7/27/2019 2013 Lect 04 RCL Circuits

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Physics 139 Lecture 04 Prof. Allen Mills Oct 04, 2013

Solving RCL circuits

For nonlinear circuits and general input time variations one may use a circuit simulator such as SPICE.http://bwrcs.eecs.berkeley.edu/Classes/IcBook/SPICE/

(I) Linear RCL circuits may be solved by writing out the set of linear differential equations and solving them.

Example 1: RC circuit. At t=0 a resistorR is connected across a capacitorCinitially charged to voltage V0. What isthe ensuing voltage as a function of time?

Ans. Kirchhoffs laws imply 0// RCVdtdV , so RCdtVdV // . Integrate to get

RCttV /)(ln and )(}/exp{)( 0 tRCtVtV , where )(t is the Heaviside (1881) stepfunction, 0)( t fort0.

Problem 8: RL circuit. At t= 0 a currentI0 is flowing around a loop composed of a resistorR

and an inductorL in series. What is the current as a function of time t>0?

Problem 9: Resonant LRC circuit connected in series in a single loop. At t=0 a resistorR andinductorL,in series, are connected across a capacitor Cinitially charged to voltage V0 (seeFig. 1). What is the ensuing voltage as a function of time? Fig. 1

The charge Q and the current I are related by IdtdQ / because I as drawn in the figure is discharging the

capacitor. By Ks voltage law dtLdIIRCQ // . Taking the time derivative

of this equation we get 0/// 22 dtILddtRdICI which we rewrite as

020 III , whereLC

10 and LR / .

(a) Show that the solution for 24

120 is (see solutions in Fig. 2)

}/sin{}exp{/

)(4

12202

1

4122

0

0

ttR

VtI .

(b) What is the solution for 24

120 ? Fig. 2

(c) Show that for critical damping, for which4

1220 / the solution is }exp{)( 2

10 ttR

VtI .

The response of more complicated circuits to a step input may be found by the method of Laplace transforms.

(II) Theorem: The response of any linear circuit to single-frequency sine wave inputs is found from its compleximpedance exactly as one solves resistive circuits. The proof uses complex analysis, i.e. the theory of functions of a

complex variable. We require the following four elements and theorems which may be familiar.

The imaginary uniti is defined by the solutions to the equation 012 x , ix .A complex number is denoted by its real and imaginary parts iyxz .

Eulers theorem: The exp. function may be expanded into its real and imaginary parts thus sincos iei .

The superposition principle: The sum of any two solutions of a linear differential equation is also a solution.Proof - Suppose the linear differential equation for our circuit is expressed using a matrix differential operator

)}({ tVD in , which,when operating on the input voltages, tells us what the output voltages are: )}({)( tVDtV inout .

In the special case that the input voltage is tVtV inin cos)(0 we will have }cos{)( 0cos tVDtV inout

In the special case that the input voltage is tVtV inin sin)(0 we will have }sin{)( 0sin tVDtV inout .

Now the differential operator works just as well on complex as on real numbers and functions.Furthermore, by the superposition principle, a linear combination of inputs will give us an output that is a linear

combination of the two solutions, i.e. of the two outputs.

Therefore if the input is the complex sum tiVtV inin sincos00 , the complex output will be

}{}sincos{)()()( 000sincos tiinininoutoutcomplexout eVDtiVtVDtiVtVtV

.

Therefore the real part of the solution for a complex inputti

ineV

0is the solution }cos{)(

0cos tVDtV inout to the

entirely real problem with the real input tVin cos0

.

0 2 4 6 8 10-0.2

0.0

0.2

0.4

0.6

0.8

.

21

1/2

1/4

1/8

1/16

I(t)R/V

0

t

2/

2=L/R

2

C

1/32

7/27/2019 2013 Lect 04 RCL Circuits

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(III)How to analyze a circuit. Throughout the circuit all the voltages and currents will in general be constant

complex amplitudes times tie .

Encountering a capacitor, the behavior VCI with a single frequency voltage tiineVV 0 implies theproportionality of the complex current and voltage amplitudes CViI .

Likewise the differential equation for the voltage across the inductor ILV implies the proportionality ofcomplex amplitudes LIiV .

The proportionality of amplitudes expressed by Ohms law is unchanged, IRV .With these three rules, IRV CViI and LIiV we are ready to solve any passive linear circuit for asteady state. We use Kirchhoffs laws and solve the implied set of linear equations in exactly the same manner asolving a resistive network with batteries and current sources.

(IV) Impedance matching the tank circuit for a pulsed NMR measurement (encountered in Phys 142L)

Example 2: We shall be concerned with impedance matching the LCR tank circuit of Fig 3a using the series

capacitor C in Fig 3b. The inspiration for this analysis is Ref [1] whichuses the circuit of Fig 3c with C-big (CB) and C-small (CS) to easily match

to a 50 drive impedance.

The impedance of a parallel LCR tank circuit Fig 3a at point a relative to

ground is the parallel combination ofCand theL in series withR:

RCiLCLiR

LiRCiZ

2

1

11)( . Fig. 3 Tank circuits

Writing this in terms of its real and imaginary parts we have

22222

2

22

2

]1[

]1[

]1][1[

]1][[)(

CRLC

LCLiR

RCiLCRCiLC

RCiLCLiRZ

The undamped resonance frequency is LC/10 and the damping rate is

RL / .

When 0 the impedance is pure real, 0220

0

1)( Z

RC

L

RCZ

.

When RCLC 021 or RCRC 202

1000 1 (in the

approximation that RC0