Download Basic Concepts - Oakland University

Exam 1 Review
Chapters 1, 2, 9
1
e12
Charge, q
q2
r12
Recall Coulomb’s Law
q1
Force F1 on charge q2 due to charge q1 is given by
q1q2
F1 
e12
2
4 0 r12
1
1
4 0
 107 c 2  8.99 109
Charge on an electron
(proton) is negative (positive)
and equal to 1.602 x 10-19 C
e12  unit vector pointing from q1 to q2
r12  distance between charge q1and q2
Unit: Newton meter2 / coulomb2
volt meter / coulomb
Note: Positive force is repulsive, negative force is attractive
2
Electric Current, i
Current (in amperes) (A) is the time rate of change of charge q
dq
i
dt
1 A = 1 C/s
t
q(t )   idt  q(t0 )
t0
t
Charge flowing past a point in the interval [t0, t] is qT (t )  t idt
0
Convention: Direction of current flow is that of positive charges,
opposite to the direction of electron flow
3
Voltage
The energy in joules (w) required to move a
charge (q) of one coulomb through an element is
1 volt (V).
dw
v
dq
1 volt = 1 joule/coulomb = 1 newton meter/coulomb
4
Power and Energy
Power (p), in watts (W), is the time rate of expending or
absorbing energy (w) in joules
dw
p
dt
dw  dw   dq 
p
      vi
dt  dq   dt 
5
Power and Energy
Change in energy from time t1 to time t2
t2
t2
t1
t1
w   pdt   vidt
Passive sign convention:
i (t )
+
p  vi
If p > 0 power is absorbed
by the element
v (t )
-
If p < 0 power is supplied
by the element
6
v
R
i
+
v
i
R
v  iR
v
i1
v  i1R
(i  i1 )
-
R
+
-
i
v
R
-
+
Ohm's Law
Units of resistance, R, is Ohms (W)
R = 0: short circuit
R   : open circuit
7
+
1
G
R
G
v
-
i
-
+
Conductance, G
Unit of G is siemens (S),
1 S = 1 A/V
i
v
G
i  Gv
i
G
v
8
Power
A resistor always dissipates energy; it transforms
electrical energy, and dissipates it in the form of heat.
Rate of energy dissipation is the instantaneous power
2
v
(t )
2
p(t )  v(t )i(t )  Ri (t ) 
0
R
2
i
(t )
2
p(t )  v(t )i(t )  Gv (t ) 
0
G
9
Elements in Series
Two or more elements are connected in series if they
carry the same current and are connected sequentially.
I
R1
V0
R2
10
Elements in Parallel
Two or more elements are connected in parallel if they
are connected to the same two nodes & consequently
have the same voltage across them.
I
I1
V
R1
I2
R2
11
Kirchoff’s Current Law (KCL)
The algebraic sum of the currents entering a
node (or a closed boundary) is zero.
N
i
n 1
n
0
where N = the number of branches connected to
the node and in = the nth current entering
(leaving) the node.
12
Sign convention: Currents entering the node are positive,
currents leaving the node are negative.
N
i
n 1
n
0
i2
i1
i5
i3
i4
i1  i2  i3  i4  i5  0
13
Kirchoff’s Current Law (KCL)
The algebraic sum of the currents entering
(or leaving) a node is zero.
Entering:
i1  i2  i3  i4  i5  0
Leaving: i1
 i2  i3  i4  i5  0
i2
i1
i5
i3
i4
The sum of the currents entering a node is
equal to the sum of the currents leaving a node.
i1  i2  i4  i3  i5
14
Kirchoff’s Voltage Law (KVL)
The algebraic sum of the voltages around
any loop is zero.
M
v
m 1
m
0
where M = the number of voltages in the loop
and vm = the mth voltage in the loop.
15
Sign convention: The sign of each voltage is the polarity of the
terminal first encountered in traveling around the loop.
I
+
R1
V1
+
A
V0
R2
The direction of travel is arbitrary.
Clockwise:
V0  V1  V2  0
V2
-
Counter-clockwise:
V2  V1  V0  0
V0  V1  V2
16
Series Resistors
I
+
R1
V1
 I  R1  R2 
+
A
V0
V0  V1  V2  IR1  IR2
R2
 IRs
V2
Rs  R1  R2
-
I
V
Rs
17
Voltage Divider
V0
V0
I

Rs R1  R2
I
R1
V1
R2
V2
A
V0
V0
V2  IR2 
R2
 R1  R2 
R2
V2 
V0
 R1  R2 
R1
Also V1 
V0
 R1  R2 
18
Parallel Resistors
I
V V
I  I1  I 2  
R1 R2
I1
R1
V
R2
1
1
1
 
Rp R1 R2
I
V
I2
Rp
1 1 
V   
 R1 R2 
V

Rp
R1 R2
Rp 
R1  R2
19
Current Division
i
+
i1
i(t)
R1
i2
R2 v(t)
-
R2
v(t )
i1 (t ) 

i (t )
R1
R1  R2
R1
v(t )
i2 (t ) 

i (t )
R2
R1  R2
R1 R2
v(t )  R p i (t ) 
i (t )
R1  R2
Current divides in inverse proportion to the resistances
20
Current Division
N resistors in parallel
1
1
1
1
 
  
Rp R1 R2
Rn
Current in
jth
branch is
v(t )  R pi (t )
v(t ) R p
i j (t ) 

i (t )
Rj
Rj
21
Source Exchange
ia '
ia
+
Rs
DC
vab
vs
-
+
vs
Rs
Rs v
ab
-
We can always replace a voltage source in series with a
resistor by a current source in parallel with the same resistor
and vice-versa.
Doing this, however, makes it impossible to directly find the
22
original source current.
Source Exchange Proof
ia '
ia
+
Rs
DC
RL vL
vs
+
vs
Rs
Rs
RL vL
-
-
RL
vL 
vs
 Rs  RL 
Rs
vs
ia ' 
 ia
 Rs  RL  Rs
vs
ia 
 Rs  RL 
RL
vL  ia ' RL 
vs
 Rs  RL 
Voltage across and current through any load are the same23
3-bit R2-R Ladder Network
KCL
2
B2
V2
IB2
IV2
1
2
V1
B1
IB1
IV1
1
2
B0
V0
IB0
2
IV0
IV 0  I B 0  IV 1
V0 B0  V0 V1  V0


2
2
1
V0  B0  V0  2V1  2V0
1
2V0  B0  V1
2
24
KCL
V1  V0 B1  V1 V2  V1


1
2
1
2
B2
V2
IB2
IV2
1
2V1  2V0  B1  V1  2V2  2V1
2
V1
B1
IB1
IV1
1
2
B0
2V0 
V0
IB0
2
IV 1  I B1  IV 2
IV0
1
B0  V1
2
1
2V1  B0  V1  B1  V1  2V2  2V1
2
1
1
2V1  B0  B1  V2
25
4
2
KCL
V2  V1 B2  V2

1
2
2
B2
V2
IB2
IV2
1
2
V1
B1
IB1
IV1
1
2
B0
V0
IB0
2
IV 2  I B 2
IV0
2V2  2V1  B2  V2
1
1
2V1  B0  B1  V2
4
2
1
1
2V2  B0  B1  V2  B2  V2
4
2
1
1
1
V2  B0  B1  B2
8
4
2
26
1
1
1
V2  B0  B1  B2
8
4
2
2
B2
V2
IB2
IV2
1
2
V1
B1
IB1
IV1
1
2
B0
V0
IB0
2
IV0
Table 11.1
Output of R2-R ladder network in Fig. 11.6
B2
0
0
0
0
1
1
1
1
B1
0
0
1
1
0
0
1
1
B0
0
1
0
1
0
1
0
1
V2
0
1/8
1/4
3/8
1/2
5/8
3/4
7/8
27
Writing the Nodal Equations by Inspection
v1
2A
i1
v3
v2
R2
R1
i3
 G1  G2

 G2
 0

R3
R4
is
G2
G2  G3  G4
G3
i5
R5
0  v1   2 
   
G3  v2    0 
  
G3  G5 
 v3   is 
•The matrix G is symmetric, gkj = gjk and all of the off-diagonal terms
are negative or zero.
The gkk terms are the sum of all conductances connected to node k.
The gkj terms are the negative sum of the conductances connected to
BOTH node k and node j.
The ik (the kth component of the vector i) = the algebraic sum of the
independent currents connected to node k, with currents entering the
28
node taken as positive.
Writing the Mesh Equations by Inspection
R2
R1
+
DC
Vs2
i1
+
R3
-
v3
+
+ v2 -
v1 -
v5 -
+
v7
-
R7
+ v6 -
R5
i3
 R1  R5  R7

 R7


 R5

0

i2
R6
DC
Vs1
i4
+
v4
-
R4
 R7
R2  R6  R7
 R5
0
0
 R6
R3  R5
0
 i1   Vs2 
   0 

 i2   
 i3   Vs1 
0

  
R4  R6  R8   i4   Vs 
 1 
0
 R6
- v +
8
R8
•The matrix R is symmetric, rkj = rjk and all of the off-diagonal terms
are negative or zero.
The rkk terms are the sum of all resistances in mesh k.
The rkj terms are the negative sum of the resistances common to
BOTH mesh k and mesh j.
The vk (the kth component of the vector v) = the algebraic sum of the
independent voltages in mesh k, with voltage rises taken as positive.
29
Turning sources off
Current source:
a
i  is
is
We replace it by a current
source where is  0
b
An open-circuit
Voltage source:
+
DC
vs
v  vs
-
We replace it by a voltage
source where vs  0
i
An short-circuit
30
Thevenin's Theorem
Thevenin’s theorem states that the two circuits given below are
equivalent as seen from the load RL that is the same in both cases.
i
i
a
a
RTh
Linear
Circuit
RL
DC
RL
VTh
b
b
Rin
Rin
VTh = Thevenin’s voltage = Vab with RL disconnected (= ) = the
open-circuit voltage = VOC
31
Thevenin's Theorem
i
i
a
a
RTh
Linear
Circuit
RL
DC
RL
VTh
b
b
Rin
Rin
RTh = Thevenin’s resistance = the input resistance with all
independent sources turned off (voltage sources replaced by short
circuits and current sources replaced by open circuits). This is the
resistance seen at the terminals ab when all independent sources are
turned off.
32
Example
W
DC
10V
W
VTh
5
RTh 

 2W
iSC 2.5
a
vOC 
W
RTh  2W
2
10V  5V  VTh
22
DC
a
VTh  5V
b
W
DC
10V
W
a
iSC 
W
b
10 2 10

 2.5A
23 4
2
3
W
b
W
W
a
RTh  1 
2 2
 2W
22
33
b
Maximum Power Transfer
In all practical cases, energy sources have non-zero internal
resistance. Thus, there are losses inherent in any real source. Also,
in most cases the aim of an energy source is to provide power to a
load. Given a circuit with a known internal resistance, what is the
resistance of the load that will result in the maximum power being
delivered to the load?
Consider the source to be modeled by its Thevenin equivalent.
i
a
RTh
DC
RL
VTh
b
34
i
a
RTh
DC
RL
VTh
b
The power delivered to the load (absorbed by RL) is
p  i RL  VTh
2
 RTh  RL 
2
RL
This power is maximum when p RL  0
p
2
3
2 
 VTh  RTh  RL   2 RL  RTh  RL    0


RL
35
p
2
3
2 
 VTh  RTh  RL   2 RL  RTh  RL    0


RL
RTh  RL  2RL
RL  RTh
Thus, maximum power transfer takes place when the resistance of
the load equals the Thevenin resistance RTh. Note also that
pmax  VTh
 RTh  RL 
2
RL
RL  RTh
pmax  VTh  2RTh  RTh  VTh 2 4 RTh
2
Thus, at best, one-half of the power is dissipated in the internal
resistance and one-half in the load.
36
Ideal Op Amp
i
i
1)
v
v
VDD
VSS  v0  VDD
+
vo
-
VSS
v0  Av  v  v 
The open-loop gain, Av, is very large, approaching infinity.
2)
i  i  0
The current into the inputs are zero.
37
Ideal Op Amp with Negative Feedback
v
+
v
-
vo
Network
Golden Rules of Op Amps:
1. The output attempts to do whatever is necessary to
make the voltage difference between the inputs zero.
2. The inputs draw no current.
38
Non-inverting Amplifier
v
vi
v
R1
+
vo
Closed-loop voltage gain
AF 
-
vo
vi
R2
vi  v  v 
R1
vo
R1  R2
vo
R2
AF   1 
vi
R1
39
Inverting Amplifier
R2
Current into op amp is zero
v  v  0
v  0 vi
ii  i

R1
R1
0  v0 v0
ii 

R2
R2
vi
ii
ii
R1
v
v
-
vo
+
vi v0

R1 R2
AF 
vo
R
 2
vi
R1
40