Firstorder transient


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1 EIE209 Basic Electronics Firstorder transient Contents Inductor and capacitor Simple RC and RL circuits Transient solutions
2 Constitutive relation An electrical element is defined by its relationship between v and i. This is called constitutive relation. In general, we write v = f (i) or i = g(v) i + v For a resistor, v = i R The constitutive relation of a resistor has no dependence upon time. 2
3 Capacitor and inductor i c C + v c The constitutive relation of a linear capacitor is: i c = C dv c dt where the proportionality constant C is capacitance (unit is farad or F) i L L The constitutive relation of a linear inductor is: + v L v L = L di L dt where the proportionality constant L is inductance (unit is henry or H) 3
4 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. 4
5 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 5
6 A simple firstorder RC circuit Let us consider a very simple dynamic circuit, which contains one capacitor. After t = 0, the circuit is closed. So, we can easily write and Thus, we have Thus, we have fi V o for t>0 If the initial condition is v C (0 + ) = 0, then A = V o. Thus, the solution is 6
7 Transient response of the RC circuit Once we have the capacitor voltage, we can find anything. Starting with We can derive the current as time constant We see the solution typically has a TRANSIENT which dies out eventually, and as t tends to, the solution settles to a steady state. 7
8 A simple firstorder 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, i L (0 ) = 0. At t = 0, the switch is opened. Current goes to R and L. We know from KCL that I o = i R + i L for t > 0, i.e., The constitutive relations give Hence, fi From the initial condition, we have i L (0 ) = 0. Continuity of the inductor current means that i L (0 + ) = i L (0 ) = 0. Hence, Thus, A = I o The solution is 8
9 Transient response of the RL circuit Starting with We can find v L (t): time constant 9
10 Observation firstorder transients First order transients are always like these: 10
11 Let s do some math 5 x(t) x(t) = 5(1 e t/t ) 6 x(t) x(t) = 1 + 5(1 e t/t ) 5 x(t) x(t) = 2 + 7(1 e t/t ) 0 t 1 0 t 0 2 t x(t) 5 x(t) = 5 e t/t x(t) 6 x(t) = e t/t x(t) 4 x(t) = e t/t 0 t 1 0 t 0 t 3 11
12 General firstorder solution NO NEED TO SOLVE ANY EQUATION, just find 1. the starting point of capacitor voltage or inductor current 2. the ending point of the time constant t 12
13 Finding t For the simple firstorder RC circuit, t = C R. For the simple firstorder RL circuit, t = L / R. The problem is Given a firstorder circuit (which may look complicated), how to find the equivalent simple RC or RL circuit. 13
14 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, shortcircuit all voltage sources and opencircuit all current sources. Then, reduce the circuit to either or R eq C eq R eq L eq Example: R 1 R 1 t = C (R 1 R 2 ) + R 2 C R 2 C R 1 R 2 C 14
15 Example 1 (boundary conditions given) Find v c (t) for t > 0 without solving any differential equation. Step 1: initial point (given) v c (0 ) = 50 V is known (but not what we want). Continuity of cap voltage guarantees that v c (0 + ) = v c (0 ) = 50 V. Step 2: final point (almost given) v c ( ) = 20 V. Step 3: time constant Answer is: v c (t) = e t/cr 50 The equivalent RC circuit is: 0 Thus, t = CR. 20 t 15
16 Example 2 (nontrivial boundary conditions) Find v 1 (t) and v 2 (t) for t > 0 without solving any differential equation. Suppose v 1 (0 ) = 5 V and v 2 (0 ) = 2 V. Problem: how to find the final voltage values. + v 1 i 1 C 1 2F t=0 R = 1Ω C 2 3F i 2 + v 2 Form 7 solution: You considered the charge transfer Dq from C1 to C2. Dq q 1 fi q 1 Dq q 2 fi q 2 + Dq Use charge balance and KVL equations to find the final voltage values. Clumsy solution! 16
17 Example 2 (elegant solution) We need not consider CHARGE! t=0 R = 1Ω 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. + v 1 i 1 C 1 2F C 2 3F i 2 + v 2 Step 2: final point (nontrivial) C1: 1 i = 2 dv 1 for all t dt C2: i 2 = 3 dv 2 for all t dt After t>0, we have i 1 = i 2, i.e., 2 dv 1 dt fi 2v 1 (t) + 3v 2 (t) = K for all t > 0. Integration constant + 3 dv 2 dt = 0 At t = 0 +, this equation means 2*5 + 3*2 = K. Thus, K = 16. Thus, 2v 1 (t) + 3v 2 (t) = 16 for t > 0. At t =, we have v 1 ( )=v 2 ( ) from KVL. Hence, 2v 1 ( )+3v 1 ( )=16 fi v 1 ( )=v 2 ( )=16/5 V. 17
18 Example 2 (elegant solution) Step 3: time constant R = 1Ω The circuit after t = 0 is + v 1 i 1 C 1 2F C 2 3F i 2 + v 2 This can be reduced to C 1 C 2 C 1 +C 2 = 6/5 F R = 1Ω The time constant is t = C 1C 2 C 1 + C 2 R = 6 5 s 18
19 Example 2 (answer) v 1 v 2 5V 16/5=3.2V 16/5=3.2V 2V t t v 1 (t ) = e 5t / 6 V v 2 (t) = (1 e5t / 6 ) V We can also find the current by i(t) = v 1 (t) v 2 (t) R = 3e 5t / 6 A 19
20 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., d 2 v c dt 2 + A dv c dt + Bv c = C In the previous example: Get the general solution. There should be n arbitrary constants for an n th order circuit. Using initial conditions, find all the arbitrary constants. 20
21 Basic question 1 Why must we choose capacitor voltage and inductor current as the variable(s) for setting up differential equations? Never try to set differential equation in terms of other kinds of variables! dv R dt + kv R = V o Answer: Capacitor voltages and inductor currents are guaranteed to be CONTINUOUS before and after the switching. So, it is always true that v C (0  ) = v C (0 + ) and i L (0  ) = i L (0 + ) 21
22 Basic question 1 Then, why capacitor voltages and inductor currents are guaranteed to be continuous? Answer: Let s try to prove it by contradiction. Suppose v c and i L are discontinuous at t = 0. That means, v C (0  ) v C (0 + ) and i L (0  ) i L (0 + ) v c or i L Now, recall the constitutive relations. Then, we have i c = C dv c dt v L = L di L and dt i C Æ and v L Æ slope = t which is not permitted in the physical world. So, capacitor voltages and inductor currents must not be discontinuous. 22
23 Basic question 2 How to get the differential equation systematically for any circuit? For simple circuits (like the simple RC and RL circuits), we can get it by an ad hoc procedure, as in the previous examples. But, if the circuit is big, it seems rather difficult! Hint: Graph theory. (See Chapter 8 of my book) 23
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