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Engineering 43. Chp 14-2 Op Amp Circuits. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]. RC OpAmp Circuits. Introduce Two Very Important Practical Circuits Based On Operational Amplifiers Recall the OpAmp. The “Ideal” Model That we Use R O = 0 - PowerPoint PPT Presentation
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[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx1
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Engineering 43
Chp 14-2Op Amp Circuits
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx2
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
RC OpAmp Circuits Introduce Two Very
Important Practical Circuits Based On Operational Amplifiers
Recall the OpAmp
The “Ideal” Model That we Use• RO = 0• Ri = ∞ • Av = ∞• BW = ∞
Consequences of Ideality• RO = 0 vO = Av(v+−v−)• Ri = ∞ i+ = i− = 0• Av = ∞ v+ = v−
• BW = ∞ OpAmp will follow the very Highest Frequency Inputs
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx3
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
RC OpAmp Ckt Integrator
KCL At v− node By Ideal OpAmp• Ri = ∞ i+ = i− = 0• Av = ∞ v+ = v− = 0
0=+v
=+
ivvdtdC
Rvv
o21
1
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx4
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
RC OpAmp Integrator cont
By the Ideal OpAmp Assumptions
Separating the Variables and Integrating Yields the Solution for vo(t)
A simple Differential Eqn
Thus the Output is a (negative) SCALED TIME INTEGRAL of the input Signal
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx5
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
RC OpAmp Ckt Differentiator
By Ideal OpAmp• v− = GND = 0V• i− = 0
KCL at v−
2i
1i
KVL
=+ iii 21 Now the KVL
01111 =++ CviRv
1R
0=+v
+1Cv
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx6
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
RC OpAmp Differentiator cont.
Recall Ideal OpAmp Assumptions• Ri = ∞ i+ = i− = 0• Av = ∞ v+ = v− = 0
Then the KCL
Recall the Capacitor Integral Law
Thus the KVL
02
121 =+=+Rviii O
=
t
C dxxiC
tv 1
+=t
dxxiC
iRtv )(1)( 11
111
Multiply Eqn by C1, then Take the Time Derivative of the new Eqn
)(111
111 t
dtdvCi
dtdiCR =+
1i
1R 2i
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx7
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
RC OpAmp Differentiator cont
In the Previous Differential Eqn use KCL to sub vO for i1• Using
Examination of this Eqn Reveals That if R1 were ZERO, Then vO would be Proportional to the TIME DERIVATIVE of the input Signal• In Practice An Ideal
Differentiator Amplifies Electrical Noise And Does Not Operate well
• The Resistor R1 Introduces a Filtering Action. – Its Value Is Kept As Small As
Possible To Approximatea Differentiator
21 R
vi O=
)(11211 t
dtdvCRv
dtdvCR o
o =+
1i
1R 200 Rv
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx8
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Aside → Electrical Noise ALL electrical signals
are corrupted by external, uncontrollable and often unmeasurable, signals. These undesired signals are referred to as NOISE
The Signal-To-Noise Ratio
Simple Model For A Noisy 1V, 60Hz Sinusoid Corrupted With One MicroVolt of 1GHz Interference
)102sin(10)120sin()( 96 ttty +=
Signal Noise
V1V10
amplitude noiseamplitude signalSN 6
===
Use an Ideal Differentiator)102cos(2000)120cos(120)( 9 ttt
dtdy +=
The SN is Degraded Due to Hi-Frequency Noise
Signal Noise
503
2000120
amplitude noiseamplitude signalSN ===
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx9
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Class Exercise Ideal Differen.
Given Input v1(t)• SAWTOOTH Wave
Let’s Turn on the Lites for 10 minutes for YOU to Differentiate
Given the IDEAL Differentiator Ckt and INPUT Signal
Find vo(t) over 0-10 ms
F 2 =
= k 1
F 2 =
= k 1
Recall the Differentiator Eqn
)(11211 t
dtdvCRv
dtdvCR o
o =+
R1 = 0; Ideal ckt
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx10
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
RC OpAmp Differentiator Ex.
Given Input v1(t)
The Slope from 0-5 mSF 2 =
= k 1
F 2 =
= k 1
sV
dtdvm 3
1
10510
==
For the Ideal Differentiator
)(112 t
dtdvCRvo =
Units Analysis
sFVQF
QsV
sQV
AV
==
===
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx11
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
RC OpAmp Differentiator cont.
Derivative Scalar PreFactor
A Similar Analysis for 5-10 mS yields the Complete vO
F 2 =
= k 1
F 2 =
= k 1
mSin 50:4520
10510102 3
3112
==
==
tVVv
SVS
dttdvCRv
o
o
sCR 36312 102F102101 ==
Apply the Prefactor Against the INput Signal Time-Derivative (slope)
InPutOutPut
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx12
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
RC OpAmp Integrator Example
Given Input v1(t)• SQUARE Wave
For the Ideal Integrator
Units Analysis Again
sFVQF
QsV
sQV
AV
==
===
μF 0.2=
=k 5
=t
ioo dxxvCR
vtv021
)(1)0()(
Vvo 0)0( =
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx13
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
RC OpAmp Integrator Ex. cont.
The Integration PreFactor
0<t<0.1 S• v1(t) = 20 mV (Const)
1
21
1000001.0
12.05
11 =
=
= S
SFkCR
Next Calculate the Area Under the Curve to Determine the Voltage Level At the Break Points
SmVSmVSvCRNote
tmVduuvvCRtvCR
o
t
oo
==
=+= 21.020)1.0(:
20)()0()(
21
012121
0.1t<0.2 S• v1(t) = –20 mV (Const)
tmVSmV
dzzvvCRtvCRt
oo
+=
+= 202
)()1.0()(1.0
12121
Integrate In Similar Fashion over• 0.2t<0.3 S• 0.3t<0.4 S
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx14
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
RC OpAmp Integrator Ex. cont.1 Apply the 1000/S PreFactor and Plot Piece-Wise
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx15
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Design Example Design an OpAmp ckt to implement in
HARDWARE this Math Relation
=t
vdyyvv0
210 25
Examine the Reln to find an
Integrator Summer
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx16
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Design Example The
Proposed Solution The by Ideal OpAmps & KCL & KVL &Superposition
=t
vdyyvv0
210 25
=t
vdyyvv0
210 25
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx17
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Design Example
3
4
21
4 2;5RR
CRRR
==
Then the Design Eqns
=t
vdyyvv0
210 25
The Ckt Eqn
TWO Eqns in FIVE unknowns
This means that we, as ckt designers, get to PICK 3 values
For 1st Cut Choose• C = 20 μF• R1 = 100 kΩ• R4 = 20 kΩ
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx18
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Design Example In the Design
Eqns
=t
vdyyvv0
210 25
=
=
kRFRk
k
2020100
205
2
2
=
=
kRRk
10
202
3
3
If the voltages are <10V, then all currents should be the in mA range, which should prevent over-heating
20μ
100k
20k
10k
20k
Then the DESIGN
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx19
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
LM741 OpAmp Schematic
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx20
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Some LM741 Specs
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx21
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
OpAmp Frequency Response The Ideal OpAmp
has infinite Band-Width so NO Matterhow FAST the inputsignals
However, REAL OpAmps Can NOT Keep up with very fast signals• The Open Loop Gain, AO, starts to degrade
with increasing input frequencies
+
ov
tvtvAtv Oo 21 =
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx22
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Gain∙BandWidth for LM741 520/100 1010 =OA
−20db/DecadeSlope
The Unity Gain Frequency, ft, is the
BandWidth Spec
100BOf
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx23
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
BandWidth Limit Implications Recall the OpAmp
based Inverting ckt
The NONideal Analysis yielded
Noting That All the R’s are Constant; Rewrite above as
For Very Large A
=
o
S
O
RRA
RK
Kvv
222
2
1
11
1
1
22
1
11lim KK
RRAKK
vv
oA
S
O =
=
=
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx24
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
BandWidth Limit Implications As Frequency
Increases the Open-Loop gain, A, declines so the Limit does NOT hold in:
If
Then the Denom in the above Eqn ≠ 1
Thus significantly smaller A DECREASES the Ideal gain
For Typical Values of the R’s the Open-Loop Gain, A, becomes important when A is on the order of about 1000
222
1RRR
A
o
dB 6010log2010 33 =
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx25
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Gain∙BandWidth for LM741 520/100 1010 =OA
Frequency significantly degrades Amplification Performance for Source Frequencies > 10 kHz
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx26
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Voltage Swing Limitations Real OpAmps Can
NOT deliver Unlimited Voltage-Magnitude Output
Recall the LM741 Spec Sheet that show a Voltage Output Swing of about ±15V• For Source Voltages
of ±20 V
If the Circuit Analysis Predicts vo of more than the Swing, the output will be “Clipped”
Consider the Inverting Circuit:
k 7.4
k 1
t 2k6.1cosV4k .15
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx27
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Vswing Clipping Since the Real
OutPut can NOT exceed 15V, the cosine wave OutPut is “Clipped Off” at the Swing Spec of 15V
0 0.5 1 1.5 2 2.5 3 3.5 4-20
-15
-10
-5
0
5
10
15
20V
oIde
al (V
)Ideal vs. Real
0 0.5 1 1.5 2 2.5 3 3.5 4-20
-15
-10
-5
0
5
10
15
20
time (mS)
VoR
eal (
V)
ENGR43_Lec14b_OpAmp_V_Swing_Plot_1204.m
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx28
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Short-Ckt Current Limitations Real OpAmps Can
NOT deliver Unlimited Current-Magnitude Output
Recall the LM741 Spec Sheet that shows an Output Short Circuit Current of about 25mA
If the Circuit Analysis Predicts io of more than This Current, the output will also be “Clipped”
Consider the Inverting Circuit:
k 7.4
k 1
t 2k6.1cosV4 105
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx29
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
CurrentSaturation
Since the Real OutPut can NOT exceed 25mA, the cosine wave OutPut is “Clipped Off” at the Short Circuit Current spec of 25mA
ENGR43_Lec14b_OpAmp_Current_Saturation_Plot_1204.m
0 0.5 1 1.5 2 2.5 3-40
-30
-20
-10
0
10
20
30
40iL
-Idea
l (m
A)
Ideal vs. Real
0 0.5 1 1.5 2 2.5 3-40
-30
-20
-10
0
10
20
30
40
time (mS)
iL-R
eal (
V)
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx30
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Slew Rate = dvo/dt For a Real OpAmp
we expect the OutPut Cannot Rise or Fall Infinitely Fast.
This Rise/Fall Speed is quantified as the “Slew Rate”, SR
Mathematically the Slew Rate limitation
The 741 Specs indicate a Slew Rate of
SRdt
dvo
SV 500,000 SROR µSV 0.5SR
==
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx31
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Slew Rate = dvo/dt If dvin/dt exceeds the
SR at any point in time, then the output will NOT be Faithful to input• The OpAmp can
NOT “Keep Up” with the Input
Consider the Example at Top Right
Then the Time Slope of the Source
tsec
10sinV5.25
sec10
sec10cosV5.2
55
=
tdtdv s
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx32
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Slew Rate = dvo/dt The Maximum value
of dvS/dt Occurs at t=0. Compare the max to the SR
Thus the source Rises & Falls Faster than the SR
When the Source Slope exceeds the SR the OpAmp Output Rises/Falls at the SR• This produces a
STRAIGHT-LINE output with a slope of the SR when the source rises/falls Faster than the SR until the OpAmp“Catches Up” withthe Ideal OutPut
t
sec10sinV5.2
5
secV 400 785
sec101V5.2
max
5
max
=
=
dtdv
dtdv
s
s
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx33
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Slew Rate = dvo/dt
Ideal,ov
Real,ov
µS t
tsec
10sinV5.25
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx34
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Full Power BandWidth The Full Power BW
is the Maximum Frequency that the OpAmp can Deliver an Undistorted Sinusoidal Signal• The Quantity, fFP, is
limited by the SLEW RATE
Determine This Metric for the LM741
The 741 has a max output, Vom, of ±12V
Applying a sinusoid to the input find at full OutPut power (Full Output Voltage)
Recall the Slew Rate tVtv omo sin=
SRdt
dvo
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx35
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Full Power BandWidth Taking d/dt of the
OpAmp running at Full Output
Thus the maximum output change-rate (slope) in magnitude
Recall ω = 2πf Setting |dvo/dt|max = to
the Slew Rate
tV
dttdv
tVtvdtd
omo
omo
cos
sin
=
=
omo
omo
Vdt
tdv
tnVdt
tdv
=
=
max
max
or
cos
SRπfVV omom ==2
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx36
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Full Power BandWidth Isolating f in the last
expression yields fFP:
From the LM741 Spec Sheet• SR = 0.5 V/µS• |Vomax|min = 12V
Then fFP:
omFP πV
SRf2
=
kHz 63.6
cycle 6316
cycle 0066.0
V 12cycle
2V/µS 5.0
LM741,
LM741,
LM741,
LM741,
==
=
=
FP
FP
FP
FP
fS
f
µSf
πf
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx37
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Full Power BandWidth Thus the 741
OpAmp can deliver UNdistorted, Full Voltage, sinusoidal Output (±12V) for input Frequencies up to about 6.63 kHz
kHz 63.6LM741, FPf
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx38
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
WhiteBoard Work
60µF
F igure P F E-3
Cx
8V+
-24V
+
-
Let’s Work These Probs
Choose C Such That
Find Energy Stored on Cx
dttvS
v So =10
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx39
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
All Done for Today
OpAmpCircuitDesign
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx40
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Bruce Mayer, PERegistered Electrical & Mechanical Engineer
Engineering 43
Appendix
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx41
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx42
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx43
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx44
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx45
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx46
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Practical Example Simple Circuit Model For a
Dynamic Random Access Memory Cell (DRAM)
Also Note the TINY Value of the Cell-State Capacitance (50x10-15 F)
Note How Undesired Current Leakage is Modeled as an I-Src
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx47
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Practical Example cont
The Criteria for a Logic “1”• Vcell >1.5 V
Now Recall that V = Q/C• Or in terms of Current
During a WRITE Cycle the Cell Cap is Charged to 3V for a Logic-1• Thus The TIME PERIOD
that the cell can HOLD the Logic-1 value
+=t
CCC dxxiC
vv0
)(1)0(
VtCIV
CtIVV
cell
leak
cell
leakcell 5.15.13
=
sA
FVtH3
12
15
105.11050
)(1050)(5.1
=
=
Now Can Calculate the DRAM “Refresh Rate”
HzmSt
fH
R 6675.111
min, ===
[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx48
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Practical Example cont.2 Consider the Cell at the
Beginning of a READ Operation fCoulfCoulfCoulQ
fCoulfFVQfCoulfFVQ
total
out
cell
8256751506754505.1
150503
=+===
==
fFCtotal 50045050 =+= Then The Output
VfF
fCoulCQV OI 65.1
500825
/ === Calc the Best-Case
Change in VI/O at the READ
When the Switch is Connected Have Caps in Parallel