Remember this Demo Our old friend, the inverter, driving another. The parasitic inductance of the wire and the gate-to-source capacitance of the MOSFET are shown [Review complex algebra appendix in Agarwal & Lang for next class]
6.002
Fall 03
2
Damped Second-Order Systems 5V
5V
50Ω
2KΩ
2KΩ
S
C A
B
+ –
large loop
Relevant circuit:
5V + –
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Fall 03
2KΩ
CGS
L B CGS
3
Observed Output
2kΩ
5 vA 0
t
vB 2kΩ
t
0
vC 0
t
Now, let’s try to speed up our inverter by closing the switch S to lower the effective resistance 6.002
Fall 03
4
Observed Output
~50Ω
5 vA 0
t
vB 50Ω
0
t
vC 0
t
Huh! 6.002
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5
In the last lecture, we started by analyzing the simpler LC circuit to build intuition
i (t )
L
vI (t )
6.002
+ –
Fall 03
C
+ v(t ) –
6
In the last lecture…
We solved d 2v 1 1 + v = vI 2 dt LC LC
For input VI
vI
0
t
And for initial conditions v(0) = 0 i(0) = 0 [ZSR]
6.002
Fall 03
7
In the last lecture… Total solution v(t ) = VI − VI cosω t o
where 1 LC
ωo =
v(t ) 2VI
LC
VI
vI
0
t i (t )
L v I (t )
6.002
+ –
Fall 03
C
+
v (t )
–
8
Today, we will close the loop on our observations in the demo by analyzing the RLC circuit R
L
vI (t ) + –
i (t ) C
+ v(t ) –
v(t ) 2VI
LC
VI
vI
0
add R
t
Damped sinusoids with R – remember demo!
See A&L Section 13.6 6.002
Fall 03
9
Let’s analyze the RLC network vA
vI (t )
L
i (t )
+ v(t ) –
R
+ –
C
Node method: vA :
1 t vA − v ∫ (vI − v A ) dt = L −∞ R vA − v dv =C R dt
v:
Recall element rules L:
vL = L t
di dt
1 vL dt = i ∫ L −∞ d 2 v R dv 1 1 + + v= vI 2 dt L dt LC LC
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C:
dvC iC = C dt
v, i state variables 10
Let’s analyze the RLC network vA
vI (t ) + –
L
i (t )
R C
+ v(t ) –
Node method: 1 t vA − v ∫ (v I − v A ) dt = L −∞ R
vA :
vA − v dv =C R dt
v:
1 d 2v ( vI − v A ) = C 2 L dt 1 d 2v ( vI − v A ) = 2 dt LC
dv v A = RC + v dt dv 1 d 2v ( vI − RC − v ) = 2 LC dt dt d 2 v R dv 1 1 + + v = vI 2 dt L dt LC LC
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Fall 03
11
Solving Recall, the method of homogeneous and particular solutions: 1
Find the particular solution.
2
Find the homogeneous solution. L 4 steps
3
The total solution is the sum of the particular and homogeneous. Use initial conditions to solve for the remaining constants.
v = vP (t ) + vH (t )
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12
Let’s solve d 2 v R dv 1 1 + + v = vI 2 dt L dt LC LC
For input VI
vI
0
t
And for initial conditions v(0) = 0 i(0) = 0 [ZSR]
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1
Particular solution d 2 vP R dvP 1 1 + + vP = VI 2 dt L dt LC LC
vP = VI
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is a solution.
14
2
Homogeneous solution
Solution to
1 d 2 vH R dvH + + vH = 0 2 dt LC dt L
Recall, vH : solution to homogeneous equation (drive set to zero)
Four-step method: A Assume solution of the form vH = Ae st , A, s = ? B
Form the characteristic equation f(s)
C Find the roots of the characteristic equation
s1 , s2
D General solution
vH = A1e s1t + A2 e s2t 6.002
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2
Homogeneous solution 1 d 2 vH R dvH + + vH = 0 2 dt LC dt L
Solution to
A Assume solution of the form vH = Ae st , A, s = ? so,
As2est +
R 1 Asest + Aest = 0 L LC characteristic equation
R 1 s + s+ =0 L LC
B
2
s + 2αs + ω 2
2
o
ωo =
=0
α=
C Roots
1 LC
R 2L
s1 = −α + α 2 − ω 2 o s2 = −α − α 2 − ω 2 o
D
General solution
vH = A1e 6.002
Fall 03
⎛⎜ −α + α 2 −ω 2 o ⎞⎟ t ⎝ ⎠
+ A2 e
⎛⎜ −α − α 2 −ω 2 o ⎞⎟ t ⎝ ⎠
16
3
Total solution v(t ) = vP (t ) + vH (t )
v(t ) = VI + A1e
⎛⎜ −α + α 2 −ω 2 o ⎞⎟ t ⎝ ⎠
+ A2 e
⎛⎜ −α − α 2 −ω 2 o ⎞⎟ t ⎝ ⎠
Find unknowns from initial conditions. v(0) = 0 : 0 = VI + A1 + A2
i (0) = 0 : dv i (t ) = C dt
( CA (− α −
) )e
= CA1 − α + α 2 − ω 2 o e 2
so,
(
α −ω 2
2
o
⎛⎜ −α + α 2 −ω 2 o ⎞⎟ t ⎝ ⎠
⎛⎜ −α − α 2 −ω 2 o ⎝
) (
+
⎞⎟ t ⎠
0 = A1 − α + α 2 − ω 2 o + A2 − α − α 2 − ω 2 o
)
Mathematically: solve for unknowns, done. 6.002
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17
Let’s stare at this a while longer… ⎛ α 2 −ω 2 o −αt ⎜⎝
v(t ) = VI + A1e e
⎞⎟ t ⎠
⎛ − α 2 −ω 2 o −αt ⎜⎝
+ A2 e e
⎞⎟ t ⎠
3 cases: α > ωo
Overdamped
v(t ) = VI + A1e
α < ωo
−α1t
v(t ) = VI + A1e e −αt
= VI + A1e e
α = ωo 6.002
+ A2 e
−α 2 t
v t
Underdamped ⎛ j ω 2 o −α 2 ⎞⎟ t −αt ⎜⎝ ⎠
= VI + K1e
VI vI
−αt
jω d t
⎛⎜ − j ω 2 −α 2 ⎞⎟ t o −αt ⎝ ⎠
+ A2 e e −αt − jωd t
+ A2e e
cosωd t + K 2e
−αt
sin ωd t
ωd = ω 2 o − α 2 e jωd t = cosωd t + j sin ωd t
Critically damped Later…
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Let’s stare at underdamped a while longer…
α < ωo
Underdamped contd…
v(t ) = VI + K1e−αt cosωd t + K 2e−αt sin ωd t v(0) = 0 : K1 = −VI
dv i (0) = 0 : i (t ) = C dt
= −CK1αe−αt cosωd t − CK 2ωd e−αt sin ωd t − CK1αe−αt sin ωd t + CK 2ωd e−αt cosωd t
0 = − K1α + K 2ωd Vα K2 = − 1
ωd
v(t ) = VI − VI e
−αt
α −αt cosωd t − VI e sin ωd t ωd
Note: For R = 0
⇒α = 0
v(t ) = VI − VI cosωot Same as LC as expected 6.002
Fall 03
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Let’s stare at underdamped a while longer…
α < ωo
Underdamped contd…
v(t ) = VI − VI e
−αt
α −αt cosωd t − VI e sin ωd t ωd
Remember, scaled sum of sines (of the same frequency) are also sines! -- Appendix B.7
ωo −αt ⎛ α v(t ) = VI − VI e cos⎜⎜ ωd t − tan −1 ωd ωd ⎝
⎞ ⎟⎟ ⎠
v(t ) 2VI
LC
VI
vI
0
6.002
add R
t Fall 03
20
α < ωo
Underdamped contd…
v(t ) = VI − VI e
−αt
α −αt cosωd t − VI e sin ωd t ωd
Remember, scaled sum of sines (of the same frequency) are also sines! -- Appendix B.7
ωo −αt ⎛ −1 α v(t ) = VI − VI e cos⎜⎜ ωd t − tan ωd ωd ⎝
⎞ ⎟⎟ ⎠
v(t ) 2VI
LC
VI
vI
0
add R
t v
α = ωo
Critically damped
underdamped criticallydamped overdamped
t
Section 13.2.3
6.002
Fall 03
21
Remember this? Closed the loop… 5 vA 0
t
vB 50Ω
0
t
vC 0
t
See example 12.9 on page 664 of the A&L textbook for inverter-pair analysis 6.002
Fall 03
22
Intuitive Analysis See Sec. 12.7 of A&L textbook ωo −αt ⎛ −1 α ⎜ v ( t ) V V e cos t tan ω = − − Underdamped I I ⎜ d ωd ωd ⎝
v(t )
e −αt
⎞ ⎟⎟ ⎠
“ringing”
VI
0
2π
t
ωd
Characteristic equation
s2 +
R 1 s+ =0 L LC
s 2 + 2αs + ω 2 o = 0
ωd : Oscillation frequency α : Governs rate of decay
ωd = ω 2 o − α 2
VI : Final value v(0) : Initial value
Q= 6.002
ωo : Quality factor (approximately 2α the number of cycles of ringing) Fall 03
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Intuitive Analysis See Sec. 12.7 of A&L textbook Ringing stops after Q cycles V I
