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Resistive Touch Screens

Dalam dokumen Buku Electric circuits -10th edition (Halaman 97-114)

Begin by analyzing the resistive grid in the x-direction. We model the resistance of the grid in the x-direction with the resistance Rx, as shown in Fig. 3.34. The x-location where the screen is touched is indicated by the arrow. The resulting voltage drop across the resistance Rxis Vx. Touching the screen effectively divides the total resistance, Rx, into two separate resistances Rxand (1 )Rx.

From the figure you can see that when the touch is on the far right side of the screen, 0, and Vx 0. Similarly, when the touch is on the far left side of the screen, 1, and Vx Vs. If the touch is in between the two edges of the screen, the value of is between 0 and 1 and the two parts of the resistance Rxform a voltage divider. We can calculate the volt- age Vxusing the equation for voltage division:

.

We can find the value of , which represents the location of the touch point with respect the far right side of the screen, by dividing the voltage

a Vx =

aRx aRx + (1 - a)Rx

Vs = aRx

Rx

Vs = aVs

a

= a =

= a =

- a a

a

Objective 6—Know when and how to use delta-to-wye equivalent circuits 3.8 Use a Y-to- transformation to find the voltage

in the circuit shown.

Answer: 35 V.

v

¢ 28

20 10

5 105

2 A v

NOTE: Also try Chapter Problems 3.60, 3.62, and 3.63.

A S S E S S M E N T P R O B L E M

across the grid resistance starting at the touch point, Vx, by the voltage applied across the entire resistive grid in the x-direction, Vs:

Now we want to use the value of to determine the x-coordinate of the touch location on the screen. Typically the screen coordinates are specified in terms of pixels (short for “picture elements”). For example, the screen of a mobile phone would be divided into a grid of pixels with pxpixels in the x-direction, and py pixels in the y-direction. Each pixel is identified by its x-location (a number between 0 and px 1) and its y-location (a number between 0 and py 1). The pixel with the location (0, 0) is in the upper left hand corner of the screen, as shown in Fig. 3.37.

-

- a

a = Vx

Vs.

(0, 0) (px1, 0)

(0, py1) (px1, py1) Figure 3.37 The pixel coordinates of a screen with pxpixels in the x-direction and pypixels in the y-direction.

(1 –α)Rx αRx

Rx

Vs Vx

Figure 3.36 The resistive touch screen grid in the x-direction.

Since represents the location of the touch point with respect to the right side of the screen, (1 ) represents the location of the touch point with respect to the left side of the screen. Therefore, the x-coordinate of the pixel corresponding to the touch point is

Note that the value of xis capped at (px 1).

Using the model of the resistive screen grid in the y-direction shown in Fig. 3.38, it is easy to show that the voltage created by a touch at the arrow is given by

Vy = bVs. - x = (1 - a)px.

a - a

Summary 75

(1 –β)Ry

βRy

Ry Vs

Vy

Figure 3.38 The resistive touch screen grid in the y-direction.

Summary

Series resistors can be combined to obtain a single equivalent resistance according to the equation

(See page 58.)

Parallel resistors can be combined to obtain a single equivalent resistance according to the equation

When just two resistors are in parallel, the equation for equivalent resistance can be simplified to give

(See pages 59–60.)

When voltage is divided between series resistors, as shown in the figure, the voltage across each resistor can be found according to the equations

(See page 61.) v2 =

R2

R1 + R2

vs. v1 =

R1

R1 + R2

vs, Req =

R1R2

R1 + R2

. 1

Req

= a

k i=1

1 Ri

= 1 R1

+ 1 R2

+ Á +

1 Rk

. Req = a

k i=1

Ri = R1 + R2 + Á + Rk.

When current is divided between parallel resistors, as shown in the figure, the current through each resistor can be found according to the equations

(See page 63.)

Voltage divisionis a circuit analysis tool that is used to find the voltage drop across a single resistance from a collection of series-connected resistances when the volt- age drop across the collection is known:

where is the voltage drop across the resistance and is the voltage drop across the series-connected resistances whose equivalent resistance is (See page 65.)

Current divisionis a circuit analysis tool that is used to find the current through a single resistance from a col- lection of parallel-connected resistances when the cur- rent into the collection is known:

Ij = Req

Rj

i,

Req. v

Rj

vj

vj = Rj

Req

v, i2 =

R1

R1 + R2

is. i1 =

R2

R1 + R2

is.

R1

R2 vs

v1

v2

is i1 R1 i2 R2

Therefore, the y-coordinate of the pixel corresponding to the touch point is

where the value of yis capped at (py 1). (See Problem 3.72.)

NOTE: Assess your understanding of the Practical Perspective by solving Chapter Problems 3.72–3.75.

- y = (1 - b)py,

Problems

Sections 3.1–3.2

3.1 a) Show that the solution of the circuit in Fig. 3.9 (see Example 3.1) satisfies Kirchhoff’s current law at junctions x and y.

b) Show that the solution of the circuit in Fig. 3.9 satisfies Kirchhoff’s voltage law around every closed loop.

3.2 a) Find the power dissipated in each resistor in the circuit shown in Fig. 3.9.

b) Find the power delivered by the 120 V source.

c) Show that the power delivered equals the power dissipated.

Figure P3.3

6 k 7 k 18 V

5 k 8 k

(a)

90 V 40 30 mA

25

300 15 35 10

(c)

100 70 90 80 50

(d) 300 200 500 27 V

1200 800

(b)

3.3 For each of the circuits shown in Fig. P3.3, a) identify the resistors connected in series, b) simplify the circuit by replacing the series-

connected resistors with equivalent resistors.

3.4 For each of the circuits shown in Fig. P3.4, a) identify the resistors connected in parallel, b) simplify the circuit by replacing the parallel-

connected resistors with equivalent resistors.

3.5 For each of the circuits shown in Fig. P3.3,

a) find the equivalent resistance seen by the source,

b) find the power developed by the source.

PSPICE MULTISIM

PSPICE MULTISIM

where is the current through the resistance and iis the current into the parallel-connected resistances whose equivalent resistance is (See page 65.)

A voltmetermeasures voltage and must be placed in par- allel with the voltage being measured. An ideal voltmeter has infinite internal resistance and thus does not alter the voltage being measured. (See page 66.)

An ammetermeasures current and must be placed in series with the current being measured. An ideal amme- ter has zero internal resistance and thus does not alter the current being measured. (See page 66.)

Digital metersand analog metershave internal resist- ance, which influences the value of the circuit variable being measured. Meters based on the d’Arsonval meter

Req.

Rj

ij movement deliberately include internal resistance as a

way to limit the current in the movement’s coil. (See page 67.)

The Wheatstone bridgecircuit is used to make precise measurements of a resistor’s value using four resistors, a dc voltage source, and a galvanometer. A Wheatstone bridge is balanced when the resistors obey Eq. 3.33, resulting in a galvanometer reading of 0 A. (See page 69.)

A circuit with three resistors connected in a configu- ration (or a configuration) can be transformed into an equivalent circuit in which the three resistors are Y con- nected (or T connected). The -to-Y transformation is given by Eqs. 3.44–3.46; the Y-to- transformation is given by Eqs. 3.47–3.49. (See page 72.)

¢

¢ p

¢

Problems 77

Figure P3.8 24

12

(a) (b) (c)

60 90 a

b

a

a b b

6 k 8 k

4 k 1200 320

480 720

2 k

5.2 k Figure P3.4

36 18 18 V

24

(a) (b)

(c) (d)

100 k 150 k 750

600

900 500 1.5 k 3 k

2 k 60 k 65 V

60 V

90 k 50 k 75 k

200 120 180 30 mA

280 210

3.6 For each of the circuits shown in Fig. P3.4, a) find the equivalent resistance seen by

the source,

b) find the power developed by the source.

3.7 a) In the circuits in Fig. P3.7(a)–(d), find the equiv- alent resistance seen by the source.

b) For each circuit find the power delivered by the source.

PSPICE MULTISIM

Figure P3.7

50 30 V

20 (a)

30 45

15 60

25

10

50 40

100

50 mA

150

500

600 300

250 750

500 1.8 k

250

300 750

2 k 1 k

20 V

1.2 k 2.5 k

1 k 80 mA

20 (b)

(d) (c)

60 20

75 60 12 30

18

3 k

3.8 Find the equivalent resistance for each of the circuits in Fig. P3.8.

Rab 3.9 Find the equivalent resistance for each of the circuits in Fig. P3.9.

Rab PSPICE

MULTISIM

PSPICE MULTISIM

3.10 a) Find an expression for the equivalent resistance of two resistors of value Rin series.

b) Find an expression for the equivalent resistance of nresistors of value Rin series.

c) Using the results of (a), design a resistive net- work with an equivalent resistance of using two resistors with the same value from Appendix H.

d) Using the results of (b), design a resistive net- work with an equivalent resistance of using a minimum number of identical resistors from Appendix H.

3.11 a) Find an expression for the equivalent resistance of two resistors of value Rin parallel.

b) Find an expression for the equivalent resistance of nresistors of value Rin parallel.

c) Using the results of (a), design a resistive net- work with an equivalent resistance of using two resistors with the same value from Appendix H.

d) Using the results of (b), design a resistive net- work with an equivalent resistance of using a minimum number of identical resistors from Appendix H.

Section 3.3

3.12 a) Calculate the no-load voltage for the voltage- divider circuit shown in Fig. P3.12.

b) Calculate the power dissipated in and c) Assume that only 0.5 W resistors are available.

The no-load voltage is to be the same as in (a).

Specify the smallest ohmic values of R1and R2. R2. R1

vo

4 kÆ 5 kÆ 4 kÆ 3 kÆ

Figure P3.12

3.13 In the voltage-divider circuit shown in Fig. P3.13, the no-load value of is 4 V. When the load resist- ance is attached across the terminals a and b, drops to 3 V. Find

Figure P3.13

3.14 The no-load voltage in the voltage-divider circuit shown in Fig. P3.14 is 8 V. The smallest load resistor that is ever connected to the divider is When the divider is loaded, is not to drop below 7.5 V.

a) Design the divider circuit to meet the specifica- tions just mentioned. Specify the numerical values of and

b) Assume the power ratings of commercially available resistors are , , , 1, and 2 W.

What power rating would you specify?

1>4 1>8 1>16 R2.

R1

vo

3.6 kÆ.

R2 RL

40

20 V

vo

a

b RL.

vo RL

vo

vo

4.7 k R1

3.3 k R2

160 V

DESIGN PROBLEM PSPICE MULTISIM

DESIGN PROBLEM

PSPICE MULTISIM

PSPICE MULTISIM

Figure P3.9

26

1 k 250 25

10

28

20

30

18 500

36

2 k 50 24 40 16 60

750 1.5 k

30 20 15 10 18

18

10

6

4 16 12

14 a

b

a

b

a

b a

b (a)

(c) (d)

(b)

Problems 79

Figure P3.14

3.15 Assume the voltage divider in Fig. P3.14 has been constructed from resistors. What is the smallest resistor from Appendix H that can be used as before one of the resistors in the divider is operat- ing at its dissipation limit?

3.16 Find the power dissipated in the resistor in the current divider circuit in Fig. P3.16.

Figure P3.16

3.17 For the current divider circuit in Fig. P3.17 calculate a) and .

b) the power dissipated in the resistor.

c) the power developed by the current source.

Figure P3.17

3.18 Specify the resistors in the current divider circuit in Fig. P3.18 to meet the following design criteria:

Figure P3.18

3.19 There is often a need to produce more than one voltage using a voltage divider. For example, the memory components of many personal computers require voltages of , 5 V, and all with respect to a common reference terminal. Select the values of , , and in the circuit in Fig. P3.19 to meet the following design requirements:

R3

R2

R1

+12 V, -12 V

ig vg

i1 R1 i2 R2 i3 R3 i4 R4

i3 = 2i2; and i4 = 4i1.

ig = 50 mA; vg = 25 V; i1 = 0.6i2; 90 6

10

10 2.4 A

20 vo

io

6 Æ vo

io

10 5 12

10 A

8 6

5 Æ

RL

1 W

R2 RL

R1

40 V

vo

a) The total power supplied to the divider circuit by the 24 V source is 80 W when the divider is unloaded.

b) The three voltages, all measured with respect to the common reference terminal, are ,

, and Figure P3.19

3.20 a) The voltage divider in Fig. P3.20(a) is loaded with the voltage divider shown in Fig. P3.20(b);

that is, a is connected to , and b is connected to Find

b) Now assume the voltage divider in Fig. P3.20(b) is connected to the voltage divider in Fig. P3.20(a) by means of a current-controlled voltage source as shown in Fig. P3.20(c). Find c) What effect does adding the dependent-voltage

source have on the operation of the voltage divider that is connected to the 380 V source?

Figure P3.20

3.21 A voltage divider like that in Fig. 3.13 is to be designed so that at no load ( ) and at full load ( ). Note that by defini- tion

a) Show that

R1 = k - a ak Ro

a 6 k 6 1.

RL= Ro

vo = avs

RL = q vo = kvs

40 k 20 k

30,000 i 30 k

180 V

10 k

i vo

(c) 30 k 180 V

a

b 10 k

i

(a)

40 k a

b

20 k

vo

(b)

vo. vo.

b¿.

a¿ 24 V

Common R1

v1

v2

v3

R2

R3

v3 = -12 V.

v2 = 5 V

v1 = 12 V

PSPICE MULTISIM

DESIGN PROBLEM PSPICE MULTISIM

DESIGN PROBLEM

PSPICE MULTISIM

DESIGN PROBLEM

and

b) Specify the numerical values of and if

, , and

c) If , specify the maximum power that will be dissipated in and

d) Assume the load resistor is accidentally short circuited. How much power is dissipated in and ?

3.22 a) Show that the current in the kth branch of the circuit in Fig. P3.22(a) is equal to the source current times the conductance of the kth branch divided by the sum of the conductances, that is,

b) Use the result derived in (a) to calculate the cur- rent in the resistor in the circuit in Fig. P3.22(b).

Figure P3.22

Section 3.4

3.23 Look at the circuit in Fig. P3.3(a).

a) Use voltage division to find the voltage across the resistor, positive at the top.

b) Use the result from part (a) and voltage division to find the voltage across the resistor, posi- tive on the left.

3.24 Look at the circuit in Fig. P3.3(d).

a) Use current division to find the current in the resistor from left to right.

b) Use the result from part (a) and current division to find the current in the resistor from top to bottom.

70 Æ 50 Æ

5 kÆ 6 kÆ

(b) 0.5

40 A 5 8 10 20 40

(a)

ig R1 R2 R3 ik Rk RN

io 5 Æ ik =

igGk

G1 + G2 + G3 + Á

+ Gk + Á + GN. ig

R2

R1 R2.

R1

vs = 60 V

Ro = 34 kÆ. a = 0.80

k = 0.85

R2 R1

R2 =

k - a a(1 - k)Ro.

3.25 Look at the circuit in Fig. P3.7(a).

a) Use voltage division to find the voltage drop across the resistor, positive at the left.

b) Using your result from (a), find the current flow- ing in the resistor from left to right.

c) Starting with your result from (b), use current division to find the current in the resistor from top to bottom.

d) Using your result from part (c), find the voltage drop across the resistor, positive at the top.

e) Starting with your result from (d), use voltage division to find the voltage drop across the resistor, positive at the top.

3.26 Attach a 450 mA current source between the termi- nals a–b in Fig. P3.9(a), with the current arrow pointing up.

a) Use current division to find the current in the 36 resistor from top to bottom.

b) Use the result from part (a) to find the voltage across the 36 resistor, positive at the top.

c) Use the result from part (b) and voltage division to find the voltage across the 18 resistor, positive at the top.

d) Use the result from part (c) and voltage division to find the voltage across the 10 resistor, positive at the top.

3.27 Attach a 6 V voltage source between the terminals a–b in Fig. P3.9(b), with the positive terminal at the top.

a) Use voltage division to find the voltage across the 4 resistor, positive at the top.

b) Use the result from part (a) to find the current in the 4 resistor from left to right.

c) Use the result from part (b) and current division to find the current in the 16 resistor from left to right.

d) Use the result from part (c) and current division to find the current in the 10 resistor from top to bottom.

e) Use the result from part (d) to find the voltage across the 10 resistor, positive at the top.

f) Use the result from part (e) and voltage division to find the voltage across the 18 resistor, posi- tive at the top.

3.28 a) Find the voltage in the circuit in Fig. P3.28 using voltage and/or current division.

vx

Æ Æ

Æ Æ Æ

Æ

Æ Æ Æ

Æ

60 Æ 50 Æ

50 Æ 25 Æ

25 Æ

PSPICE MULTISIM

PSPICE MULTISIM

Problems 81

b) Replace the 18 V source with a general voltage source equal to Assume is positive at the upper terminal. Find as a function of

Figure P3.28

3.29 Find in the circuit in Fig. P3.29 using voltage and/or current division.

Figure P3.29

3.30 Find and in the circuit in Fig. P3.30 using volt- age and/or current division.

Figure P3.30

3.31 For the circuit in Fig. P3.31, find and then use cur- rent division to find .

Figure P3.31

3.32 For the circuit in Fig. P3.32, calculate and using current division.

i2

i1 io

ig

v1

3 V

90 60

30 75

40 150

v2

v2

v1

2 k 10 k

vo

18 mA

15 k

3 k

12 k 4 k

vo

vx

2 k

6 k

9 k

3 k

18 V

Vs. vx

Vs

Vs.

Figure P3.32

3.33 A d’Arsonval ammeter is shown in Fig. P3.33.

a) Calculate the value of the shunt resistor,RA, to give a full-scale current reading of 5 A.

b) How much resistance is added to a circuit when the 5 A ammeter in part (a) is inserted to meas- ure current?

c) Calculate the value of the shunt resistor,RA, to give a full-scale current reading of 100 mA.

d) How much resistance is added to a circuit when the 100 mA ammeter in part (c) is inserted to measure current?

Figure P3.33

3.34 A shunt resistor and a 50 mV, 1 mA d’Arsonval movement are used to build a 5 A ammeter. A resistance of is placed across the terminals of the ammeter. What is the new full-scale range of the ammeter?

3.35 A d’Arsonval movement is rated at 2 mA and 200 mV. Assume 1 W precision resistors are avail- able to use as shunts. What is the largest full-scale- reading ammeter that can be designed using a single resistor? Explain.

3.36 a) Show for the ammeter circuit in Fig. P3.36 that the current in the d’Arsonval movement is always of the current being measured.

b) What would the fraction be if the 100 mV, 2 mA movement were used in a 5 A ammeter?

c) Would you expect a uniform scale on a dc d’Arsonval ammeter?

1>25th 20 mÆ 150 mV

3 mA

Ammeter RA

8

250 mA 60 30 20

80 4

i1 i2

PSPICE MULTISIM

PSPICE MULTISIM

PSPICE MULTISIM

DESIGN PROBLEM

20 15

5 6

13 12 125 V

2

ig

io

Figure P3.36

3.37 A d’Arsonval voltmeter is shown in Fig. P3.37. Find the value of for each of the following full-scale readings: (a) 50 V, (b) 5 V, (c) 250 mV, and (d) 25 mV.

Figure P3.37

3.38 Suppose the d’Arsonval voltmeter described in Problem 3.37 is used to measure the voltage across the resistor in Fig. P3.38.

a) What will the voltmeter read?

b) Find the percentage of error in the voltmeter reading if

Figure P3.38

3.39 The ammeter in the circuit in Fig. P3.39 has a resist- ance of Using the definition of the percent- age error in a meter reading found in Problem 3.38, what is the percentage of error in the reading of this ammeter?

Figure P3.39

60

20 10

Ammeter 50 V

0.1 Æ.

15 45

io

50 mA

% error = ¢measured value

true value -1≤ * 100.

45 Æ

20 mV 1 mA Voltmeter Rv

Rv

100 mV, 2 mA

(25/12) imeas

im

3.40 The ammeter described in Problem 3.39 is used to measure the current in the circuit in Fig. P3.38.What is the percentage of error in the measured value?

3.41 The elements in the circuit in Fig. 2.24 have the follow-

ing values: , , ,

, , , and

a) Calculate the value of in microamperes.

b) Assume that a digital multimeter, when used as a dc ammeter, has a resistance of If the meter is inserted between terminals b and 2 to measure the current , what will the meter read?

c) Using the calculated value of in (a) as the cor- rect value, what is the percentage of error in the measurement?

3.42 You have been told that the dc voltage of a power supply is about 350 V.When you go to the instrument room to get a dc voltmeter to measure the power supply voltage, you find that there are only two dc voltmeters available. One voltmeter is rated 300 V full scale and has a sensitivity of The other voltmeter is rated 150 V full scale and has a sensitivity of (Hint:you can find the effective resistance of a voltmeter by multiplying its rated full-scale voltage and its sensitivity.)

a) How can you use the two voltmeters to check the power supply voltage?

b) What is the maximum voltage that can be measured?

c) If the power supply voltage is 320 V, what will each voltmeter read?

3.43 Assume that in addition to the two voltmeters described in Problem 3.42, a precision resis- tor is also available. The resistor is connected in series with the series-connected voltmeters. This circuit is then connected across the terminals of the power supply. The reading on the 300 V meter is 205.2 V and the reading on the 150 V meter is 136.8 V. What is the voltage of the power supply?

3.44 The voltmeter shown in Fig. P3.44(a) has a full- scale reading of 500 V. The meter movement is rated 100 mV and 0.5 mA. What is the percentage of error in the meter reading if it is used to measure the voltage in the circuit of Fig. P3.44(b)?

Figure P3.44

100 mV

0.5 mA Common

500 V

50 k 250 k 10 mA

v

(b) (a)

Rm

v

50 kÆ 50 kÆ 1200 Æ>V.

900 Æ>V.

iB

iB

1 kÆ. iB

39.

b = 0.6 V V0 =

7.5 V VCC=

0.2 kÆ RE =

0.82 kÆ RC=

80 kÆ R2 =

20 kÆ R1 =

io

PSPICE MULTISIM

Problems 83

3.45 The voltage-divider circuit shown in Fig. P3.45 is designed so that the no-load output voltage is of the input voltage. A d’Arsonval voltmeter having a sensitivity of and a full-scale rating of 200 V is used to check the operation of the circuit.

a) What will the voltmeter read if it is placed across the 180 V source?

b) What will the voltmeter read if it is placed across the resistor?

c) What will the voltmeter read if it is placed across the resistor?

d) Will the voltmeter readings obtained in parts (b) and (c) add to the reading recorded in part (a)?

Explain why or why not.

Figure P3.45

3.46 Assume in designing the multirange voltmeter shown in Fig. P3.46 that you ignore the resistance of the meter movement.

a) Specify the values of , , and

b) For each of the three ranges, calculate the percent- age of error that this design strategy produces.

Figure P3.46

3.47 The circuit model of a dc voltage source is shown in Fig. P3.47. The following voltage measurements are made at the terminals of the source: (1) With the terminals of the source open, the voltage is meas- ured at 50 mV, and (2) with a resistor con- nected to the terminals, the voltage is measured at 48.75 mV. All measurements are made with a digi- tal voltmeter that has a meter resistance of 10 MÆ.

15 MÆ 100 V

Common

50 mV 2 mA R1

1 V

R3 10 V

R2

R3. R2

R1 180 V

20 k

70 k vo

20 kÆ

70 kÆ

100 Æ>V

7>9ths

a) What is the internal voltage of the source ( ) in millivolts?

b) What is the internal resistance of the source ( ) in kilo-ohms?

Figure P3.47

3.48 Design a d’Arsonval voltmeter that will have the three voltage ranges shown in Fig. P3.48.

a) Specify the values of , , and

b) Assume that a resistor is connected between the 150 V terminal and the common terminal. The voltmeter is then connected to an unknown voltage using the common terminal and the 300 V terminal. The voltmeter reads 288 V. What is the unknown voltage?

c) What is the maximum voltage the voltmeter in (b) can measure?

Figure P3.48

3.49 A resistor is connected from the 200 V ter- minal to the common terminal of a dual-scale volt- meter, as shown in Fig. P3.49(a). This modified voltmeter is then used to measure the voltage across the resistor in the circuit in Fig. P3.49(b).

a) What is the reading on the 500 V scale of the meter?

b) What is the percentage of error in the measured voltage?

360 kÆ 600 kÆ

30 V

Common 50 mV

1 mA

150 V 300 V R3

R2

R1

750 kÆ

R3. R2

R1

vs Terminals of

the source Rs

Rs

vs

DESIGN PROBLEM

DESIGN PROBLEM

Dalam dokumen Buku Electric circuits -10th edition (Halaman 97-114)