1 SECOND ORDER PSpice Example

2 Example 1 (i)Find v(t) for t > 0. (ii) Find i(t) for t >0. v i 24 V R = 6  1 H 0.25 F  t = 0 The switch has been closed for a long time and it is opened at t = 0.

3 Example 1 : Solution (i)Find v(t) for t > 0. (ii) Find i(t) for t >0. v i 24 V R = 6  1 H 0.25 F  t = 0 The switch has been closed for a long time and it is opened at t = 0.

4 1. Draw the circuit for t = 0 - This statement means that the circuit is in steady state at t = 0 -. Therefore, C is opened and L is shorted. “The switch has been closed for a long time and it is open at t = 0” 24 V 6  11 i(0) = A v (0) = V + - Find i(0) and v(0) Steps to solve this problem

5 2. Draw the circuit for t = 0 + This is a starting point for the circuit to experience transient. Therefore, C is not opened and L is not shorted. We know that i(0 - ) = i(0 + ) and v(0 - ) = v(0 + ) 24 V 6  i(0 + ) = A 0.25 F + - v(0 + ) = V Find dv(0 + )/dt or/and di(0 + )/dt 1 H

6 24 V 6  i(0 + ) = A 0.25 F + - v(0 + ) = V Find dv(0 + )/dt or/and di(0 + )/dt 1 H + - v L (0 + ) (3.429)+v L (0 + ) = 0 v L (0 + ) = 0 KVL around the loop

7 3. Draw the circuit for t = ∞ At t = ∞ the circuit has reached the steady state again. Therefore, C is open and L is shorted. 24 V 6  + - Find v(∞) or/and i(∞) v(∞) = 24 V i(∞) = 0

8 4. Draw the source free circuit for t >0 1H 6  + - Voltage source is shorted and current source is opened. i v 0.25F Generally for this step, we have to find the differential equation for the source free circuit, then its characteristic equation. Since the circuit is a series RLC, we can directly write down its characteristic equation, and determine the type of the response.

9 For second order circuit:For second order circuit, its characteristics equation is General form Series RLC circuit rad/s the response is overdamped the roots are real and distinct

10 5. Write the general solution for the circuit for t > V 6  i 0.25 F + - v 1H or

11 6. Find A 1 and A 2 24 V 6  i 0.25 F + - v 1H 1 2

12 7. Find other circuit quantities for t > 0. 24V 6  i 0.25 F + - v 1H

13 Verification Using PSpice 24V 6  i 0.25 F + - v 1H h 2h E1h G2h 1  Hand Calculation Verification: Example 1 **Overdamped response: Case 1.Param R=6 L=1 C=0.25 ************************************************************************************** E1h 1h 0 value={ *EXP(-0.764*TIME)+0.448*EXP(-5.236*TIME)} R1h 1h 0 1 ; V(1h) = output voltage ************************************************************************************** G2h 0 2h value={4.015*EXP(-0.764*TIME)-0.586*EXP(-5.236*TIME)} R2h 2h 0 1 ; V(2h) = output current *************************************************************************************** V1 1 0 DC 24V R1 1 2 {R} L1 2 3 {L} IC=3.429A C1 3 0 {C} IC=3.429V.Tran 5m 5s 0 5m UIC.Probe V(1h) V(3).Probe I(R2h) I(L1).End

14 Verification Using PSpice 24V 6  i 0.25 F + - v 1H h 2h E1h G2h 1  To plug-in v(t) obtained from hand calculation in PSpice, we use E element. The E element is a dependent voltage source. We can write the equation in its value field. The 1  resistor connected to E element is just to satisfy PSpice requirement. What we want is the voltage across E element, V(1h) R2h R1h To plug-in i(t) obtained from hand calculation in PSpice, we use G element. The G element is a dependent current source. We can write the equation in its value field. The 1  resistor connected to G element is just to satisfy PSpice requirement. What we want is the current of the G element; it is also the current through R2h, I(R2h)