First-order Transient-Basic Electronics-Lecture 02-Electronic and Information Engineering, Lecture notes of Basic Electronics

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EIE209 Basic Electronics

First-order transient

Contents

• Inductor and capacitor
• Simple RC and RL circuits
• Transient solutions Prof. C.K. Tse: First Order Transient

Prof. C.K. Tse: First Order Transient

Constitutive relation

® An electrical element is defined by its relationship between v and i.

This is called constitutive relation. In general, we write

® For a resistor,

® v = i R

® The constitutive relation of a resistor has no dependence upon time.

v = f ( i ) or i = g( v)

+ v –

i

Prof. C.K. Tse: First Order Transient What happens if a circuit has C and/or L? ® The circuit becomes dynamic. That means: ® Its behaviour is a function of time. ® Its behaviour is described by a (set of) differential equation(s). ® It has a transient response as well as a steady state.

Prof. C.K. Tse: First Order Transient Resistive circuits have no transient

® Consider a resistive circuit.

® When the switch is turned on,

the voltage across R becomes V

immediately (in zero time).

® v = V = i R for all t > 0 ® i = V / R for all t > 0

Prof. C.K. Tse: First Order Transient Transient response of the RC circuit ®Once we have the capacitor voltage, we can find anything. ®Starting with ®We can derive the current as ®We see the solution typically has a TRANSIENT which dies out eventually, and as t tends to ∞, the solution settles to a steady state. time constant

Prof. C.K. Tse: First Order Transient

A simple first-order RL circuit

®Consider a RL circuit. ®Before t = 0, the switch is closed (turned on). Current goes through the switch and nothing goes to R and L. Initially, iL(0–) = 0. ®At t = 0, the switch is opened. Current goes to R and L. ®We know from KCL that Io = iR + iL for t > 0, i.e., ®The constitutive relations give ®Hence, ® fi ®The solution is From the initial condition, we have iL(0–) = 0. Continuity of the inductor current means that iL(0+) = iL(0–) = 0. Hence, A = – Io Thus,

Prof. C.K. Tse: First Order Transient Observation — first-order transients ® First order transients are always like these:

Prof. C.K. Tse: First Order Transient Let’s do some math 0 5 x( t) t x( t) = 5(1 – e–^ t/t) 0 5 x( t) t x( t) = 5 e–^ t/t 0 6 x( t) t 1 x( t) = 1 + 5(1 – e–^ t/t) x( t) = 1 + 5 e–^ t/t 0 5 x( t) t

• 2 x( t) = – 2 + 7(1 – e–^ t/t) 0 4 x( t) t
• 3 0 6 x( t) t 1 x( t) = – 3 + 7 e–^ t/t

Prof. C.K. Tse: First Order Transient Finding t For the simple first-order RC circuit, t = C R. For the simple first-order RL circuit, t = L / R. The problem is Given a first-order circuit (which may look complicated), how to find the equivalent simple RC or RL circuit.

Prof. C.K. Tse: First Order Transient A quick way to find t Since the time constant is independent of the sources, we first of all set all sources to zero. That means, short-circuit all voltage sources and open- circuit all current sources. Then, reduce the circuit to

• R 1 R 2 C R 1 R 2 C R 1 || R 2 C either (^) Req Ceq or Req Leq Example: t = C ( R 1 || R 2 )

Prof. C.K. Tse: First Order Transient Example 2 (non-trivial boundary conditions) Find v 1 ( t) and v 2 ( t) for t > 0 without solving any differential equation.

v 1

• v 2

C 1 2F C 2 3F i 1 R = 1Ω t= i 2 Suppose v 1 (0–) = 5 V and v 2 (0–) = 2 V. Problem: how to find the final voltage values. Form 7 solution: You considered the charge transfer D q from C1 to C2. ++++++ ++++

D q q 1 fi q 1 – D q q 2 fi q 2 + D q Use charge balance and KVL equations to find the final voltage values. Clumsy solution!

Prof. C.K. Tse: First Order Transient Example 2 (elegant solution) We need not consider CHARGE! Step 1: initial point (given) v 1 (0–) = 5 V and v 2 (0–) = 2 V are known. Continuity of cap voltage guarantees that v 1 (0+) = 5 V and v 2 (0+) = 2 V. Step 2: final point (non-trivial) C1: for all t C2: for all t After t>0, we have i 1 = – i 2 , i.e.,

v 1

• v 2

C 1 2F C 2 3F i 1 R = 1Ω t= i 1 = 2 dv 1 d t i 2 = 3 dv 2 d t 2 dv 1 dt

• 3 dv 2 dt = 0 fi 2 v 1 ( t) + 3 v 2 ( t) = K for all t > 0. i 2 Integration constant At t = 0+, this equation means 25 + 32 = K. Thus, K = 16. Thus, 2 v 1 ( t) + 3 v 2 ( t) = 16 for t > 0. At t =∞, we have v 1 (∞)= v 2 (∞) from KVL. Hence, 2 v 1 (∞)+3 v 1 (∞)= fi v 1 (∞)= v 2 (∞)=16/5 V.

Prof. C.K. Tse: First Order Transient Example 2 (answer) 5V 16/5=3.2V v 1 t

2V

16/5=3.2V

v 2 t v 1 ( t ) = 3 .2 + 1.8 e

• 5 t / 6 V v 2 ( t) = 2 + 1 .2( 1 - e - 5 t / 6

) V

We can also find the current by i( t) =

v 1 ( t) - v 2 ( t)

R

= 3 e

• 5 t / 6 A

Prof. C.K. Tse: First Order Transient

General procedure

® Set up the differential equation(s) for the circuit in terms of capacitor voltage(s) or inductor current(s). ® The rest is just Form 7 Applied Math! ® E.g., ® Get the general solution. ® There should be n arbitrary constants for an nth-order circuit. ® Using initial conditions, find all the arbitrary constants.

d

2

vc

d t

2

+ A

dvc

dt

+ Bvc = C

In the previous example: