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Electronics Fundamentals�Week 1 - Lecture 2�
Mark Bocko �
Topics:�• Charge, Coulomb’s Law �• Current �• Electric fields�• Voltage�• Resistors, Ohm’s Law �• Kirchoff’s Laws �• Series and parallel resistors�• Voltage and current dividers�
Electric Charge�
k = 8.988 x 109 Nt-m2/Coul2 �
1 Newton is the force required to accelerate 1 kg by 1 m/sec2 �
F =maNewton’s 2nd Law �
m = 1 kg 1 Newton �
F�
1 m/sec2 �
a�
q1 � q2 �r�
F�
“Pith” balls�
F = kq1q2
r 2Newtons�
Coulomb’s Law (1783)�
Charge can be + or – � Like charges repel� Opposite charges attract �
For:� q1 = q2 = 1 Coul� r = 1 meter� F = 8.988 x 109 Nt �
That’s a huge Force! �Enough to levitate about 5000, 200 ton locomotives! �
Problem: Coulomb’s Law �When you are unpacking objects packed in styrofoam “peanuts” the peanuts usually stick to everything! How many electrons would there need to be on a styrofoam peanut of size 1 cm3 to be picked up by your hand from a distance of 5 cm? Assume that the styrofoam peanut and your hand both have the same amount of charge on them, but with opposite signs. (the density of styrofoam is about 0.035 gram/cm3 and the Earth’s gravitational acceleration is g = 9.8 m/sec2. �
1 Coulomb is a lot of charge! "��1897 – J.J. Thompson discovered that charge comes in
“corpuscles” – electrons
1908 -‐ Millikan measured the charge of a single electron
1 electron = 1.602 x 10-‐19 Coulomb 1 Coulomb = 6.242 x 1018 electrons
Answers: ��a) 6.2 x 1018 electrons�b) 6.1 x 1010 electrons�c) 9.8 x -9 electrons�d) 1 electron �
+ q�
5 cm�
mg �
Coulomb �Force�
Gravity �Force�- q�
Electrical Current �For our purposes it’s not important that charge is made up of discrete particles – we will treat charge like a fluid that is infinitesimally divisible.�
-q à surplus of electrons�+q à deficit of electrons�
I �1 Ampere = 1 Coulomb/sec�
+ current corresponds to positive charge moving in the direction indicated by arrow �
+ q Positive current �
-‐ q Negative current �
+ q Negative current �
-‐ q Positive current �
Positive direction �
Another look at Coulomb’s Law …�
q1 � q2 �r�
F�
F =q1 kq2
r 2
⎛
⎝⎜⎞
⎠⎟=q1E2
F = kq1q2
r 2
F = q1E2 �
E2 = kq2
r 2where�
Electric Field: E�
q1 � q2 �E2 � Electric field lines
are directed radially from the charge q2 �
q2 �q1 �E1 � F =q2 k
q1
r 2
⎛
⎝⎜⎞
⎠⎟=q2E1
F = q2E1 = q1E2 �
+Q � -Q �
E�q�
Electric Field between two charged plates�
Electric field points �from +Q to -Q �
F = qE�To move the charge against the electric field
force requires that we do work.�
W = F x d�Work
“Energy” Force Distance mg sin θ θ
d
W = mg sin x d θ
E�+q�
d
F +q�
W = qE x d�= positive à so work is done by the electric field on the charge (energy is added)�
E�-q�
d
F -q�
W = -qE x d�= negative à so work must be done to move the charge �
E�+q�
d
F +q�
E�-q�
d
F -q�
Electric Potential (Voltage)�
The charge is moving in the direction of the E-field force. �àThe E field does work on the charge (adds energy). �à So we say that the charge moves from a higher to a lower electric potential.�
The charge is moving against the direction of the E-field force. �àWork must be done to move the charge against the force (spend energy). �à So we say that the charge moves from a lower to a higher electric potential.�
q�
V2�
q�
V1 �∆V = V2 – V1 �
Work done by the Electric field to bring a charge from potenQal V2 to V1
W = q∆V = q(V2 – V1)�
+Q � -Q �
So, if q is positive and V2 > V1 the field does work (adds energy) to the charge.�
If q is positive and V2 < V1 we have to do work to move the charge.�
Conclusions:�
q�
V2
V1
V2
V1
Higher potenQal
Lower potenQal
q
Higher potenQal
Lower potenQal
Only the difference in potential matters – we can set the zero of voltage any place we wish. �
So rather than writing ∆V we just use V for the voltage (potential) difference between two points.�
Ohm’s Law �Resistor�
R �
I �+ � - �
V �
V = IR �
I = V/R �or�
For a fixed voltage: �" "higher resistance à less current �
p2 p1 large flow resistance
p2 p1 small flow resistance
Analogy: Fluid flowing through a pipe or a straw: �
Small flow
Large flow
A first Simple Circuit �Introduce the battery �
" "à source of voltage (potential difference)�
V
+
-‐
V
+
-‐
V +
-‐
I R
+
-‐ a
A first simple circuit: how much current flows through R?�
I = V / R
Voltage across R is V so by Ohm’s law
Kirchoff’s Voltage Law: The sum of the voltages in going around a closed path in a circuit is zero.
Vnn∑ = 0
“Solving” the circuit with KVL �
V +
-‐
I R
+
-‐ a
Vnn∑ = 0
Start at the point a and proceed clockwise.�Voltage increases by V (go from – to +)�Then voltage drops across resistor (+ to -) ��
V – IR = 0�
V = IR à I = V/R �
This is a lot of formality just to retrieve Ohm’s Law! ��
A slightly more complicated circuit �(voltage divider)�
V +
-‐
I R1 +
-‐
a
R2 +
-‐
b
Find the current I �Find the voltage at b ��
Vb =VR2
R1 +R2
Voltage divider: ��Special Cases:�"R1 = R2 ; Vb = ½ V �
� "R2 >> R1 ; Vb à V ��
Rseries = R1 +R2Series resistors:��
Volume Control Circuit �
I Signal�
Output �
Max Volume�
Min Volume�
Logarithmic vs. linear potentiometer�
hap://www.bcae1.com/potenQo.htm
Why do we use logarithmic pots for volume control?�
Human perception of loudness is logarithmic.�
Loudness perception �is subjective! �
To double perceived loudness the sound pressure level must increase by a factor of about 2.8, which is about (9 decibels). �
Poten. Posi*on
Voltage Increase
1 1 (ref)
2 2.8
4 7.8
8 22
dB (Volts )= 20log VVref
⎛
⎝⎜
⎞
⎠⎟ V =Vref 10
dBV20
0 5 10 15 20 25 30 35
0 2 4 6 8 10 12
PotenQometer posiQon
Outpu
t
Kirchoff’s Current Law �
The sum of all currents entering or leaving a node of a circuit is zero.�
i1
i2
i3 Pick a consistent convention: � current flowing into node is positive� current flowing out of node is negative�
i1 + i2 – i3 = 0�
Charge is neither destroyed or created in a circuit.��What goes in must come out.�
Current divider�
KVL Loop 1: ��
V
i1
R1
a
R2
b
i2 i
1
(1) V – i1R1 = 0 à i1 = V/R1 ��
KVL Outer Loop: ��
(2) V – i2R2 = 0 à i2 = V/R2 ��
KCL @ node a: ��
(3) i – i1 – i2 = 0��
Find i, i1, i2�
R// =R1R2
R1 +R2
Parallel Combination R1, R2�
�
i – V/R1 – V/R2 = 0�
i = V(1/R1 + 1/R2) = V/R//�
1/R// = 1/R1 + 1/R2 à 1/R// = (R2 + R1)/R1R2 �
(1) V – i1R1 = 0 à i1 = V/R1 �(2) V – i2R2 = 0 à i2 = V/R2 �(3) i – i1 – i2 = 0�
Current divider cont.�
V
i1
R1
a
R2
b
i2 i
1
2
1
12V
2
10
1 1
i
i1 i2
i3
i4
i5
a b
Using KCL and KVL in a more complex problem. ��(Nodal analysis)�
Find all of the currents (as indicated) in the above circuit and then find the voltage at points a and b.�