W1 L2 Slides Final

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.�