Starting with We can derive the current as We see the solution typically has a

Starting with we can derive the current as we see the

This preview shows page 7 - 16 out of 23 pages.

¨ 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
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Prof. C.K. Tse: Dynamic circuits—Transient 8 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, 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, ¨ Þ ¨ The solution is 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, A = – I o Thus,
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Prof. C.K. Tse: Dynamic circuits—Transient 9 Transient response of the RL circuit ¨ Starting with ¨ We can find v L ( t ): time constant
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Prof. C.K. Tse: Dynamic circuits—Transient 10 Observation — first-order transients ¨ First order transients are always like these:
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Prof. C.K. Tse: Dynamic circuits—Transient 11 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
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Prof. C.K. Tse: Dynamic circuits—Transient 12 General first-order solution NO NEED TO SOLVE ANY EQUATION, just find 1. the starting point of capacitor voltage or inductor current 2. the ending point of ………… ………. ……. ………. ………. 3. the time constant t
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Prof. C.K. Tse: Dynamic circuits—Transient 13 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.
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Prof. C.K. Tse: Dynamic circuits—Transient 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, 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 C R 1 || R 2 C eq R eq R eq L eq either or Example: t = C ( R 1 || R 2 )
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Prof. C.K. Tse: Dynamic circuits—Transient
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  • Summer '16
  • Martin Chow
  • RC circuit, RL circuit, Prof. C.K. Tse

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