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Nano‐Process and Measurement (Lecture 6)
Prof. Dong‐Weon LeeMEMS & Nanotechnology Laboratory
School of Mechanical Systems EngineeringChonnam National University
Learning Objectives1. Write differential equations for circuits containing inductors and
capacitors.2. Determine the DC steady state solution of circuits containing
inductors and capacitors.3. Write the differential equation of first order circuits in standard
form and determine the complete solution of first order circuits excited by switched DC sources.
4. Write the differential equation of second order circuits in standard form and determine the complete solution of second order circuits excited by switched DC sources.
5. Understand analogies between electrical circuits and hydraulic, thermal and mechanical systems.
Systematic methodology for the solution of first‐ and second‐ circuits excited by switched DC sources
Nano‐Process and Measurement
Transient analysis
Chapter 5
Transient analysis is to describe the behavior of a voltage or a current during the transitionbetween two distinct steady‐state conditions on/off 때마다발생
The response of a first‐order circuit to a switched DC source decaying exponential and rising exponential
Nano‐Process and Measurement
DE for circuits containing Ls & Cs: Summary
Chapter 5
Any circuit containing a single energy storage element
RLC circuit
First‐order system equation
time constant DC gain
natural frequency DC gain
damping ratio
Nano‐Process and Measurement
Differential equation of circuit
Known quantitiesR1=10kRL=50; C=0.1F; L=10mH
FindDifferential equation in IL(t)
Chapter 5
KCL
KVL두개의에너지원을가지고있으므로회로를완벽하게표현하기위해서는두개의식이필요
vc에대한 2차미분방정식은?
Nano‐Process and Measurement
L & C를포함하는 DC 정상상태해: initial and final conditions
DC steady‐state solution: a circuit connected to a DC source for a very long time
Ks : DC gain(직류이득)
Chapter 5
general form
From Ex. 5.1
과도해를구하는데매우유용하게이용
KCL
Nano‐Process and Measurement
L & C를포함하는 DC 정상상태해: initial and final conditions (Cont.)
DC steady‐state solution: a circuit connected to a DC source for a very long time
Chapter 5
Second‐order circuit
Capacitor currents and inductor voltages become zero in the DC steady state.
Initial condition: before a switch is first activatedFinal condition: a long time after a switch has been activated
From Ex. 5.2
Nano‐Process and Measurement
Initial and final conditions
Chapter 5
Known quantitiesR1=1k; R2=5k; R3=3.33k; L=0.1H; V1=12V; V2=4V
FindInductor current IL(t) just before the switch is opened
Steady state: all capacitors behave as open circuits doall inductors behave as short circuits do
Va
가정: the switch has been closed for a long time
Nano‐Process and Measurement
Transient response of first‐order circuits
1. switch 변화가 발생하기 전과 과도응답이 소멸된 후의 정상상태 응답2. C의 전압과 L의 전류의 연속성을 이용하여 초기값을 규정3. switch가 위치를 바꾼 직 후에 대한 DE 작성. 식을 표준형(5.8)으로간소화 한다.4. 회로의 시상수를 구함: capacitive type ; inductive type 5. 완전해 작성
Methodology: first transient response
1차 시스템: 어떤 형태에 에너지를 저장하거나 저장된 에너지를 방출할 수있는 능력을 갖는 시스템 RC 또는 RL system
과도응답: (1) 과도전의 정상상태 응답(2) 회로가 새로운 여진에 적응하는 동안의 과도응답(3) 과도 끝의 정상상태 응답
Chapter 5
Nano‐Process and Measurement
1차 회로의 자연응답
자연응답
‐ natural response: general solution강제함수를 0으로 놓은 상태(homogeneous)
‐ forced response: particular solution강제함수를 고려한 상태(non‐homogeneous)
‐ complete solution: 자연응답과 강제응답의 합 (초기조건 필요)
강제응답
완전응답
Chapter 5
Transient response of first‐order circuits
Excitation frequency
초기값이며완전응답이결정되어야만결정가능
t=0에서스위치가 on된 DC 강제함수
In response to a constant DC excitation
초기조건, 최종조건, 시상수로 1차시스템해석
t=0에서의경우를계산하여 α계산
Nano‐Process and Measurement
Complete response for first‐order circuit
Chapter 5
Known quantities: Initial capacitor voltage; battery voltage; resistor and capacitor values
FindCapacitor voltage as a function of time vc(t) for all t
t=0‐
1k
470 mFStep 1. Steady‐state response
Step 2. Initial condition
Step 3. Writing differential equation
Step 4. Time constant
Step 5. Complete solution
Nano‐Process and Measurement
First‐order circuit의완전응답
Chapter 5
Steady‐state and transient component Natural and forced response
First‐order circuits의해를구하는데필요한모든단계를기술
Nano‐Process and Measurement
Energy storage in capacitors and inductors
Chapter 5
Current와관련
Voltage와관련
00
)(0
ccc
c
Bccc
vdt
dvRCR
dtdv
Cv
VvRiv
00
)(0
LL
LL
BLLL
idtdi
RLRi
dtdiL
IiRiv
Nano‐Process and Measurement
Starting transient of DC motor
Chapter 5
Known quantities: Initial motor current; battery voltage; resistor and inductor values
FindInductor current as a function of time iL(t) for all t
4
0.1H
50VStep 1. Steady‐state response
Step 2. Initial condition
Step 3. Writing differential equation
Step 5. Complete solution
Step 4. Time constant
가정: the switch has been opened for a long time
Nano‐Process and Measurement
Starting transient of DC motor (Cont.)
Chapter 5
Note: In practice it is not a good idea to place a switch in series with an inductor 스위치 open의경우 inductor 전류는즉시변하여 vL(t)가무한대로접근
Steady‐state and transient component Natural and forced response
Nano‐Process and Measurement
First‐order circuit의등가회로표현: a more involved RC circuit
Chapter 5
T_EC N_EC
cv
i
Nano‐Process and Measurement
1차과도의풀이에서 TEC 사용
Chapter 5
Step 1. Steady‐state response
Step 2. Initial condition
Step 3. Writing differential equation
Known quantities: Battery voltage; resistor and capacitor values
Find: The switch has been opened for a very long time. At t=0 the switch closes, and then at t = 50 ms the switch open againCapacitor voltage as a function of time vc(t) for all t
1kW
25mF
0.5k
Part 1.
cvKVL
VT
Nano‐Process and Measurement
1차과도의풀이에서 TEC 사용 (Cont.)
Chapter 5
Step 5. Complete solution
Step 4. Time constant
Part 2. No external forcing function
1. the initial condition () at the time when the switch is opened.2. the time constant of the decay is now the time constant of the switch open, = (R2+R3)C
Note: The circuit has two different time constant
cvci
cT
ccT
ccTc v
CRdtdv
vRdt
dvCvRi 100
Nano‐Process and Measurement
DC 모터의과도꺼짐(turn‐off) :전시간에대한모터전압결정
Chapter 5
가정: the switch has been closed for a long time
Inductor short
Turn‐off 전의전압
Figure 5. 33 참조
sm RRL
Inductive kick : combustion engine ignition
LL
Lm iLR
dtdivRi 0 KVL
Nano‐Process and Measurement
DC motor (turn‐off): determine the motor voltage for all time
An inductor is capable of producing a momentary voltage that is much higher than the voltage of the power source that supplied the current to create its magnetic field.
Automotive spark ignition or Neon bulb
Inductive kick
Nano‐Process and Measurement
Focus on measurements: Coaxial cable pulse response
Chapter 5
Known quantities: Cable length, resistance and capacitance; voltage pulse amplitude and time duration
Find: Cable voltage as a function of time
10 m C=1mF/m
150W
0.2W/m
Nano‐Process and Measurement
Focus on measurements: Coaxial cable pulse response (Cont.)
Chapter 5
Part 1. 0 < t < 1 ms: when the switch closes
vc at t = 1 ms: 4.93 V
Assumption: The short voltage pulse is applied to the cable at t = 0. Assume zero initial conditions.
cv
150
0.2/m
Nano‐Process and Measurement
Focus on measurements: Coaxial cable pulse response (Cont.)
Chapter 5
Part 2. t > 1 ms: the switch opened again
off = RLC
cvci
Nano‐Process and Measurement
2차 회로의 과도 응답: 2차 미분방정식 유도
KCL
KVL
1
2
3
Inductor voltage
is(t) ic(t)
Nano‐Process and Measurement
2차 회로의 과도 응답: 2차 미분방정식 유도
Q. 1. Why is the differential equation expressed in term of iL(t)? [Why not vC(t)]Q. 2. Why did not use equation 5.38 in deriving equation 5.44?
A. 1. iL(t)을알고있으면회로의어떤전압이나전류도유도가능A. 2. 5.44가주어진회로를표현하는유일한미분식이아님
iL(t) vC(t)로대체- forcing function이등가전압의미분형미분방정식의선택은회로의응용에따라달라짐: 유일한방법은없음
IS: 양변미분
Nano‐Process and Measurement
2차 회로의 과도 응답: 2차 미분방정식 유도
vout(t) = vC(t) vout(t) = RTiout(t)
Find a better choice for network analysis
Nano‐Process and Measurement
Examples: Second-order circuits (Source free)
dtdvCii c
CL
Nano‐Process and Measurement
2차 회로의 transient response: 2차 미분방정식 해
natural frequency DC gain
damping ratio
1. 최종값이 1에수렴2. Response가 6s 주기로 oscillation3. Oscillation이시간의흐름에따라 decay
1. DC gain Ks = 1, steady state x(t) = k(t)2. 진동주기: n = 1 T = 2/n = 2 6.28s3. 감쇠비는 damping ratio 에의해결정
Response of switched second-order system with Ks =1, n = 1 and = 0.1
모든미분항 ‘0’
Nano‐Process and Measurement
2차 회로의 transient response: 2차 미분방정식 해 (Cont.)
Response of switched second-order system with Ks =1, n = 1 and ranging from 0.2 to 4car breaking system
Nano‐Process and Measurement
Analogy between electrical and mechanical systems
Mechanical systems
n=2fn= mk
Electrical systems
n=2fn= LC1
= mkc
21
C
= LCR
2
주파수응답을결정하는 parameters
Frequency response is the measure of any system's response at the output to a signal of varying frequency (but constant amplitude) at its input.
C
Nano‐Process and Measurement
Natural response of a second-order circuit
변수: s (공학수학에서는 )
Case 1: 두개의실근 > 1 이상의경우에만발생 over damping을초래(Ex. An overdamped door-closer will take longer to close the door than a critically damped door close)Case 2: 실수중근 = 1의경우에만발생 critical damping을초래(Ex. An example of critical damping is the door-closer seen on many hinged doors in public buildings)Case 3: 공액인복소수 < 1의경우에만발생 under damping을초래(Ex. The system will oscillate at the damped frequency)
http://en.wikipedia.org/wiki/Damping
c
Nano‐Process and Measurement
Natural response of a second-order circuit (Cont.)
Natural responses of several second-order circuits
j term 추가 1=2
공학수학: 2.2 상수계수를 갖는 2계 제차미분방정식
Nano‐Process and Measurement
Forced response
Forced response
The derivative term become zero in response to a constant DC excitation
The forcing function f(t)is equal to a constant F
Complete response
- Some of the two response- Three casesoverdampedcritically dampedunderdamped
Nano‐Process and Measurement
Natural response of a second-order circuitKnown quantitiesR1=8k; R2=8k; C=10F; L=1H; VS=?V
FindNatural response of D.E. of inductor current IL(t)
Ks = 1, n = 316.2 rad/s and = 0.04
두개의초기조건을알면두상수를구할수있음
Natural response
: in case of under damped system
R=R1R2=4k
by TEC
KCL
임계감소가 되는 R 값은?
“We set source equal to zero”
Nano‐Process and Measurement
Second-order transient response
1. Switch 변화가 발생하기 전(t=0-)과 과도응답이 소멸된 후의 정상상태응답(t )
2. C의 전압과 L의 전류의 연속성을 이용하여 초기값을 규정3. switch가 위치를 바꾼 직 후에 대한 DE 작성. 식을 표준형(5.49)으로 간소화한다.4. 회로의 parameter를 구함: n and 5. 완전해 작성 > 1, =0 and <1
Methodology: Second-order transient response
> 1
=0
<1
6. 두 상수를 구하기 위해 초기값 대입
Nano‐Process and Measurement
Complete response of a second-order circuit
FindComplete response of D.E. of inductor current IL(t)Step 1. Steady-state responsea. before the switch closes, the current in the circuit must be zero
Step 2. Initial condition
Step 3. Writing differential equation
Assumptions: vC(0) = 5V
iL(0-) = 0A and vC(0-) = 5V
Step 4. n and
iL() = 0A and vC() = 25V
vC(0) = 5V
Over-damped system
KVL
Nano‐Process and Measurement
Complete response of second-order circuits (Cont.)Step 5. Complete solution
Step 6. Solve for the constants
iL() = 0A : forced response
Nano‐Process and Measurement
Critically damped second-order circuit
FindComplete response of D.E. of capacitor voltage VC(t)Step 1. Steady-state responsea. before the switch closes, inductor current in the circuit must be zero
Step 2. Initial condition
Step 3. Writing differential equation (KCL)
Assumptions: none
iL(0-) = 0A and vC(0-) = 0VIS = 5A; R = 500 ; C = 2 F; L = 2 H
iL() = 5A and vC() = 0V
iL(0-) = iL(0+) = 0A vC(0-) = vC(0+) = 0V
미분방정식의 변수 이므로 두 개의초기조건은 vC(0+)와 dvC(0+)/dt
0
KCL
Nano‐Process and Measurement
Step 6. Solve for the constants
Step 4. n and
2차시스템은 critically damped system
xF = iLF = iL() = 0A
첫번째조건
표준형과비교
Step 5. Complete solution
두번째조건
Critically damped second-order circuit (Cont.)Nano‐Process and Measurement
Underdamped second-order circuit
FindComplete response of D.E. of inductor current IL(t)Step 1. Steady-state responsea. before the switch closes, inductor current in the circuit must be zero
Step 2. Initial condition
Step 3. Writing differential equation
Assumptions: vC(0) = 2V
iL(0-) = 0A and vC(0-) = 2VVS = 12V; R = 200 ; C = 10 F; L = 0.5 H
iL() = 0A and vC() = 12V
iL(0-) = iL(0+) = 0AvC(0-) = vC(0+) = 2V
diL(0+)/dt by KVL
0
iL=iC
KVL
Nano‐Process and Measurement
Step 5. Complete solution
Step 6. Solve for the constants
Step 4. n and
2차시스템은 underdamped system
xF = iLF = iL() = 0A
첫번째조건
Underdamped second-order circuit (Cont.)Nano‐Process and Measurement
Step 6. Solve for the constants (Cont.)
두번째조건
By Euler’s identity
Underdamped second-order circuit (Cont.)Nano‐Process and Measurement
Maximum value of v(t) for t>0
VS = VR + VL + VC
L: short and C: open
dtdv
Cii cCL
Nano‐Process and Measurement
Inductor current
iiRiC
Characteristic polynomial로부터 U.D system
KVL
KCL
iR
iC
Nano‐Process and Measurement
Time determination
ii1
i2
)0(. 21 iiicf
)3.( 11 ViVcf
V1
KCL
KVL
다른 표현
Nano‐Process and Measurement
Summary1. L과 C가포함된전기회로의수학적modeling: differential equation 필요등가회로의표현기법(테브난등가) 및 L과 C의미분형적분형이용
2. DC 정상상태의해: 미분항이 ‘0’으로표현됨 L은 short circuit으로 C는 open circuit로동작
3. 1차시스템회로를미분방정식의표준형으로작성 complete solution = steady-state + transient : time constant와 gain
4. 2차시스템회로를미분방정식의표준형으로작성 : DC source에의한여진 complete solution : DC gain, eigen frequency and damping ratio
systemElectrical input Electrical output
First or second systems(RL, RC or RLC를포함하는전기회로)
time과 frequency에따른 electrical output의특성
Nano‐Process and Measurement