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AC circuit analysis Procedures to solve a problem Identify the sinusoidal and note the excitation frequency. Covert the source(s) to phasor form Represent each circuit element by its impedance Solve the resulting phasor circuit using previous learnt analysis tools Convert the (phasor form) answer to its time domain equivalent. Ex. 4.16, p180

AC circuit analysis

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AC circuit analysis. Procedures to solve a problem Identify the sinusoidal and note the excitation frequency. Covert the source(s) to phasor form Represent each circuit element by its impedance Solve the resulting phasor circuit using previous learnt analysis tools - PowerPoint PPT Presentation

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Page 1: AC circuit analysis

AC circuit analysis

• Procedures to solve a problem– Identify the sinusoidal and note the excitation frequency.– Covert the source(s) to phasor form– Represent each circuit element by its impedance– Solve the resulting phasor circuit using previous learnt analysis

tools– Convert the (phasor form) answer to its time domain equivalent.

Ex. 4.16, p180

Page 2: AC circuit analysis

Example

)451000cos(205.0

05.005.0100

55

1010055

)901000cos(05.0

)(9005.005.0100

5

100

05

05)100100()100(

010)100(100100

100

10010101000/1

2

2

2

1

1

21

21

6

ti

jj

ji

jij

ti

Ajjj

i

ijjij

ijij

jZ

jjZ

L

c

)631000cos(11.0

)05.0

1.0tan1000cos(1.005.0

1.005.005.005.005.0

122

21

t

t

jjjiiiC

a. Calculate the current through the capacitor and inductor.

b). If 10V and 5V batteries are replace by V1=10Vcos(1000t)

and V2=5Vcos(1000t), respectively, calculate the current

through the capacitor.

j

jj

j

jj

j

j

i

jj

j

jj

j

i

05.005.0

10000

1000500500

0100

100100100

5100

10100100

05.010000

500

0100

100100100

05

10010

2

1

Page 3: AC circuit analysis

Charging a Capacitor

RC

t

etv 1 RC

t

eRdt

tCdv

dt

dqi

0.63

t

v(t)

t=RC= t

i

/R

t=RC=

0.63 /R

Time constant (): time needs to charge a capacitor to 63% of its full charge.

The larger the RC, the longer it takes to charge a capacitor.

The larger the R value, the smaller the current is in the circuit.The larger the C value, the more the charge the capacitor can hold

Page 4: AC circuit analysis

Discharging a Capacitor

i=0

Vc=++++ ----

t = 0

i Vc=++++ ----

RC

t

RC

t

eR

i

etv

t

vC(t)

Page 5: AC circuit analysis

Examplea) A young MacGyver enthusiast is attempting to design a simple switched RC circuit to use as a fuse timer. The child has a 5 F capacitor and one AAA cell with an emf of 1.5 V and an internal resistance of 0.6 ohm. If the fuse will ignite when the capacitor is charged to a voltage of 1.0 V, how much time does the youngster have to vacate the premises?

b). With a never ending enthusiasm for adding batteries to a circuit, the youngster connects a fresh 9 V lithium battery as shown. Now how much time expires after switch closure until the fuse is ignited?

Page 6: AC circuit analysis

Examplea) A young MacGyver enthusiast is attempting to design a simple switched RC circuit to use as a fuse timer. The child has a 5 F capacitor and one AAA cell with an emf of 1.5 V and an internal resistance of 0.6 ohm. If the fuse will ignite when the capacitor is charged to a voltage of 1.0 V, how much time does the youngster have to vacate the premises?

b). With a never ending enthusiasm for adding batteries to a circuit, the youngster connects a fresh 9 V lithium battery as shown. Now how much time expires after switch closure until the fuse is ignited?

sec3.3

5.1

0.11ln56.0

1ln

1

S

C

RC

t

SC

V

VRCt

eVV

Page 7: AC circuit analysis

•The average ac power (Pav) is the power dissipated on the load resistor. • 0cos1, dependent on the complex load.• ideal power factor: cos=1, Z=R, pure resistive load

AC Power

RI

ZI

VItP

2

2

~

cos~

cos~~

)(

tIti

tVtv

p

p

cos)(

cos)(

rms value

Average power

Page 8: AC circuit analysis

Review cont.: Complex Power

jQP

ZVZIIVS

av */

~*

~*

~~ 22

• real power Pav: power absorbed by the load resistance.• Q (volt-amperes reactive, VAR): exchange of energy between the source and the reactive part of the load. No net power is gained or lost during the process.• S : compute by measuring the rms load voltage and currents without regard for the phase angle.• if Q<0, the load is capacitive, Q>0, the load is inductive

Instantaneous power p(t)

Pay attention to complex conjugate

Page 9: AC circuit analysis

Topic 2: Digital Circuit: Combinational Logic

• Logic operation• Real problem to truth table• Karnaugh Map:

– Box “1” or box “0”• Largest supercell possible• 2n ones or zeros in each supercell• Edges of Karnaugh map are connected• Finish all ones or zeros• Doesn’t matter (“d” or “x”) can be considered as either “1” or

“0”.

Page 10: AC circuit analysis

Digital Circuit Review: Sequential Logic• Flip Flips• Timing diagram

1. When CLK signal arrives (rising edge or falling edge), FF will have outputs (Q and Q’) depending on the input (ex. D, or J, K). At this stage, ignore combinational logic if there exist in the circuit.

2. After finishing the output (Q and Q’), then work on the combinational logic, which typically determines the inputs (ex. D, or J, K) which will determine the output (Q and Q’) at next CLK signal

• Sequential circuit design: State Map1. Construct a state map.2. Convert the state map to truth map. Note: have to include all

combination. Ex. If there are three outputs, Q0, Q1, and Q2, then there are 8 states (combinations) that have to be listed. Some of them may be listed as “d”.

3. Prepare the inputs such that the outputs (Q0, Q1, and Q2) at next state will follow the state map.

4. Convert the truth map to Karnaugh map: the inputs of FF (ex, D, or J, K) is the results in Karnuagh map, i.e. the value of the inputs of FF goes into cells. The outputs of FF is the inputs in the Karnaugh map.

Page 11: AC circuit analysis

Example: Combinational Logic

• Problem 2 of Exam 2:

  CD  00   01   11   10 

AB\  

 00     1  0 0   1 

 01     1   1  1   0 

 11     1   1  1   0 

 10     1   0   0  1 

Page 12: AC circuit analysis

Example

Determine the state diagram for this two-bit counter.

How would you reconfigure the counter to reverse the sequence determined above? (Show all work!)