Download Lecture 7

Lecture #7 Graphical analysis,
Capacitors
•Finish graphical analysis
•Capacitors
•Circuits with Capacitors
•Next week, we will start exploring semiconductor
materials (chapter 2).
Reading:
Malvino chapter 2 (semiconductors)
9/15/2004
EE 42 fall 2004 lecture 7
Power of Load-Line Method
We have a circuit containing a two-terminal non-linear element
“NLE”, and some linear components.
The entire linear part of the circuit can be replaced by an equivalent,
for example the Thevenin equivalent. This equivalent circuit consists
of a voltage source in series with a resistor. (Just like the example
we just worked!).
So if we replace the entire linear part of the circuit by its Thevenin
equivalent our new circuit consists of (1) a non-linear element, and (2)
a simple resistor and voltage source in series.
If we are happy with the accuracy of graphical solutions, then we just
graph the I vs V of the NLE and the I vs V of the resistor plus voltage
source on the same axes. The intersection of the two graphs is the
solution. (Just like the problem on page 6)
9/15/2004
EE 42 fall 2004 lecture 7
The Load-Line Method
We have a circuit containing a two-terminal non-linear element
“NLE”, and some linear components.
First replace the entire linear part of the circuit by its Thevenin
equivalent (which is a resistor in series with a voltage source). We will
learn how to do this in Lecture 11.
Then define I and V at the NLE terminals (typically associated signs)
D
Nonlinear
9mA
element
S
9/15/2004
D
250K 1M
+
-
1V
N
L
E
EE 42 fall 2004 lecture 7
ID
+
VDS + 200K
2V
S
-
Example of Load-Line method (con’t)
Given the graphical properties of two terminal non-linear circuit
(i.e. the graph of a two terminal device)
And have this connected to a linear
(Thévenin) circuit
D
N
L
E
Whose I-V can also be graphed
on the same axes (“load line”)
S
9/15/2004
200K
+
-
2V
The intersection gives circuit solution
ID (mA)
D ID
VDS+
+
VDS
S
ID
10
The solution !
200K
+
-
2V
EE 42 fall 2004 lecture 7
1
2
VDS (V)
Load-Line method
The method is graphical, and therefore approximate
But if we use equations instead of graphs, it could be accurate
It can also be use to find solutions to circuits with three terminal
nonlinear devices (like transistors), which we will do in a later
lecture
ID (mA)
D ID
10
The solution !
200K
S
9/15/2004
+
-
2V
EE 42 fall 2004 lecture 7
1
2
VDS (V)
Power Calculation Review
Power is calculated the same way for linear and non-linear
elements and circuits.
For any circuit or element the dc power is I X V and, if associated signs
are used, represents heating (or charging) for positive power or
extraction of energy for negative signs.
For example in the last example the NLE has a power of +1V X 5mA or
5mW . It is absorbing power. The rest of the circuit has a power of - 1V
X 5mA or - 5mW. It is delivering the 5mW to the NLE.
So what it the power absorbed by the 200K resistor?
Answer: I X V is + 5mA X (5mA X 200K) = 5mW. Then the voltage
source must be supplying a total of 10mW. Can you show this?
9/15/2004
EE 42 fall 2004 lecture 7
Storing charge
• If you try to store charge on a piece of
metal, like a Van de Graaff generator, the
work required to put more and more
charge on the piece of metal gets very
large, and thus the voltage gets high fast.
+
+
+
+
+
+
9/15/2004
EE 42 fall 2004 lecture 7
+
Capacitor
• If you want to store more charge, at less
voltage, then you can put two plates of
metal close together, and store positive
charge on one, and negative charge on
the other.
------------++++++++++++
9/15/2004
EE 42 fall 2004 lecture 7
Capacitance
• The voltage is proportional to
the charge which is stored on
the plates
• +Q on one
• -Q on the other
• Current is the flow of charge per
unit time
• Since charge is proportional to
voltage, we define a
proportionality constant C called
the Capacitance
9/15/2004
EE 42 fall 2004 lecture 7
Charge
I
Time
dQ
I
dt
Q  CV
dV
I C
dt
Capacitance of parallel plates
• The bigger the area of two plates which
are close together, the more charge they
can hold for a given voltage
(proportionally)
• The further apart, the bigger the voltage is
for the same charge. (also proportional)
Area
Q
V
Separation
Area
C

Separation
The variable ε is a property of the material, called the dielectric constant
9/15/2004
EE 42 fall 2004 lecture 7
Capacitors
• A battery is a complex device, chemical
reactions generate current (charges) as needed
to maintain the voltage
• A capacitor is a much simpler device, it just
stores charge by keeping positive charges close
to the negative charges on separate conductors
• As charge is added, the repulsion and therefore
the voltage increases
• As charge is removed, the voltage drops
9/15/2004
EE 42 fall 2004 lecture 7
CAPACITORS
+V 
|(
C
i(t)
capacitance is defined by
dV
iC
dt
9/15/2004
dV i
So

dt C
EE 42 fall 2004 lecture 7
CAPACITORS IN SERIES
+ V 1  + V2 
|(
|(
C1
C2
i(t)
i  C1
+ Veq 
|(
Equivalent to
q
Equivalent capacitance defined by
dVeq
d(V1  V2 )
Veq  V1  V2 and i  C eq
 C eq
dt
dt
dV1
dV
 C2 2
dt
dt
dV1
i
So

,
dt C1
dV2
i

,
dt
C2
Clearly, Ceq 
1
1
9/15/2004
i(t)
Ce
1

C1 C2

C1C2
C1  C2
so
dVeq
dt
 i(
1
1
i
 )
C1 C 2
C eq
CAPACITORS IN SERIES
EE 42 fall 2004 lecture 7
CAPACITORS IN PARALLEL
+
C2
|(
C1
|(
i(t)
i( t )  C1
V

Clearly, Ceq  C1  C2
9/15/2004
i(t)
Ceq
|(
Equivalent capacitance defined by
dV
i  C eq
dt
dV
dV
 C2
dt
dt
+
V(t)

CAPACITORS IN PARALLEL
EE 42 fall 2004 lecture 7
Charging a Capacitor with a constant current
+ V(t)

|(
C
i
dV(t) i

dt
C
t
t
dV(t)
i
0 dt dt  0 C dt
i
i t
V(t)   dt 
C
C
0
t
9/15/2004
EE 42 fall 2004 lecture 7
Discharging a Capacitor through a resistor
 V(t)
+
i
C
i
R
dV(t)
i(t)
V(t)


dt
C
RC
This is an elementary differential equation, whose
solution is the exponential:
V (t )  V0 e
9/15/2004
 t /
Since:
d t /
1 t /
e
 e
dt

EE 42 fall 2004 lecture 7
Voltage vs time for an RC
discharge
1.2
Voltage
1
0.8
0.6
0.4
0.2
0
0
1
2
3
Time
9/15/2004
EE 42 fall 2004 lecture 7
4