EE 221 Midterm Solutions -...

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EE 221 Midterm Solutions October 30, 2002 1 1) Figure 1 a) Determine and sketch the Thevenin’s equivalent circuit for the output terminals A and B. Solution First one should realize that R 3 , R 5 , and R 7 along with I 2 and V 2 have no bearing on the final solution. Thus the circuit to determine R TH is: R TH = R2 + R4//R6 = 300Circuit to determine V TH is: V A = I 1 x R 2 = 50 mA x 260 = 13 V and V V R R R V V B 10 60 120 120 15 6 4 4 1 = + = + =

Transcript of EE 221 Midterm Solutions -...

Page 1: EE 221 Midterm Solutions - IEEEsites.ieee.org/sb-usask/files/2011/09/ee-221-2002-t1-midterm.1.pdf · EE 221 Midterm Solutions October 30, 2002 1 1) Figure 1 a) Determine and sketch

EE 221 Midterm Solutions October 30, 2002

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1)

Figure 1

a) Determine and sketch the Thevenin’s equivalent circuit for the output terminals A and B. Solution First one should realize that R3, R5, and R7 along with I2 and V2 have no bearing on the final solution. Thus the circuit to determine RTH is:

RTH = R2 + R4//R6 = 300Ω Circuit to determine VTH is:

VA = I1 x R2 = 50 mA x 260 Ω = 13 V and VVRR

RVVB 1060120

1201564

41 =

Ω+ΩΩ=

+=

Page 2: EE 221 Midterm Solutions - IEEEsites.ieee.org/sb-usask/files/2011/09/ee-221-2002-t1-midterm.1.pdf · EE 221 Midterm Solutions October 30, 2002 1 1) Figure 1 a) Determine and sketch

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VAB = VTH = VA – VB = 13 V – 10 V = 3 V

Thevenin Equivalent Circuit. b) At what value of load resistance will 600 µA flow through the load? RLoad = _4.7 kΩ_

Ω=Ω−=−=∴+

= kA

VRA

VRRR

VA THTH

LoadLoadTH

TH 7.4300600

3600

600µµ

µ

c) At what value of load resistance is maximum power transferred? RLoad = __300 Ω__ Maximum Power is transferred when RLoad = RTH = 300 Ω d) What would be the minimum load resistance if this Thevenin’s equivalent circuit was to be a stiff voltage supply? RLoad(min) = ___30 kΩ__ From the definition of a stiff supply! RLoad(min) = 100 x RS. In this case RS is the Thevenin resistance (RTH). Hence RLoad(min) = 100 x 300 Ω = 30 kΩ e) What is the Norton’s equivalent circuit? (Provide a circuit sketch as well)

RN = RTH = 300 Ω and mAVRVI

TH

THN 10

3003 =

Ω==

Norton Equivalent Circuit.

Page 3: EE 221 Midterm Solutions - IEEEsites.ieee.org/sb-usask/files/2011/09/ee-221-2002-t1-midterm.1.pdf · EE 221 Midterm Solutions October 30, 2002 1 1) Figure 1 a) Determine and sketch

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2) (Use the second approximation of the diode!)

Figure 2

NOTE: There is an error on the diagram, both D1 and D2 should be flipped so that the anode is connected to ground. With the above configuration the output would be zero as there is no ground connection for current flow. a) What is the peak or maximum voltage across the load? VLoad = __20 V_ For a primary transformer voltage of 120 Vac gives a secondary voltage of 20 Vac means that

the transformer turns ratio is, 620

120 =VacVac . Thus for a primary voltage of 90.8 Vac the secondary

voltage is PVVacVacVac 4.212133.15133.1568.90 ≅×==

Realizing that this is a bridge rectifier and the fact that we are using the second approximation of the diode gives a load voltage of 21.4 V – 1.4 V = 20 V. b) What is the peak to peak ripple voltage at the load? Vripple = __0.250 VP-P__

ARV

ILoad

LoadLoad 2.0(max) == and since this is a bridge rectifier circuit as well means that the rectified

frequency is double the input frequency of 60 Hz. Hence the time between charging for the

capacitor is sec3.8120

1 mHz

=

So to find the change in charge is as follows; mCmAQ 6.1sec3.82.0 =×=∆

PPPP VmVmF

mCCQV

VQ

−− ≅==∆=∆∴∆∆= 250.09.249

67.66.1eCapacitanc

Page 4: EE 221 Midterm Solutions - IEEEsites.ieee.org/sb-usask/files/2011/09/ee-221-2002-t1-midterm.1.pdf · EE 221 Midterm Solutions October 30, 2002 1 1) Figure 1 a) Determine and sketch

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-22

-20

-18

-16

-14

-12

-10-8-6-4-20246810121416182022

030

6090

120

150

180

210

240

270

300

330

360

390

420

450

480

510

540

570

600

630

660

690

720

Voltage

VAVB

With

out F

ilter C

ap.

VB W

ith F

ilter C

ap.

Vripp

le ~

0.250

Vp-

p

21.4

V

c) What rectifier configuration is this supply? Bridge Rectifier (well supposed to be ) d) Using the graph on the next page, sketch the waveform at point ‘A’. Also on the same graph sketch the resulting waveform at point ‘B’ both with and without the capacitor in circuit. (You must indicate which waveform is which along with voltage levels.) Please try to be neat! A sloppy diagram may lead to a misinterpretation and lost marks.

Page 5: EE 221 Midterm Solutions - IEEEsites.ieee.org/sb-usask/files/2011/09/ee-221-2002-t1-midterm.1.pdf · EE 221 Midterm Solutions October 30, 2002 1 1) Figure 1 a) Determine and sketch

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3) (Use the second approximation of the diode!) Sketch the waveform at each terminal, A, B, C, and D on the supplied graphs. Please try to be neat! A sloppy diagram may lead to a misinterpretation and lost marks.

Figure 3

Circuit break down: D1 and C3 = Peak Detector 1. Sets up a bias for Clamper 1. C2 and D4 = Clamper 1. D3 and C5 = Peak Detector 2. Sets up a bias for Clamper 2. C1 and D2 = Clamper 2. D5 and C4 = Peak Detector 3.

Page 6: EE 221 Midterm Solutions - IEEEsites.ieee.org/sb-usask/files/2011/09/ee-221-2002-t1-midterm.1.pdf · EE 221 Midterm Solutions October 30, 2002 1 1) Figure 1 a) Determine and sketch

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Question 3 Continued

-2-1.9-1.8-1.7-1.6-1.5-1.4-1.3-1.2-1.1

-1-0.9-0.8-0.7-0.6-0.5-0.4-0.3-0.2-0.1

00.10.20.30.40.50.60.70.80.9

11.11.21.31.41.51.61.71.81.9

2

0 30 60 90 120 150 180 210 240 270 300 330 360 390 420 450 480 510 540 570 600 630 660 690 720

Vs VA

-3-2.8-2.6-2.4-2.2

-2-1.8-1.6-1.4-1.2

-1-0.8-0.6-0.4-0.2

00.20.40.60.8

11.21.41.61.8

22.22.42.62.8

3

0 30 60 90 120 150 180 210 240 270 300 330 360 390 420 450 480 510 540 570 600 630 660 690 720

Vs VB

Graph 1 for VA

Graph 2 for VB

Page 7: EE 221 Midterm Solutions - IEEEsites.ieee.org/sb-usask/files/2011/09/ee-221-2002-t1-midterm.1.pdf · EE 221 Midterm Solutions October 30, 2002 1 1) Figure 1 a) Determine and sketch

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Question 3 Continued

-3-2.8-2.6-2.4-2.2

-2-1.8-1.6-1.4-1.2

-1-0.8-0.6-0.4-0.2

00.20.40.60.8

11.21.41.61.8

22.22.42.62.8

3

0 30 60 90 120 150 180 210 240 270 300 330 360 390 420 450 480 510 540 570 600 630 660 690 720

Vs VB VC

-0.7 V

0.9 V

-3-2.8-2.6-2.4-2.2

-2-1.8-1.6-1.4-1.2

-1-0.8-0.6-0.4-0.2

00.20.40.60.8

11.21.41.61.8

22.22.42.62.8

3

0 30 60 90 120 150 180 210 240 270 300 330 360 390 420 450 480 510 540 570 600 630 660 690 720

Vs VD

1.5 V

Graph 3 for VC

Graph 4 for VD

Page 8: EE 221 Midterm Solutions - IEEEsites.ieee.org/sb-usask/files/2011/09/ee-221-2002-t1-midterm.1.pdf · EE 221 Midterm Solutions October 30, 2002 1 1) Figure 1 a) Determine and sketch

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1 2

4 0.7 ( 2 ) 2.65V V VI mAR R

− − −= =+ 1 1 2 0.65V IR V V= − =

-5.5

-5

-4.5

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

5.5

0 30 60 90 120 150 180 210 240 270 300 330 360 390 420 450 480 510 540 570 600 630 660 690 720

Vin Vout

1.35 V

4) (Use the second approximation of the diode!) With the given Vin sketch Vout on the same graph. Please try to be neat! A sloppy diagram may lead to a misinterpretation and lost marks.

Figure 4

Because there are supplies attached means that the diodes will have a forward biased region for an input waveform. To determine what this region is and how the output will look is as follows. First let’s determine a relationship between Vin and Vout. This can be done by choosing an intermediate point such as V1. Vout = V1 + 0.7 V and V1 = Vin – 0.7 V Vout = Vin – 0.7 V +0.7 V But this is not for all cases of Vin! The upper end will be limited by the +4 V supply. In other words for an input signal greater than 4 V D2 will be reversed biased. To determine the lower limit we have to find what V1 is when no input signal is applied. In this condition current will flow from the +4 V supply through R2, D2 and R1 to the –2 V supply.

But we are interested in Vout = V1 + 0.7 V = 1.35 V therefore the lower limit is 1.35 V.

Page 9: EE 221 Midterm Solutions - IEEEsites.ieee.org/sb-usask/files/2011/09/ee-221-2002-t1-midterm.1.pdf · EE 221 Midterm Solutions October 30, 2002 1 1) Figure 1 a) Determine and sketch

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5) (Assume β = 100, VBE = 0.7 V and VCE(sat) = 0.3 V)

Figure 5

Find the following for the above transistor Schmitt trigger. a) Determine the: - Upper Trip Point? UTP = __10 V_ - Lower Trip Point? LTP = __4 V_ - Hysteresis Voltage? VH = __6 V_

VVk

kVRR

RUTP CC 10122001

1

54

4 =Ω+Ω

Ω=+

=

VVkk

kVRR

RLTP CC 41212

1

42

4 =Ω+Ω

Ω=+

=

VH = UTP – LTP = 10V – 4V = 6V

Page 10: EE 221 Midterm Solutions - IEEEsites.ieee.org/sb-usask/files/2011/09/ee-221-2002-t1-midterm.1.pdf · EE 221 Midterm Solutions October 30, 2002 1 1) Figure 1 a) Determine and sketch

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Question 5 Continued b) It is determined that a capacitor can be used in order to increase the switching time.

On the circuit diagram sketch where this capacitor is to be connected. See CS on diagram (Figure 5)!

c) If this was a 621 pF capacitor, what would the maximum switching frequency be? Fmax = _700 kHz_

max

1 3.2 F

treR

tC re == R = R3//R2

max32

32

32

32

max

3.23.2

1

FRRRR

RRRR

FC +=

+

=∴

kHzkHzpFkk

kkCRR

RRF 70013.700

621223.222

3.2 32

32max ≅=

×Ω×Ω×Ω+Ω=

+=⇒

d) List two advantages the transistor Schmitt Trigger has over a Basic Transistor Switch. 1) Different trip points for turn on (UTP) and turn off (LTP). 2) The output signal tracks the input signal. i.e. When the input is low the output is low

and when the input is high the output is high.

Page 11: EE 221 Midterm Solutions - IEEEsites.ieee.org/sb-usask/files/2011/09/ee-221-2002-t1-midterm.1.pdf · EE 221 Midterm Solutions October 30, 2002 1 1) Figure 1 a) Determine and sketch

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6) (Assume β = 100, VBE = 0.7 V, VCE(sat) = 0.3 V, and at room temperature (300 oK))

Figure 6

a) Draw the DC load line and determine the Q point. (Use the supplied transistor curve and label the X and Y axis’ values) ICQ = _1.02 mA_ VCEQ = _5.42 V_

VCE (V)

IC (mA)

2 4 6 8 10 12 14 16 18 20

1.41

1.20

1.00

0.80

0.60

0.40

0.20

1.40

Q

5.42 V

1.02

Page 12: EE 221 Midterm Solutions - IEEEsites.ieee.org/sb-usask/files/2011/09/ee-221-2002-t1-midterm.1.pdf · EE 221 Midterm Solutions October 30, 2002 1 1) Figure 1 a) Determine and sketch

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Question 6 Continued Page for Q point calculations

mAmAkk

VRRR

VI CC

C 41.1406.105.217512

20

543(max) ≅=

Ω+Ω+Ω=

++=

Vkk

kVRR

RVV CCBQ 3317

32021

2 =Ω+Ω

Ω=+

=

VVVVVV BEBQEQ 3.27.03 =−=−=

mAk

VRR

VI EQ

EQ 034.105.2175

3.2

54

=Ω+Ω

=+

=

mAmAmAII EQCQ 02.1024.1034.11100

1001

≅=+

=+

β

VkmAVRIVV CQCCCQ 72.712024.1203 =Ω×−=−=

VVVVVV EQCQCEQ 42.53.272.7 =−=−=

Page 13: EE 221 Midterm Solutions - IEEEsites.ieee.org/sb-usask/files/2011/09/ee-221-2002-t1-midterm.1.pdf · EE 221 Midterm Solutions October 30, 2002 1 1) Figure 1 a) Determine and sketch

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Question 6 Continued b) Determine the input impedance (Zin), output impedance (Zout) and voltage gain (Av).

Zin = __2.26 kΩ__ Zout = _12 kΩ__ Av = __- 60 _

Ω== kRZout 123 Zin = R1//R2//β( er ' + R4)

EQ

T

IVer =' V

qkTVT 02585.0== @ room temp. (300oK). Ω≅=∴ 25

034.102585.0'

mAVer

Zin = 17 kΩ//3 kΩ//100(25 Ω + 175 Ω) = 2.26 kΩ

6017525

12' 4

3 −=Ω+Ω

Ω−=+

−= k

RerR

AV

Page 14: EE 221 Midterm Solutions - IEEEsites.ieee.org/sb-usask/files/2011/09/ee-221-2002-t1-midterm.1.pdf · EE 221 Midterm Solutions October 30, 2002 1 1) Figure 1 a) Determine and sketch

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Question 6 Continued c) Sketch the T and π transistor ac models for this circuit. (Remember to label all components)

Page 15: EE 221 Midterm Solutions - IEEEsites.ieee.org/sb-usask/files/2011/09/ee-221-2002-t1-midterm.1.pdf · EE 221 Midterm Solutions October 30, 2002 1 1) Figure 1 a) Determine and sketch

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Question 6 Continued d) Sketch the DC transistor model for this circuit. (Remember to label all components)

e) Is this a stiff voltage divider bias? (You must show proof to support your answer)

R1//R2 = Ω=Ω+ΩΩ×Ω k

kkkk 55.2

317317

Ω=Ω+Ω=+= kkRRRE 225.205.217554

Definition of a stiff voltage divider bias is R1//R2 < 0.01 β RE Since β = 100 then 0.01 β = 1 and 2.55 kΩ is NOT less than 2.225 kΩ Therefore this is NOT a stiff voltage divider bias.