Download 2nd Order Circuits

Boundary Conditions
Objective of Lecture
 Demonstrate how to determine the boundary
conditions on the voltages and currents in a 2nd order
circuit.
 These boundary conditions will be used when
calculating the transient response of the circuit.
nd
2
Order Circuits
 A second order differential equation is required to
solve for the voltage across or the current flowing
through a component.
 The circuit will contain at least one resistor and the
equivalent of two energy storage elements

2 capacitors, 2 inductors, or a capacitor and an inductor
Boundary Conditions
 Steady state
 For step response functions u(t- to) for all times between
t = +/- ∞ except for some time period after t = to


Capacitors are opens
Inductors are short circuits
 During the transition at the step t = to
 Voltage across a capacitor is continuous

vC(to +) = vC (to -)
 Current through an inductor is continuous

iL(to +) = iL(to -)
Initial Condition
 Redraw the circuit at t < to
 Determine the value of all voltage and current sources
at t< to
 Make the appropriate substitutions for the energy
storage devices.
 Substitute an open circuit (∞W resistor) for all
capacitors.

Note: IC(t < to ) = 0A.
 Substitute an short circuit (0W resistor) for all inductors.
 Note: VL(t < to ) = 0V.
 Calculate VC(t < to ) and IL(t < to ).
Final Condition
 Redraw the circuit at t =∞ s
 Determine the value of all voltage and current sources
at t =∞ s
 Make the appropriate substitutions for the energy
storage devices.
 Substitute an open circuit (∞W resistor) for all
capacitors.

Note: iC(t =∞ s) = 0A.
 Substitute an short circuit (0W resistor) for all inductors.
 Note: vL(t =∞ s) = 0V.
 Calculate vC(t =∞ s) and iL(t =∞ s).
Example #1
Example #1 (con’t)
 Initial Condition: The circuit is:
Example #1 (con’t)
∞W
Example #1 (con’t)
∞W
iL (-∞) = iL (to-) = 0A
vL (-∞) = vL (to-) = 0V
iC (-∞) = iC (to-) = 0A
vC (-∞) = vC (to-) = [R2/(R1+R2)]V
Example #1 (con’t)
 Final Condition: The switch opens,
 which removes V1 and R1 from the circuit.
Example #1 (con’t)
 The energy stored in the inductor and capacitor will be
dissipated through R2 and R3 as t increased from t= to.
Example 1 (con’t)
 At time t = ∞s, the energy stored in the inductor and
in the capacitor will be completely released to the
circuit.
∞W
Example #1 (con’t)
∞W
iL (∞s) = 0A
vL (∞s) = 0V
iC (∞s) = 0A
vC (∞s) = 0V
Example #1 (con’t)
For to < t << ∞s
iL (t) ≠ 0
vL t) ≠ 0
iC (t) ≠ 0
vC (t) ≠ 0
Electronic Response
 Draw the circuits when t < to and t = ∞s for the
following circuit:
Example #2 (con’t)
Example #2 (con’t)
iL (-∞s) = 0.3mA
vL (-∞s) = 0V
iC (-∞s) = 0A
vC (-∞s) = 3.5V
Example #2 (con’t)
iL (∞s) = 0A
vL (∞s) = 0V
iC (∞s) = 0A
vC (∞s) = 5V
Example #3
Example #3 (con’t)
iL1 (-∞s) = -1mA
vL1 (-∞s) = 0V
iL2 (-∞s) = 1mA
vL1 (-∞s) = 0V
Example #3 (con’t)
Example #3 (con’t)
iL1 (∞s) = -1mA
vL1 (∞s) = 0V
iL2 (∞s) = 1.4mA
vL2(∞s) = 0V
Example #4
Example #4 (con’t)
iL1 (-∞s) =- 1 mA
vL1 (-∞s) = 0V
iC1 (-∞s) = viC2 (-∞s) = 0A
vC1 (-∞s) = vC2 (-∞s) = 4V
Example # 4 (con’t)
iL1 (∞s) = 0mA
vL1 (∞s) = 0V
vC1 (∞s) = vC2 (∞s) = 1V
iC1 (∞s) = iC2 (∞s) = 0A
Summary
 Calculation of the initial and final conditions for 2nd order
circuits requires:
 Knowledge of the magnitude of the voltage and/or current sources
in the circuit before and after a step function transition.
 In steady state (t < to and t = ∞s), replace energy storage devices.


Capacitors are opens circuits => iC = oA
Inductors are short circuits => vL = oA
 Calculate the voltage across the capacitor and the current through
the inductor.
 During the transition at the step t = to
 Voltage across a capacitor is continuous

vC(to +) = vC (to -)
 Current through an inductor is continuous

iL(to +) = iL(to -)