SECTION 5:
CAPACITANCE & INDUCTANCE
MAE 2055 – Mechetronics I
2
K. Webb
Capacitance
MAE 2055 – Mechetronics I
Fluid Capacitor
3
Consider the following scenario:

Two hemi-spherical metal shells

Separated by impermeable elastic
membrane and bolted together

Modulus of elasticity is a function of
thickness and material

Incompressible fluid

External pumps control
pressure/flow rate
Total mass inside the device is constant


Mass on each side will vary as the
membrane deforms
K. Webb
MAE 2055 – Mechetronics I
Fluid Capacitor - Equilibrium
4

Equal pressure on both sides
P1  P2
P  P1  P2  0

No fluid flow
m1  m 2  0


Membrane does not
deform
Equal volume and mass on each side
m
m1  m 2 
2
K. Webb
MAE 2055 – Mechetronics I
Fluid Capacitor – P1 > P2
5

Greater pressure on side 1

Membrane deforms

More fluid on side 1 than on side 2

Mass differential
proportional to:
P1  P2 , P  0
m1  m 2

Pressure differential

Membrane properties
m  P ,  , A

Energy is stored in the membrane
K. Webb
MAE 2055 – Mechetronics I
Fluid Capacitor – P1 < P2
6

Lower pressure on side 1
P1  P2

Membrane deforms

Less fluid on side 1 than on side 2
m1  m 2

Total mass remains const.

Still no steady-state fluid flow
(transient, yes – steady-state, no)
m1  m 2  m
m1  m 2  0
K. Webb
MAE 2055 – Mechetronics I
Fluid Capacitor – Forced Flow Rate
7

Constant flow rate forced into side 1

Incompressible, so flows are equal
and opposite
m1  0
m 1  m 2

Total mass still remains
constant
m1  m 2  m

Mass on each side is linearly
proportional to time
m 1 (t )  m 1  t C 1
m 2 (t )  m 2  t C 2  m 1  t C 2
K. Webb
MAE 2055 – Mechetronics I
Fluid Capacitor – Capacitance
8

Define a relation between ΔP and Δm
m
C 
P

C is capacitance

Differential mass stored
per differential pressure

Intrinsic device property

Proportional to physical parameters:

Membrane area

Modulus of elasticity
K. Webb
MAE 2055 – Mechetronics I
Fluid Capacitor – DC vs. AC
9

In steady-state (DC), no fluid flows
m1  m 2  0

Consider sinusoidal ΔP (AC):
P  P cos(t )

Mass flow rate into/out of
the device is prop. to
capacitance and to the time
rate of change of the
differential pressure:
m 1  m 2  C
K. Webb
dP
 C  P sin(t )
dt
MAE 2055 – Mechetronics I
Fluid Capacitor – Time-Varying ΔP
10

Fluid flowing out equals fluid flowing in
m 1  m 2




Not the same fluid, because it can’t
permeate the membrane, but fluid
appears to flow through the device
This is a displacement flow
Results from displacement
of the membrane
The faster ΔP changes, the higher
the mass flow rate
m 

The larger the capacitance the higher the mass
flow rate
m C
K. Webb
MAE 2055 – Mechetronics I
ΔP Can’t Change Instantaneously
11

A given ΔP corresponds to a particular
membrane displacement

Forces must balance

Membrane can’t instantaneously jump
from one displacement to another

Step-change in
displacement is impossible

Step-change in pressure is
impossible – would require
an infinite flow rate

Pressure across a fluid capacitor
cannot change instantaneously
K. Webb
MAE 2055 – Mechetronics I
Capacitor
12



Capacitors are components
that store energy in the form
of an electric field
A charge differential is
stored on parallel metal
plates (electrodes) separated
by a dielectric (insulator)
Schematic symbol:
Dielectric
E-field lines
Negative
Charge
-Q
V
V+
-
C

Units of capacitance are
Farads(F)
K. Webb
Positive
Charge
+Q
Metal plates
MAE 2055 – Mechetronics I
Capacitors – charge and voltage
13

Capacitance is the ratio of
charge stored on the
electrodes to the voltage
across the capacitor
Q
C
V

Dielectric
E-field lines
Positive
Charge
+Q
Negative
Charge
-Q
Voltage across a capacitor
is proportional to the
stored charge and
inversely proportional to
the capacitance
K. Webb
Q
V
C
+
V
MAE 2055 – Mechetronics I
Parallel Plate Capacitor
14
Capacitance of an ideal parallel
plate capacitor
C
A
d
ε: dielectric constant of the
material between the plates
A:
area of the plates
Capacitance is maximized by
using high dielectric constant
material and large, very closely
spaced plates
K. Webb
MAE 2055 – Mechetronics I
Capacitors – I-V characteristics
15


Current through a capacitor is equal to the capacitance
times the time derivative of the voltage
Voltage across a capacitor is given by the integral of
current with respect to time
i
+
C
K. Webb
v
-
dv
i t   C
dt
1
v t    idt
C
Current is greater
for rapidly varying
voltages.
For a constant (DC)
current, voltage increases
linearly with time
No current flows
for DC voltages.
I t
v t  
C
MAE 2055 – Mechetronics I
Voltage Change Across a Capacitor
16



Voltage across a capacitor cannot change instantaneously
Current through a capacitor can change instantaneously,
but voltage is the integral of the current
Analogous to temperature of and heat flow into a large
thermal mass – heat flow may change instantaneously, but
temperature will not
1
lim V  lim
t 0
t 0 C
K. Webb

t 0 t
t0
idt  0
MAE 2055 – Mechetronics I
Capacitor – open circuit at DC
17

Current through a capacitor is proportional to the
time rate of change of the voltage across the
capacitor
dv
iC

dt
A DC voltage does not change with time, so
dv
 0 , and i  0
dt

A capacitor is an open circuit for DC signals
K. Webb
MAE 2055 – Mechetronics I
Capacitors In Parallel
18
The total charge on the two
parallel capacitors is
Q  Q1  Q 2  C 1V  C 2V
Q  C 1  C 2 V
Q  C eqV
Parallel capacitors behave like
an equivalent capacitance of
C eq  C 1 C 2
Capacitances in parallel add
K. Webb
MAE 2055 – Mechetronics I
Capacitors In Series
19



Charge +Q flows onto the top plate of C1
+Q must flow out the bottom plate of C1
onto the top plate of C2
+Q flows out the bottom plate of C2
Total voltage across the series combination is
V V 1 V 2 
 C C 2  Q
Q Q

Q  1

C1 C 2
 C 1C 2  C eq
Series capacitors add similar to parallel resistors
 C 1C 2 
C eq  

 C 1 C 2 
K. Webb
MAE 2055 – Mechetronics I
Constant Current Onto a Capacitor
20


Current steps from 0A
to a constant value of
I0 A at t=t0
Capacitor charges and
v(t) increases linearly
while i(t) = I0
K. Webb
𝐼0 𝑡 − 𝑡0
𝑣 𝑡 =
,
𝐶
𝑡 ≥ 𝑡0
slope 
I0
C
MAE 2055 – Mechetronics I
21
K. Webb
Inductance
MAE 2055 – Mechetronics I
Fluid Inductor
22
Consider the following scenario:

Lossless pipe

Heavy paddle-wheel or turbine in the flow mounted on a frictionless
bearing – moment of inertia, I
Incompressible flow: m 1  m 2


External pumps control
pressure/flow rate

Paddle-wheel
rotates at the
same rate as
the flow
K. Webb
MAE 2055 – Mechetronics I
Fluid Inductor – Constant Flow Rate
23

Constant flow rate
m 1  m 2  C

Constant paddle-wheel angular velocity

Acceleration is zero – net applied force is
zero
Frictionless bearing –
no force required to
maintain rotation


Pressure differential is
zero
P  P1  P2  0
K. Webb
MAE 2055 – Mechetronics I
Fluid Inductor – Constant ΔP
24

Constant pressure differential
P  P1  P2  C

Constant applied torque

Constant angular acceleration

Flow rate increases linearly with time
P
m 1 (t )  m 2 (t ) 
 t C 1
L

L is inductance


Intrinsic device
property
Moment of inertia
L I
K. Webb
MAE 2055 – Mechetronics I
Fluid Inductor – Increasing Flow
25

Flow rate initially zero – paddle-wheel initially at rest

Apply pressure differential to increase flow

Torque, resulting from ΔP, required to accelerate paddle-wheel from rest

Any non-zero applied ΔP results in
increasing flow rate.

The change in flow rate is
proportional to ΔP:
dm
 P
dt
K. Webb
MAE 2055 – Mechetronics I
Fluid Inductor – DC vs. AC
26

In steady state (DC) there is no pressure differential across the device
P  P1  P2  0

Consider a sinusoidal (AC) flow rate: m 1  m 2  M cos(t )

Paddle wheel oscillates with the flow

Sinusoidal acceleration requires
sinusoidal torque (ΔP)
P  L

dm
 L  M sin(t )
dt
ΔP is proportional to
frequency and to
inductance
K. Webb
MAE 2055 – Mechetronics I
Flow Rate Can’t Change Instantaneously
27

A given flow rate corresponds to a particular angular velocity of the flywheel

Angular velocity can’t change instantaneously from one value to another

Must accelerate continuously from one
velocity to another

Step change in flow rate would require
an infinite differential pressure

Flow rate through a
fluid inductor cannot
change
instantaneously
K. Webb
MAE 2055 – Mechetronics I
Inductance
28



Inductance is an electrical
property that resists changes
in electrical current
Electrical current induces a
magnetic field around the
conductor in which it flows –
electromagnetic induction
As the current changes, the
changing magnetic field
induces a voltage opposing
the changing current –
Faraday’s Law of Induction
K. Webb
MAE 2055 – Mechetronics I
Inductors
29




Inductors are electrical
components that store energy
in a magnetic field
Inductors are coils of wire –
often wrapped around a
magnetic core
Magnetic fields from adjacent
coils sum – inductance is
proportional to the number of
turns of wire
Schematic symbol:
L

Units of inductance are
Henries (H)
K. Webb
MAE 2055 – Mechetronics I
Inductors – I-V characteristics
30

Voltage across an inductor is equal to the
inductance times the time derivative of the
current through the inductor
i
L
+
v
-
K. Webb
di
v t   L
dt
Voltage across the inductor
is greater for rapidly varying
voltages.
For DC currents there is no
voltage across the inductor –
it is just a wire at DC.
1
i t    vdt
L
For a constant (DC) voltage,
current increases linearly
with time
V t
i
L
MAE 2055 – Mechetronics I
Current Change Through an Inductor
31


Current through an inductor cannot change
instantaneously
Voltage across an inductor can change instantaneously,
but current is the integral of the voltage
1 t 0 t
lim i  lim
vdt

0
t 0
t 0 L t 0
K. Webb
MAE 2055 – Mechetronics I
Inductor – short circuit at DC
32

Voltage across an inductor is proportional to the
time rate of change of the current through the
inductor
di
v L

dt
A DC current does not change with time, so
di
 0, and v  0
dt

An inductor is a short circuit for DC signals
K. Webb
MAE 2055 – Mechetronics I
Inductors In Series
33


Voltage across the series combination is
the sum of the individual voltages
Current through each inductor is equal
Total voltage across the series combination is
v  v1  v 2  L1
di
di
di
 L 2  L1  L 2 
dt
dt
dt
Inductances in series add
Leq  L1  L 2
K. Webb
MAE 2055 – Mechetronics I
Inductors In Parallel
34
Equal voltage across each inductor
di1
di 2
v  L1
 L2
dt
dt
Time derivative of the total current is
1 1
di di1 di 2 1
1
   v  v    v
dt dt dt L1 L 2  L1 L 2 
Parallel inductors behave like an
equivalent inductance of
1
1 1 
L1L 2
Leq     
L1  L 2
 L1 L 2 
K. Webb
MAE 2055 – Mechetronics I
35
K. Webb
RC Circuits
MAE 2055 – Mechetronics I
Step Response
36


The manner in which a linear system (not necessarily
electrical) responds to a step function input – its step
response – is a useful way to characterize that system
The unit step function, or Heaviside step function, is
0, t  0
u (t )  
1, t  0

To characterize an electrical network a voltage step
can be applied as an input
K. Webb
MAE 2055 – Mechetronics I
RC Circuit – voltage step response
37
Step response of this RC circuit is the output voltage, vo(t), when the
input, vi(t), is a voltage step
We want to find out how Vo(t) gets
from vo(0) = 0V to vo(∞) = 1V
What happens in this region?
vs(t) = 1VDC for t >> 0,
C is an open circuit to DC, so
vo(t) = 1V for t >> 0
vs(t) = 0V for t < 0, so
vo(t) = 0V for t < 0
K. Webb
MAE 2055 – Mechetronics I
RC Circuit – voltage step response
38
To determine the step response of the RC circuit, we must determine
what happens in the transition region
Substituting in the expression for i(t) and vs(t)
dvo
vo t   u t   R C
dt
rearranging
Output voltage is related to the
input voltage by Ohm’s Law
vo t   vs t   i t  R
Current, i(t), is the current through
the capacitor
dvo
i t   C
dt
K. Webb
dvo
1
1

vo t  
u t 
dt R C
RC
This is a first-order linear ODE, which we will now
solve for vo(t).
Note that even though this circuit is described by a
differential equation (due to the presence of the
capacitor), that equation was derived by a simple
application of KVL and Ohm’s Law.
MAE 2055 – Mechetronics I
RC Circuit – voltage step response
39
Solving the ODE that describes the RC circuit
The governing differential equation
dvo
1
1

vo t  
u t 
dt R C
RC
The characteristic equation is

1
0
RC
includes a source term or forcing function
to the right of the equal sign. In other
words it is a non-homogeneous ODE, so
the solution will be the sum of the
homogeneous and particular solutions.
which has a single real root at
First, solving the homogeneous part:
The homogeneous solution is:
The homogeneous equation describes the
circuit’s natural response, and is obtained
by setting the forcing function to zero:
dvo
1

vo t   0
dt R C
K. Webb

1
1

RC

where τ = RC is the circuit time constant.
voh t   A e t  A e

t
RC
 Ae

t

where A is a constant that will be determined
by the initial and final conditions after we
determine the particular part of the solution.
MAE 2055 – Mechetronics I
RC Circuit – voltage step response
40
Solving the ODE that describes the RC circuit - (cont’d)
Next, determine the particular solution:
The particular solution accounts for the
effect of the input or forcing function on
the response of the system. It can be
determined by the method of
undetermined coefficients.
For a step input (constant for t ≥ 0) the
particular solution will be of the form:
vop t   B
where, again, B is a constant that will be
determined by the initial and final
conditions of the circuit.
K. Webb
The general solution to the non-homogeneous
differential equation is the sum of the
homogeneous and particular solutions:
vo t   vop t   voh t   A e t /  B
The next step is to apply the initial and final
conditions to determine the unknown
coefficients, A and B.
You determine the initial and final conditions
through basic circuit analysis and based on
your knowledge of both the input signal and
how capacitors behave, specifically that the
voltage across a capacitor cannot change
instantaneously.
MAE 2055 – Mechetronics I
RC Circuit – voltage step response
41
Solving the ODE that describes the RC circuit - (cont’d)
Application of the initial and final
conditions:
The initial condition is the value of the
output (the voltage across the capacitor
in this case) at t = 0+ (i.e. at t=0, just after
the input step has transitioned*):
VI  vo (0 )
The final condition is the value of the
output (the voltage across the capacitor
in this case) as t→∞ (i.e. after the circuit
has reached steady-state):
VF  vo ()
* Note that the 0+ notation is actually unnecessary, because the
unit step is defined such that u(0) = 1. It is used here simply to
stress that point.
K. Webb
Applying the initial condition:
vo  0   A e 0  B  A  B V I
A V I  B
Applying the final condition:
vo     A e   B V F
B V F
The constant coefficient values are
B V F  1V
A V I V F  0V  1V  1V
MAE 2055 – Mechetronics I
RC Circuit – voltage step response
42
Having determined the value of the coefficients A, B, and C, we can now plot the output
voltage, which is given by:
t
vo t   1  e   1  e
t
RC
Note that the time axis is normalized
to the time constant, τ.
• Vo(t) rises slowly to its final value.
• It cannot change instantaneously.
• Time constant, τ, determines the rate
at which C charges and vo(t) rises
K. Webb
MAE 2055 – Mechetronics I
RC Circuit – general response
43
The expression just derived applies to an RC circuit with an initial output voltage of 0V and a
unit (1V) step input. Prior to application of the initial and final conditions, we also derived a
general solution to the capacitor voltage:
vo (t ) V F  (V I V F )e
t /
where: V  v (0)
I
o
V F  vo ()
This is a general-form solution for the step response of an RC circuit with any initial and final
conditions, and in fact for the step response of a first-order system of any kind – not just
electrical circuits (assuming a finite steady-state value). We will see that this solution also
applies to the current in an RC circuit, as well as the voltages and currents in an RL circuit.
Further generalizing the solution, we have:
where: Y  y (0)
y (t ) Y F  (Y I Y F )e
t /
I
Y F  y ( )
where y(t) is any circuit value (e.g. voltage or current) of interest.
K. Webb
MAE 2055 – Mechetronics I
RC Circuit – time constant
44
The RC circuit is characterized by its time constant
  RC
• τ has units of seconds
• As R or C increase, τ increases
• Larger R or C  circuit charges
more slowly
The RC time constant, τ , determines
the initial slope of Vo(t)
dvo
dt

t 0
V F V I

V F V I

RC
At t=7τ, vo(t) has reached
99.9% of its final value
vo(t) reaches 63% of its final
value after one time constant
vo   V I  0.63  (V F V I )
where VF is the final value of vo(t), VI is
the initial value and VF – VI is the
magnitude of the input voltage step.
K. Webb
MAE 2055 – Mechetronics I
RC Circuit Response – an example
45
Consider the response of the RC circuit to a negative-going input voltage step
• vs(t) is initially at 1V
• Capacitor and vo(t) charged to 1V for t < 0, and
vo(t) cannot change instantaneously, so VI = 1V.
• Input steps down to 0V at t = 0. Capacitor
eventually discharges to 0V, so VF = 1V.
Vo(t) reaches 63% of its final
value after one time
constant
t
vo t  V F  V I V F e 
t
vo t   e 
K. Webb
MAE 2055 – Mechetronics I
RC Circuit – current step response
46
Now consider the current through the circuit driven by a positive step
• At t=0, vs(t) steps instantaneously to 1V
• vo(t) cannot change instantaneously, so
the voltage across the resistor at t=0 is
vs  0   vo  0   1V
• According to Ohm’s Law, current will
flow at t=0
vs  0   vo  0  1V
i (0) 

 II
R
R
• As t → ∞, the capacitor will charge up to
1V, and current will no longer flow
I F  i ()  0
K. Webb
MAE 2055 – Mechetronics I
RC Circuit – current step response
47
The current step response is given by the general-form solution as
t
i t   I F  I I  I F e 
where,
II 
1V
, and I F  0
R
The current step response then
simplifies to
t
i t   e 
• Initial current flows due to the
voltage across R at t=0
• Current decays toward zero with RC
time constant, τ
K. Webb
i(t) is zero for t<0.
It jumps instantaneously to
i(0) = 1V/R = 1A (for unit step
input and R = 1Ω)
(for R = 1Ω)
i(t) decays to 63% of its final
value after one time constant
MAE 2055 – Mechetronics I
48
K. Webb
RL Circuits
MAE 2055 – Mechetronics I
RL Circuit – current step response
49
We’ll first look at the step response of the current , i(t), through an RL
circuit when the input, vs(t), is a voltage step
We want to find out how i(t) gets
from i(0) = 0A to i(∞) = 1V/R
What happens in this region?
vs(t) = 1VDC for t >> 0,
L is a short circuit for DC, so
i(t) = 1V/R for t >> 0
vs(t) = 0V for t < 0, so
i(t) = 0A for t < 0
K. Webb
MAE 2055 – Mechetronics I
RL Circuit – current step response
50
To determine the step response of the RL circuit, we must determine what
happens in the transition region between II and IF
Substituting in the expressions for vo(t) and vi(t)
i t  
rearranging
Current is related to the input and
output voltages by Ohm’s Law
i t  
vs t   vo t 
R
Output voltage, vo(t), is
di
vo t   L
dt
K. Webb
1
L di
u t  
R
R dt
di R
1
 i t   u t 
dt L
L
This is a first-order linear ODE, whose general
solution* will be given by
i (t )  I F  (I I  I F )e t /
where τ = L/R is the circuit time constant
* See the RC circuit step response notes for a detailed solution to this ODE
MAE 2055 – Mechetronics I
RL Circuit – current step response
51
The next step in determining the current step response of the RL circuit is
to determine the initial and final conditions
At t=0, the input, vs(t) steps instantaneously to 1V,
but the current through the inductor cannot
change instantaneously, so
I I  i  0   i t  0   0A
As t → ∞, the circuit reaches steady state (i.e. DC).
The inductor looks like a short circuit at DC, so
vo(t) → 0V. The steady-state current is then
For t<0, before the input step
vo t  0   vs t  0   0V
and the current through the circuit, i(t), is
the current through the resistor, which is
given by Ohm’s Law as
i t  0  
K. Webb
vs t  0   vo t  0 
R
 0A
I F  i  
vs     vo   
R
1V

R
Having determined the initial and final conditions,
we can now apply them to the general solution and
plot the resulting expression.
MAE 2055 – Mechetronics I
RL Circuit – current step response
52
We can now plot the step response of the RL circuit
Note that the time axis is normalized
to the time constant, τ.
Applying the initial and final conditions
to the general solution:
t
1V 
i t  
1  e 
R 
 1V

 R
t

L
1  e R






• i(t) rises slowly to its final value.
• It cannot change instantaneously.
• Time constant, τ, determines the rate
at which i(t) rises to its final value
K. Webb
for R = 1Ω
MAE 2055 – Mechetronics I
RL Circuit – time constant
53
The RC circuit is characterized by its time constant
• τ has units of seconds
L

R
• As L increases or R decreases, τ increases
• Larger τ → current reaches its steadystate value more slowly
The time constant, τ , determines the
initial slope of i(t)
di
dt

t 0
IF II


IF II
L
R
At t=7τ, i(t) has reached
99.9% of its final value
i(t) reaches 63% of its final
value after one time constant
i    I I  0.63  (I F  I I )
where If is the final value of i(t), and Ii
is the initial value of i(t)
K. Webb
MAE 2055 – Mechetronics I
RL Circuit – voltage step response
54
Now consider the step response of the voltage across the inductor
• At t=0, vs(t) steps instantaneously to 1V
• For t<0, the current through the inductor was
zero, and the current through the inductor
cannot change instantaneously, so
i (0)  i t  0   0A 
vs (0)  vo (0)
R
• Which can only be satisfied if the output
voltage steps up instantaneously at t=0,
following the input voltage:
V I  vo (0)  vs (0)  1V
• As t → ∞, the current reaches steady state and
the inductor behaves like a short circuit:
V F  vo ()  0V
K. Webb
MAE 2055 – Mechetronics I
RL Circuit – voltage step response
55
The voltage step response is given by the general-form solution as
t
vo t  V F  V I V F e 
Where,
V I  1V , and V F  0A
vo(t) is zero for t < 0.
It jumps instantaneously to
vo(0) = vi(0) = 1V
The voltage step response then
simplifies to
t
vo t   e 
• Voltage jumps up instantaneously,
following vs(t)
• Voltage decays toward 0V according
to the time constant, τ
K. Webb
vo(t) decays to 63% of its final
value after one time constant
MAE 2055 – Mechetronics I
RL Circuit Response – an example
56
Consider the voltage step response of the RL circuit to a 1V to 0.2V input step
•
•
•
•
•
vs(t) is initially at 1V
Inductor is a short circuit for DC, so vo(t<0)=0V
Input steps down to 0.2V at t=0
Output voltages steps down with the input, VI=v0(0)=-0.8V
Inductor is a short for DC so, VF=0V
t
vo t  V F  V I V F e 
t
vo t   0.8e 
K. Webb
Vo(t) reaches 63% of its final
value after one time
constant
MAE 2055 – Mechetronics I