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ECEN3714 Network Analysis
Lecture #27
21 March 2016
Dr. George Scheets
www.okstate.edu/elec-eng/scheets/ecen3714
Read 15.4
 Problems 15.8, 9, & 12
 Exam #2 Friday

13 – 15
 EXCEPT Convolution
 Chapter
ECEN3714 Network Analysis
Lecture #28
23 March 2016
Dr. George Scheets
www.okstate.edu/elec-eng/scheets/ecen3714
Problems: Old Exam #2
 Exam #2 this coming Friday, 25 March

13 – 15
 EXCEPT Convolution
 Chapter

HKN Test Review
 Thursday,
24 March, CLB 101, 5:30 pm
Op Amp Characteristics
vp(t)
Zin
vn(t)
+Vcc
+
vout(t) =
Av(vp(t)-vn(t))
Av
-
-Vcc

Zin?
 In

Hopamp(f) f3dB?
 In

M ohms
XX or XXX MHz
Voltage gain Av?
 On
order of 104 - 106
Op Amps: No Feedback
+Vcc
+
vin(t)
vout(t) =
Av(vp(t)-vn(t))
Av
-
-Vcc

Output likely to hit rails
 Unless

tiny voltage
Use: Comparator
 Compares
two voltages
 Yields binary output
Op Amps: Positive Feedback
& No Negative Feedback
0v
0v5v
+
Av
vin(t)
-

vout(t)
0v
Output likely to hit rails
 May
get stuck there
Use: None
 Suppose |Vcc| = 15 v

Op Amps: Positive Feedback
10 v if feedback & input resistors =
5v
+
Av
vin(t)
-

15 v
Output likely to hit rails
 May

vout(t)
get stuck there
Suppose |Vcc| = 15 v
Op Amps: Positive Feedback
7.5 v if feedback & input resistors =
0v
+
Av
vin(t)
-

15 v
Output likely to hit rails
 May

vout(t)
get stuck there
Suppose |Vcc| = 15 v
Op Amps: Positive Feedback
≈ 0 v if feedback & input resistors =
-15 v
+
Av
vin(t)
-

15 v
Output likely to hit rails
 May

vout(t)
get stuck there
Suppose |Vcc| = 15 v
Op Amps: Negative Feedback
-
vin(t)
0v
Av
vout(t)
+
Safe to assume vp(t) = vn(t)
 Safe to assume no current enters Op Amp

 If
low Z outside paths exist
Op Amps: Negative Feedback
-
vin(t)
0v
vout(t)
+
OpAmp feedback makes these two =
 Analyze This

Op Amps: Negative Feedback
0v
0v5v
vin(t)
-
Av
+
vout(t) =
Av[v+(t) – v-(t)]
0v
Stable System
 LTI so long as don't hit Power
Supply Limits & get Clipped Output
 Suppose |Vcc| = 15 v

Op Amps: Negative Feedback
(Actual)
0.0004999 v
5v
vin(t)
-
Av
+

vout(t) =
10,000[v+(t) – v-(t)]
-4.999 v
Suppose…
OpAmp Voltage Gain = 10,000
OpAmp Input Impedance = ∞
External elements are Resistors of R Ω
Op Amps: Negative Feedback
(Ideal)
0.0+ v
5v
vin(t)
-
Av
+

vout(t)
-5.0 v
Suppose…
OpAmp Voltage Gain = ∞External elements are Resistors of R Ω
V+ = V-
Op Amps: Output Load
-
vin(t)
Av
+
vout(t)
Zload
Ideally, load does not effect characteristics
 Practically, load may effect characteristics

 If
Op Amp output can't source or sink
enough current
OpAmps: External Resistors
1.020a
-.1020v
1.020a
Note:
+&feedback
is OK
1Ω
|Vcc| = 25v
Av = 10,000
Zin = 1 MΩ
-
5v
5Ω
+
-11/49*5v = -1.122v
-.1020v
1Ω
.1020a
10 Ω
.1020a
Generally not too
small: X or XX
Ohms.
Current flows too
large.
OpAmps: External Resistors
< 10 μa
10 μa
1 MΩ
|Vcc| = 25v
Av = 10,000
Zin = 1 MΩ
5 MΩ
Generally not too
large: X or XX
M Ohms.
+
1 MΩ
10 MΩ
Current flows
in/out inputs
cannot be ignored.
OpAmps? External Resistors
1.020 ma
-.1020v
1.020 ma
10K Ω
|Vcc| = 25v
Av = 10,000
Zin = 1 MΩ
-
5v
5K Ω
+
-11/49*5v = -1.122v
-.1020v
1K Ω
102.0 μa
10K Ω
102.0 μa
Just Right if in
X or XX
K Ohm range
OpAmps (Analysis Assumptions)

Is there negative feedback?
 Generally,
must be "Yes" to be LTI
 Yes? Then V+ = V|Vcc| = 25v
Av = 10,000
vin
5Ω
1Ω
1Ω
+
vout
10 Ω
OpAmps (Analysis Assumptions)

If LTI, analyze circuit with OpAmp removed.
 No
current enters OpAmp inputs
 Current can enter/exit OpAmp outputs
|Vcc| = 25v
Av = 10,000
vin
5Ω
1Ω
1Ω
+
vin
1Ω
-
5Ω
vout
10 Ω
OpAmp
negative
feedback
makes
V- = V+
1Ω
+
vout
10 Ω
Analyze this circuit.
Ideal OpAmps

Default Assumption on Quiz or Test
…
unless specifically stated otherwise
Voltage Gain = ∞
 Output can source or sink any current
 Input impedance = ∞

 No
current enter or exits inputs
OpAmp BW = ∞
 V+ = V- if negative feedback
 Output doesn't hit Power Supply rails

2nd Order Low Pass Filter
(Two back-to-back 1st order active filters)
|H(ω)|
3dB break point changes.
1
0.707
1st order
2nd order
1
ω