Transcript
Page 1: Dynamic Behavior of Electrical Networks

Dynamic Behavior of Electrical Networks

Carlos Coimbra April 14, 2014

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Overview of Assignments Due in Lab. Sections

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Grade sheet used for your lab reports Item Maximum ScoreTitle and format of report 5Abstract 5Introduction 5Theory 10Experimental Procedures 10Data and Results 15Discussion and Analysis 20Conclusions 5Error analysis (can be in Discussion) 10Figures, Tables and References 5Raw Data Summaries (Appendix) 5Overall impression 5

Important point: If you do not have enough time to finish the lab, that is OK. We prefer quality over quantity.

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Week Date Description Instructor

1 3/31 Class overview, Introduction to circuits Cattolica, Coimbra

2 4/7 A/D conversion, sampling rates, error analysis, and lab report writing, LabView

Cattolica, Mailo

3 4/14 Filters , frequency analysis, LabView Coimbra, Mailo

4 4/21 Operational amplifiers: one and two stage Cattolica, Mailo

5 4/28 Measurement of temperature and heat transfer. Lab View

Coimbra, Mailo Midterm Exam I

6 5/5 Pressure transducers and accelerometers LabView

Coimbra, Mailo

7 5/12

Measurement of strain and force, beam vibration. LabView

Coimbra, Mailo

8 5/19 Introduction to Position Control. LabView Cattolica, Mailo

9 5/26 Make Up Labs/Review Lab Practical Final No lecture - Holiday 10 6/2 Lab Practical Final Examination Cattolica, Midterm II

Lecture Schedule 2014

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What is due this week?

•  Lab Report from week 2

•  LabVIEW VI from week 2

•  Pre-lab for week 3 (summary of what you will be doing in the lab.+ answers to questions at the start of the lab handout)

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Objectives of Week 3 lab.

•  To investigate the response of a first order RC circuit to step and sine wave inputs, and to determine the value of an unknown capacitance, C, from this response.

•  To investigate the use of RC circuits in filtering signals (both low pass and high pass filters).

•  Application of low pass filter

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Basic concepts: Kirchoff’s Current Law

– The current flowing into a node is equal to the sum of the individual currents leaving the node (Principle of charge conservation)

I1I2

I3

I1 = I2 + I3

In

n∑ = 0

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Kirchoff's Voltage Law

The voltage drop around any closed circuit is equal to zero

0= (Vb −Va)+(Vc −Vb)+ Vd −Vc( )+(Va −Vd)

a

b c

d

(Va −Vd )

Vd −Vc( )

(Vc −Vb)

(Vb −Va)

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Voltage divider

•  A most useful circuit, allowing for voltage control

Vout =Vin

Rbottom

Rbottom +Rtop

⎝⎜

⎠⎟

Vin −Vout

Rtop

= I =Vout − 0Rbottom

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Capacitors

•  An electrical device which stores charge

•  The magnitude of the charge stored (Q) is directly proportional to the potential difference (V) between the plates.

–  Q=CV –  Units of C = farad = coulombs/volt

•  The current flow through a capacitor is:

I = dQ

dt=C dV

dt

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Capacitors

10-6 Coulomb of opposite charge on 10-7 farad capacitor V = Q/C = 10 Volts

•  Capacitors commonly found in electronic devices typically range from 10-6 to 10-12 farad.

•  10-12 farad is a “picofarad” or pF

~ 6.2 x 1012 electrons lab capacitor + + + + + +

+

+

- - - - - - -

+ C = coulombs/volt = farad

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RC circuit (in series, DC)

V0- IR - Q/C = 0 (Kirchoff’s 2nd)

V0

This system would charge with switch at “a” and discharge with switch at “b”.

dQdt

=V0

R−

QRC

= −1

RCQ −CV0( )

Q =CV0 1− e

−t

RC⎛

⎝⎜

⎠⎟

When t = 0, Q = 0 When t → ∞, Q = CVo http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=31

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0.86Vo

τ 2τ

RC circuit - charging

Charge rises exponentially with time, current decreases exponentially with time

Q =CV0 1− e−

tRC

⎝⎜

⎠⎟ ⇒V(t) =Vo 1− e

−t

RC⎛

⎝⎜

⎠⎟

I = dQdt

=V0

Re−

tRC = I0e

−t

RC

0.63Vo

When t = RC = τ = characteristic time

(τ = RC)

http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=31

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•  Note that voltage rises to ~ 6.3 V in 1 ms. •  τ = RC = 0.001 sec = characteristic time.

V(τ ) = 10(1− e−1) = 6.32V

Illustration of the time constant

e−tRC = e−1

when t = RC

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RC circuit- discharging

http://www.phy.ntnu.edu.tw/java/rc/rc.html

Units of RC

I =−dQdt

=VR

C =QV⇒Q =CV

−dQdt

= −C dVdt

−CdVdt

=VR

dVV

=−dtRC

ln VVo

⎝⎜

⎠⎟ =

−tRC

RC = ohm x farad = volt/(coulomb/sec ) x

coulombs/volt = sec

When t = RC = τ V = Vo/e

V = 0.37Vo V =Vo exp −t

RC⎛

⎝⎜⎞

⎠⎟

V = 0.37Vo

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•  Charging capacitor –  V(t) = Vo(1 - e-Δt/RC) –  When Δt = RC

•  1 - e-1 = 0.63 •  Thus when V(t) = 0.63Vo –  Δt = RC

•  Discharging capacitor –  V(t) = Voe-Δt/RC –  When Δt = RC

•  e-Δt/RC = 0.37 •  Thus, when V(t) = 0.37Vo –  Δt = RC

Δt

0.37 Vo VO

LTA

GE

Δt

0.63Vo

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Capacitive circuit phase angle depends on both capacitance and resistance

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Impedance

•  Describes a measure of opposition to an alternating current – Z = V/I

•  Describes amplitudes and relative phases •  Complex quantity •  Impedance can be

– A resistor – A capacitor – An inductor

Z = Z eiθ

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Application of complex algebra to impedance

•  If we take the voltage V and the current I as complex quantities with frequency ω and phases ϕ and φ

V = |V|ei(ωt+φ) I = |I|ei(ωt+ψ) •  The complex impedance is:

Z = VI=

VI

eiθ

Z = Z eiθ

θ = φ - ψ

⎢Z ⎢ θ Z

Resistor R 0 R

Capacitor 1/ωC -π/2 1/iωC

Inductor ωL π/2 iωL

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A

Frequency response

•  The most useful input waveform is the sine wave –  Vinput = Asin(ωt) where ω = 2πf

•  The output voltage has the same frequency ω for any linear circuit although its amplitude and phase may change –  Voutput = Bsin(ωt + φ)

Vi Vo

B

volta

ge

φ

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•  Consider when ω = 1/RC •  A decibel is one tenth as large as a bel •  For example, if we have10 times larger

signal, it is 20 dB

dB ≡ 20log10

A2

A1

V0

Vi

=12

G = 20log10

12

⎝⎜⎞

⎠⎟Gain = -3 dB (at which power is halved) Gain = 0.707

Decibels

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Filters

•  A filter is a device that impedes the passage of signals whose frequencies fall within a band called the stop band

•  It permits frequencies in the pass band through relatively unchanged

•  In signal processing, to remove unwanted parts of the signal such as random noise

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Low pass filter – a RC circuit

•  Passes low frequencies Measure voltage across capacitor

Vout−capacitor

Vin

=Gain = 1

(ωRC)2 +1

1iωC

R +1

iωC

⎜⎜⎜

⎟⎟⎟

1iωC

R −1

iωC

⎜⎜⎜

⎟⎟⎟

*

=1

1+ iωRC⎛

⎝⎜⎞

⎠⎟1

1− iωRC⎛

⎝⎜⎞

⎠⎟

•  Vinput = R + Zc (Zc = 1/iωC) •  Voutput = Zc

•  For an RC circuit, (Vo/Vi)2 =

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•  When ωRC << 1 – Low frequency signals are

passed through •  When ωRC >> 1

– The output voltage becomes attenuated •  Eventually → 0

•  When ωRC = 1 – ω = 1/RC

Low pass filter characteristics

Vo

Vi

=1

ωRC

Vo

Vi

=12

ω = 1/RC gain

-3dB frequency

0.707

1

Vo

Vi

= 1

Vout−capacitor

Vin

=Gain = 1

(ωRC)2 +1

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High pass filter •  Passes high frequencies

– A RC circuit •  Measure voltage across resistor

Vout

Vin

=IR

IR +I

iωC

=iωRC

iωRC +1Vout −resistorVin

=Gain = ωRCω 2R 2C 2 +1

As ω → ∞, Gain → 1 As ω → 0, Gain → 0 "-3 dB frequency" is frequency where Vo/Vi = 1/√2 Gain = 1/√2 "-3 dB frequency" occurs when ω = 1/RC

0

1

f-3dB

Vo

Vi

12

Swap resistor with capacitor

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THIS WEEK IN THE LAB

NOTE: All the answers to the lab quiz questions are in this lecture and

in the laboratory handout

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Part 1: Finding the value of the capacitor & -3 dB frequency of a RC Filter

No

- Use 1 kHz, 5 Vp-p square wave - Measure the voltage across capacitor - Set the frequency to 60 Hz (input and output equal)

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V = Vo[1 – e-Δt/RC] Δ t = τ = RC when V = 0.63 Vp-p

V = 0.63 Vp-p

Δ t = τ

Determining C

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Question 2

2a: Compare calculated capacitance with value given in class (mine was about 0.1 µF), include plots (via Data Capture) of oscilloscope screens showing voltage and time cursor lines used in determining time constant. Plots should be clear, include notes so that they are self-explanatory. 2b: What units (seconds, microseconds, etc.) must time be in if R is in ohms and C is in farads and e-t/RC is dimensionless?

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Another way to calculate capacitance - from frequency

Gain =

Voutput−capacitor

Vinput

=1

1+ (RCω)2

"-3dB frequency" occurs when

Gain =12

ω = 2π f =1

RC

Calculate C from determining the frequency corresponding to a gain of 1/√2

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Determining the -3 dB frequency

•  On the scope, look at Vo and Vi

•  Vary the frequency until Vo = 1/√2Vi = 0.707Vi

•  -3.01 = 20log(1/√2) –  Vo/Vi = 1/√2

Note - input and output frequencies are the same but phase is shifted.

f = 164.6 Hz= -3dB frequency Then C = 1/(2πfR)

Ch. 1 (Vi) Ch. 2 (Vo)

4.99 V 3.53 V

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Frequency response of a low pass filter

Gain decreases with frequency; note 3 dB point

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Plot frequency response (in Excel)

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Question 3

What is a linear circuit?

(a) Include plot of oscilloscope screen used to determine -3dB frequency and indicate on the plot the input and output voltages and frequency.

(b) Does the filtering effect the frequency, i.e. is the output frequency different than the input frequency, i.e. is this a linear circuit?

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High pass filter

20log

Vo

Vi

⎝⎜

⎠⎟ dB

f/f-3dB

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Application of low pass filter

•  A common application for filters is to reduce the effect of unwanted i n t e r f e r e n c e o r “n o i s e” w h e n measuring an electrical signal in the presence of other electrical equipment.

•  Devices like motors, switches, and air conditioning equipment produce electromagnetic radiation which can influence the sensitive measurement.

•  This produces unwanted variation which obscures the true value of the

signal to be measured.

A low-pass filter can reduce the unwanted noise and improve the quality of the signal.

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Next week

•  Operational amplifiers and applications


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