Electronics 5

© 2012 Dirar Abu-Saymeh. All rights reserved. CME 241 – Electronics 1 Dirar Abu-Saymeh

CME 241 – Electronics 1 Dirar Abu-Saymeh

Module 5

Transistor Bias Circuits Slides adapted from the set of slides provided by the author of textbook

Electronic Devices, Thomas Floyd, 9th Ed.


Summary

A transistor amplifies base current linearly only in the active region

Linear Amplification


Summary

Q-Point is the DC operating point of the circuit (IC & VCE)

Q-Point Calculation


Summary

At IB= 200 µA è IC= 20 mA

Q-Point Calculation

VCE =VCC − ICRC = 10V − (20mA)(220Ω) = 5.6V


Summary


Q-Point Calculation



Summary


Q-Point Calculation



Summary

This is the equation of a straight line: What is the Slope, x-intercept & y-intercept?

DC Load Line IC = VCC −VCERC

= − 1RC

⎛⎝⎜

⎞⎠⎟VCE +

VCCRC

Slope = -1/RC , x-intercept @ VCE = VCC , and y-intercept @ IC = VCC/RC = IC(sat)


Summary

Transistor Characteristics


Summary

Collector Circuit impact


Summary

Base Circuit Impact


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 mA. When the input causes the base current to vary between 200 mA and 400 mA, 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



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 will be clipped because it is too close to cutoff.



Summary

Or, the bias has established a high Q- point. As a result, the signal will be clipped because it is too close to saturation.



Summary

Or, the bias was established properly, but the input signal to the transistor is too high. As a result, the signal will be clipped because the output reaches cutoff and saturation.


Summary

a)  Determine Q-point b)  Draw DC load line c)  Find maximum value of peak base

current for linear operation

Example βDC = 200


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

+VCC

RCR1

RER2

βDC = 200

27 kΩ

12 kΩ

+15 V

680 Ω

1.2 kΩ

4.62 V

What is the collector current IC and VCE?

IC is almost equal to IE:

IC = IE = 5.76 mA 3.92 V

VC =VCC − IC RC = 15− (5.76m)(1.2k) = 8.09V

VCE =VC −VE = 8.09− 3.92 = 4.17V


Summary

To consider the impact of the transistor load on the bias: If RIN(BASE) > 10R2 , then we have stiff divider.

DC Input Resistance

RIN (BASE ) =VBIB

= VBIE / βDC

= βDCVBIE



Summary

+VCC

RCR1

RER2

βDC = 200

27 kΩ

12 kΩ

+15 V

680 Ω

1.2 kΩ

4.62 V

What is the DC input resistance if the transistor?

3.92 V

RIN ( BASE ) >10R2

RIN (BASE ) =βDCVBIE

= (200)(4.62)5.76m

= 160kΩ


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

RTH

+VCC

R E

+VTH + –

IB

+

+

–

–IE

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

RTH

+VCC

R E

+VTH + –

IB

+

+

–

–IE

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

If RTH/β << RE => Stiff Divider


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

RE680 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

Find IC and VEC

Example


Summary

Transistors have a wide range of β due to manufacturing variations. When a transistor fails and you need to replace it with another one, it is important that the Q-Point does not change even if the new replacement transistor has a different β. => Bias Stability

Bias Stability


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

IE =

−1 V −VEE

RE

= −1 V − (−15 V)7.5 kΩ

= 1.87 mA


Summary

KVL on base-emitter loop:

Emitter Bias Accurate Analysis V

V

CC

EE

RC

RE

RB68 kΩ

+15 V

−15 V

7.5 kΩ

3.9 kΩ

VEE +VRB +VBE +VRE = 0VEE + IBRB +VBE + IERE = 0

IE =−VEE −VBERE + RB / βDC

IE =15 − 0.7

7.5k + 68k /100= 1.75mA

VCE =VCC − ICRC − IERE −VEE = 10.05V

βDC=100


Summary

What happens if β changes to 200

Emitter Bias Accurate Analysis V

V

CC

EE

RC

RE

RB68 kΩ

+15 V

−15 V

7.5 kΩ

3.9 kΩ

IE =−VEE −VBERE + RB / βDC

IE =15 − 0.7

7.5k + 68k / 200= 1.82mA

VCE =VCC − ICRC − IERE −VEE = 9.25V

%ΔVCE =9.25 −10.0510.05

100% = −8% %ΔIC = 1.82 −1.751.75

100% = 4%

βDC=200


Summary

Base Bias

RCRB

+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− −= = =

Ω25.5 mA

RC

RB

+VCC

560 kΩ

+15 V

1.8 kΩ


Summary

Base Bias

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

RC

RB

+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 = 200: IC = β IB = 200( ) 25.5 µA( ) = 5.1 mA

VCE =VCC − ICRC

= 15 V − 5.1 mA( ) 1.8 kΩ( ) = 5.82 V

%ΔVCE =5.82 −10.410.4

100% = −44% %ΔIC = 5.1− 2.552.55

100% = 100%


Summary

Emitter-Feedback Bias

R

R

C

E

RB

+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:

IE =VCC −VBE

RE +RB

βDC

Negative feedback: When IC tries to increase, this increases VE and VB so IB decreases causing IC to decrease.


Summary

Emitter-Feedback Bias R

R

C

E

RB

+VCC

RC=560Ω, RB=330kΩ, RE=1kΩ, βDC=100, VCC=12V

IE = 12− 0.7

1k + 330k / 100= 2.63mA

VCE =VCC − ICRC − IERE = 12 − (2.63m)(560 +1k) = 7.9V

IE = 12− 0.7

1k + 330k / 200= 4.26mA

VCE =VCC − ICRC − IERE = 12 − (4.26m)(560 +1k) = 5.35V

%ΔVCE =5.35 − 7.97.9

100% = −32% %ΔIC = 4.26 − 2.632.63

100% = 62%

If βDC changes to 200:


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.

CC BEC

BC

DCβ

V VI RR

−=+

+VCC

RCRB

VC ≅VCC − ICRC

IB = (VC −VBE ) / RB

ICβDC

= VCC − ICRC −VBERB


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

RCRB

330 kΩ

1.8 kΩ

+ 15 V

Compare IC for the case when β = 100 with the case when β = 200.

2.80 mA

When β = 200,

IC =VCC −VBE

RC +RB

βDC

= 15 V − 0.7 V1.8 kΩ+ 330 kΩ

200= 4.14 mA


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 kW

b. 6.67 kW

c. 10 kW

d. 30 kW

+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

RE680 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

RCRB

+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

Engineering

Electronics 5