ELECTRIC CIRCUITS
ECSE-2010
Spring 2003
Class 13
ASSIGNMENTS DUE
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Today (Tuesday/Wednesday):
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Will do Experiment #5 in Class (EP-5)
Activity 13-1 (In Class)
Thursday:
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Experiment #4 Report Due
Will do Experiment #6 in Class (EP-6)
Activity 14-1 (In Class)
Next Monday:
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No Classes – President’s Day
Next Tuesday:
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Monday’s Class – All Sections Meet Tuesday
EXAM I
• Expect to have Exam I graded
and returned in class next
week Tuesday
REVIEW
• Circuits with C & L:
• iC = C dvC/dt; vL = L diL/dt
• DC Steady State: d/dt = 0 => iCSS = 0; vLSS = 0
• DC Steady State: C => Open Circuit;
L
=> Open Circuit
• vC and iL cannot change instantaneously
• Electrical energy is stored in C and L
• C and L can interchange electrical energy
with the circuit
• Circuits become far more interesting
CAPACITANCE
ic

dvc
ic  C
dt
vc
C [Farads]

d
In DC Steady State;
0
dt
iCSS  0  Open Circuit
INDUCTANCE
iL

vL

di L
vL  L
dt
L [Henries]
d
In DC Steady State;
0
dt
vLSS  0  Short Circuit
SWITCHED CIRCUITS
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Circuits that Contain Switches
Switches Open or Close at t = t0
to = Switching Time
Often choose to = 0
Want to Find i’s and v’s in Circuit
Before and After Switching Occurs
• i(to-), v(t0-); i(to+), v(t0+)
• Initial Conditions of Circuit
INITIAL CONDITIONS
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C’s and L’s Store Electrical Energy
vC Cannot Change Instantaneously
iL Cannot Change Instantaneously
In DC Steady State; C => Open Circuit
In DC Steady State; L => Short Circuit
Use to Find i(to-), v(t0-); i(to+), v(t0+)
Let’s do an Example
EXAMPLE
i1
Switch Opens at t  0
 v1 
12 V
2
 v3 
i2
2
 4
1 F
v2

Assume Switch has been Closed
for a long time before t  0
i3

iC
vC

Find Initial Conditions
i's and v's at t  0 and t  0 +
EXAMPLE
At t  0  :
i1 DC Steady State
 v1  Switch Closed
2
12
i1 (0 )  i 2 (0 ) 
3A
22
v1 (0 )  v 2 (0 )  3x2  6 V


 v3 
iC
i2
2
12 V
i3
 4 
v C C 
v2
 Open Ckt

i C (0 )  0 i3 (0 )  0
v3 (0 )  0
vC (0 )  v 2 (0 )  v3 (0  )  6 V
EXAMPLE

At t  0 :
i1
i1 (0 )  0  v1 (0 )
 v1  Switch Open
2
i2
2
12 V
i3  i C
 v3 
 4
v 1 F
2


vC

6
i 2 (0 )  i3 (0 )  i C (0 ) 
 1 A vC (0 )  v C (0 )  6 V
42
v 2 (0 )  2x1  2 V
v3 (0 )  4x(  1)  4 V



EXAMPLE
i3  0 A
Initial Conditions
t  0
i1  0 A
i2  1 A
i3  1 A
iC  0 A
i C  1 A
v1  6 V
v1  0 V
v2  6 V
v2  2 V
v3  0 V
v3  4 V
vC  6 V
vC  6 V
t  0
i1  3 A
i2  3 A
1ST ORDER SWITCHED DC
CIRCUITS
 Will Look at 1 Order Circuits (Circuits with
1 C or 1 L) with Switched DC Inputs Tomorrow
 Will Use Initial Conditions to Help Us Solve the
st
st
1 Order Differential Equation Relating the
Output to the Input
 Today We Will Look at a 1st Order Circuit using
PSpice
ACTIVITY 13-1
R

100 V
20 nF
vC

ACTIVITY 13-1
• Charge a 20 nF Capacitor to 100 V
thru a Variable Resistor, Rvar:
• Let’s Use a Switch that Closes at t = 0
• Rvar = 250k, 500k, 1 M
• Circuit File Has Been Run:
• C:/Files/Desktop/CE-Studio/Circuits/act_52.dat
• But Let’s Practice Using Schematics
and Take a Quick Look
PSPICE WITH C AND L
• To Describe C and L in Schematics:
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Capacitor: Use Part Named c
Inductor: Use Part Named L
Doubleclick on C or L
Set Value
Set Initial Conditions in Volts and Amps
vC(0+) and iL(0+)
ACTIVITY 13-1
Circuit File
v 1 0 dc 100
R 1 2 {R}
C 2 0 20n ic=0
.param R=250k
.step param R list 250k 500k 1meg
.tran .1 .1 uic
.probe
.end
ACTIVITY 13-1
Print Graphs of vC vs. time
Fill in Table for Activity 13-1
Hand In for Grading
PSPICE TRANSIENTS
• Transient Analysis – Schematics:
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Click on Setup Analysis
Choose Transient
Doubleclick on Transient
Choose Print Step = .1
Choose Final Time = .1
Save
Click Simulate
VARIABLE RESISTOR
• Schematics:
• Choose Part = Rvar:
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Place in Circuit
Doubleclick on Rvar
Set Value = {Rvar}
Change SET to 1.0
• Choose Part = Param
• Doubleclick on Param
• Set Name1 = Rvar, Value1 = 250k
VARIABLE RESISTOR
• Click on Setup Analysis:
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Select Parametric
Select Global Parameter
Select List
Set Variable Name = Rvar
Set List of Values = 250k 500k 1meg
PSPICE WITH C AND L
• To Describe C and L in Circuit File:
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C2 4 5 1n IC=3
L4 4 5 3m IC=4u
Capacitor Named C2 (Must Start Name with C)
Inductor Named L4 (Must Start Name with L)
Positive Terminal = Node 4
Negative Terminal = Node 5
Value of C = 1 nanoFarad
Value of L = 3 milliHenries
vC(0+) = 3 Volts; iL(0+) = 4 microAmps
PSPICE TRANSIENTS
• Transient Analysis – Circuit File:
• .tran Statement
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General Form: .tran tp tf uic
Start time for Analysis is always t = 0
Print Step Size = tp; End time = tf
Usually Choose tp = tf
uic = use initial conditions
Must Specify Initial Conditions when
describe C and L
VARIABLE RESISTOR
• Circuit File:
• R 1 2 {R}:
• Need a .param Statement:
• .param R = 250k
• Need a .step Statement
• .step param R list 250k 500k 1meg
OP AMPS WITH R’s AND C’s
• Can Make Very Useful Circuits by
using Capacitors in Op Amp Circuits
• Replace RF with C in an Inverting
Voltage Amplifier:
• Replace R1 with C in an Inverting
Voltage Amplifier:
OP AMP INTEGRATOR
i1
C
 vC 
R1
vin
i1
vin  0
i1 
R1
0
v out
1
v out  0  v C    i1dt
C
1
 vout  
vin dt

R1C
OP AMP DIFFERENTIATOR
RF
i1
vin
 vF 
C
i1
vout
0
d(vin  0)
i1  C
dt
vout  0  vF  i1R F
 vout
dvin
 R FC
dt
OP AMPS WITH R’s AND C’s
• Replace RF with C in an Inverting
Voltage Amplifier:
• i1 = vin / R1 = ic
• vout = 0 - vc = - vc
 vout
1

vin dt  Op Amp Integrator

R1C
• Interchange R and C:
 vout
dvin
 R FC
 Op Amp Differentiator
dt
EXPERIMENT 5
Use 2 100k's in Parallel
Or Use Digital Pot
Op Amp Differentiator
Cu
i1
i1
50 k
LM741
v out
vin
Use to Measure Cu
EXPERIMENT 5
• Use Cu in Plastic Box; Unknown C
• Use Function Generator for vin:
• Measure vin with Scope, not FG
• Reading of FG is seldom correct
• Use 741 Op Amp:
• Must supply + 5 V and - 5 V for Op Amp:
• Set voltages independently
• Handle wires carefully
EXPERIMENT 5
• Op Amp Differentiator:
• Input Voltage = Sinusoid:
vin  A sin t
dvin
i1  Cu
 ACu cos t
dt
i1 flows through R F
vout  i1R F  AR FCu cos t
 Provides method for measuring Cu
EXPERIMENT 5
Step 1: At 2 kHz, set amplitude of vin such that vR  1 V (rms)
as measured on SCOPE
Step 2: When "repeating" measurements of Step 1
at lower frequencies, do NOT try to readjust the input
to get vR  1 V (rms)
At lower frequencies, the GAIN of the circuit is too low
You cannot make vR  1 V (rms)
Just choose a value of vin that ensures that you are
operating with a sinusoid in, and a sinusoid out
741 PIN LAYOUT
VDC
1, 5 Used for Offsets
vp
v out
vn
Note Pin Layout
VDC
OP AMP PIN LAYOUT
VDC v out
8 7 6 5
Note Indentation
For 741
1
2
vn
3
4
v p VDC
MOUNTING OP AMPS
8
7
1
2
8
Straddle Seam
in Protoboard
1
6
5
3
4
7
2
6
3
5
4