Chapter 7 Response of First-order RL and RC Circuits
7.1-2 The Natural Response of RL and RC Circuits 7.3 The Step Response of RL and RC Circuits 7.4 A General Solution for Step and Natural Responses 7.5 Sequential Switching 7.6 Unbounded Response
1 Overview
Ch9-10 discuss “steady-state response” of linear circuits to “sinusoidal sources”. The math treatment is the same as the “dc response” except for introducing “phasors” and “impedances” in the algebraic equations.
From now on, we will discuss “transient response” of linear circuits to “step sources” (Ch7-8) and general “time-varying sources” (Ch12-13). The math treatment involves with differential equations and Laplace transform.
2 First-order circuit
A circuit that can be simplified to a Thévenin (or Norton) equivalent connected to either a single equivalent inductor or capacitor.
In Ch7, the source is either none (natural response) or step source.
3 Key points
Why an RC or RL circuit is charged or discharged as an exponential function of time?
Why the charging and discharging speed of an RC or RL circuit is determined by RC or L/R?
What could happen when an energy-storing element (C or L) is connected to a circuit with dependent source?
4 Section 7.1, 7.2 The Natural Response of RL and RC Circuits
1. Differential equation & solution of a discharging RL circuit 2. Time constant 3. Discharging RC circuit
5 What is natural response?
It describes the “discharging” of inductors or capacitors via a circuit of no dependent source. No external source is involved, thus termed as “natural” response. The effect will vanish as t. The interval within which the natural response matters depends on the element parameters.
6 Circuit model of a discharging RL circuit
Consider the following circuit model:
For t < 0, the inductor L is short and carries a
current Is, while R0 and R carry no current.
For t > 0, the inductor current decreases and the energy is dissipated via R.
7 Ordinary differential equation (ODE), initial condition (IC)
For t > 0, the circuit reduces to:
IC depends on initial energy of the inductor: )0()0( 0 IIii s
By KVL, we got a first-order ODE for i(t): d L tRiti .0)()( dt where L, R are independent of both i, and t. 8 Solving the loop current
d ;)0(:IC ,0)()( :ODE L tRiti ,0)()( IIi ;)0(:IC dt 0 s di R dtRidiL ,0)()( dt, i L ti )( id R t ti )( R t , ln itd t , i )0( i L 0 i )0( L 0 ti )( R iti ln)0(ln)(ln t, I0 L
t L )( 0eIti , where …time constant R 9 Time constant describes the discharging speed
The loop current i(t) will drop to e-1 (37%) of its
initial value I0 within one time constant . It will
be <0.01I0 after elapsing 5. If i(t) is approximated by a linear function, it will vanish in one time constant.
37.0 I0
10 Solutions of the voltage, power, and energy
The voltage across R (or L) is: ,0 t ,0for tRitv )()( …abrupt change at t=0. t 0 , teRI .0for
The instantaneous power dissipated in R is:
2 22 t τ )()( 0 , teRIRtitp .0for
The energy dissipated in the resistor R is: t t 2 2t τ )( 0 tdeRItdtpw 0 0 2t τ 2 0 ,1 00 ,2 tLIwew .0for initial energy stored in L 11 Example 7.2: Discharge of parallel inductors (1)
Q: Find i1(t), i2(t), i3(t), and the energies w1, w2
stored in L1, L2 in steady state (t).
For t<0: (1) L1, L2 are short, and (2) no current flows through any of the 4 resistors, i1 8)0( ,A i2 4)0( ,A i3 ,0)0( 2 2 w1 5()0( 8H)( 1602A) w2 1602)4)(20()0(,J .J 12 Example 7.2: Solving the equivalent RL circuit (2)
For t>0, switch is open, the initial energy stored in the 2 inductors is dissipated via the 4 resistors. The equivalent circuit becomes:
I0=(8+4)=12 A Req =
v(0-)=0 8 Leq =(5H//20H)=4 H
The solutions to i(t), v(t) are:
t 2t Leq 4 )( 0 12eeIti ,A 5.0 ,s 2t Req 8 96)()( etRitv V. 13 Example 7.2: Solving the inductor currents (3)
The two inductor currents i1(t), i2(t) can be calculated by v(t):
V 96e2t V
t t 1 1 2t 2t iti 11 )0()( tdtv 8)( tde 6.96.196 e A. 0 0 L1 5
t t 1 1 2t 2t 2 iti 2 )0()( tdtv 4)( 96 tde 4.26.1 e A. 0 0 L2 20
14 Example 7.2: Solutions in steady state (4)
Since 2t 1 ti e 1.66.96.1)( A, 2t 2 ti 4.26.1)( e 1.6 A, the two inductors form a closed current loop!
The energies stored in the two inductors are: 1 tw )( 5( 6.1)(H 2 4.6)A J, 1 2 1 tw )( 2 6.25)6.1)(20( J, 2 2 which is ~10% of the initial energy in total. 15 Example 7.2: Solving the resistor current (5)
By current division, i3(t) = 0.6i4(t), while i4(t) can be calculated by v(t):
i4
tv )( 96e2t ti )( 6.9 e2t A. 4 )10//15(4 10 3 )( 76.5)( etiti 2t A. 3 5 4 16 Circuit model of a discharging RC circuit
Consider the following circuit model:
For t<0, C is open and biased by a voltage Vg,
while R1 and R carry no current.
For t>0, the capacitor voltage decreases and the energy is dissipated via R.
17 ODE & IC
For t>0, the circuit reduces to:
IC depends on initial energy of the capacitor: )0()0( 0 VVvv g
By KCL, we got a first-order ODE for v(t): d tv )( C tv )( .0 dt R where C, R are independent of both v, and t. 18 Solving the parallel voltage
d tv )( )( :ODE C tv )( ,0 VVv ;)0(:IC dt R 0 g
t )( 0eVtv , where RC …time constant
Reducing R (loss) and parasitic C is critical for high- speed circuits.
19 Solutions of the current, power, and energy
The loop current is: tv )( ,0 t ,0for ti )( …abrupt change at t=0. R t 0 , teRV .0for
The instantaneous power dissipated in R is: 2 tv )( V 2 tp )( 0 2t τ , te .0for R R
The energy dissipated in the resistor R is: 2 t 2t τ CV0 0 ,1)( wewtdtpw 0 , t .0for 0 2 initial energy stored in C 20 Procedures to get natural response of RL, RC circuits
1. Find the equivalent circuit.
2. Find the initial conditions: initial current I0 through the equivalent inductor, or initial voltage
V0 across the equivalent capacitor. 3. Find the time constant of the circuit by the values of the equivalent R, L, C: , RCRL ;or 4. Directly write down the solutions: t t 0 ,)( 0eVtveIti .)(
21 Section 7.3 The Step Response of RL and RC Circuits
1. Charging an RC circuit 2. Charging an RL circuit
22 What is step response?
The response of a circuit to the sudden application of a constant voltage or current source, describing the charging behavior of the circuit.
Step (charging) response and natural (discharging) response show how the signal in a digital circuit switches between Low and High with time.
23 ODE and IC of a charging RC circuit
IC depends on initial energy of the capacitor: )0()0( Vvv 0
Derive the governing ODE by KCL:
tv )( d dv 1 I C tv ),( Vv , s R dt dt f
where , sf RIVRC are the time constant and final (steady-state) parallel voltage. 24 Solving the parallel voltage, branch currents
dv 1 dv 1 Vv f , dt, dt Vv f tv )( vd 1 t )( Vtv t )( Vtv td ln, f , f et , V 0 0 s RIv 0 VV f 0 VV f
t )( f 0 f eVVVtv .
The charging and discharging processes have the same speed (same time constant =RC). The branch currents through C and R are: d V tv )( Cti )()( Itv 0 t , tie )( , t .0for C dt s R R R 25 Example 7.6 (1)
Q: Find vo(t), io(t) for t 0.
For t <0, the switch is connected to Terminal 1 for long, the capacitor is an open circuit: k 60 k v 40()0( V) 30 ,V i .0)0( o 60)20( k o
26 Example 7.6 (2)
At t 0+, the “charging circuit” with two terminals 2 and G can be reduced to a Norton equivalent:
G
vv )0()0( + o o v V 30 V 0 0 I s 5.1 mA 27 Example 7.6 (3)
The time constant and the final capacitor voltage of the charging circuit are: RC 40( 25.0)(k 10)F ,ms
sf RIV 5.1( 40)(mA 60)k .V
t 100t o )( f 0 f eVVVtv 9060 e .V
t 100t o )( s 0 eRVIti 25.2 e .mA
At the time of switching, the capacitor voltage is continuous: v o 30)0( v 0 )9060()0(V V ,
while the current io jumps from 0 to -2.25 mA. 28
Charging an RL circuit
IC depends on initial energy of the inductor: )0()0( Iii 0 d LtRiV ),()( )( eIIItiti t , s dt f 0 f
L Vs where , I f . R R
The charging and discharging processes have the same speed (same time constant =L/R). 29 Section 7.5 Sequential Switching
30 What is and how to solve sequential switching?
Sequential switching means switching occurs n (2) times. It is analyzed by dividing the process into n+1 time intervals. Each of them corresponds to a specific circuit. Initial conditions of a particular interval are determined from the solution of the preceding interval. Inductive currents and capacitive voltages are particularly important for they cannot change abruptly. Laplace transform (Ch12) can solve it easily.
31 Example 7.12: Charging and discharging a capacitor (1)
Q: v (t)=? for t ≥ 0.
t =0 vV 0)0( t =15 ms IC: 0
For 0 t 15 ms, the 400-V source charges the
capacitor via the 100-k resistor, Vf = 400 V, = RC =10 ms, and the capacitor voltage is: t 100t 1 f 0 f eVVVtv 400400)()( e V. 32 Example 7.12 (2)
For t>15 ms, the capacitor is disconnected from the 400-V source and is discharged via the 50-
k resistor, V0 = v1(15 ms) = 310.75 V, Vf = 0, = RC = 5 ms. The capacitor voltage is:
tt 0 )( t )015.0(200 2 )( 0eVtv 75.310 e V.
33 Section 7.6 Unbounded Response
34 Definition and reason of unbounded response
An unbounded response means the voltages or currents increase with time without limit. It occurs when the Thévenin resistance is
negative (RTh <0), which is possible when the first-order circuit contains dependent sources.
35 Example 7.13: (1)
Q: vo(t)=? for t≥0.
For t>0, the capacitor seems to “discharge” (not really, to be discussed) via a circuit with a current-controlled current source, which can be represented by a Thévenin equivalent. 36 Example 7.13 (2)
Since there is no independent source, VTh =0,
while RTh can be determined by the test source method:
vvv i T T 7 T , T 10 20k k 20 k
ivR TTTh -5 .0k 37 Example 7.13 (3)
For t0, the equivalent circuit and governing differential equation become:
0 vV o 10)0( V, RC 25 0,ms t 40t ..grow without limit. o )( 0 10eeVtv V.
38 Why the voltage is unbounded?
Since 10-k, 20-k resistors are in parallel,
i10k =2i, the capacitor is actually charged (not
discharged) by a current of 4i!
Charging effect will increase vo, which will in turn
increase the charging current (i = vo/20 k) and
vo itself. The positive feedback makes vo soaring.
6i
2i
39 Lesson for circuit designers & device fabrication engineers
Undesired interconnection between a capacitor and a sub-circuit with dependent source (e.g. transistor) could be catastrophic!
40 Key points
Why an RC or RL circuit is charged or discharged as an exponential function of time?
Why the charging and discharging speed of an RC or RL circuit is determined by RC or L/R?
What could happen when an energy-storing element (C or L) is connected to a circuit with dependent source?
41