ELECTRIC CIRCUITSECSE-2010Spring 2003

Class 13

ASSIGNMENTS DUE• Today (Tuesday/Wednesday):

• Will do Experiment #5 in Class (EP-5)• Activity 13-1 (In Class)

• Thursday:• Experiment #4 Report Due• Will do Experiment #6 in Class (EP-6)• Activity 14-1 (In Class)

• Next Monday:• No Classes – President’s Day

• Next Tuesday:• 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

cc

dvi Cdt

cv

ci

C [Farads]

CSS

dIn DC Steady State; 0dt

i 0 Open Circuit

INDUCTANCE

LL

div Ldt

L [Henries]Lv

Li

LSS

dIn DC Steady State; 0dt

v 0 Short Circuit

SWITCHED CIRCUITS

• 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• 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

12 V2 4

2 1 F

Cv

1iSwitch Opens at t 0

Assume Switch has been Closed for a long time before t 0

2v

1 v

2i

3 v 3i

+

Find Initial Conditions

i's and v's at t 0 and t 0

Ci

EXAMPLE

12 V2

2 4

At t 0 :3i

3v (0 ) 0

3 v 1i

2i

1 212i (0 ) i (0 ) 3 A

2 2

1 v

1 2v (0 ) v (0 ) 3x2 6 V

2v

Cv

C 2 3v (0 ) v (0 ) v (0 ) 6 V

COpen Ckt

Switch ClosedCi

Ci (0 ) 0 3i (0 ) 0

DC Steady State

EXAMPLE

12 V2

2 4

Cv

At t 0 :

C Cv (0 ) v (0 ) 6 V

1 F

1 1i (0 ) 0 v (0 )

2i

3 Ci i

2 3 C6i (0 ) i (0 ) i (0 ) 1 A

4 2

2v

3 v

2v (0 ) 2x1 2 V 3v (0 ) 4x( 1) 4 V

1 v 1i

Switch Open

EXAMPLEInitial Conditions

t 0

1

2

3

C

1

2

3

C

i 3 Ai 3 Ai 0 Ai 0 A v 6 Vv 6 Vv 0 Vv 6 V

t 0

1

2

3

C

1

2

3

C

i 0 Ai 1 Ai 1 Ai 1 A v 0 Vv 2 Vv 4 Vv 6 V

1ST ORDER SWITCHED DC CIRCUITS

st

st

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

1 Order Differential Equation Relating the

st

Output to the Input

Today We Will Look at a 1 Order Circuit using PSpice

ACTIVITY 13-1

100 V

R

20 nF Cv

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_5-2.dat

• But Let’s Practice Using Schematics and Take a Quick Look

PSPICE WITH C AND L

• To Describe C and L in Schematics:• 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 13-1Circuit Filev 1 0 dc 100R 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

CPrint Graphs of v vs. timeFill in Table for Activity 13-1Hand In for Grading

PSPICE TRANSIENTS

• Transient Analysis – Schematics:• 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:

• 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:• 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:

• 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 • 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:

out in

1

1v v dt R C

inv1R

C

outv

OP AMP INTEGRATOR

01i

1i

in1

1

v 0iR

C v

out C 11v 0 v i dtC

inout F

dvv R C dt

outv inv C

FR OP AMP DIFFERENTIATOR

01i

in1

d(v 0)i Cdt

1i F v

out F 1 Fv 0 v i R

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

• Interchange R and C:

out in1

1v v dt Op Amp IntegratorR C

inout F

dvv R C Op Amp Differentiatordt

EXPERIMENT 5

inv

outv

uC50 k

Op Amp Differentiator

1i

1i

LM741

uUse to Measure C

Use 2 100k's in ParallelOr Use Digital Pot

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:

inv A sin t

in1 u u

dvi C AC cos tdt

1 Fi flows through R

out 1 F F uv i R AR C cos t

uProvides method for measuring C

EXPERIMENT 5in RStep 1: At 2 kHz, set amplitude of v such that v 1 V (rms)

as measured on SCOPE

R

Step 2: When "repeating" measurements of Step 1 at lower frequencies, do NOT try to readjust the input to get v 1 V (rms) At lower frequencies, the GAIN of the

R

in

circuit is too low You cannot make v 1 V (rms) Just choose a value of v that ensures that you are operating with a sinusoid in, and a sinusoid out

741 PIN LAYOUT

pv

nv

outv

DCV

DCV

Note Pin Layout

1, 5 Used for Offsets

OP AMP PIN LAYOUT

1 2 3 4

5678

Note Indentation

pvnv

outvDCV

DCV

For 741

MOUNTING OP AMPS

1 2 3 4

8 7 6 5

1 2 3 4

5678

Straddle Seamin Protoboard