ac sources

26
AC Sources AC Circuits •Often, electrical signals look like sines or cosines •AC power, Radio/TV signals, Audio •Sine and cosine look nearly identical •They are related by a phase shift •Cosine wave is advanced by /2 (90 degrees) compared to sine •We will always treat source as sine wave m ax m ax sin cos V V t V V t 1 2 cos sin t t Frequency, period, angular frequency related 1 f T 2 f •This symbol denotes an arbitrary AC source

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AC Circuits. AC Sources. Often, electrical signals look like sines or cosines AC power, Radio/TV signals, Audio Sine and cosine look nearly identical They are related by a phase shift Cosine wave is advanced by /2 (90 degrees) compared to sine We will always treat source as sine wave. - PowerPoint PPT Presentation

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Page 1: AC Sources

AC SourcesAC Circuits

•Often, electrical signals look like sines or cosines•AC power, Radio/TV signals, Audio

•Sine and cosine look nearly identical•They are related by a phase shift•Cosine wave is advanced by /2 (90 degrees) compared to sine

•We will always treat source as sine wave

max

max

sin

cos

V V t

V V t

12cos sint t

Frequency, period, angular frequency related

1f T 2 f

•This symbol denotes an arbitrary AC source

Page 2: AC Sources

RMS Voltage•There are two ways to describe the amplitude

•Maximum voltage Vmax is an overstatement•Average voltage is zero

•Root-mean-square (RMS) voltage is probably the best way•Plot (V)2, find average value, take square root

•We can do something similar with current

max sinV V t

2rmsV V 2 2

max sinV t 21max2 V

maxrms

2

VV

House current (US) is at f = 60 Hz and

Vrms = 120 V Vmax = 170 V

maxrms

2

II

Page 3: AC Sources

Amplitude, Frequency, and Phase ShiftWe will describe any sort of wave in terms of three quantities:•The amplitude A is how big it gets

•To determine it graphically, measure the peak of the wave •The frequency is how many times it repeats per second

•To determine it graphically, measure the period T•Frequency f = 1/T and angular frequency = 2f

•The phase shift is how much it is shifted earlier/latercompared to basic sine wave

•Let t0 be when it crosses the origin while rising•The phase shift ist0 (radians)

sinA t

1f T

0t

2 f

0.02 sT

1

50 Hz

100 s

f

1.60 VA

100 0.005

12

0 0.005 st

Page 4: AC Sources

Our Goal•Feed AC source through an arbitrary circuit

•Resistors, capacitors, inductors, or combinations of them•We will always assume the incoming wave has zero phase shift

maxV max sinV V t

•We want to find current as a function of time

•For these components, can show angular frequency is the same•We still need to find amplitude Imax and phase shift for current•Also want instantaneous power and average power consumed

•Generally, maximum current will be proportional to maximum voltage•Call the ratio the impedance, Z

?

max sinI I t

I V P

max maxI V Z

Page 5: AC Sources

Degrees vs. Radians

•All my calculations will be done in radians•Degrees are very commonly used as well•But the formulas look different

•Probably best to set your calculator on radians and leave it there

max sin 360I I ft

max sinI I t max sin 2I ft 0t

0360 ft

Page 6: AC Sources

Resistors

max 170 V

60 Hz

V

f

R = 1.4 k

•Can find the current from Ohm’s Law

VI

R

max sin

Vt

R

•The current is in phase with the voltage

VoltageCurrent

max sinV V t max sinI I t

max max

0

I V R

Impedance vector:•A vector showing relationship between voltage and current

•Length, R is the ratio•Direction is to the right,representing the phase shift of zero

1.4 k

rms rmsI V R

Page 7: AC Sources

Power in Resistors

•We want to know•Instantaneous power•Average Power

R = 1.4 k

I V P 2max max sinV I t

2 2max sinRI tP

21max2 RIP

2 2max sinRI t

2rmsRIP

max 170 V

60 Hz

V

f

Page 8: AC Sources

Capacitors

•Charge of capacitor is proportional to voltage•Current is derivative of charge

C= 2.0 F max sinV V t Q C V

dQI

dt d V

Cdt

•Current leads voltage by /2•We say there is a –/2 phase shift:

max cosC V t

1CX

C

1max 2sinC V t

max 170 V

60 Hz

V

f

maxmax

12

C

VI

X

Impedance vector:•Define the impedance* for a capacitor as:

•Make a vector out of it•Length XC

•Pointing down for = –½

1.3 k2

*We will ignore the term “reactance”

max sinI I t

Page 9: AC Sources

Power in CapacitorsC= 2.0 F

We want to know•Instantaneous Power•Average Power

I V P max max sin cosV I t t

•Power flows into and out of capacitor•No net power is consumed by capacitor

Only resistors contribute to the average power consumed

0P

max 170 V

60 Hz

V

f

Page 10: AC Sources

Capacitors and Resistors Combined

•Capacitors and resistors both limit the current – they both have impedance•Resistors: same impedance at all frequencies•Capacitors: more impedance at low frequencies

1CX C

max max

max max C

I V R

I V X

Page 11: AC Sources

Inductors•Voltage is proportional to change in current

•Integrate this equation

L= 4.0 HdI

Ldt

E max sinV t

•Current lags voltage by /2•We say there is a +/2 phase shift

LX L

max 170 V

60 Hz

V

f

max max

12

LI V X

Impedance vector:•Define the impedance for an inductor as:•Make a vector out of it

•Length XL

•Pointing up for = +½ 1.5

k

2

max cosV

LI t

max cosV

I tL

max 1

2sinV

tL

Page 12: AC Sources

Impedance TableResistor Capacitor Inductor

Impedance R

Phase 0

VectorDirection

right down up

1CX

C LX L

12 1

2

max max

max max

max max

C

L

I V R

I V X

I V X

Page 13: AC Sources

Adding Impedances Graphically•Suppose we have 2+ items in series

•Resistors, Capacitors, Inductors•We can get the total impedance and phase shift by adding the impedances graphically

1.4 k

The impedance and phase shift of two components in series can be found by adding

the vector sum of the two separate impedances

1.4 k

Z = 2.8 k

•Each impedance is represented by a 1.4 k vector pointing to the right•The length of the combination is 2.8 k

•The total impedance is denoted Z•The total arrow is to the right, so phase shift is 0

maxmax

VI

Z

2.8 k

0

Z

1.4 k 1.4 k

60 Hz170 V

rmsrms

VI

Z

Page 14: AC Sources

Adding Impedances Graphically (2)We can add different types of components as well•The resistor is 1.4 k to the right•The capacitor is 1.3 k down•The total is 1.9 k down-right

1.4 k1.4 k 2.0 F

60 Hz170 V

11330 CX

C

1.3 k

1.9 k

•The current is then the voltage over the impedance

•The phase can be found from the diagram

1330tan

1400

0.759 rad

2 2CZ X R 2 21.4 1.3 k 1.9 k

maxmax

VI

Z

170 V

1930

max 88.0 mAI

max sinI I t

Page 15: AC Sources

RLC AC CircuitsRC

f

Vmax

L•For this rather general circuit, find

•Impedance•Phase shift

1. Find the angular frequency 2. Find the impedance of the capacitor and inductor3. Find the total impedance Z4. Find the phase shift 5. Find the current6. Find the average power consumed

1CX

C LX L

R XL

XC

Z

2 f

22L CZ R X X

1tan L CX X

R

maxmax

VI

Z

max sinI I t

•Current•Average power

Page 16: AC Sources

Power in RLC AC Circuits

RC

f

Vmax

L6. Find the average power consumed

max sinI I t Only resistors

contribute to the average power

consumed

2R I RP

maxmax

VI

Z

2 2max sinRI t

2 2max sinRI t P 21

max2 RI

maxrms

2

II

2rmsRIP

•Most power is delivered to resistor when Imax is maximized•When Impedance is minimized

•The “resistor” might well represent some useful device•Like a speaker for a stereo

2rms2

RV

Z P

Page 17: AC Sources

Frequency and RLC Circuits

RC

f

Vmax

L•Impedance tends to be dominated bywhichever component has largest impedance

•At low frequencies, that’s the capacitor•At high frequencies, that’s the inductor•At intermediate, that’s the resistor

•If the circuit includes a capacitor, it blocks low frequencies•If the circuit includes an inductor, it blocks high frequencies

High pass filter Low pass filter

Page 18: AC Sources

A Sample Circuit

10 1000 F

0.1 mH1C

L

X C

X L

2 f

•What frequencies make it through the capacitor?

CR X1

RC

1

RC

•What frequencies make it through the inductor?

LR X R L R

L

16 Hzf

16 kHzf

•These inequalities compatible if:

1 R

RC L

2L R C

f

Vmax = 5 V

Page 19: AC Sources

A Sample Circuit (2)

10 1000 F

f

Vmax = 5 V

0.1 mH1C

L

X C

X L

2 f

Power

Phase Shift

•At low frequencies, blocked by capacitor•At high frequencies, blocked by inductor•At intermediate, power goes to resistor•Frequencies from about 16 Hz–16 kHz get through

•Close to perfect for an audio system

What happens if L > R2C ?

Page 20: AC Sources

The Narrow Band Filter

2.0 2.1 pF

f

Vmax = 1 mV

1.54 H1C

L

X C

X L

2 f

•At resonant frequency, capacitor and inductor cancel •Perfect for picking up WFDD

Page 21: AC Sources

Types of RLC CircuitsHigh Pass Filter•Lets frequencies through if > 1/RC

Low Pass Filter•Lets frequencies through if < R/L

RLC circuit•If R2 > L/C, it is a combination of Low and High pass filter•If R2 < L/C it only lets a narrowrange of frequencies through•The smaller R2C/L, the narrower it is

0

1

CL

Page 22: AC Sources

Comments on Phase Shifts

1tan L CX X

R

max sinI I t

•The phase shift represents how the timing of the current compares to the timing of the voltage•When it is positive, the current lags the voltage

•It rises/falls/peaks later•When it is negative, the current leads the voltage

•It rises/falls/peaks earlier

max sinV V t

0t

Page 23: AC Sources

Transformers

•Let two inductors share the same volume•You can (should) give them an iron core too

•The EMF’s can be calculated from the flux

N1 tu

rns N

2 tu

rns

B

1

1B

d

dt

E 1

Bd

Ndt

2

2 2B B

d dN

dt dt

E

1 2

1 2N N

E E

•The magnetic flux must be changing•Only works for AC

Page 24: AC Sources

What Transformers are Good For•Their main purpose in life is to change the voltage

1 2

1 2N N

E E

N1 =

500

N2 =

5000

V

1 = 120 V V

2 = ?

2120 V

500 5000

V

2

5000 120 V

500V 1200 V

•Voltage can increases, does that mean power increases?•When you increase voltage, you decrease current

•In an ideal transformer, the product is conserved

I V P1 1 2 2I V I V

A 120 V AC source is fed into a transformer, with N1 = 500 turns on the primary coil, and N2 = 5000 turns on the

secondary. What is the voltage out of the transformer?

Realistic transformers are 80-95% efficient

Page 25: AC Sources

Transformers and Power Transmission

•Why transmit at 10 kV, instead of 500 V or 120 V?•Transmission wires are long – they have a lot of resistance•By using a step up transformer, we increase the voltageand decrease the current•Power lost for a resistor is:•You then step it down so youdon’t kill the customer

Generator500 V

Transmission Line 100 kV

House Current 120 V

12 1

2

VI I

V

22rmsI RP=

2

2 11rms

2

VI R

V

Page 26: AC Sources

Power Supplies

•What if we need a different voltage for a specific device?•Use a transformer

•What if we want direct current?•A diode is a device that only lets current through one direction

•What if we don’t like the ripples•Capacitors store charge from cycle to cycle

120 V AC

21 V AC

To devices20 V

ripply DC

20 Vsmooth

DC