting Configuration
0oID
vv
A Assume gain is
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1 20 IDv v v virtual short
Infinite input impedance
Non-Inverting Configuration
1 0i
1 10o
F a
v v vTherefore
R R
1 a Fo
a
v R Rv
R
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Virtual short
Non-Inverting Configuration
2 1Iv v v
a FI o
a
R Rv v
R
1o F
I a
v RG
v R
1 F
a
RG
R
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The Buffer Stage
0, 1 1FF
a
RIf R G
R
Although voltage gain is low, current gain can be quite high. Buffer stage can be used to interface between processors and switches.
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The Voltage Follower
- Unity gain amplifier- 100% negative feedback
inR
0outR
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No feedback with feedback descriptionA(f) Ani(f) GainAMBOa AMBni Midband gainf2oa f2ni 3 dB freq ptGBWoa GBWni Gain-BW prod
Frequency Response – Non-Inverting
1MBoa
MBniMBoa
AA
A F
a
a F
RF
R R
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Frequency Response – Non-Inverting
1,MBoaIf A F
11a F F
MBnia a
R R RA
F R R
( )1ni
AA f
AF
2
( )1 /
MBoa
oa
AA f
jf f
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Frequency Response – Non-Inverting
2
1( )
1 11
MBoani
MBoa
MBoa oa
AA f
fA F jA F f
1MBoa
MBniMBoa
AA
A F
2 2 1ni oa MBoaf f A F
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2
( )1 /
MBnini
ni
AA f
jf f
Frequency Response – Non-Inverting
2 2 11
MBoani MBni ni oa MBoa
MBoa
AGBW A f f A F
A F
2 2ni MBni ni MBoa oa oaGBW A f A f GBW
Gain-Bandwidth product is constant
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Midband voltage gain is reduced from AMBoa to AMBni
The upper 3-dB frequency will be greater than that of the op amp by the same factor of gain reduction.
If the low-frequency gain of the op amp is AMBoa= 200,000 and with resistors AMBni = 40, the gain is reduced by a factor of 5,000. If the basic 3-db frequency is 5 Hz, the noninverting 3dB frequency will be 25 kHz.
Frequency Response – Non-Inverting
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1(1 )F
iF
ARA
R A R
Frequency Response – Inverting OP Amp
2
Using1 /
MBoa
oa
AA
jf f
1 1 2(1 ) /MBoa F
iMBoa F F oa
A RA
R A R j R R f f
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1 1Using ( 1) MBoa MBoaR A R A
Frequency Response – Inverting OP Amp
1
2 1 1
1
11 /[ ]
Fi
oa MBoa F
RA
fR jf A R R R
and neglecting FR
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1
FMBi
RA
R
Frequency Response – Inverting OP Amp
2 2 1 11 /[ ]i oa MBoa Ff f A R R R
12 2
1 1
1 FMBi i oa MBoa
F
R RA f f A
R R R
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2 21
1 FMBi i oa MBoa
F
RA f f A
R R
Frequency Response – Inverting OP Amp
2 2MBi i oa MBoaA f f A
if RF >> R1
2
( )1 /
MBii
i
AA f
jf f
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Frequency Response – Inverting OP Amp
if RF >> R1 gain-bandwidth is constant
if RF ~ R1 ,
2 21
F
i MBi i MBoa oaF
RGBW A f A f
R R
1
F
i oaF
RGBW GBW
R R
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Frequency Response – Inverting OP Amp
12
1 1
1 oaFi oa
F MBi F
GBW RRf GBW
R R A R R
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10200
0.05MBoA
ExampleDesign an amplifier to couple a microphone to a resistive load. The microphone generates a peak output of 50 mV for a typical voice input level and has a 10-kW output impedance. The output voltage across the 2-kW load is to have a peak value of 10 V. the bandwidth of the voltage gain should be at least 40 kHz. If the GBW of the op amp used is 3106 Hz, calculate the bandwidth of the final design
6
2
3 1015
200ni oa
niMBni MBni
GBW GBWf kHz
A A
The midband voltage gain is:
For a single noninverting stage with this gain, the upper corner frequency is:
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1
21
1 10FR
R 2
22
1 20FR
R
This value of BW will not work need 2 stages
Example (cont’)
Pick first stage gain 10; second stage gain 20. We must then have:
and
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6
2 2
3 10150
20nif kHz
6
2 1
3 10300
10nif kHz
Choose R21 and R22 arbitrarily and use above equations to extract RF1 and RF2; we get:
RF1= 18 kW, RF2 = 38 kW
Next, find 3-dB bandwidth of each stage by dividing respective gains into GBWoa or GBWni
Example (cont’)
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5 5
10 20( )
1 /3 10 1 /1.5 10oA jf jf
2 22 210 10
2 1 19 10 2.25 10
o of f
2 126of kHz
The overall gain is:
At 3dB point, magnitude squared of denominator must be 2
From which
Example (cont’)