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1st order transientsin time domain
Electrical circuits 2 – Lecture 6 / 7
Pavel Máša
XE31EO2 - Pavel Máša - Lectures 6, 7
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• If in the circuit occur:– Source is connected
– Source is disconnected
– Circuit element is connected or disconnected (R, L, C, …)
– The value of some circuit parameter changes (R, L, C, amplification, …)
The values of voltages and currents in the circuit have to be changed
The change is not instantaneous
The change will be called transientThe transients are quite common in ordinary life; whenever some physical system is deflected from its equilibrium
position, and force, which cause such deflection stops its action – pendulum swinging, swing swinging, spring oscillations, car shock absorbers oscillations, …
How to compute waveforms of transients?
What affects transients?
How many forms it could have?
INTRODUCTION
XE31EO2 - Pavel Máša - Lectures 6, 7
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• With the ideal resistor, or ideal controlled source, where the relation between voltage and current is just multiplication by a constant, any change is instantaneous
• Just phenomena, related to energy supply takes nonzero time – if such phenomena had been instantaneous, the sources would deliver infinite instantaneous power
To make the transient in the circuit happen, it has to contain energy storing circuit elements – C, L
Order of the transients / circuits• Each circuit may be described by set of an integral‐differential equations
• By elimination of circuit variables (by successive differentiation of primary circuit equations and by substitution of circuit variables and its derivatives into other equations) we obtain nth order differential equation of selected circuit variable
where– constants are combinations of R, L, C, M parameters of passive circuit elements and K, R, G, H
parameters of controlled sources
– is linear combination of independent voltage sources and its derivatives
The order of resulting differential equation (transient) is at most equivalent to the number of energy storing circuit elements (L, C) –number of incompatible inductors and capacitors
an
dny
dtn+ an¡1
dn¡1y
dtn¡1+ ¢ ¢ ¢ + a1
dy
dt+ a0y = x(t)an
dny
dtn+ an¡1
dn¡1y
dtn¡1+ ¢ ¢ ¢ + a1
dy
dt+ a0y = x(t)
a0; a1; : : : ana0; a1; : : : an
x(t)x(t)
Both resistors and capacitorsmay be joined
each in one circuit element
⇒XE31EO2 - Pavel Máša - Lectures 6, 7
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1. Find energetic initial conditions (steady state before the change)
2. Elimination of circuit variables
3. Find general solution of homogeneous differential equation
preferably by solution of characteristic equation
the form of general solution is different depending on kind of roots of characteristic equation:
GENERAL PROCEDURE IN TIME DOMAIN
an
dny
dtn+ an¡1
dn¡1y
dtn¡1+ ¢ ¢ ¢ + a1
dy
dt+ a0y = 0an
dny
dtn+ an¡1
dn¡1y
dtn¡1+ ¢ ¢ ¢ + a1
dy
dt+ a0y = 0
y0(t)y0(t)
an¸n + an¡1¸
n¡1 + ¢ ¢ ¢+ a1¸ + a0 = 0an¸n + an¡1¸
n¡1 + ¢ ¢ ¢+ a1¸ + a0 = 0
roots
Distinct real
Repeated real (multiplicity m)
Complex conjugated
y0(t) =
nXk=1
Ake¸kty0(t) =
nXk=1
Ake¸kt
y0m(t) =¡A1 + A2t + A3t
2 + ¢ ¢ ¢+ Amtm¡1¢e¸ty0m(t) =
¡A1 + A2t + A3t
2 + ¢ ¢ ¢+ Amtm¡1¢e¸t
¸1;2 = ¡®§ j!¸1;2 = ¡®§ j!
K1e¸1t + K2e
¸2t = e¡®t (A sin !t + B cos !t) =
= D sin (!t + Ã)
K1e¸1t + K2e
¸2t = e¡®t (A sin !t + B cos !t) =
= D sin (!t + Ã)
Form does not depend on the kind of exciting sources, just on circuit elements
λ < 0 (asymptotically) stable circuit (when passive, then always)XE31EO2 - Pavel Máša - Lectures 6, 7
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4. Complete solution of transient has also particular solution
the form of particular solution depends on the kind of exciting sources (it is steady state after the change)
5. How to find constants Ak (detailed description and examples in the lecture 2nd order transients)
– N ‐ 1 × differentiate complete solution (where n is circuit order)
– We need to know solution of the equation at certain time – and we know it – energetic initial conditions! But, it give us just one equation…
– N ‐ 1 × differentiate primary circuit equations, substitute t = 0 and energetic initial conditions – we obtain mathematical initial conditions
– Finally solve system of linear equations with n unknown variables Ak
source
‐ 0 0DC X0 Y0
AC
periodical
x(t)x(t) yp(t)yp(t)
Xm sin(!t + ')Xm sin(!t + ') Ym sin(!t + Ã)Ym sin(!t + Ã)
X0 +1P
k=1
Xmk sin(k!t + 'k)X0 +1P
k=1
Xmk sin(k!t + 'k) Y0 +1P
k=1
Ymk sin(k!t + Ãk)Y0 +1P
k=1
Ymk sin(k!t + Ãk)
yp(t)yp(t)
y(t) = yo(t) + yp(t)y(t) = yo(t) + yp(t)
XE31EO2 - Pavel Máša - Lectures 6, 7
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1ST ORDER CIRCUIT WITH CAPACITOR
Connection of voltage source to the capacitor without initial energy
Intuitive description of circuit operation:• In the moment, when we connect voltage source, the voltage across resistor is U, and
the passing current is . The current charge the capacitor.
• In the time interval Δt the charge q = I Δt is delivered to the capacitor, so the voltage across capacitor rise of
• But the voltage across resistor decrease of ΔU, so the current decrease, and the charge rate also decrease. It continues till the capacitor is fully charged.
I = URI = UR
¢U = qC = I¢t
C¢U = qC = I¢t
C
Mathematical solution:1. Initial condition – in this example we suppose circuit without accumulated energy ⇒2. Circuit equation – nodal analysis
to find general solution mesh analysis is also possible, but the initial condition is not continuous
uc(0) = 0uc(0) = 0
Cduc(t)
dt+
uc(t)¡ U
R= 0 ) RC
duc(t)
dt+ uc(t) = UC
duc(t)
dt+
uc(t)¡ U
R= 0 ) RC
duc(t)
dt+ uc(t) = U
3. Characteristic equation and its solution by method of variation of parameters→ general solution
RC ¸ + 1 = 0 ) ¸ =¡1
RCRC ¸ + 1 = 0 ) ¸ =
¡1
RC
uco = Ke¸t = Ke¡tRCuco = Ke¸t = Ke¡tRC
4. Steady state in the circuit after the transient dies away→ particular solutiona) DC source
ucp = uc(1) = Uucp = uc(1) = UXE31EO2 - Pavel Máša - Lectures 6, 7
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5. Substitute t = 0 into complete solution→ initial condition and calculate K
uc(0) = Ke0 + U = K + U ) 0 = K + U ) K = ¡Uuc(0) = Ke0 + U = K + U ) 0 = K + U ) K = ¡U
uc(t) = U³1¡ e
¡tRC
´uc(t) = U
³1¡ e
¡tRC
´We define time constant of the circuit
¿ =¡1
¸= RC¿ =
¡1
¸= RC
Theoretically, the dying away of uco(t) lasts infinitely long, however:
time % of steady state
τ 63.213τ 95.025τ 99.33
• Usually, we suppose, the substantial part of transient is finished at time t = 3τ• Time constant is essential parameter, limiting maximum frequency of amplifiers, busses and
other circuits• Time constant is given by intersection of a tangent line at the origin with steady state value
XE31EO2 - Pavel Máša - Lectures 6, 7
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4. Steady state in the circuit after transient dies away → particular solutiona) Sine wave source
Compare transient solution in the RC circuit with parameters R = 850 Ω, C = 1 μF witha) DC excitation U = 90 Vb) AC excitation u(t) = 90 sin(2¼ 5000 t) Vu(t) = 90 sin(2¼ 5000 t) V
Time constant ¿ = RC = 850 ¢ 10¡6 = 0:85 ms ) ¸ =¡1
0:00085= ¡1176:47¿ = RC = 850 ¢ 10¡6 = 0:85 ms ) ¸ =
¡1
0:00085= ¡1176:47
uc(t) = Ke¡1176:47t + up(t)uc(t) = Ke¡1176:47t + up(t)Complete solution
DC up(t) = Uup(t) = U uc(t) = 90(1¡ e¡1176:47t) Vuc(t) = 90(1¡ e¡1176:47t) V
AC bUC = bU 1j!C
R + 1j!C
= bU 1
1 + j!RC= 90
1
1 + j ¢ 2¼ ¢ 5000 ¢ 850 ¢ 10¡6= 3:37e¡1:53jbUC = bU 1
j!C
R + 1j!C
= bU 1
1 + j!RC= 90
1
1 + j ¢ 2¼ ¢ 5000 ¢ 850 ¢ 10¡6= 3:37e¡1:53j
up(t) = 3:37 sin(31416t¡ 1:53) [V]up(t) = 3:37 sin(31416t¡ 1:53) [V]
0 = K + 3:37 sin(¡1:53) ) K:= 3:370 = K + 3:37 sin(¡1:53) ) K:= 3:37
Now we set t = 0
uc(t) = 3:37e¡1176:47t + 3:37 sin(31416 t¡ 1:53) Vuc(t) = 3:37e¡1176:47t + 3:37 sin(31416 t¡ 1:53) V
XE31EO2 - Pavel Máša - Lectures 6, 7
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• Maximum amplitude of the waveform is almost twice compared with steady state• Exponential behavior is response of the circuit to the voltage source connection; it is characteristic
property of the circuit, the source waveform does not affect this behavior• Initial amplitude of this exponential function depend on initial conditions (initial voltage of the
capacitor and the voltage of the source at the moment of source connection – in case of sine wave any voltage between h¡Um;Umih¡Um;Umi
XE31EO2 - Pavel Máša - Lectures 6, 7
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Disconnection of voltage source from the capacitor
Capacitor is initially charged, initial condition is (generally) nonzerosuppose steady state before the source is disconnected
DC uc(0) = Uuc(0) = U up = 0up = 0
AC u(t) = 3:37 sin(31416t¡ 1:53) Vu(t) = 3:37 sin(31416t¡ 1:53) Vt < 0t < 0
t = 0t = 0 uc(0) = 3:37 sin(¡1:53):= ¡3:37 Vuc(0) = 3:37 sin(¡1:53):= ¡3:37 V
t !1t !1 up = 0up = 0
90 = K + 090 = K + 0 uc(t) = 90 e¡1176:47t Vuc(t) = 90 e¡1176:47t V
uc(t) = ¡3:37 e¡1176:47t Vuc(t) = ¡3:37 e¡1176:47t V
-1 0 1 2 3 4x 10-3
-4
-2
0
2
4
t[s]
u c(t)
-1 0 1 2 3 4
x 10-3
0
50
100
t[s]
u c(t)
DC AC
XE31EO2 - Pavel Máša - Lectures 6, 7
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uc(t) = [uC(0+)¡ up(0)] e¡t¿ + up(t)uc(t) = [uC(0+)¡ up(0)] e¡t¿ + up(t)
Initial voltage of capacitor before source connection / disconnection
Steady state voltage across capacitor in the moment, when the change occurs – in AC affected by phase shift
Steady state – 0, DC, AC, or periodical steady state
General solution in RC circuit
More complex circuit with one capacitor
Circuit, from the point of view of capacitor’s terminals will be replaced by Thévenin’s
equivalent circuitThe solution is then same as above
⇒
XE31EO2 - Pavel Máša - Lectures 6, 7
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Example:Circuit in the figure below is supplied from the rectangular voltage source with maximum
voltage 2V. As the valid value of logic 1 will be considered voltage greater than 75% of maximum voltage. Let at the most 20 % of half period of the clock pulse is the allowed rising time. What is maximum possible clock frequency of the bus?
V1
R1
100
C1 4p
u(t) = U (1¡ e¡t¿ )
t = ¡¿ ln(1¡ u(t)
U) = ¡4 ¢ 10¡10 ¢ ln(1¡ 1:5
2) = 554 ps
¿ = RC = 100 ¢ 4 ¢ 10¡12 = 400 ps
f =1
2 ¢ 5 ¢ t = 180:38 MHzf =1
2 ¢ 5 ¢ t = 180:38 MHz
XE31EO2 - Pavel Máša - Lectures 6, 7
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Example:USB bus in full speed operation mode (12 Mb/s) has the voltage 1.8 V. The bus (wave)
impedance is 90 Ω. Maximum allowed loading capacitance is 18 pF. Maximum allowed rising time is10 ns (from 0,45 to 1,35 V). Satisfy the bus this specification?
• Maximum allowed loading capacitance of high speed bus (240 MHz, 480 Mb/s) is 14 pF. Is the bus able to operate with voltage excitation?
NO, time constant is too long
• Is the bus able to operate with current excitation 17.78 mA?
YES, and it does – the capacitor is charged faster by current source
from 0.45 V to 1.35 V it rises in t = ¡1:65¢ln 1:35¡ 1:8
0:45¡ 1:8= 1:78 nsfrom 0.45 V to 1.35 V it rises in t = ¡1:65¢ln 1:35¡ 1:8
0:45¡ 1:8= 1:78 ns
YES; ¿ = 1:62 nsYES; ¿ = 1:62 ns
q = I ¢ tq = I ¢ t U =q
C=
I ¢ tC
U =q
C=
I ¢ tC
t =CU
I=
14 ¢ 10¡12 ¢ 0:917:78 ¢ 10¡3
= 0:7 nst =CU
I=
14 ¢ 10¡12 ¢ 0:917:78 ¢ 10¡3
= 0:7 ns
XE31EO2 - Pavel Máša - Lectures 6, 7
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Circuit, where we compute different circuit variable from voltage across capacitor
u2(t)u2(t)
In circuit on the figure compute the waveform of voltage u2(t)when the switcher is switched on / off
Solution:The waveform of the voltage across resistor R3 is given by waveform of passing current i2(t) – the same current, passing the capacitor.First, we compute the waveform of voltage across the capacitor.
Switching on:uc(0) = 0 Vuc(0) = 0 V
Ui = up = UR2
R1 + R2= 15
2000
1000 + 2000= 10 VUi = up = U
R2
R1 + R2= 15
2000
1000 + 2000= 10 V
i2(t)i2(t)i1(t)i1(t)
uc(0)uc(0)
i2(t) = 0 no voltage drop across R3, UR2 = UcUR2 = Uc
Ri = R3 +R1 ¢R2
R1 + R2= 1000 +
2000
3 ¢ 106= 1666:6 ÐRi = R3 +
R1 ¢R2
R1 + R2= 1000 +
2000
3 ¢ 106= 1666:6 Ð
¿ = RiC = 1:6 ms¿ = RiC = 1:6 ms
uc(t) = [0¡ 10] e¡600 t + 10uc(t) = [0¡ 10] e¡600 t + 10
i2(t) = Cduc(t)
dt= 10¡6
£¡10 ¢ (¡600)e¡600 t
¤= 6e¡600 t mAi2(t) = C
duc(t)
dt= 10¡6
£¡10 ¢ (¡600)e¡600 t
¤= 6e¡600 t mA
u2(t) = R3 ¢ i2(t) = 6 e¡600 t Vu2(t) = R3 ¢ i2(t) = 6 e¡600 t VXE31EO2 - Pavel Máša - Lectures 6, 7
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Switching off:
up = 0 Vup = 0 V
uc(0) = UR2
R1 + R2= 15
2000
1000 + 2000= 10 Vuc(0) = U
R2
R1 + R2= 15
2000
1000 + 2000= 10 V
Ri = R3 + R2 = 1000 + 2000 = 3000 ÐRi = R3 + R2 = 1000 + 2000 = 3000 Ð
¿ = RiC = 3 ms¿ = RiC = 3 ms
uc(t) = [10¡ 0] e¡333:3 t + 0uc(t) = [10¡ 0] e¡333:3 t + 0
i2(t) = Cduc(t)
dt= 10¡6
h10 ¢ (333:3)e¡333:3 t
i= 3:3 e¡333:3 t mAi2(t) = C
duc(t)
dt= 10¡6
h10 ¢ (333:3)e¡333:3 t
i= 3:3 e¡333:3 t mA
u2(t) = R3 ¢ i2(t) = 3:3 e¡333:3 t Vu2(t) = R3 ¢ i2(t) = 3:3 e¡333:3 t V
XE31EO2 - Pavel Máša - Lectures 6, 7
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RC circuit containing controlled source
ux(t)ux(t)Uv = K UxUv = K Ux
In the circuit in the figure compute waveform of the voltage u2(t)When the voltage source U is connected
1. Thévenin’s equivalent circuitUiUi
UvUvUxUx
Ui + K Ux ¡ Ux = 0
¡U + 0 + Ux = 0
)Ui = U (1¡K)
Ui + K Ux ¡ Ux = 0
¡U + 0 + Ux = 0
)Ui = U (1¡K)
IkIk
Ik =U
R) Ri =
Ui
Ik=
U(1¡K)UR
= R(1¡K)Ik =U
R) Ri =
Ui
Ik=
U(1¡K)UR
= R(1¡K)
uc(t) = [0¡ U(1¡K)] e¡t
RC(1¡K) + U(1¡K) = U(1¡K)h1¡ e
¡tRC(1¡K)
iuc(t) = [0¡ U(1¡K)] e
¡tRC(1¡K) + U(1¡K) = U(1¡K)
h1¡ e
¡tRC(1¡K)
i¿ = RiC = RC(1¡K)¿ = RiC = RC(1¡K)
ux(t) = uc(t)+K ux(t) ) ux(t)¢(1¡K) = uc(t) ) ux(t) = U³1¡ e
¡tRC(1¡K)
´ux(t) = uc(t)+K ux(t) ) ux(t)¢(1¡K) = uc(t) ) ux(t) = U
³1¡ e
¡tRC(1¡K)
´XE31EO2 - Pavel Máša - Lectures 6, 7
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ux(t)ux(t)Uv = K UxUv = K Ux
uc(0) = 0 Vuc(0) = 0 V
ux(t)¡ U
R+ C
d
dt[ux(t)¡Kux(t)] = 0
ux(t)¡ U
R+ C
d
dt[ux(t)¡Kux(t)] = 0
RC(1¡K)¸ + 1 = 0 ) ¸ =¡1
RC(1¡K)RC(1¡K)¸ + 1 = 0 ) ¸ =
¡1
RC(1¡K)
¡U + R ¢ 0 + uxp = 0 ) uxp = ux(1) = U¡U + R ¢ 0 + uxp = 0 ) uxp = ux(1) = U
uc(0) + K ux(0)¡ ux(0) = 0 ) ux(0) = 0 Vuc(0) + K ux(0)¡ ux(0) = 0 ) ux(0) = 0 V
2. Direct solution
ux(t) = [0¡ U ] e¡t
RC(1¡K) + U = U³1¡ e
¡tRC(1¡K)
´ux(t) = [0¡ U ] e
¡tRC(1¡K) + U = U
³1¡ e
¡tRC(1¡K)
´
We have to compute initial condition using circuit equations
XE31EO2 - Pavel Máša - Lectures 6, 7
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Periodical rectangular waveform – integrating circuit
¿ ¿ T¿ ¿ T ¿ = 1 ms¿ = 1 ms
T = 20 msT = 20 ms
In this case each change of excitation voltage will be considered as the origin of new transient– the transient voltage at the end of first half of the period is the initial condition of subsequent transientin this case may be omitted
0 0.01 0.02 0.03 0.04 0.05 0.06
0
0.5
1
0 0.01 0.02 0.03 0.04 0.05 0.06
0
0.5
1
Periodical steady state:
u1(t) = 0:5+
1Xk=1
2
(2k ¡ 1)¼sin(2k¡1)!0t ) U01 = 0:5, U1k =
2
(2k ¡ 1)¼u1(t) = 0:5+
1Xk=1
2
(2k ¡ 1)¼sin(2k¡1)!0t ) U01 = 0:5, U1k =
2
(2k ¡ 1)¼
U02 = 0:5, U2k = U1k1
1 + j(2k ¡ 1)!0RC=
2
(2k ¡ 1)¼¢ 1
1 + j(2k ¡ 1)0:1¼U02 = 0:5, U2k = U1k
1
1 + j(2k ¡ 1)!0RC=
2
(2k ¡ 1)¼¢ 1
1 + j(2k ¡ 1)0:1¼
R = 1 kÐ; C = 1 ¹FR = 1 kÐ; C = 1 ¹F
!0 = 100¼!0 = 100¼
u2(t) = 0:5+ 0:607 sin(314:16t¡ 0:304)+0:154 sin(942:48t¡ 0:756) +0:068 sin(1570:79t¡ 1:004)+ ¢ ¢ ¢u2(t) = 0:5+ 0:607 sin(314:16t¡ 0:304)+0:154 sin(942:48t¡ 0:756) +0:068 sin(1570:79t¡ 1:004)+ ¢ ¢ ¢
Transient:
t 2 (0; 0:1) u2(t) = 1¡ e¡1000tt 2 (0; 0:1) u2(t) = 1¡ e¡1000t
u2(0:01) = 0:9999546 Vu2(0:01) = 0:9999546 V
t 2 (0:1; 0:2) u2(t) = 0:9999546 e¡1000(t¡0:1)t 2 (0:1; 0:2) u2(t) = 0:9999546 e¡1000(t¡0:1)
u2(0:02) = 0:0000454 Vu2(0:02) = 0:0000454 V
t 2 (0:2; 0:3) u2(t) = 1¡ 0:9999546 e¡1000(t¡0:2)t 2 (0:2; 0:3) u2(t) = 1¡ 0:9999546 e¡1000(t¡0:2)
Um = 1 VUm = 1 V
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¿ À T¿ À T ¿ = 1 ms¿ = 1 ms
T = 0:2 msT = 0:2 ms
R = 1 kÐ; C = 1 ¹FR = 1 kÐ; C = 1 ¹F
!0 = 10000¼!0 = 10000¼
0 1 2 3 4 5 6x 10-4
0
0.5
1
0 1 2 3 4 5 6x 10-4
0
0.05
0.1
0.15 Periodical steady state:
Periodical steady state does not resolve transient part,but just steady state part – it is equal to the meanvalue of rectangular waveform
Exponential trend is equivalent to the „slow“ transient with time constant τ
u1(t) =1X
k=¡1U1ke
jk!0t ) U01 =Um ¢ t0
T= 0:12, U1k =
Um
¡jk2¼
£e¡jk!0t0 ¡ 1
¤=
1
¡jk2¼
£e¡jk0:24¼ ¡ 1
¤u1(t) =
1Xk=¡1
U1kejk!0t ) U01 =
Um ¢ t0T
= 0:12, U1k =Um
¡jk2¼
£e¡jk!0t0 ¡ 1
¤=
1
¡jk2¼
£e¡jk0:24¼ ¡ 1
¤
U02 = U01, U2k = U1k1
1 + jk!0RC=
1
¡jk2¼
£e¡jk0:24¼ ¡ 1
¤¢ 1
1 + jk10¼k = ¡1; : : :¡1; 1 : : :1U02 = U01, U2k = U1k
1
1 + jk!0RC=
1
¡jk2¼
£e¡jk0:24¼ ¡ 1
¤¢ 1
1 + jk10¼k = ¡1; : : :¡1; 1 : : :1
XE31EO2 - Pavel Máša - Lectures 6, 7
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24 μs 176 μsu1
u2
u(24 ¹s) = u2 = (u1 ¡ 1) e¡0:024 + 1u(24 ¹s) = u2 = (u1 ¡ 1) e¡0:024 + 1
u(176 ¹s) = u1 = u2 e¡0:176u(176 ¹s) = u1 = u2 e¡0:176
uc(0) = u1; up = 1; t = 24 ¹suc(0) = u1; up = 1; t = 24 ¹s
Steady state – how from the transient equation find limit voltages:
1st part – increasing exponential
2nd part – decreasing exponentialuc(0) = u2; up = 0; t = 176 ¹suc(0) = u2; up = 0; t = 176 ¹s
Set of equations24e¡0:024 ¡1
¡1 e¡0:176
35 ¢ "u1
u2
#=
"e¡0:024 ¡ 1
0
#) u1 = 0:109711
u2 = 0:130824
24e¡0:024 ¡1
¡1 e¡0:176
35 ¢ "u1
u2
#=
"e¡0:024 ¡ 1
0
#) u1 = 0:109711
u2 = 0:130824
t 2 (0; 24 ¹s) u2(t) = 1¡ e¡1000tt 2 (0; 24 ¹s) u2(t) = 1¡ e¡1000t u2(24¹s) = 0:0237143 Vu2(24¹s) = 0:0237143 V
t 2 (24 ¹s; 200 ¹s) u2(t) = 0:0237143 e¡1000(t¡0:000024)t 2 (24 ¹s; 200 ¹s) u2(t) = 0:0237143 e¡1000(t¡0:000024) u2(200 ¹s) = 0:01988723 Vu2(200 ¹s) = 0:01988723 V
t 2 (200 ¹s; 224 ¹s) u2(t) = 1¡ 0:9801127 e¡1000(t¡0:0002)t 2 (200 ¹s; 224 ¹s) u2(t) = 1¡ 0:9801127 e¡1000(t¡0:0002) u2(224 ¹s) = 0:04312991 Vu2(224 ¹s) = 0:04312991 V: : :: : :
XE31EO2 - Pavel Máša - Lectures 6, 7
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Zero mean valueUmax = 1Umin = ‐1
Negative mean valueUmax = 1Umin = ‐1
Positive mean valueUmax = 1Umin = 0
Positive mean valueUmax = 1Umin = ‐1
XE31EO2 - Pavel Máša - Lectures 6, 7
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Periodical rectangular waveform – derivative circuit¿ = 1 ms¿ = 1 ms
T = 20 msT = 20 ms
R = 1 kÐ; C = 1 ¹FR = 1 kÐ; C = 1 ¹F
!0 = 100¼!0 = 100¼
Um = 1 VUm = 1 V
0 0.01 0.02 0.03 0.04 0.05 0.06
0
0.5
1
0 0.01 0.02 0.03 0.04 0.05 0.06-1
0
1
RC circuit exhibits „derivative“ properties, only when ¿ ¿ T¿ ¿ T
u1(t) = 0:5 +1X
k=1
2
(2k ¡ 1)¼sin(2k ¡ 1)!0tu1(t) = 0:5 +
1Xk=1
2
(2k ¡ 1)¼sin(2k ¡ 1)!0t
U01 = 0:5, U1k =2
(2k ¡ 1)¼U01 = 0:5, U1k =
2
(2k ¡ 1)¼
¿ ¿ T¿ ¿ T
¿ >T
2¿ >
T
2
U02 = 0
U2k = U1kj(2k ¡ 1)!0RC
1 + j(2k ¡ 1)!0RC
=2
(2k ¡ 1)¼¢ j(2k ¡ 1)0:1¼
1 + j(2k ¡ 1)0:1¼
U02 = 0
U2k = U1kj(2k ¡ 1)!0RC
1 + j(2k ¡ 1)!0RC
=2
(2k ¡ 1)¼¢ j(2k ¡ 1)0:1¼
1 + j(2k ¡ 1)0:1¼
uc(t) = [uc(0)¡ ucp(0)] e¡t¿ + ucp(t)uc(t) = [uc(0)¡ ucp(0)] e¡t¿ + ucp(t)
i(t) = Cduc(t)
dt=¡C
¿[uc(0)¡ ucp(0)] e
¡t¿i(t) = C
duc(t)
dt=¡C
¿[uc(0)¡ ucp(0)] e
¡t¿
u2(t) = Ri(t) =¡RC
¿[uc(0)¡ ucp(0)] e
¡t¿
= [ucp(0) ¡ uc(0)] e¡t¿ = uR(0) e
¡t¿
u2(t) = Ri(t) =¡RC
¿[uc(0)¡ ucp(0)] e
¡t¿
= [ucp(0) ¡ uc(0)] e¡t¿ = uR(0) e
¡t¿
ucp(0)ucp(0) is particular solution on capacitor – alternatively 1 V (first half period) and 0V
uR(0)uR(0) positive on rising edge, negative on falling edge uR(0)n = §1 ¨ uR
¡T2
¢n¡1
uR(0)n = §1 ¨ uR
¡T2
¢n¡1 XE31EO2 - Pavel Máša - Lectures 6, 7
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¿ À T¿ À T Derivative circuit
T = 0:2 msT = 0:2 ms¿ = 1 ms¿ = 1 ms t0 = 176 ¹st0 = 176 ¹s
• The voltage is almost rectangular, if the duration of voltage values are: Um is t0 and 0 is T – t0, than the waveform has boundary voltages and
• The magnitude is still Um
• „Slow“ exponential transient is boundary waveform
• The waveform of the voltage across capacitor is the same as voltage across capacitor in the integrating circuit above (the sum of both voltages must be in each time instant Um).
Umt0TUmt0T ¡Um(1¡ t0
T )¡Um(1¡ t0T )
XE31EO2 - Pavel Máša - Lectures 6, 7
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1ST ORDER CIRCUIT WITH INDUCTOR
• Energetic initial condition – current passing inductor at t = 0. Even if we would compute voltage across inductor, it is more suitable to choose current as the initial condition – it is continuous.
• ⇒ circuit equation –mesh analysis
Ri(t) + Ldi(t)
dt¡ u(t) = 0Ri(t) + L
di(t)
dt¡ u(t) = 0
L
R
di(t)
dt+ i(t) =
u(t)
R
L
R
di(t)
dt+ i(t) =
u(t)
R
L
R¸ + 1 = 0 ) ¸ =
¡R
L
L
R¸ + 1 = 0 ) ¸ =
¡R
L
iLo(t) = Ke¸t = Ke¡tRLiLo(t) = Ke¸t = Ke¡tRL
iL(t) = [iL(0+)¡ ip(0)] e¡t¿ + ip(t)iL(t) = [iL(0+)¡ ip(0)] e¡t¿ + ip(t)
¿ =L
R¿ =
L
R
• To find the solution, method of variation of parameters is most suitable
))• Compute steady state in the circuit after transient dies away• Setting t = 0 compute integration constant K
XE31EO2 - Pavel Máša - Lectures 6, 7
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Voltage across inductor
uL(t)uL(t)iL(0)iL(0)
R1 = 1 kÐ; R2 = 1 Ð; L = 0:5 H; U = 1 VR1 = 1 kÐ; R2 = 1 Ð; L = 0:5 H; U = 1 V
Energetic initial condition is current⇒ First we find current, then voltage
Switcher is switched on:
iL(0) = 0; ip =U
R2=
1
1= 1 AiL(0) = 0; ip =
U
R2=
1
1= 1 ASteady state when t < 0 and t > 0:
Time constant: ¿ =L
R2=
0:5
1= 0:5 s¿ =
L
R2=
0:5
1= 0:5 s
R2 i(t) + Ldi(t)
dt¡ U = 0 ) R2 + L¸ = 0 ) ¸ =
¡R2
LR2 i(t) + L
di(t)
dt¡ U = 0 ) R2 + L¸ = 0 ) ¸ =
¡R2
LEquation and its solution:
t = 0:
i(t) = Ke¸t + ipi(t) = Ke¸t + ip
iL(0) = K + ip ) K = iL(0)¡ ipiL(0) = K + ip ) K = iL(0)¡ ip
i(t) = 1¡ e¡2ti(t) = 1¡ e¡2tCurrent solution:
Voltage solution: uL(t) = Ldi(t)
dt= L
d
dt
¡1¡ e¡2t
¢= e¡2tuL(t) = L
di(t)
dt= L
d
dt
¡1¡ e¡2t
¢= e¡2t
i(t)i(t) uL(t)uL(t)
t [s]t [s] t [s]t [s]
I [A]I [A]U [V ]U [V ]
XE31EO2 - Pavel Máša - Lectures 6, 7
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Switcher is switched off:
iL(0) =U
R2=
1
1= 1 A; ip = 0iL(0) =
U
R2=
1
1= 1 A; ip = 0Steady states when t < 0 and t > 0:
Time constant: ¿ =L
R1 + R2=
0:5
1001= 499:5 ¹s¿ =
L
R1 + R2=
0:5
1001= 499:5 ¹s
(R1+R2) i(t)+Ldi(t)
dt= 0 ) R1+R2+L¸ = 0 ) ¸ = ¡R1 + R2
L(R1+R2) i(t)+L
di(t)
dt= 0 ) R1+R2+L¸ = 0 ) ¸ = ¡R1 + R2
LEquation and its solution:
i(t) = e¡2002ti(t) = e¡2002tCurrent solution:
Voltage solution: uL(t) = Ldi(t)
dt= L
d
dt
¡e¡2002t
¢= ¡1001 e¡2002t VuL(t) = L
di(t)
dt= L
d
dt
¡e¡2002t
¢= ¡1001 e¡2002t V
i(t)i(t)
uL(t)uL(t)t [ms]t [ms]
I [A]I [A]
U [V ]U [V ]
t [ms]t [ms]
Current is continuous variable – voltage rise on such value, in order that the current passes arbitrary resistivity (theoretically even infinite voltage) spark / arc if we just disconnect inductor from the source!!! XE31EO2 - Pavel Máša - Lectures 6, 7
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Voltage across inductor – overvoltage protection
Resistor in parallel with the coil
R2 all the time the circuit is switched on resistor dissipates power = unwanted losses; the larger the
resistivity is, dissipated power is smaller, but at once the overvoltage when the source is disconnected is larger; large coils have large dissipated power on this resistor!!!
Resistor in parallel with the switcher
Such connection, of course, could not be used to safe disconnect the coil from the source; so it is applicable
merely to reduce current passing the coil
Capacitor in parallel with the coil
Effective overvoltage protection, capacitor correctspower factor as well, accumulated energy dissipates on
heat in the coil winding (R1)
Diode in parallel with the coil
Applicable only to DC circuits – e.g. relay winding protection; correct selection of the diode is important (limit current), limiting current may be reduced by
resistorXE31EO2 - Pavel Máša - Lectures 6, 7
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