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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


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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)


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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


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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


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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


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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


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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

⇒


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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


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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


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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


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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


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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


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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: : :: : :


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Zero mean valueUmax = 1Umin = ‐1

Negative mean valueUmax = 1Umin = ‐1

Positive mean valueUmax = 1Umin = 0

Positive mean valueUmax = 1Umin = ‐1


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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

XE31EO2 -

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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 )


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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


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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 ]


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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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