Exam 1
-
q=−1.6∗10−19 → proton charge=−q
-
I=dQ
dt 1A=1C
sI is positive current
-
R=ρL
Aρ=resistivity=1
G=1
conductivity
-
Ohm
'
s Law :V=IR
-
RMS=Root Mean Square=
√
average
(
x2
(
t
)
)
o
P
avg
=
(
I
RMS
)
(V
RMS
)
o
RMSAC =Peak ÷
√
2
-
E=
[
1
T
∫
0
T
P
(
t
)
dt
]
T=¿E=P
avg
∗T Eis ∈Joules (J)
o
P=dE
dt ∧E
(
t
0
)
=
∫
0
t
0
P
(
t
)
dt
-
KVL→
∑
voltage drops=0=
∑
voltage rises → applies for aloop
-
KCL →∑I entering=0=∑I leaving → applies at a node
-
Series:R
eq
=
∑
R
n
∥:G
eq
=
∑
G
n
where G=1
R
-
Voltage Divider Rule :V
j
=R
j
R
eq
V
ab
-
Current Divider Rule:Ij=Gj
Geq
I=
1
Rj
1
Req
I
-
I−V Characteristics
o
I=f
(
V
)
o
For an ideal resistor
I=V
R
o
For an ideal voltage source, slope of line is ∞
o
For an ideal current source, slope of line is 0
Exam 2
-
Semiconductor Diode
-
Exponential diode model:
o
I
d
=I
s
(
e
qV d
kT
−1
)
o
k=1.38∗10
−23
J
K
(
Boltzman Constant
)
o
q=1.38∗10
−19
C T =temperature∈Kelvin
o
Note: Doesn’t work in breakdown region
