Chapter 7. First and second order transient circuits

Contents 1. Introduction 2. First order circuits 3. Second order circuits 4. Application examples

1. Introduction Camera flash discharge :

Analysis

2. First order transient circuits Solution to 1st order differential equation : f (t) = 0 → homogeneous equation f (t) ≠ 0 → inhomogeneous equation xh(t) or xc(t) → homogeneous or complementary solution xp(t) → inhomogeneous or particular solution

Particular solution : For polynomial functions f (t) , x(t) should also be a polynomial function of the order of equal or lower degree.

Homogeneous solution : Integrating both sides with respect to the arguments, The constant K2 can be found, if the value of x(t) is known at one instant of time.

General solution : 1st order differential equation K1 : steady-state solution: x(t) → K1 as t→∞ when the second term becomes negligible. τ : time constant

Simple RC circuit

Simple RL circuit

Example 7.1 Consider the circuit shown in Fig. 7.4a. Assuming that the switch has been in position 1 for a long time, at time t=0 the switch is moved to position 2. We wish to calculate the current i(t) for t > 0. Initial condition :

Suppose

Example 7.2 The switch in the network in Fig. 7.5a opens at t=0. Let us find the output voltage vo(t) for t > 0. Initial condition : Thevenin’s equivalent

Example 7.3 Consider the circuit shown in Fig. 7.6a. The circuit is in steady state prior to time t=0, when the switch is closed. Let us calculate the current i(t) for t > 0. Initial condition :

Problem-Solving Strategy Step 1. We assume a solution for the variable x(t) of the form Step 2. Assuming that the original circuit has reached steady state before a switch was thrown. Solve for the voltage across the capacitor, υc(0-) or the current through the Inductor iL(0-), prior to switch action. Step 3. Recall that voltage across a capacitor and the current flowing through an inductor cannot change in zero time after the switch is changed. Solve for the initial value of the variable x(0+) using υc(0+) = υc(0-), iL(0+) = iL(0-). Step 4. Assuming that steady state has been reached after the switches are thrown, draw the equivalent circuit, valid for t >5τ by replacing the capacitor by an open circuit or the inductor by a short circuit.

Step 5. Since the time constant for all voltages and currents in the circuit will be the same, it can be obtained by reducing the entire circuit to a simple series circuit containing a voltage source, resistor, and a storage element (i.e., capacitor or inductor) by forming a simple Thévenin equivalent circuit at the terminals of the storage element. This Thévenin equivalent circuit is obtained by looking into the circuit from the terminals of the storage element. Step 6. Using the results of steps 3, 4, and 5, we can evaluate the constants.

Example 7.4 The circuit shown in Fig. 7.8a is assumed to have been in a steady-state condition prior to switch closure at t=0. We wish to calculate the voltage v(t) for t>0. Initial condition (before the switch action)

Right after the switch action : (t = 0+) Applying Kirchhoff current law to node υ1,

Steady state solution :

Pulse response

Unit step function

Example 7.6 Let us determine the expression for the voltage υo(t). Since the source is zero for all negative time, the initial conditions for the network are zero 9 0.3

7.3 Second order transient circuit Integrating both sides,

Solution to 2nd order differential equation : Homogeneous differential equation : Standard form : ζ : damping ratio ω0 : natural frequency Trial solution : Characteristic equation :

Characteristic equation : Homogeneous solution : Initial condition

(1) |ζ|>1 : over damped (2) |ζ|=1 : critically damped (3) |ζ|<1 : under-damped

Derivation for the critically damped case :

Typical unit step responses with ζ changed (1) |ζ|>1 : over damped (2) |ζ|=1 : critically damped (3) |ζ|<1 : under-damped (ζ is proportional to R.)

Example 7.7 Let us assume that the initial conditions on the storage elements are iL(0) = -1 A and υC(0) = 4 V. Let us find the node voltage v(t) and the inductor current. ( R=2, C=1/5F, L=5 H) Applying Kirchhoff current law to node υ(t), Trial solution : Characteristic equation :

General solution : Initial condition for υ(0) Initial conditions for are used to determine unknown coefficients. Initial conditions can be extracted from KCL on node υ

Example 7.8 The series RLC circuit shown in Fig. 7.18 has the following parameters: C=0.04 F, L=1 H, R=6 , iL(0) = 4 A, and υC(0) = -4 V. Let us determine the expression for both the current and the capacitor voltage. Applying Kirchhoff voltage law to the loop, Trial solution : Characteristic equation :

Initial condition for i(0) : Initial conditions for

Example 7.9 Determine the current i(t) and the voltage υ(t) in the figure. Using KVL : Using KCL :

Trial solution : Characteristic equation : Initial condition for υ(0) Initial conditions for

Example 7.11 Let us determine the output voltage υ(t) for t>0.

Particular solution : Homogeneous solution :

Initial conditions : Before the switch action: Using KCL :

Application example 7.13 Consider the high-voltage pulse generator circuit shown in Fig. 7.26. This circuit is capable of producing high-voltage pulses from a small dc voltage. Let’s see if this circuit can produce an output voltage peak of 500 V every 2 ms, that is, 500 times per second. Switch in position 1 :

Switch in position 2 :

Application example 7.15 Consider the simple RL circuit, shown in the figure, which forms the basis for essentially every dc power supply in the world. The switch opens at t=0. Vs and R have been chosen simply to create a 1-A current in the inductor prior to switching. Let us find the peak voltage across the inductor and across the switch. Before the switch action : After the switch action :

Inductor voltage : (inductive kick) Switch voltage : At the peak inductor voltage is negative infinity! This voltage level is caused by the attempt to disrupt the inductor current instantaneously. The peak switch voltage jumps up to positive infinity. This phenomenon is called inductive kick, and it is the nemesis of power supply designers.

Circuit design to avoid inductive kick Before the switch action : After the switch action :

Set arbitrarily Initial conditions :

Application example 7.16 DC voltage sources or power supplies are fed from AC outlets on walls. The ac waveform is converted to a quasi-dc voltage by an inexpensive ac–dc converter whose output contains remnants of the ac input and is unregulated. A higher quality dc output is created by a switching dc–dc converter. Of the several versions of dc–dc converters. We will focus on a topology called the boost converter.

Assuming that Vo is nearly constant, The boost converter with switch settings for time intervals (a) ton and (b) toff. (I0 is the initial current at the beginning of each switching cycle) State #1 : Assuming that Vo is nearly constant, State #2 :

State #1 : State #2 :

Application example 7.20 : automobile ignition system The voltage source represents the standard 12-V battery. The inductor is the ignition coil. The inductor’s internal resistance is modeled by the resistor. The switch is the keyed ignition switch. Requirement: Optimum starter operation requires an over-damped response for iL(t) that reaches at least 1 [A] within 100 ms after switching and remains above 1 [A] for between 1~and 1.5 s. let us find a value for the capacitor that will produce such a current waveform.

Requirement: 1~1.5s 100 ms Characteristic equation : The values of s1 and s2 should be set to satisfy the requirements.

Comparing the two expressions, we see that Derived constraints: Derived constraints:

Design : (1) Let us arbitrarily choose s1=3 and s2=17, which satisfies the condition. Success Fail 0.5s 1.5s (2) Let us arbitrarily choose s1=1 and s2=19, which satisfies the condition.

Application example 7.21 : Defibrillator A defibrillator is a device that is used to stop heart fibrillations—erratic uncoordinated quivering of the heart muscle fibers—by delivering an electric shock to the heart. Its key feature, as shown in the figure, is its voltage waveform. A simplified circuit diagram that is capable of producing the Lown waveform. Let us find the necessary values for the inductor and capacitor. Lown waveform The Lown waveform is oscillatory. → Under-damped (ζ <1). The waveform starts from zero voltage.

Characteristic eq. for a series RLC circuit. The period = 10 [ms]. From the ratio of voltage peaks Natural frequency :