barcode reader using vb net source code Figure 551 Underdamped circuit voltage response in Software

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Figure 551 Underdamped circuit voltage response
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Focus on Computer-Aided Tools: The Matlab m- le containing the numerical analysis and plotting commands for this example may be found in the CD that accompanies this book An EWB simulation is also enclosed
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EXAMPLE 513 Transient Response of Automotive Ignition Circuit
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The circuit shown in Figure 552 is a simpli ed but realistic representation of an automotive ignition system The circuit includes an automotive battery, a transformer4 (ignition coil), a capacitor (known as condenser in old-fashioned automotive parlance) and a switch The switch is usually an electronic switch (eg, a transistor see 9), and can be treated as an ideal switch The circuit on the left represents the ignition circuit immediately after the electronic switch has closed, following a spark discharge Thus, one can assume that no energy is stored in the inductor prior to the switch closing, say at t = 0 Furthermore, no energy is stored in the capacitor, as the short
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4 Transformers
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are discussed more formally in s 7 amd 17; the operation of the transformer in an ignition coil will be explained ad hoc in this example
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Circuits
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N2 = 100 N1 N1 + VB C switch closed i spark plug N2 LP, RP VB + i C N1
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N2 = 100 N1 N2 LP , RP
spark plug switch closed
circuit (closed switch) across it would have dissipated any charge in the capacitor The primary winding of the ignition coil (left-hand-side inductor) is then given a suitable length of time to build up stored energy, and then the switch opens, say at t = t, leading to a rapid voltage buildup across the secondary winding of the coil (right-hand-side inductor) The voltage rises to a very high value because of two effects: the inductive voltage kick described in Examples 53 and 58, and the voltage multiplying effect of the transformer The result is a very short high-voltage transient (reaching thousands of volts), which causes a spark to be generated across the spark plug
Solution
Known Quantities: Battery voltage, resistor, capacitor, inductor values Find: The ignition coil current, i(t), and the open circuit voltage across the spark plug,
vOC (t) 10 F; Lp = 5 mH
Schematics, Diagrams, Circuits, and Given Data: VB = 12 V; RP = 2
;C =
Assumptions: The switch has been open for a long time, and closes at t = 0 The switch opens again at t = t Analysis: With no energy stored in either the inductor or the capacitor, the action of
closing the switch will create a closed circuit comprising the battery, VB , the coil primary inductance, LP , and the coil primary resistance, RP The inductor current will therefore rise exponentially to a nal value equal to VB /RP , as described in the following equation: VB VB 3 1 e t/ = 1 e Rp t/L = 6 1 e t/25 10 0<t < t iL (t) = RP RP We know from Example 54 that the energy storage element will acquire approximately 90 percent of its energy in 3 time constants; let s assume that the switch remains closed for 5 time constants; ie, t = 125 ms Thus, at t = t, the inductor current will be equal to VB 1 e 5 / = 6 1 e 5 = 596 A iL ( t) = RP that is, the current reaches 99 percent of its nal value in 5 time constants Now, when the switch opens at t = t, we are faced with a series RLC circuit similar to that of Example 513 The inductor current at this time is 596 A, and the capacitor voltage is zero, because a short circuit (the closed switch) had been placed across
5
Transient Analysis
the capacitor The differential equation describing the circuit for t > VB vL (t) vR (t) vC (t) = 0 VB L diL 1 iL R dt C
t
t is given below
iL dt = 0
1 R diL 1 dVB d 2 iL + iL = =0 t> t + 2 dt L dt LC L dt Next, we solve for the roots of the characteristic polynomial: 1 R =0 s2 + s + L LC s1,2 = L 1 2R 2 R L
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