Inductors
The inductor is typically a coil of wire (a solenoid).
The inductance is the ability of inductor to resisting any change of
electric current through the coil.
The voltage drop across the inductor in
response to the current change:
∂I
V =L
∂t
The coefficient L is called “inductance”
The circuit symbol for inductor
L
The circuit
diagram symbol
for inductance
The unit for inductance is Henry
∂I
V =L
∂t
 ∂I 
L =V  
 ∂t 
1 H is the inductance of the coil that generates
the voltage of 1V when the current change
rate is 1 A per 1 s:
1H = 1V ×1s / 1A
Factors affecting the Inductance
A- the cross-section area
l – the length
Greater number of turns N, results in greater inductance
(stronger magnetic field)
Greater coil area A results in greater inductance (stronger magnetic field)
N 2 µ0 A
µ0 is the magnetic permittivity of the vacuum, µ0 = 4π ×10-7 H/m
L=
l
Factors affecting the Inductance
Certain materials increase the magnetic field.
This increase is described by a factor called magnetic permittivity µ
The inductance of a coil having N turns, the area A, the length l and the
magnetic permittivity µ of the filling material:
N 2 µ µ0 A
L=
l
Energy stored in inductor
VL
L
VB
R
L× I
WL =
2
2
Parallel connection of inductors
V = VB
IT
I1
I2
L2
L1
V = L1
∂I1
∂t
V = L2
∂I 2
∂t
The voltage V1 = V2; The total current IT = I1 + I2.
V ∂I1
=
;
L1 ∂ t
V ∂I 2
=
;
L2 ∂ t
∂I ∂I
V V
+
= 1+ 2;
L1 L2 ∂ t ∂ t
  1 1
V = 1 /  +
  L1 L2
1/Lpar= 1/L1 + 1/L2;
 1 1
V +
 L1 L2
  ∂I T
;

 ∂ t
L1 L2
L par =
L1 + L2
 ∂ ( I1 + I 2 ) ∂IT
;
=
=
∂t
∂t

Series connection of inductors
V = VB
L1
V1
L2
V2
Inductors L1 and L2 are connected in series:
the current I through L1 and L2 is the same;
the total voltage drop VT = V1 + V2.
V1 = L1
∂I
∂t
V2 = L2
From the KVL, VB = VTot= V1 + V2.
 ∂I
∂ I
∂I
∂I
 = (L1 + L2 )
VTot =  L1
+ L2
= LS
;
∂t 
∂t
∂t
 ∂t
The equivalent series inductance,
LS= L1 + L2
∂I
∂t
Transients in Inductors
∂I
VL = L
∂t
VL
L
VB
R
When the switch is turned on, dI/dt >0,
hence VL >0.
The inductor acts a dynamic (transient)
resistance.
The current through inductor cannot change instantly:
Initial condition: IL(0+) = IL (0-) = 0
Transients in Inductors
∂I
VL = L
∂t
VL
L
VB
R
When the switch is turned off, dI/dt <0,
hence VL < 0.
The inductor acts as a dynamic current
source (trying to maintain the same
current as before switching).
The current through inductor cannot change instantly:
Initial condition: IL(0+) = IL (0-) = VB/R
Transient current in R-L circuit
VL
L
VB
R
(
)
i (t ) = I DC × 1- e-Rt/L ;
where I DC =VB / R;
v L (t ) =VB e-Rt/L ;
Characteristic time of the RL circuit:
τRL = L/R
General form of step response of R-L circuit
VL
L
VS
R
R-L circuit
iL = iLF + ( iL0 − iLF ) e− t / τ
τ = L/R
iL0 is the inductor current right before the commutation;
iLF is the inductor current after all the transient processes are over.
R is the total resistance connected to the inductor after commutation
(al the sources are zeroed to find the equivalent total resistance)
General form of step response of R-C and R-L circuits
C
R
VS
L
VS
R
R-L circuit
R-C circuit
General solution for the step response:
x = xF + ( x0 − xF ) e− t / τ
RC-circuit: x = vC
vC = vCF + ( vC 0 − vCF ) e
τ = RC
x0 = initial value;
xF = final value;
RL-circuit: x = iL
−t /τ
iL = iLF + ( iL0 − iLF ) e− t / τ
τ = L/R
Example 1
Delayed alarm circuit
S1
R=10 k
VB=4.5 V
S2
C=0.5 mF
Sensor switch: S1.
Electronic switch S2 triggers the alarm system when the voltage across it
exceeds the preset threshold value VT.
R = 10k; C = 0.5 mF; VB = 4.5 V.
The required time delay between the switch S1 turn-on and triggering switch
S2 must be tt = 3s.
What threshold voltage VT must the switch S2 be tuned to?
Example problem
The switch in the circuit shown in Fig. 7.28 has been open for a
long time. At t = 0 the switch is closed.
Find the expression for i at t>0
i = iF + ( i0 − iF ) e − t / τ
i0 = 20V/(1Ω+3Ω) = 5 A
iF = 20V/1Ω = 20 A
τ = L/R = 80 mH / 1Ω = 80 ms.
i = 20 −15 e
− t / 0.08
A
Electronic photo flash circuit