Download Inductor Discharging

Lesson 13:
Inductor Transient Analysis
1
Learning Objectives
• Calculate inductor voltage and current as a function of time.
• Explain inductor DC characteristics.
• Calculate inductor energy stored.
2
Inductor Charging
• Inductor: Oppose Changes (Choking effects).
• Inductor is initially fully discharged:
− acts like a open circuit.
• When the switch is initially closed, the changing current
across the inductor immediately induces a voltage that
opposes that change, which keeps the current near zero:
vL  E
3
Inductor Charging Equations
• As current (iL) across the inductor builds up, the voltage
across the R1 resistor increases.
iL (t )  I 0 (1  e  tR1 / L )
vL (t )  E (e  t / )
I0 
E
R1
• Voltages and currents in a the inductor charging circuit
change exponentially over time.
4
Power/Energy
• Similarly to capacitance; the energy stored in an
inductor can be calculated by:
1
W  * L * iL 2
2
5
Steady State Conditions
• When the circuit is at steady state:
− The voltage and current reach their final values and stop changing.
• There is no change in current in the circuit, so the inductor
has zero voltage induced across it. Inductor current will be
steady:
E
100V
iL 

 100mA
R1 1000
• Inductor then looks like a short circuit.
6
The Time Constant
• Rate at which an inductor charges depends on R and L, which
is called the TIME CONSTANT:
 ch arg ing
• Transients can be considered
to last for five time constants.
7
L

R1
Example Problem 1
In the circuit below, the switch is initially open and conditions are at
steady-state.
After the switch is shut, determine:
a)
b)
c)
how long it will take for the inductor to reach a steady-state condition
(>99% of final current).
Write the equation for the VL(t) & iL(t). Sketch the transient.
Find the Energy stored in the Inductor.
 ch arg ing 
I0 
L
4H

 2m sec
R1 2k 
E 50V

 25mA
R1 2k 
iL (t )  I o (1  e  t / )  25mA(1  e  t /2 ms )
vL (t )  E (e  t / )  50V (e  t /2 ms )
W
1
* L * iL 2  0.5* 4 H * (25mA) 2  125mJ
2
8
Interrupting Current in an
Inductive Circuit
• When switch opens in an RL circuit:
− Energy is released in a short time.
− This may create a large voltage.
− Induced voltage is called an inductive kick.
• Opening of inductive circuit may cause voltage
spikes in thousands of volts range.
9
Interrupting a Circuit
• Switch flashovers are generally undesirable.
− They can be controlled with proper engineering design.
• These large voltages can be useful.
− Such as in automotive ignition systems.
10
Inductor Discharging
• Assume an Inductor is initially fully charged with a constant
100 mA (IO) current flow. It acts like a short circuit…
• When the switch is opened, the inductor will immediately
induce a voltage to keep the 100 mA current constant.
− KVL can be used to calculate this induced voltage.
− Notice the polarity of the induced voltage!
vL  vR1  vR 2  1100V
vR1  I 0 R1  100mA 1000   100V
vR 2  I 0 R2  100mA 10000   1000V
11
vL  vR1  vR 2  1100V
Inductor Discharging
• As stored energy is released, the induced voltage across the
inductor drops.
• This makes the voltage drop across the resistor drop, so
current in the circuit drops.
iL (t )  I 0e
 t  R1  R2  / L
vL (t )  -E (e  t / ) -50V (e  t /2 ms )
12
Inductor Discharging Equations
• Voltages and currents in a discharging circuit also change
exponentially over time.
13
The Time Constant
• Rate at which an inductor discharges depends on R and L,
which is called the TIME CONSTANT:
 disch arg ing
L

 R1  R2 
• Transients can be considered
to last for five time constants.
14
Example Problem 2
•
The circuit shown below has been in operation with the switch shut for a
long time. The switch opens at time t = 0, determine:
a)
b)
How long it will take for the inductor to discharge.
Write the discharge equation for the VL, iL,.
 disch arg ing 
L
4H

 800  s
 R1  R2  (2k   3k  )
 steadystate  5*  5*800  s  4000 s
I0 
E 50V

 25mA
R1 2k 
iL (t )  I o (e  t / )  25mA(e  t /800  s )
vL  vR1  vR2  ( I 0 * R1 )  ( I 0 * R2 )
vL  25mA * 2k   25mA *3k   125V
15
vL (t )  -E (e  t / ) -125V (e  t /800  s )
QUESTIONS?
16