# Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Transients Analysis.

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Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Transients Analysis

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Solution to First Order Differential Equation Consider the general Equation Let the initial condition be x(t = 0) = x( 0 ), then we solve the differential equation: The complete solution consists of two parts: the homogeneous solution (natural solution) the particular solution (forced solution)

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING The Natural Response Consider the general Equation Setting the excitation f (t) equal to zero, It is called the natural response.

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING The Forced Response Consider the general Equation Setting the excitation f (t) equal to F, a constant for t  0 It is called the forced response.

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING The Complete Response Consider the general Equation The complete response is: the natural response + the forced response Solve for , The Complete solution: called transient response called steady state response

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING WHAT IS TRANSIENT RESPONSE Figure 5.1

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Figu re 5.2, 5.3 Circuit with switched DC excitation A general model of the transient analysis problem

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING In general, any circuit containing energy storage element Figure 5.5, 5.6

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Figure 5.9, 5.10 (a) Circuit at t = 0 (b) Same circuit a long time after the switch is closed The capacitor acts as open circuit for the steady state condition (a long time after the switch is closed).

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING (a) Circuit for t = 0 (b) Same circuit a long time before the switch is opened The inductor acts as short circuit for the steady state condition (a long time after the switch is closed).

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Why there is a transient response? The voltage across a capacitor cannot be changed instantaneously. The current across an inductor cannot be changed instantaneously.

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING REASON

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Transients Analysis 1. Solve first-order RC or RL circuits. 2. Understand the concepts of transient response and steady-state response. 3. Relate the transient response of first- order circuits to the time constant.

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Transients The solution of the differential equation represents are response of the circuit. It is called natural response. The response must eventually die out, and therefore referred to as transient response. (source free response)

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Discharge of a Capacitance through a Resistance icic iRiR Solving the above equation with the initial condition V c (0) = V i

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Discharge of a Capacitance through a Resistance

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Exponential decay waveform RC is called the time constant. At time constant, t =RC, the voltage is 36.8% of the initial voltage. Exponential rising waveform RC is called the time constant. At time constant, the voltage is 63.2% of the initial voltage.

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING RC CIRCUIT for t = 0 -, i(t) = 0 u(t) is voltage-step function

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Step Function for t = 0 -, u(t) = 0 for t = 0 +, u(t) = 1 u(t) is voltage-step function Vu(t) is a voltage – step function scaled with a constant of V. for example, if V is 5, then for t ≥ 0 5u(t) is 5, and for t < 0, 5u(t) is 0.

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING RC CIRCUIT Solving the differential equation for Vc. Complete response = natural response + forced response

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Complete Response Complete response = natural response + forced response Natural response (source free response) is due to the initial condition Forced response is the due to the external excitation.

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Figure 5.17, 5.18 Complete response is the sum of natural and forced responses of the circuit Forced response (Steady State Response) v c (∞) = 12V Natural response (Source Free Response) v c (t) = Ae -t/RC due to v c (0+) = 5 Complete Solution

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Circuit Analysis for RC Circuit Apply KCL v s is the external excitation applied.

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Example Initial condition Vc(0) = 0V time constant is 10 -3

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Example Initial condition Vc(0) = 0V and

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Energy stored in capacitor If the zero-energy reference is selected at t o, implying that the capacitor voltage is also zero at that instant, then

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Power dissipation in the resistor is: p R = V 2 /R = (V o 2 /R) e -2 t /RC RC CIRCUIT Total energy turned into heat in the resistor

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING RL CIRCUITS Initial condition i(t = 0) = I o

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING RL CIRCUITS Initial condition i(t = 0) = I o

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING RL CIRCUIT Power dissipation in the resistor is: p R = i 2 R = I o 2 e -2Rt/L R Total energy turned into heat in the resistor It is expected as the energy stored in the inductor is

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING RL CIRCUIT where L/R is the time constant

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING DC STEADY STATE The steps in determining the forced response for RL or RC circuits with dc sources are: 1. Replace capacitances with open circuits. 2. Replace inductances with short circuits. 3.Determine the time constant by finding Thévenin equivalent resistance that the energy element facing 4. As  = RC or L/R

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Example

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Solution Find i L (0 - ), the initial condition, i L (0 - ) = 20 A Find R th that the inductor connected, that is, Thévenin equivalent resistance R eq = 2+10//40 = 2+8 =10 , time constant = L/R eq = 2/10 = 0.2 sec i L (∞) = 0 i L (t) = A e –t/0.2 +0 = A e –5t, i L (0)=20  A=20, i L (t) = 20 e –5t

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING iLiL

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING After having been closed for a long time, the switch in the circuit of Fig. 8.41 is opened at t = 0. 1. Find i L (t) for t > 0. 2. Find t1 if i L (t1) =0.5i L (0). Solution 1. Determine the initial condition of i L for t  0 2. Determine the Thévenin equivalent resistance that the inductor is facing. 3. Determine the steady state response of i L

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING 2. Find t1 if i L (t1) =0.5i L (0). 1. Find i L (t) for t > 0.

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING 1.Find v c (t) for all time in the circuit 2.At what time is v c =0.1v c (0) ? Initial condition Final steady state response Time constant = RC R eq = 24  = 24 × 0.333 × 10 -3 = 8 ms

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Find v c (t) for all time in the circuit At what time is v c =0.1v c (0) ?

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Determine i(t) and v(t) Initial condition v(0 - )=20V, i(0 - )=0.1A Steady state response v(∞) = 0, i(∞) = 0 Time constant = 50×20×10-6 = 1000×10-6 i(t) = 0 as short circuit after t >0

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING Determine i L (t) and v x (t) Initial condition i L (0 - ) = 24/(10+50)=0.4A, v x (0 - ) = 0.4×50 =20V v x (0 + ) = v L (0 + ) ×30/(80)×50=7.5 V Steady state response i L (∞) = 0 A, v x (∞) = 0 V Time constant = 25×10 -3 /(50//30)=25×10 -4 /18.75=1.33×10 -3 s and i L (t) = 0.4e -750t

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING After being closed for a long time, the switch in the circuit is opened at t = 0. Find v C (t) for t > 0. Calculate values for i A (-100 µs) and i A (100 µs). Initial condition v c (0 - )=10V, Steady state response v c (∞) = 0 i A (-100 µs) = i A (0 - ) =10/200 = 0.05 A, i A (∞) = 0 Time constant = [10 + (50//200) ]×1×10 -6 s= 5×10 -5 s

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING After being closed for a long time, the switch in the circuit is opened at t = 0. Find v C (t) for t > 0. Calculate values for i A (-100 µs) and i A (100 µs).

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING After being in the configuration shown for a long time, the switch is opened at t = 0. Determine values for (a)i s (0 - ); (b) i x (0 - ); (c) i x (0+); (d) i s (0+); (e) i x (0.4 s).

Department of Electronic Engineering BASIC ELECTRONIC ENGINEERING After being in the configuration shown for a long time, the switch is opened at t = 0. Before the switch is opened, the cap. acts as an open circuit. After the switch is opened, it is isolated from the rest of the circuit; simply in series with the (75+25)  resistor. Time constant = 100×10×10 -3 s = 1s Finish it as a tutorial question.