exam for Electronic Engineering
Jocelyn yang 
EE1301Session16-Capacitors-Class.pdf
2/28/2018
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EE 1301: MODERN ELECTRONIC TECHNOLOGY
SESSION #16: CAPACITORS 02/26/2018
Instructor: Joseph Cleveland, Ph.D. Email: [email protected]
Thought for the Day
The pessimist complains about the wind; the optimist expects it to change; the realist adjusts the sails.
William Arthur Ward
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Session Topics
• Capacitors
• Exponential rise/decay – Time constant measurement – 36.79% = 1/e
• Computer clock rate and heat generation – Moving charges very fast = high current
• Getting ready for lab #5
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Capacitors
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Capacitor Structure
• Capacitors are electronic devices that store charge separation without chemical action
• It consists of two metal plates separated by a dielectric (insulator)
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Capacitance C • Proportional to area A • Inversely proportional
to separation d
Capacitor Usage
• We find capacitors in most electronic devices: – Computers
– Mobile phones
– Radios
– TVs
– DVD/CD players
– Digital cameras
So, how do capacitors function to make them so useful?
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Charge Separation
• If the + terminal of a battery is connected to one plate and the – terminal to the other, electrons are moved from the +plate to the plate.
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I The battery pumps from one plate to the other
VB electric field
The separated charge creates a voltage drop across the capacitor
e-
Charge Separation …
• Disconnect the battery. Then the capacitor maintains the charge separation!
• It becomes a battery with no chemicals. • It stores energy. • It stores information (“1”) - memory
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Unit for C: When 1-volt is applied to a 1Farad capacitor, it separates 1 Coulomb of from 1 Coulomb of +charge. That is:
electric field
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Why Your Laptop Gets Hot
Laptop Heat Generation
• If you want to store a “1”, charge the capacitor
• If you want to store a “0” discharge the capacitor
• Charging and discharging current produces heat
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VB
R
V0 Q0
a “0”
VB
R VVB QCVBa “1”
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Laptop Heat Generation …
• Suppose , . ,
• Current if switching time is 1 nsec = 10-9 sec. .
.
• Heat in the resistor
. .
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. .
VB
R
Laptop Heat Generation …
• If this occurs for 10 million 01 or 1 changes the total heat dissipated is
. ! • Now double the rate of change so the
switching time is 0.5 nsec = 0.5x10-9 sec.
• The current doubles in size so the power increases by 4X:
Power20W • The clock rate on most laptops is less than
2.5x109 cycles/sec (time=0.4 nsec) for this reason
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Exponential Decay
This week’s lab topic
Charge Transfer Process
• Now, let’s look at what happens as the capacitor builds this charge separation
• Start with the circuit below
• Close the switch then find the initial current.
• Repeat this every 0.5 seconds
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VB=10 V
5 C = 0.5 F
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Charge Transfer Process …
• At the instant the switch is closed
• The initial current is
.
• During the next 0.5 sec, this current transfers about Q 2 A x 0.5 s = 1 Coul
of electrons to the bottom plate. • Then, . ⁄ .
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VB=10 V
5 C = 0.5 F
Charge Transfer Process …
• At 0.5 sec we have
• The instantaneous current value then is
.
• During the next 0.5 sec, this current transfers about
Q 1.6 A x 0.5 s = 0.8 Coul of electrons.
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VB=10 V
5 C = 0.5 F
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Charge Transfer Process …
• The total charge transferred after 1 sec is:
1.0 Coul + 0.8 Coul = 1.8 Coul
• The capacitor voltage becomes . . ⁄ .
• So, after 1 sec we have
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VB=10 V
5 C = 0.5 F
.
.
Charge Transfer Process …
• After 1.5 sec we have
• Repeat this process to find – Instantaneous current
– Charge transferred
– Total charge
– New capacitor voltage
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VB=10 V
5 C = 0.5 F
.
.
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Charge Transfer Process …
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0.0
2.0
4.0
6.0
8.0
10.0
12.0
0 2 4 6 8 10 12
C a p a ci to r V o lt a g e
Time (sec)
Charging a Capacitor
Voltage rises to 63% of final value
Time constant
Saturates at
10 1 ⁄“Famous” exponential rise curve:
Discharge Process
• Suppose we take a fully “charged” capacitor and discharge it through a resistor
• The initial current is I=2A, which reduces Q, which reduces V, which reduces I, …
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5 electric field
C = 0.5 F
.
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Discharge Process …
• Capacitor after 1 sec
Charge removed: 2 A x 1 sec = 2 Coul
• Current:
.
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5 electric field
C = 0.5 F
.
Discharge Process …
• Capacitor another 1 sec
Charge removed: 1.2 A x 1 sec = 1.2 Coul
• Current: .
.
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5 electric field .
C = 0.5 F
.
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Discharge Process …
• Capacitor another 1 sec
Charge removed: 0.72 A x 1 sec = 0.72 Coul
• Current: .
.
• etc. 2/26/2018 EE1301 Modern Electronic Technology 23
5 electric field .
C = 0.5 F
.
Discharge Process …
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0.0
2.0
4.0
6.0
8.0
10.0
12.0
0.0 2.0 4.0 6.0 8.0 10.0 12.0
C a p a ci to r V o lt a g e
Time (sec)
Capacitor Discharge
37% of initial value
=2.5 s
V .⁄
The rate of change of V is proportional to V.
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Charge the capacitor by holding down the key, then release the key and measure the capacitor voltage for decay through 100K and 470K resistors.
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Write a “1” then watch it decay to a “0” when connected to 470 K and then 100 K
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Lab #5 Data Entry
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Name Example
Partner None
Date May 25, 2015
Time Cons4 1 minutes
Measured Values Calculated values Least squares fit
Time (sec.Time (minV(t) 470K MV(t) 100K|V(t) 470K CV(t) 100K|470K 100K||470K
0 0.00 9.00 9.00 9.00 9.00 0.00 0.00
10 0.17 8.59 6.96 8.63 7.62 0.00 0.43
30 0.50 7.83 4.17 7.94 5.46 0.01 1.66
60 1.00 6.82 1.93 7.01 3.31 0.04 1.91
90 1.50 5.93 0.90 6.19 2.01 0.07 1.23
120 2.00 5.16 0.41 5.46 1.22 0.09 0.65
180 3.00 3.91 0.09 4.25 0.45 0.12 0.13
240 4.00 2.96 0.02 3.31 0.16 0.12 0.02
300 5.00 2.24 0.00 2.58 0.06 0.11 0.00
360 6.00 1.70 0.00 2.01 0.02 0.09 0.00
420 7.00 1.29 0.00 1.56 0.01 0.08 0.00
480 8.00 0.98 0.00 1.22 0.00 0.06 0.00
540 9.00 0.74 0.00 0.95 0.00 0.04 0.00
600 10.00 0.56 0.00 0.74 0.00 0.03 0.00
660 11.00 0.42 0.00 0.58 0.00 0.02 0.00
720 12.00 0.32 0.00 0.45 0.00 0.02 0.00
780 13.00 0.24 0.00 0.35 0.00 0.01 0.00
840 14.00 0.18 0.00 0.27 0.00 0.01 0.00
900 15.00 0.14 0.00 0.21 0.00 0.01 0.00
Least‐squares error sum = 0.92 6.04
Lab #5 Data Plot
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0
1
2
3
4
5
6
7
8
9
10
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
v (t ) (v o lt s)
time (minutes)
Lab 5 Data: v(t) vs. time
V(t) 470K Meas.
V(t) 470K Calc.
V(t) 100K||470K Meas.
V(t) 100K||470K Calc.
37% value
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End of Session
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