ENGN 2211 Electronic Circuits and Devices Problem Set #8 ...

[Pages:9]ANU

AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering

ENGN 2211 Electronic Circuits and Devices Problem Set #8 BJT CE Amplifier Circuits

Q1

Consider the common-emitter BJT amplifier circuit shown in Figure 1.

Assume VCC = 15 V, = 150, VBE = 0.7 V, RE = 1 k, RC = 4.7 k, R1 = 47 k, R2 = 10 k, RL = 47 k, Rs = 100 .

+VCC

ENGN 2211

R1

Rs

C1

vs

vin

R2

RC C2

RE CE

vo RL

Figure 1: The circuit for Question 1.

(a) Determine the Q-point. (b) Sketch the DC load-line. What is the maximum (peak to peak) output voltage swing available in this amplifier. (c) Draw the AC equivalent circuit and determine the AC model parameters. (d) Derive expressions for Rin, Rout , Avoc, Av, Ai, G. (e) Find Rin, Rout , Avoc, Av, Ai, G. (f) Find the output voltage waveform if vs = 10 ? 10-3 sin(25000t). Sketch the source and output voltage waveforms. (g) Determine whether clipping will take place if vs = 25 ? 10-3 sin(25000t).

Problem Set #8

page 1

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Q2

Consider the common-emitter BJT amplifier circuit shown in Figure 2.

Assume VCC = 15 V, = 150, VBE = 0.7 V, RE = 2.7 k, RC = 4.7 k, R1 = 47 k, R2 = 10 k, RL = 47 k, Rs = 100 .

+VCC

ENGN 2211

R1

Rs

C1

vs

vin

R2

RC C2

RE CE

vo RL

Figure 2: The circuit for Question 2.

(a) Determine the Q-point. (b) Sketch the DC load-line. What is the maximum (peak to peak) output voltage swing available in this amplifier. (c) Draw the AC equivalent circuit and determine the AC model parameters. (d) Derive expressions for Rin, Rout , Avoc, Av, Ai, G. (e) Find Rin, Rout , Avoc, Av, Ai, G. (f) Find the output voltage waveform if vs = 10 ? 10-3 sin(25000t). Sketch the source and output voltage waveforms.

Problem Set #8

page 2

ANU

AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering

ENGN 2211 Electronic Circuits and Devices Problem Set #8 Solution

Q1

Complete Solution

Given that VCC = 15 V, = 150, VBE = 0.7 V, RE = 1 k, RC = 4.7 k, R1 = 47 k, R2 = 10 k, RL = 47 k, Rs = 100 .

+VCC

R1

Rs

C1

vs

vin

R2

RC C2

RE CE

vo RL

ENGN 2211

(a)

Analyzing the DC Voltage-divider bias circuit, we have

VT H

=

R1

R2 + R2

VCC

10k

=

(15) = 2.63 V

10k + 47k

RT H

=

R2R1 R1 + R2

= (10k)(47k) = 8.2456 k 10k + 47k

IB

=

VT H - VBE RT H + ( + 1)RE

2.63 - 0.7

=

= 12.12 ?A

8.2456k + (151)(1k)

IC = IB

= (150)(12.12?) = 1.8179 mA

IE = ( + 1)IB = (151)(12.12?) = 1.83 mA

VE = IE RE = (1.83m)(1k) = 1.83 V

VC = VCC - ICRC = 15 - (1.8179m)(4.7k) = 6.456 V

VCE = VCC - ICRC - IE RE = 15 - (1.8179m)(4.7k) - (1.83m)(1k) = 4.626 V

Problem Set #8

page 3

ANU As IB > 0 and VCE > 0.2 V, the transistor is in active region of operation.

The Q-point lies at ICQ = 1.8179 mA

VCEQ = 4.626 V

ENGN 2211

(b)

For ideal cut-off

VCE(o f f ) = VCC = 15 V

For ideal saturation

IC(sat)

=

VCC

15 = = 2.63 mA

RC + RE 5.7k

The plot of DC load line is shown in figure below

3 Ideal Saturation

2.5

Current IC (mA)

2 Q-point

1.5

1

0.5 0 0

5

10

Voltage VCE (V)

Ideal Cut-off 15

We see that the Q-point lies closer to saturation (VCE = 0.2 V) than cut-off (VCE = 15 V). Hence the maximum available peak to peak output voltage swing = 2(VCEQ - 0.2) = 8.852 V.

(c)

Replacing the capacitors by short circuits and VCC by virtual AC ground, the AC equivalent circuit is

Rs

vs

vin

R1

R2

vo RC RL

Problem Set #8

page 4

ANU Replacing the transistor by the small-signal AC equivalent circuit, we have

Rs

iin

ib

ic

B

C

vs

vin

RB vbe r ib

RC

ENGN 2211

io

vo

RL

E

E

The AC Model parameters are

26 mV

re =

IEQ

= 26 = 14.207 1.83

r = ( + 1)re

= (151)(14.207) = 2.1453 k

RB = R1||R2 = 8.2456 k

RL = RL||RC = (47 k)||(4.7 k) = 4.27 k

(d)

For derivations, please see Lecture 13.

(e)

The BJT CE amplifier parameters are

Rin = RB||r = (8.2456 k)||(2.1453 k) = 1.7024 k

Ro = RC

= 4.7 k

Avoc

=

- RC r

= (4.7 k)(150) = -328.62 2.1453 k

Av

=

- RL r

(4.27 k)(150)

=

= -298.56

2.1453 k

Ai

=

Av

Rin RL

1.7024 k

= (-298.56)

= -10.81

47 k

G = AiAv

= (-10.81)(-298.56) = 3228.69

Problem Set #8

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(f)

Finding the equation for output voltage with load, we have

vs = 10 ? 10-3 sin(25000t)

vin

=

Rs

Rin + Rin

vs

1.7024 k = 100 + 1.7024 k vs

= 0.9445vs

vo = Avvin

= (-298.56)(0.9445 vs) = (-298.56)(0.9445)(10 ? 10-3 sin(25000t))

= -2.82 sin(25000t)

The required peak to peak output voltage swing = 2(2.82)=5.64 V. The maximum available peak to peak output voltage swing = 8.852 V > 5.64 V. Hence no clipping will take place.

ENGN 2211

The -ve sign indicates that output voltage is 180 out of phase with input voltage (inverting amplifier).

The

time

period

is

T

=

1 5000

=

0.2

ms.

The sketch of input and output voltages is shown in figures below:- (Note y-axis has different units in the two

figures)

Voltage vs (mV) Voltage vo (V)

10

5

0

-5

-10 0

0.1

0.2

0.3

0.4

Time t (ms)

Figure 3: Source voltage vs(t).

3

X: 0.00015

2

Y: 2.824

1

0

-1

-2

-3

0

0.1

0.2

0.3

0.4

Time t (ms)

Figure 4: Output voltage vo(t).

(g)

For vs = 25 ? 10-3 sin(25000t), we have

vo = Avvin = (-298.56)(0.9445 vs) = (-298.56)(0.9445)(25 ? 10-3 sin(25000t)) = -7.05 sin(25000t)

The required peak to peak output voltage swing = 2(7.05)=14.1 V. However the maximum available peak to peak output voltage swing = 8.852 V < 14.1 V. Hence clipping will take place.

See ProbSet08_Q1.sch.

Problem Set #8

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Q2

Solution

Given that VCC = 15 V, = 150, VBE = 0.7 V, RE = 2.7 k, RC = 4.7 k, R1 = 47 k, R2 = 10 k, RL = 47 k, Rs = 100 .

+VCC

R1

Rs

C1

vs

vin

R2

RC C2

RE CE

vo RL

ENGN 2211

(a)

Analyzing the DC Voltage-divider bias circuit, we have

VTH = 2.63 V RTH = 8.2456 k

IB = 4.64 ?A IC = 0.696 mA IE = 0.7006 mA VC = 11.72 V VE = 1.89 V VCE = 9.837 V

As IB > 0 and VCE > 0.2 V, the transistor is in active region of operation.

The Q-point lies at

ICQ = 0.696 mA VCEQ = 9.837 V

Problem Set #8

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(b)

For ideal saturation and cut-off

VCE(o f f ) = 15 V IC(sat) = 2.027 mA

The plot of DC load line is shown in figure below

ENGN 2211

3

2.5

Current IC (mA)

2 Ideal Saturation

1.5

1 0.5

0 0

Q-point

5

10

Voltage VCE (V)

Ideal Cut-off 15

We see that the Q-point lies closer to cut-off (VCE = 15 V) than saturation (VCE = 0.2 V). Hence the maximum available peak to peak output voltage swing = 2(VCC -VCEQ) = 10.34 V.

(c)

The AC Model parameters are re = 37.11 r = 5.6037 k RB = 8.2456 k RL = 4.27 k

(d)

For derivations, please see Lecture 13.

(e)

The BJT CE amplifier parameters are

Rin = 3.336 k Ro = 4.7 k Avoc = -125.81 Av = -114.3 Ai = -8.11 G = 926.97

Problem Set #8

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