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Electronics Fundamentals 8th editionFloyd/Buchla

© 2010 Pearson Education, Upper SaddleRiver, NJ 07458. All Rights Reserved.

chapter 6

electronics fundamentalscircuits, devices, and applications

THOMAS L. FLOYDDAVID M. BUCHLA


Chapter 6


Most practical circuits havecombinations of series and parallelcomponents.

Identifying series-parallel relationships

Components that are connected inseries will share a common path.

Components that are connected inparallel will be connected acrossthe same two nodes.

1 2


Chapter 6


SummarySummary

Most practical circuits have various combinations ofseries and parallel components. You can frequentlysimplify analysis by combining series and parallelcomponents.

Combination circuits

An important analysis method is to form an equivalentcircuit. An equivalent circuit is one that hascharacteristics that are electrically the same asanother circuit but is generally simpler.


Chapter 6


Equivalent circuits

R1

R2R2

1

1.0 k

1.0 k

For example:

is equivalent to R11

2.0 k

There are no electrical measurements that candistinguish the boxes.


Chapter 6


Another example:

There are no electrical measurements that candistinguish the boxes.

R R1 2

1.0 k 1.0 kR

1,2

500

is equivalent to

Equivalent circuits


Chapter 6


R

R1

R2R2

31.0 k

4.7 k2.7 k

is equivalent to

R R1,2 3

4.7 k3.7 k

R1,2,3

2.07 k

is equivalent to

There are no electricalmeasurements that candistinguish between thethree boxes.


Chapter 6


Kirchhoff’s voltage law and Kirchhoff’s current lawcan be applied to any circuit, including combinationcircuits.

R5100

R3330

R2470

R1270

VS5.0 V

R4

100

R6

100Start/Finish

R5100

R3330

R2470

R1270

VS5.0 V

R4

100

R6

100Start/Finish

So willthis path!

For example,applying KVL, thepath shown willhave a sum of 0 V.


Chapter 6


I+

+

26.5 mA

I+ 18.5 mA

I+ 8.0 mA

R5100 Ω

R3330 Ω

R2470 Ω

R1270 Ω

VS5.0 V

R4

100 Ω

R6

100 Ω

A− −

−

Kirchoff’s current law can also be applied to the samecircuit. What are the readings for node A?


Chapter 6


Tabulating current, resistance, voltage and power is auseful way to summarize parameters. Solve for theunknown quantities in the circuit shown.

I1= R1= 270 Ω V1= P1=

I2= R2= 330 Ω V2= P2=

I3= R3= 470 Ω V3= P3=

IT= RT= VS= 10 V PT=

4.18 V

4.18 V

5.82 V

21.6 mA

8.9 mA

12.7 mA

21.6 mA 126 mW

53.1 mW

37.2 mW

216 mW

R1

R3

470 Ω

270 Ω

R2

330 Ω

VS +10 V

464 Ω

Combination circuits


Chapter 6


Kirchhoff’s laws can be appliedas a check on the answer.

I1= R1= 270 Ω V1= P1=

I2= R2= 330 Ω V2= P2=

I3= R3= 470 Ω V3= P3=

IT= RT= VS= 10 V PT=

4.18 V

4.18 V

5.82 V

21.6 mA

8.9 mA

12.7 mA

21.6 mA 126 mW

53.1 mW

37.2 mW

216 mW464 Ω

R1

R3

470 Ω

270 Ω

R2

330 Ω

VS +10 V

equal to the sum of the branch currents in R2 and R3.Notice that the current in R1 is

The sum of the voltages around the outside loop is zero.


Chapter 6


Loaded voltage divider

A voltage-divider with a resistive load is a combinationalcircuit and the voltage divider is said to be loaded. Theloading reduces the total resistance from node A to ground.

The voltage-divider equationwas developed for a seriescircuit. Recall that the outputvoltage is given by

A

22 S

T

RV V

R

R1

R2 R3

+


Chapter 6


Loaded voltage dividerA

What is the voltageacross R3?

Form an equivalent series circuit by combining R2 andR3; then apply the voltage-divider formula to theequivalent circuit:

2,33 2,3 S

1 2,3

38715 V

330 387

RV V V

R R

+15 V R1

R2 R3

330 Ω

470 Ω 2.2 kΩ

VS=

2,3 2 3 470 2.2 k = 387R R R

8.10 V


Chapter 6


Stiff voltage divider

A stiff voltage-divider is one inwhich the loaded voltage nearlythe same as the no-load voltage.To accomplish this, the loadcurrent must be small comparedto the bleeder current (or RL is large compared to the divider resistors).

R1

R2 RL

VS

If R1 = R2 = 1.0 kΩ, what value of RL will make the divider astiff voltage divider? What fraction of the unloaded voltage isthe loaded voltage?RL > 10 R2; RL should be 10 kΩ or greater. For a 10 kΩ load,

2 LL S S S

1 2 L

|| 0.91 k0.476

|| 1.0 k 0.91 k

R RV V V V

R R R

This is 95% of theunloaded voltage.


Chapter 6


Loading effect ofa voltmeter

All measurements affect the quantity being measured. Avoltmeter has internal resistance, which can change theresistance of the circuit under test. In this case, a 1 MΩinternal resistance of the meter accounts for the readings.

Assume VS = 10 V, but themeter reads only 4.04 Vwhen it is across either R1

or R2.

R1

470 kΩ

R2

47 k0 Ω

VS +10 V

+ 10 V

Can you explain what is happening?

+4.04 V

+4.04 V


Chapter 6


Wheatstone bridge

The Wheatstone bridge consistsof a dc voltage source and fourresistive arms forming twovoltage dividers. The output istaken between the dividers.Frequently, one of the bridgeresistors is adjustable.When the bridge is balanced, the output voltage is

-

+

R1R3

R4R2

VSOutput

zero,and the products of resistances in the opposite diagonalarms are equal.


Chapter 6


Summary

Wheatstone bridge

-

+

R1R3

R4R2

VSOutput

Example: What is thevalue of R2 if the bridgeis balanced?

470 Ω 330 Ω

270 Ω

12 V384 Ω


Chapter 6


Thevenin’s theorem states that any two-terminal,resistive circuit can be replaced with a simpleequivalent circuit when viewed from two outputterminals. The equivalent circuit is:

Thevenin’s theorem

VTH

RTH


Chapter 6


VTH

RTH

VTH is defined as


RTH is defined as

the open circuit voltage between the twooutput terminals of a circuit.

the total resistance appearing betweenthe two output terminals when all sources have beenreplaced by their internal resistances.


Chapter 6



R

R

1

R2R2 L

VSVS

12 V10 k

68 k27 k

Output terminals

What is the Thevenin voltage for the circuit? 8.76 V

What is the Thevenin resistance for the circuit? 7.30 kΩ

Remember, theload resistorhas no affect onthe Theveninparameters.


Chapter 6


Maximum power transfer

The maximum power is transferred from a source to aload when the load resistance is equal to the internalsource resistance.

The maximum power transfer theorem assumes thesource voltage and resistance are fixed.

RS

RL

VS +


Chapter 6


Maximum power transfer

What is the power delivered to the matching load?

The voltage to theload is 5.0 V. Thepower delivered is

RS

RL

VS + 50 Ω

50 Ω10 V

22

LL

5.0 V= 0.5 W

50

VP

R


Chapter 6


Superposition theoremThe superposition theorem is a way to determine currentsand voltages in a linear circuit that has multiple sources bytaking one source at a time and algebraically summing theresults.

What does theammeter read forI2? (See next slidefor the method andthe answer).

+-

-

+

-

+

R1 R3

R2

I2

VS2VS1

12 V

2.7 k 6.8 k

6.8 k

18 V


Chapter 6


6.10 kΩ

What does the ammeterread for I2?

1.97 mA 0.98 mA

8.73 kΩ 2.06 mA

+-

-

+

-

+

R1 R3

R2

I2

VS2VS1

12 V

2.7 k 6.8 k

6.8 k

18 V

0.58 mA

1.56 mA

Source 1: RT(S1)= I1= I2=

Source 2: RT(S2)= I3= I2=

Both sources I2=

Set up a table ofpertinent informationand solve for eachquantity listed:

The total current is the algebraic sum.

+ -

-

+

R1 R3

R2

I2VS1

12 V

2.7 kΩ 6.8 kΩ

6.8 kΩ

+ -

-

+

R1 R3

R2

I2

VS2

2.7 kΩ 6.8 kΩ

6.8 kΩ

18 V+-

-

+

-

+

R1 R3

R2

I2

VS2VS1

12 V

2.7 k 6.8 k

6.8 k

18 V1.56 mA


Chapter 6


Quiz

1. Two circuits that are equivalent have the same

a. number of components

b. response to an electrical stimulus

c. internal power dissipation

d. all of the above


Chapter 6


Quiz

2. If a series equivalent circuit is drawn for a complexcircuit, the equivalent circuit can be analyzed with

a. the voltage divider theorem

b. Kirchhoff’s voltage law

c. both of the above

d. none of the above


Chapter 6


Quiz

3. For the circuit shown,

a. R1 is in series with R2

b. R1 is in parallel with R2

c. R2 is in series with R3

d. R2 is in parallel with R3

-

+

R1

R3

R2

VS


Chapter 6


Quiz

4. For the circuit shown,

a. R1 is in series with R2

b. R4 is in parallel with R1

c. R2 is in parallel with R3


-

+R1

R4

R3

R2

VS


Chapter 6


Quiz

5. A signal generator has an output voltage of 2.0 V with noload. When a 600 Ω load is connected to it, the outputdrops to 1.0 V. The Thevenin resistance of the generator is

a. 300 Ω

b. 600 Ω

c. 900 Ω

d. 1200 Ω.


Chapter 6


Quiz

6. For the circuit shown, Kirchhoff's voltage law

a. applies only to the outside loop

b. applies only to the A junction.

c. can be applied to any closed path.

d. does not apply.R1

R3

470 Ω

270 Ω

R2

330 Ω

VS +10 V A


Chapter 6


Quiz

7. The effect of changing a measured quantity due toconnecting an instrument to a circuit is called

a. loading

b. clipping

c. distortion

d. loss of precision


Chapter 6


Quiz

8. An unbalanced Wheatstone bridge has the voltagesshown. The voltage across R4 is

a. 4.0 V

b. 5.0 V

c. 6.0 V

d. 7.0 V

-

+

RL

R1R3

R4R2

VS

12 V

1.0 V

7.0 V+ -


Chapter 6


Quiz

9. Assume R2 is adjusted until the Wheatstone bridge isbalanced. At this point, the voltage across R4 is measuredand found to be 5.0 V. The voltage across R1 will be

a. 4.0 V

b. 5.0 V

c. 6.0 V

d. 7.0 V

-

+

RL

R1R3

R4R2

VS

12 V

5.0 V

+ -


Chapter 6


Quiz

10. Maximum power is transferred from a fixed sourcewhen

a. the load resistor is ½ the source resistance

b. the load resistor is equal to the source resistance

c. the load resistor is twice the source resistance



Chapter 6


Quiz

Answers:

1. b

2. c

3. d

4. d

5. b

6. c

7. a

8. a

9. d

10. b

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