Pulse and Alternating Currents
Direct current (DC) is the electrical current flowing in a constant
direction, and possessing a voltage with constant polarity.
DC is the kind of electricity made by a battery.
Current
DC: I = const
VB
+
-
RL
Time
Current
Pulse
VB
+
-
RL
Time
+
-
Pulse Currents
Transient currents can be found in any circuit that is switched on or
off, e.g. logical circuits, power modulators etc.
RL
Curren
t
In circuit diagrams, the circle with the waveform line inside is the
generic symbol for any time-dependent voltage source
Time
Alternating Current (AC):
both the magnitude and the direction change with time
In circuit diagrams, the circle with the “=“ is often used to for DC sources
(batteries)
The circle with the wavy line inside is the generic symbol for any AC
voltage source.
How can we produce AC sources and why do we need them?
Flexible AC generators - Alternators
An AC voltage will be produced across the wire coils as that shaft is rotated, in
accordance with Faraday's Law of electromagnetic induction
Hydroelectric Power Stations produce AC voltage
A flux of water
rotates the coils
of the AC
generators and
produces AC
voltage
Wireless communications, Radio, TV, Audio
Radio, TV, Audio, Cell phones, WiFi, etc. - all use the AC waveforms
Electric circuit for AC or Pulse application contain
a variety of components, not just resistors
Boost converter: converts 1.5 V voltage of
the battery into 3 – 5 V voltage to pump the
LED (Light Emitting Diode)
Batteries, resistors,
diodes, transistors, capacitors, inductors
Electronic Photo- Flash circuit (used in digital cameras and cell
phones)
Capacitors
Typical capacitor consists of two metal plates separated by an
insulating gap.
It is quite obvious that the DC current cannot flow through the
capacitor.
When the capacitor is connected to the battery, the plate potentials
become equal to the corresponding potentials of the battery terminals.
The capacitor plates thus accumulate the charge.
The voltage between the plates and the charge accumulated on each of
the plates are related
(positive charge on the top plate and negative charge on the bottom plate):
Q = C ×V
C is called a capacitance
For a parallel plate capacitor, shown above
C=
εA
d
A is the area of the plate, d is the distance between the plates and
ε is the dielectric permittivity of the material between the plates
(for different plate configurations the expressions for the capacitance are different)
The circuit symbol for capacitors.
C
Capacitor symbol
used in electrical
circuits
The capacitance is measured in Farads.
The Farad definition comes from the relation
Q = C ×V
or
C =Q V
1C
1F = ;
1V
1 Farad (F) is the capacitance of a capacitor that accumulates the
charge of 1 Coulomb when the voltage between its plates is 1 Volt
Example problem 1
If the voltage of 10 V applied to the capacitor induces 25 mC charge on
its plates, what is the capacitance of the capacitor in Farads?
Q = C ×V
[Q] = C; [C] = F; [V] = V
0
of
5
120
Timed response
Electric current through capacitor
VB
When the switch is OFF, there is no charge on
the capacitor, and the voltage between the plates
is zero.
The current through the capacitor and in the
whole open circuit is zero
Electric current through capacitor
VB
VC
After the switch is ON for a long time, the voltage on the
capacitor
VC = VB
The charge on the capacitor, QC = C × VC
The current through the capacitor and in the whole open circuit is
zero, because the capacitor presents the break in the circuit.
(!) Note: “The charge on the capacitor Q” means the charge on one of the plates.
The charges on the two plates always have equal values and opposite signs.
Capacitor charging during the
transient Positive charge transfer
VB
VC
Negative charge transfer
After the switch was turned ON, during the transient the charge
on the capacitor changed from zero to QC.
Hence there was the current during the transient.
Q=0
Switch turned ON at t=0
QC = C × VC
Charge changes with time
electric current flows.
Time
Electric current through capacitor
VB
VC
VC
I
Time
VB
Time
Capacitor current in high frequency
circuits
VB
I
Time
Electric current through capacitor
When the voltage changes, the current flows “through” a
capacitor:
Q = C ×V
∂Q
∂V
=C
I=
∂t
∂t
There is no actual electron drift trough the gap between the
plates.
However, the charges on one plate induce the charges on
the other plate so that there the current flows in the circuit.
Example
The voltage across the capacitor changes at the rate of 1V per 1 µs.
What capacitance is needed to get the current of 1 mA flowing
through the capacitor?
∂Q
∂V
I=
=C
∂t
∂t
dV/dt = 1V/10-6 (V/s) = 106 V/s;
C = I/(dV/dt) = 1 mA/(106 V/s) = 10-9 F
Example problem 2
The voltage across 0.5 nF capacitor changes with time as shown.
What is the current through the capacitor (in µA, micro-Amperes)?
25
∂Q
∂V
I=
=C
∂t
∂t
Voltage, V
20
15
10
5
0
0
2
4
6
8
10
12
Time, ms
0
of
5
180
Timed response
Capacitance of a parallel-plate capacitor with the air gap
W
d
L
A
C = ε0 ×
d
The capacitance of a parallel plate capacitor is proportional to the plates
area (for rectangular plates, A = W × L)
and inversely proportional to the distance between the plates, d.
The parameter ε0 is the dielectric permittivity of the vacuum (the air).
ε0 = 8.85 ×10-12 F/m.
The capacitance can be significantly increased by filling the
gap between the plates with so-called dielectric materials.
+Q
+Q
Air
-Q
Dielectric
-Q
A
Cd = ε 0
d
For different dielectrics,
εd = 1 … several thousands
A
Cd = ε d ε 0
d
Cd = C0 ×ε d
Dielectric permittivity effect on capacitance
C=
Material
ε d ⋅ ε0 A
d
ε
Dielectric permittivity, d
Air
Paper
Wood
Glass
Distilled water
Barium Strontium Titanite
1
2 – 2.5
3.3
4.9 – 7.5
80
7500
Note: sometimes the expressions for the capacitance use
so-called “total dielectric permittivity”: ε
C=
εA
d
= εd ε 0
Example
What should be the plate size of a square parallel plate air-gap capacitor
that has the capacitance C = 1F and plate spacing, d = 1 mm
C=
ε0 A
d
A=
ε0 = 8.85 ×10-12 F/m
d ×C
ε0
Unit conversion: d = 1 mm = 10-3 m
A=
d ×C
ε0
10− 3 m × 1 F
109 2
8
2
=
=
m
=
1
.
12
×
10
m
8.85 ×10−12 F / m 8.85
L=W ≈ 104 m
1F is a HUGE capacitance
Commercial capacitors
C=
20 mF
capacitor
ε d ⋅ ε0 A
d
Integrated Capacitor fabrication:
The process starts
with the insulating
substrate
First metal deposition
(bottom plate)
Substrate (Ceramic or semiconductor)
Substrate (Ceramic or semiconductor)
Dielectric deposition
Substrate (Ceramic or semiconductor)
Second metal deposition
(top plate)
Substrate (Ceramic or semiconductor)
Integrated Capacitor fabrication:
Substrate (Ceramic or semiconductor)
Top view
C1
C2
Example problem 3
A parallel plate capacitor has the plate size of 500 µm × 200 µm.
The plate are separated by 2 µm thick dielectric film with relative dielectric
permittivity εd = 10.
What is the capacitance of this capacitor?
C=
ε d ⋅ ε0 A
ε0 = 8.85 ×10-12 F/m
d
0
of
5
180
Timed response