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Electronic Devices, 9th edition Thomas L. Floyd

Electronic Devices Ninth Edition

Floyd

Chapter 5



The DC Operating Point

Bias establishes the operating point (Q-point) of a transistor amplifier; the ac signal moves above and below this point.

Summary

For this example, the dc base current is 300 µA. When the input causes the base current to vary between 200 µA and 400 µA, the collector current varies between 20 mA and 40 mA.

0VCE (V)

400 A

IC (mA)

300 A = IBQ

200 A

A

B

Q

1.2 3.4 5.6VCEQ

ICQ

Vce

IbIc

20

30

40 µ

µ

µ

Load line



The DC Operating Point

A signal that swings outside the active region will be clipped.

Summary

For example, the bias has established a low Q- point. As a result, the signal is will be clipped because it is too close to cutoff.

VCCVCE

Cutoff

Q

I BQ

IC

ICQ

Cutoff 0

Vce

VCEQ

Inputsignal



Voltage-Divider Bias

A practical way to establish a Q-point is to form a voltage-divider from VCC.

Summary

R1 and R2 are selected to establish VB. If the divider is stiff, IB is small compared to I2. Then,

+VCC

RCR1

RER2

2B CC

1 2

RV VR R

≈ +

+VCC

RCR1

RER2

βDC = 200

27 kΩ

12 kΩ

+15 V

680 Ω

1.2 kΩ

Determine the base voltage for the circuit.

( )

2B CC

1 2

12 k 15 V27 k 12 k

RV VR R

= +

Ω = + = Ω + Ω 4.62 V

I2

IB




Summary

+VCC

RCR1

RER2

βDC = 200

27 kΩ

12 kΩ

+15 V

680 Ω

1.2 kΩ

4.62 V

What is the emitter voltage, VE, and current, IE?

VE is one diode drop less than VB:

VE = 4.62 V – 0.7 V = 3.92 V 3.92 V

Applying Ohm’s law:

EE

E

3.92 V680

VIR

= = =Ω

5.76 mA



Summary


The unloaded voltage divider approximation for VB gives reasonable results. A more exact solution is to Thevenize the input circuit. +VCC

RCR1

RER2

βDC = 200

27 kΩ

12 kΩ

+15 V

680 Ω

1.2 kΩ

VTH = VB(no load)

= 4.62 V

RTH = R1||R2 = = 8.31 kΩ

The Thevenin input circuit can be drawn

R C

R TH

+VCC

R E

+V TH + –

IB

+

+

–

–I E

VBE

8.31 kΩ

680 Ω

1.2 kΩ

4.62 V

+15 V

βDC = 200



Summary


Now write KVL around the base emitter circuit and solve for IE.

R C

R TH

+VCC

R E

+V TH + –

IB

+

+

–

–I E

VBE

8.31 kΩ

680 Ω

1.2 kΩ

4.62 V

+15 V

βDC = 200

TH B TH BE E EV I R V I R= + +

TH BEE

THE

DCβ

V VI RR

−=

+

Substituting and solving,

E4.62 V 0.7 V

8.31 k680 + 200I −

= =ΩΩ

5.43 mA

and VE = IERE = (5.43 mA)(0.68 kΩ)

= 3.69 V



Summary


Multisim allows you to do a quick check of your result.



Summary

Voltage-Divider Bias A pnp transistor can be biased from either a positive or negative supply. Notice that (b) and (c) are the same circuit; both with a positive supply.

+

+

V V

V

EEEE

EE

R

RR

2

22 1

11

R

RR

R

RR

C

CC R

RR

E

EE

−

(a) (b) (c)



Summary


Determine IE for the pnp circuit. Assume a stiff voltage divider (no loading effect).

+VEE

R 2

1R RC1.2 kΩ

R E680 Ω

27 kΩ

12 kΩ

+15 V

( )

1B EE

1 2

27 k 15.0 V 10.4 V27 k 12 k

RV VR R

= +

Ω = + = Ω + Ω

E B BE 10.4 V 0.7 V = 11.1 VV V V= + = +

EE EE

E

15.0 V 11.1 V680

V VIR

− −= = =

Ω5.74 mA

10.4 V 11.1 V



Summary

Emitter Bias

Emitter bias has excellent stability but requires both a positive and a negative source.

Assuming that VE is −1 V, what is IE?

V

V

CC

EE

RC

RE

RB68 kΩ

+15 V

−15 V

7.5 kΩ

3.9 kΩ

For troubleshooting analysis, assume that VE for an npn transistor is about −1 V.

−1 V

EEE

E

1 V 15 V ( 1 V)7.5 k

VIR

− − − − −= = =

Ω−1.87 mA



Summary

Emitter Bias

A check with Multisim shows that the assumption for troubleshooting purposes is reasonable.

For detailed analysis work, you can include the effect of βDC. In this case,

EEE

BE

DC

1 V

β

VI RR

− −=

+



Summary

Base Bias

RC

R B

+VCC

Base bias is used in switching circuits because of its simplicity, but not widely used in linear applications because the Q-point is β dependent. Base current is derived from the collector supply through a large base resistor.

What is IB?

CCB

B

0.7 V 15 V 0.7 V560 k

VIR− −

= = =Ω

25.5 µA

RC

R B

+VCC

560 kΩ

+15 V

1.8 kΩ



Summary

Base Bias

Compare VCE for the case where β = 100 and β = 200.

RC

R B

+VCC

560 kΩ

+15 V

1.8 kΩ

( )( )C Bβ 100 25.5 μA 2.55 mAI I= = =

10.4 V

For β = 100:

( )( )CE CC C C

15 V 2.55 mA 1.8 kV V I R= −

= − Ω =

For β = 300: ( )( )C Bβ 300 25.5 μA 7.65 mAI I= = =

( )( )CE CC C C

15 V 7.65 mA 1.8 kV V I R= −

= − Ω = 1.23 V



Summary

Emitter-Feedback Bias

R

R

C

E

R B

+VCC

An emitter resistor changes base bias into emitter-feedback bias, which is more predictable. The emitter resistor is a form of negative feedback. The equation for emitter current is found by writing KVL around the base circuit. The result is:

CC BEE

EE

DCβ

V VI RR

−=

+



Summary

Collector-Feedback Bias

Collector feedback bias uses another form of negative feedback to increase stability. Instead of returning the base resistor to VCC, it is returned to the collector. The equation for collector current is found by writing KVL around the base circuit. The result is

CC BEC

BC

DCβ

V VI RR

−=

+

+VCC

RCRB



Summary

Collector-Feedback Bias

When β = 100,

CC BEC

BC

DC

15 V 0.7 V330 k1.8 k 100β

V VI RR

− −= = =

ΩΩ ++

+VCC

RCR B

330 kΩ

1.8 kΩ

+ 15 V

Compare IC for the case when β = 100 with the case when β = 300.

2.80 mA

When β = 300,

CC BEC

BC

DC

15 V 0.7 V330 k1.8 k 300β

V VI RR

− −= = =

ΩΩ ++4.93 mA



Electronic Devices Ninth Edition

Floyd

Chapter 6



AC Quantities

AC quantities are indicated with a italic subscript; rms values are assumed unless otherwise stated.

Summary

VceVce

Vce

vce

VCE

00

t

Vce

rmsavg

V

The figure shows an example of a specific waveform for the collector-emitter voltage. Notice the DC component is VCE and the ac component is Vce.

Resistance is also identified with a lower case subscript when analyzed from an ac standpoint.



Linear Amplifier

A linear amplifier produces an replica of the input signal at the output.

Summary

For the amplifier shown, notice that the voltage waveform is inverted between the input and output but has the same shape.

RC

+VCC

R1

RER2 RL

C2

Vb

IbRs

Ic

ICQ

Vce

VCEQ

Vs

C1IBQ

VBQ



AC Load Line

Operation of the linear amplifier can be illustrated using an ac load line.

Summary

The ac load line is different than the dc load line because a capacitor looks open to dc but effectively acts as a short to ac. Thus the collector resistor appears to be in parallel with the load resistor.

VCEQ

0 VCE

I b

IC

Ic

Q

Vce

ICQ

I BQ



Transistor AC Model

The five resistance parameters (r-parameters) can be used for detailed analysis of a BJT circuit. For most analysis work, the simplified r-parameters give good results.

Summary

βacIb

C

B

E

βacIb

C

B

E

Ib ′re

′re

The simplified r-parameters are shown in relation to the transistor model. An important r-parameter is re'. It appears as a small ac resistance between the base and emitter.

'

E

25 mVer I

=



The Common-Emitter Amplifier

In the common-emitter (CE) amplifier, the input signal is applied to the base and the inverted output is taken from the collector. The emitter is common to ac signals.

Summary

R2

RE

R1

Vin

RC

VCC

Vout

RL

C1

C2

C3




Summary

R2

RE

R1

RC

VCC

RL

C1

C2

C3

68 kΩ

27 kΩ2.2 kΩ

3.9 kΩ

3.9 kΩ1.0 Fµ

10 Fµ

100 Fµ

+15 V

What is re' for the CE amplifier? Assume stiff voltage-divider bias.

B27 k 15 V =

68 k 27 kV Ω = Ω + Ω

4.26 V

VE = 4.26 V – 0.7 V = 3.56 V

EE

E

3.56 V2.2 k

VIR

= = =Ω

1.62 mA

'

E

25 mV 25 mV1.62 mAer I

= = = 15.4 Ω




Summary

R2

RE

R1

RC

VCC

RL

C1

C2

C3

68 kΩ

27 kΩ2.2 kΩ

3.9 kΩ

3.9 kΩ1.0 Fµ

10 Fµ

100 Fµ

+15 V

What is the gain of the amplifier?

C' ' out c L

vin e e

V R R RAV r r

= = =

127

Notice that the ac resistance of the collector circuit is RC||RL.

3.9 k 3.9 k15.4 vA Ω Ω

= =Ω

The gain will be a little lower if the input loading effect is accounted for.




Summary

Greater gain stability can be achieved by adding a swamping resistor to the emitter circuit of the CE amplifier. The gain will be lower as a result.

R 2

R

R

E2

E1

R1

RC

VCC

RL

C1

C2

C3

68 kΩ

27 kΩ2.2 kΩ

3.9 kΩ

3.9 kΩ

1.0 Fµ

10 Fµ

100 Fµ

+15 V

33 Ω

What is the gain with the addition of the swamping resistor? (Ignore the small effect on re'.)

C' '

E1 E1

out c Lv

in e e

V R R RAV r R r R

= = =+ +

38.2 3.9 k 3.9 k15.4 33 vA Ω Ω

= =Ω + Ω




Summary

Multisim is a good way to check your calculation. For an input of 10 mVpp, the output is 378 mVpp as shown on the oscilloscope display for the swamped CE amplifier.

input

output




Summary

In addition to gain stability, swamping has the advantage of increasing the ac input resistance of the amplifier. For this amplifier, Rin(tot) is given by

R 2

R

R

E2

E1

R1

RC

VCC

RL

C1

C2

C3

68 kΩ

27 kΩ2.2 kΩ

3.9 kΩ

3.9 kΩ

1.0 Fµ

10 Fµ

100 Fµ

+15 V

33 Ω

Rin(tot) = R1||R2||βac(re' + RE1)

What is Rin(tot) for the amplifier if βac = 200?

Rin(tot) = R1||R2||βac(re' + RE1)

= 68 kΩ||27 kΩ||200(15.4 Ω + 33 Ω)

= 6.45 kΩ



The Common-Collector Amplifier

Summary

The common-collector amplifier (emitter-follower) has a voltage gain of approximately 1, but can have high input resistance and current gain. The input is applied to the base and taken from the emitter.

+VCC

R1

R2RE RL

Vin

Iin Vout

C1

C2




Summary

The power gain is the ratio of the power delivered to the input resistance divided by the power dissipated in the load. This is approximately equal to the current gain. That is, Ap ≈Ai.

Vin

Vout

C1

R1

VCC

R2

RLRE

C2

You can also write power gain as a ratio of resistances:

2

( )22

( )

( ) ( )1

Lin totL L

p vinin L

in tot

in tot in tot

L L

VRP RA A

VP RR

R RR R

= = =

≅ =

The next slide is an example…




Summary

Calculate the power gain to the load for the CC amplifier using a ratio of resistances. Assume Av = 1 and βac = 200. Use re' = 2 Ω.

Rin(tot) = R1||R2||βac(re' + RE||RL)

= 39 kΩ||220 kΩ||200(2 Ω + 500 Ω)

= 24.9 kΩ Vin

Vout

C1

R1

VCC

R2

RLRE

C20.22 Fµ

3.3 Fµ

+15 V

39 kΩ

220 kΩ1.0 kΩ 1.0 kΩ

RL = 1.0 kΩ

( ) 24.9 k1.0 k

in totp

L

RA

RΩ

= = =Ω

24.9




Summary

The input voltage-divider in the previous example is not “rock-solid” but the overall power gain is good. A “rock solid” stiff voltage-divider is not always the best design. Can you spot the problem illustrated here? Rin(tot) = R1||R2||βac(re' + RE||RL)

= 10 kΩ||10 kΩ||200(25 Ω + 3.0 kΩ)

= 4.96 kΩ RL = 10 kΩ

( ) 4.96 k10 k

in totp

L

RA

RΩ

= = =Ω

0.496!

Vin

Vout

C1

R1

VCC

R2

RLRE

C2

10 kΩ

10 kΩ10 kΩ

4.3 kΩ

+10 V

β = 200

The problem is the power gain is less than 1!



The Darlington Pair

Summary

A Darlington pair is two transistors connected as shown. The two transistors act as one “super β” transistor. Darlington transistors are available in a single package. Notice there are two diode drops from base to emitter. VCC

R1

C2

R2

RE RL

Vin

Q2

Q1

VCC

RC C1

Vout

CE Amplifier Darlington CC amplifier Load



The Sziklai Pair

Summary

Another high β pair is the Sziklai pair (sometimes called a complementary Darlington), in which a pnp and npn transistor are connected as shown. This configuration has the advantage of only one diode drop between base and emitter.

+VCC

RE

βDC1

βDC2

IB1

IC1 IE2

Vin

IC1 is βDC1 x IB1 and is equal to IB2

IE2 is approximately equal to βDC2 x IC1 Therefore, IE2 ≈ βDC1βDC2IB1

The DC currents are:

What is the relation between IE2 and IB1?



The CB Amplifier

Summary

The common-base (CB) amplifier is used in applications where a low input impedance is acceptable. It does not invert the signal, an advantage for higher frequencies as you will see later when you study the Miller effect.

C2 forces the base to be at ac ground.

What is the purpose of C2?

+VCC

R1

RC

RE

RL

R2

C2

C3

C1

Vin

Vout



Multistage Amplifiers

Summary

To improve amplifier performance, stages are often cascaded where the output of one drives another. This an example of a two-stage direct-coupled amplifier in which the input and output signals are capacitively coupled.

Q1

Q2

VCC+12 V

Vin

Vout

VS

100 mV1.0 kHz

pp

RE3

RE1

RE2

RCR1

R2

C1

C2

C3

RL

1.0 kΩ

1.0 µF

47 µF

100 Ω

330 Ω

2N3904

330 Ω

10 µF330 Ω2N3906

10 kΩ

4.7 kΩ



Differential Amplifiers

Summary

A differential amplifier (diff-amp) has two inputs. It amplifies the difference in the two input voltages. This circuit is widely used as the input stage to operational amplifiers. Differential-mode inputs are illustrated.

+VCC

RC1 RC2

Q1 Q2

–VEE

RE

1 2

21

Vout1

Vin1 Vin2

Vout2



Differential Amplifiers

Summary

The same amplifier as in the last slide now is shown with common-mode inputs. Diff-amps tend to reject common-mode signals, which are usually due to noise. Ideally, the outputs are zero with common-mode inputs.

+VCC

RC1 RC2

Q1 Q2

–VEE

RE

1 2

21

Vout1

Vin1 Vin2

Vout2



Key Terms

Q-point

DC load line

Linear region

Stiff voltage divider

Feedback

The dc operating (bias) point of an amplifier specified by voltage and current values.

A straight line plot of IC and VCE for a transistor circuit.

The region of operation along the load line between saturation and cutoff.

A voltage divider for which loading effects can be ignored.

The process of returning a portion of a circuit’s output back to the input in such a way as to oppose or aid a change in the output.



Quiz

1. A signal that swings outside the active area will be

a. clamped

b. clipped

c. unstable

d. all of the above



Quiz

2. A stiff voltage divider is one in which

a. there is no load current

b. divider current is small compared to load current

c. the load is connected directly to the source voltage

d. loading effects can be ignored



Quiz

3. Assuming a stiff voltage-divider for the circuit shown, the emitter voltage is

a. 4.3 V

b. 5.7 V

c. 6.8 V

d. 9.3 V

+VCC

RCR1

RER2

βDC = 200

20 kΩ

10 kΩ

+15 V

1.2 kΩ

1.8 kΩ



Quiz

4. For the circuit shown, the dc load line will intersect the y-axis at

a. 5.0 mA

b. 10.0 mA

c. 15.0 mA

d. none of the above

+VCC

RCR1

RER2

βDC = 200

20 kΩ

10 kΩ

+15 V

1.2 kΩ

1.8 kΩ



Quiz

5. If you Thevenize the input voltage divider, the Thevenin resistance is

a. 5.0 kΩ

b. 6.67 kΩ

c. 10 kΩ

d. 30 kΩ

+VCC

RCR1

RER2

βDC = 200

20 kΩ

10 kΩ

+15 V

1.2 kΩ

1.8 kΩ



Quiz

6. For the circuit shown, the emitter voltage is

a. less than the base voltage

b. less than the collector voltage

c. both of the above

d. none of the above

+VEE

R 2

1R RC1.2 kΩ

R E680 Ω

27 kΩ

12 kΩ

+15 V



Quiz

7. Emitter bias

a. is not good for linear circuits

b. uses a voltage-divider on the input

c. requires dual power supplies

d. all of the above



Quiz

8. With the emitter bias shown, a reasonable assumption for troubleshooting work is that the

a. base voltage = +1 V

b. emitter voltage = +5 V

c. emitter voltage = −1 V

d. collector voltage = VCC

V

V

CC

EE

RC

RE

RB68 kΩ

+15 V

−15 V

7.5 kΩ

3.9 kΩ



Quiz

9. The circuit shown is an example of

a. base bias

b. collector-feedback bias

c. emitter bias

d. voltage-divider bias

RC

R B

+VCC



Quiz

10. The circuit shown is an example of

a. base bias

b. collector-feedback bias

c. emitter bias

d. voltage-divider bias

+VCC

RCRB



Quiz

Answers:

1. b

2. d

3. a

4. a

5. b

6. d

7. c

8. c

9. a

10. b

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