Ideal linear inductor

Inductor is an energy storing device like Capacitor. Inductor stores energy as

magnetic flux around it. Stored energy can be later recovered.

(Tip: inductor and capacitor are complementary to each other in the sense that if you

build circuit with one kind of element (say inductor) and develop relations for it, then

with little piece of modification it can be developed for other complementary element

(capacitor) as well. E.g. if you build a circuit with one element with voltage source then

the same kind of circuit can be built for other element with current source)

The amount of energy stored by inductor depends on the inductance of inductor

Inductance:

Let Number of turns=N

Magnetic permeability of core = µ

Let area of cross section of core=A

Circumference of core = l

λ = total flux linked with wire coil

Let inductance = L

Then L= µ * N2* A/ l

The permeability of the vacuum is µ0 = 4 * * 10 -7. This is also called the

magnetic constant.

Relative permeability:

Magnetic materials such as manganese-zinc ferrite have a higher permeability

than the vacuum.

The ratio of the permeability of material to the permeability of vacuum is called

the relative permeability of the material.

Area = A (green)

Circumference= l (dashed line)

For example, the relative permeability of one kind of manganese-zinc ferrite is

about 640.

Electrical symbol of inductor is given as:

The magnetic flux through the coil is given as

λ = L * i

Where unit for λ is Webbers, unit for L (inductance) is Henry and unit for i

(current) is Ampere.

Inductors are linear as they obey both the property of homogeneity and

superposition. (Short cut to proof: two equations below are linear, as derivative

and integral are linear operators)

v = d/dt (λ) (from Maxwell’s equation)

v = d/dt (L * i) (since λ=L * i)

Now we assume for simplicity that L (inductance) is constant w.r.t voltage and

time.

v=L * (d/dt (i))

di = 1/L * (v * dt)

∫

Energy storage in inductor:

λ = L * i

v = L * di/ dt

Since power= voltage * current

P = v * i

P = L * di/dt * i (since v = L * d/dt(i))

P = d/dt(L * i2/2) A

+

v

-

i

L

+

v

-

L

i

Since power is rate of change of energy w.r.t time

Power= Energy/ time

So Energy = E = 1/2 * (L * i2)

Inductor is a memory device: The behavior (output) of the inductor not only

depends on the set of inputs but also on its previous state, so it is a memory

device.

P = d/dt (E) where P = Power and E = Energy

If we integrate the above equation in P between two times, we get the energy

change in the inductor over the time period

∫ ( )

= ∫ ( ) ( ( )

(P=L*i* di/dt)

( ( )) ( ( ))

Inductor and a voltage source:

Let the voltage source v(t), a square wave of magnitude V, with the time period T.

V=d/dt (λ) from Maxwell’s equation

λ Eqn.β

Where i(0), is the current when time t=0

V

v(t)

t 0 T

t 0

i(0)

T

λ =L*i

∆λ=L*∆i

∆i=∆λ/L

∆i=V*T/L

(eqn.β)

(In practical, inductor wire has resistance, so it has power dissipation. So there is

always parasitic resistance within inductor which causes power dissipation and

current decay)

(Note: inductors like to hold current the same and for continuous input changes

and for non-infinite jumps, inductor current changes slowly, i.e. inductor current

is always continuous)

As we see in the diagram that current value changes slowly, so if we use a current

source to make a sudden and large change in current, inductor will act as

“instantaneous open to external change” and as a result can damage other

components in the circuit.

Analyzing an RL circuit:

(The reason to analyze this kind of a circuit is that, if we have linear elements in the

circuit with the independent current and voltage sources, we make the rest of circuit as

Norton equivalent and analyze our element of interest out of it (which is inductor in this

case). It makes Analysis easy)

We have a current source iI(t) in parallel with a resistor R and inductor L.

Now we apply node method to analyze the circuit

At node 1:

-iI + vL/ R + iL = 0 (since vL=L*d/dt(iL))

L/ R * d/dt(iL) + iL = iI

iI(t) R L

iL(t) iR

+

vL(t)

-

1

(L/R) has units of time and this parameter (L/R) is time constant ‘Ʈ’ for (‘RL’)

circuit

An example of RL circuit with Current source:

Let iL(t)=iL(0) at the time, t=0

Let iI(t)=II

iL(0)= I0

Now from the above equation of ‘RL’ circuit, we get

L/ R * d/dt(iL)+iL= II Eq.1

(WE will use Method of Homogenous and Particular Solutions to solve this differential

equation. It comprises of the following steps

1. Find the particular solution

2. Find the homogenous solution

3. The total solution is the sum of particular and homogenous solution. Then use the

initial conditions to solve for the remaining constant)

1. Particular solution:

L/ R * d/dt(iLp)+ iLp = II

iLp is the any solution that satisfies the above equation

Let iLp =II (guess)

Putting this value in the above equation

L/ R * d/dt(II) + II = II

II = II (since d/dt(II)=0 as II is constant w.r.t time)

So ILp = II

2. ILH: solution to homogenous equation by setting drive (II) equal to zero

L/ R * d/dt(iLH) + iLH = 0

Let iLH = A * e st

Putting this value in the above equation

L/ R * d/dt(A * e st) + A * e

st = 0

L/ R * s * A * e st + A * e

st = 0

L/ R * s + 1 = 0

S = - R/ L

So iLH = A * e –(R/L)*t

Since ‘L/R’ is time constant ‘Ʈ’

iLH = A * e -t/ Ʈ

3. Total solution:

iL = iLp + iLH

putting the value of iLp and iLH in the above equation, we get

iL = II + A * e –(R/L) * t

Eq.2

Now find the value of ‘A’ using the initial condition

iL = I0 at t=0

So I0 = II + A * e –(R/L)* 0

I0 = II + A

So A = I0 – II

So Eq.2 becomes

iL = II + (I0 – II) * e –(R/L) * t

Plot solution:

iI(t) = II

iL(0) = I0

iL = II + (I0 – II) * e –(R/L) * t

(where iL is actually iL(t)

corresponding to t)

After long period of time the potential difference across the inductor becomes

zero and whole current supplied by source passes through the inductor

Transient analysis of ‘RL’ circuit with voltage Source:

I0

II

iL(t)

0 Ʈ=L/R t

vs(t)= step input

The input waveform VS is assumed to be a voltage supplied at t=0 and the

inductor current is assumed to be zero just before the step input i.e. iL = 0 at t<0

Now using the node method at point 1:

(vL - VS)/ R + iL = 0

VL -VS + iL * R = 0

L * d/dt(iL) + iL * R = 0

The homogeneous equation is

L * d/dt(iLH) + iLH * R = 0 Eqn.3

Assume a solution of the form

iLH = A * e st

So L * S * A * e st + R * A * e

st = 0 (from Eqn.3)

L*s + R = 0

S= - R/ L

Thus homogenous solution is thus

iLH = A * e –(R/L) * t

The particular equation is given as

0 t -t

vS(t)

VS

t -t 0

VS

vS(t)

iLp * R + L d/dt(iLP) = VS

Drive is step input which is constant for large t, it is appropriate to assume the

solution of the form

iLp = K

Substituting the value in the above equation

K * R = V (since VS = V for large time, t)

K = V/R

iLp = V/ R

Thus the complete solution is of sum of homogeneous and particular

solution

iL = V/ R + A * e –(R/L) * t

The initial condition together with continuity condition, can now be applied

to evaluate A.

Since inductor voltage cannot be infinite in circuit.

So d/dt (i) must be finite and hence inductor current must be continuous

Therefore at t=0, iL = 0 (given)

V/ R + A = 0

A = - V/R

For t>0

iL = V/ R * (1 – e –(R/L) * t

)

The voltage across the inductor is

vL = L * d/dt(iL) = V * e –(R/L) * t

Combination of inductors:

1. Series Combination:

Consider the series combination of two inductors, we assume that neither

inductor carry a current at the time of their connection

Since the two inductors share the common current

i(t) = λ1(t)/ L1 + λ2(t)/ L2

Now using the KVL, we get

v(t) = v1(t) + v2(t)

Since we know that

λ(t) = ∫ ( )

The above equation yield

λ(t) = λ1(t) + λ2(t)

Finally, since the effective inductance L of two series inductors is λ/I, it

follows that

L= λ(t)/ i(t) = L1 + L2

i.e. L = L1 + L2

2. Parallel combination:

Since the two inductors share a common voltage, it follows that they share a

common flux linkage λ.

λ(t) = λ1(t) = λ2(t) = ∫ ( )

Thus from above equation

λ(t) = L1 * i1(t) = L2 * i2(t)

Now using KCL, we observe that

i(t) = i1(t) + i2(t)

Since the effective inductance L of the two parallel inductors is λ/i

1/ L = i(t)/ λ(t)= 1/L1 + 1/ L2

1/ L = 1/ L1 + 1/ L2