chapter 17.ppt

Electronics Fundamentals 8th edition Floyd/Buchla

© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.

chapter 17

electronics fundamentalscircuits, devices, and applications

THOMAS L. FLOYDDAVID M. BUCHLA


Chapter 17Chapter 17


Summary

Bipolar junction transistors (BJTs)

The BJT is a transistor with three regions and two pn junctions. The regions are named the emitter, the base, and the collector and each is connected to a lead.There are two types of BJTs – npn and pnp.

n

np

p

pn

E (Emitter)

B (Base)

C (Collector)

E

B

C

Separating the regions

are two junctions.

Base-Collector junction

Base-Emitter junction




Summary

BJT biasing

For normal operation, the base-emitter junction is forward-biased and the base collector junction is reverse-biased.

npn

BE forward- biased

BC reverse- biased

For the npn transistor, this condition requires that the base is more positive than the emitter and the collector is more positive than the base.

+

+

For the pnp transistor, this condition requires that the base is more negative than the emitter and the collector is more negative than the base.

pnp

+

+




Summary

BJT currents

A small base current (IB) is able to control a larger collector current (IC). Some important current relationships for a BJT are:

E C BI I I

C DC EαI I

C DC BβI II

I

I

IB

IE

IC




Summary

Voltage-divider bias

Because the base current is small, the approximation2

B CC1 2

RV V

R R

is useful for calculating the base voltage.

R1

R2

RC

RE

VB

VE

After calculating VB, you can find VE by subtracting 0.7 V for VBE.

Next, calculate IE by applying Ohm’s law to RE:

C EI IThen apply the approximation

Finally, you can find the collector voltage from C CC C CV V I R

VC

EE

E

VI

R




Summary

Voltage-divider bias

Calculate VB, VE, and VC for the circuit.

2B CC

1 2

6.8 k15 V =

27 k + 6.8 k

RV V

R R

R1

R2

RC

RE

VE = VB 0.7 V =

C E 2.32 mAI I

C CC C C 15 V 2.32 mA 2.2 kV V I R

EE

E

2.32 V 2.32 mA

1.0 k

VI

R

27 k

6.8 k 1.0 k

2.2 k

+15 V

2N3904

3.02 V

2.32 V

9.90 V




Summary

Collector characteristic curves

The collector characteristic curves are a family of curves that show how collector current varies with the collector-emittervoltage for a given IB.

IC

VCE0

IB6

IB5

IB4

IB3

IB2

IB1

IB = 0

The saturation region occurs when the base-emitter and the base-collector junctions are both forward biased.

The curves are divided into three regions:The active regionactive region is after the saturation region. This is the region for operation of class-A operation.

The breakdownbreakdown regionregion is after the active region and is is characterized by rapid increase in collector current. Operation in this region may destroy the transistor.




Summary

Load lines

A load line is an IV curve that represents the response of a circuit that is external to a specified load. For example, the load line for the Thevenin circuit can be found by calculating the two end points: the current with a shorted load, and the output voltage with no load.

+12 V

2.0 k

00

4 8 12

2

4

6

I (mA)

V (V)

ISL = 6.0 mA

VSL = 0 V

INL = 0 mA

VNL = 12 V

Load line




Summary

Load lines

The IV response for any load will intersect the load line and enables you to read the load current and load voltage directly from the graph.

+12 V

2.0 k

00

4 8 12

2

4

6

I (mA)

V (V)

Read the load current and load voltage from the graph if a 3.0 k resistor is the load.

3.0 kRL =

IV curve for 3.0 k resistor

VL = 7.2 V IL = 2.4 mA

Q-point




Summary

Load lines

The load line concept can be extended to a transistor circuit. For example, if the transistor is connected as a load, the transistor characteristic

+12 V

2.0 k

00

4 8 12

2

4

6

I (mA)

V (V)

curve and the base current establish the Q-point.




Summary

Load lines

Load lines can illustrate the operating conditions for a transistor circuit.

00

4 8 12

2

4

6

I (mA)

V (V)

If you add a transistor load to the last circuit, the base current will establish the Q-point. Assume the base current is represented by the blue line.

+12 V

2.0 k

For this base current, the Q-point is:

Assume the IV curves are as shown:

The load voltage (VCE) and current (IC) can be read from the graph.




Summary

Load lines

For the transistor, assume the base current is established at 10 A by the bias circuit. Show the Q-point and read the value of VCE and IC.

00

4 8 12

2

4

6

IC (mA)

VCE (V)

+12 V

2.0 k

Bias circuitIB = 5.0 A

IB = 10 A

IB = 15 A

IB = 20 A

IB = 25 A

The Q-point is the intersection of the load line with the 10 A base current.

VCE = 7.0 V; IC = 2.4 mA




Summary

Signal operation

When a signal is applied to a transistor circuit, the output can have a larger amplitude because the small base current controls a larger collector current.

00

4 8 12

2

4

6

IC (mA)

VCE (V)

IB = 5.0 A

IB = 10 A

IB = 15 A

IB = 20 A

IB = 25 A For the load line and characteristic curves from the last example (Q-point shown) assume IB varies between 5.0 A and 15 A due to the input signal. What is the change in the collector current?

Reading the collector current, IC varies from 1.2 mA to 3.8 mA.

The operation along the load line is shown in red.




Summary

CE amplifier

In a common-emitter amplifier, the input signal is applied to the base and the output is taken from the collector. The signal is larger but inverted at the output.

R1

R2

RC

REInput coupling capacitor

Output coupling capacitor

Bypass capacitor

C3

C2

C1

VCC




Summary

Voltage gain of a CE amplifier

Calculate the voltage gain of the CE amplifier. The dc conditions were calculated earlier; IE was found to be 2.32 mA.

Coutv

in e

V RA

V r

R1

R2

RC

RE C3

C2

C1

VCC = +15 V

27 k

6.8 k 1.0 k

2.2 kE

25 mV 25 mV10.8

2.32 mAer I

2042.2 F

1.0 F

100 F

2.2 k

10.8

Sometimes the gain will be shown with a negative sign to indicate phase inversion.

2N3904




Summary

R1

R2

RC

RE C3

C2

C1

VCC = +15 V

27 k

6.8 k 1.0 k

2.2 k

2.2 F1.0 F

100 F

Input resistance of a CE amplifier

The input resistance of a CE amplifier is an ac resistance that includes the bias resistors and the resistance of the emitter circuit as seen by the base.Because IB << IE, the emitter resistance appears to be much larger when viewed from the base circuit. The factor is (ac+1), which is approximately equal to ac.Using this approximation, Rin(tot) = R1||R2||acre.

2N3904




Summary

R1

R2

RC

RE C3

C2

C1

VCC = +15 V

27 k

6.8 k 1.0 k

2.2 k

2.2 F1.0 F

100 F

Input resistance of a CE amplifier

Rin(tot) = R1||R2||acre

Calculate the input resistance of the CE amplifier. The transistor is a 2N3904 with an average ac of 200. The value of re was found previously to be 10.8 . Thus, acre = 2.16 k.

2N3904= 27 k||6.8 k||2.16 k= 1.55 k

Notice that the input resistance of this configuration is dependent on the value of ac, which can vary.




Summary

CC amplifier

In a common-collector amplifier, the input signal is applied to the base and the output is taken from the emitter. There is no voltage gain, but there is power gain.

R1

R2 RE

C1

VCCThe output voltage is nearly the same as the input; there is no phase reversal as in the CE amplifier.

The input resistance is larger than in the equivalent CE amplifier because the emitter resistor is not bypassed.




Summary

CC amplifier

R1

R2 RE

C1

VCC = +15 V

22 k

27 k1.0 k

10 F

2N3904

Calculate re and Rin(tot) for the CC amplifier. Use = 200.

2B CC

1 2

27 k15 V = 8.26 V

22 k + 27 k

RV V

R R

E B 0.7 V = 7.76 V V V

E

25 mV 25 mV

7.76 mAer I

EE

E

7.76 V7.76 mA

1.0 k

VI

R

Rin(tot) = R1||R2||ac(re+ RE)

= 22 k||27 k|| 200 (1.0 k) = 9.15 k

Because re is small compared to RE, it has almost no affect on Rin(tot).

3.2




Summary

Class B amplifiers

R1

R2

RL

C1

VCC

The class B amplifier is more efficient than the class A amplifier. It is widely used in power amplifiers.

C2

Q1

Q2

D1

D2

The bias method shown avoids cross-over distortion by bringing the transistors just above cutoff using diodes.

The amplifier shown uses complementary transistors – one is an npn and the other is a pnp.

C3




Summary

The BJT as a switch

BJTs are used in switching applications when it is necessary to provide current drive to a load.

In cutoff, the input voltage is too small to forward-bias the transistor. The output (collector) voltage will be equal to VCC.

In switching applications, the transistor is either in cutoff or in saturation. RC

VCC VCC

RC

VOUTIIN = 0 = VCC

IIN > IC(sat)/DC

= 0 V

When IIN is sufficient to saturate the transistor, the transistor acts like a closed switch. The output is near 0 V.




Summary

The FET

The field-effect transistor (FET) is a voltage controlled device where gate voltage controls drain current. There are two types of FETs – the JFET and the MOSFET.

JFETs have a conductive channel with a source and drain connection on the ends. Channel current is controlled by the gate voltage.

p

n

S (Source)

G (Gate)

D (Drain)

S

G

D

n

n

p p

The gate is always operated with reverse bias on the pn junction formed between the gate and the channel. As the reverse bias is increased, the channel current decreases. n-channel JFET p-channel JFET




Summary

The FET

The MOSFET differs from the JFET in that it has an insulated gate instead of a pn junction between the gate and channel.Like JFETs, MOSFETs have a conductive channel with the source and drain connections on it.

p

S (Source)

G (Gate)

D (Drain)

S

G

D

n

n

p

p-channel MOSFET

n-channel MOSFET

ChannelSubstrate

Channel current is controlled by the gate voltage. The required gate voltage depends on the type of MOSFET.




Summary

The FET

In addition to the channel designation, MOSFETs are subdivided into two types – depletion mode (D-mode) or enhancement mode (E-mode).The D-MOSFET has a physical channel which can be enhanced or depleted with bias. For this reason, the D-MOSFET can be operated with either negative bias (D-mode) or positive bias (E-mode).

The E-mode MOSFET has no physical channel. It can only be operated with positive bias (E-mode). Positive bias induces a channel and enables conduction as shown here with a p-channel device.

p

S

G

D

nn

p

E-MOSFET with biasE-MOSFET

n

D

G

S

induced channel

n




Summary

JFET biasing

JFETs are depletion mode devices – they must be operated such that the gate-source junction is reverse biased.The simplest way to bias a JFET is to use a small resistor is in series with the source and a high value resistor from the gate to ground. The voltage drop across the source resistor essentially reverse biases the gate-source junction.

n-channel p-channel

RG

+VDD VDD

RG RSRS

RD RD

VG = 0 V VG = 0 V

+VS VS

Because of the reverse-biased junction, there is almost no current in RG. Thus, VG = 0 V.




Summary

D-MOSFET biasing

D-MOSFETs can be operated in either depletion mode or in enhancement-mode. For this reason, they can be biased with various bias circuits.

The simplest bias method for a D-MOSFET is called zero bias. In this method, the source is connected directly to ground and the gate is connected to ground through a high value resistor.

n-channel D-MOSFET with zero bias

RG

+VDD

RD

VG = 0 V

Only n-channel D-MOSFETs are available, so this is the only type shown.




Summary

E-MOSFET biasing

E-MOSFETs can use bias circuits similar to BJTs but larger value resistors are normally selected because of the very high input resistance.

The bias voltage is normally set to make the gate more positive than the source by an amount exceeding VGS(th).

Drain-feedback bias Voltage-divider bias

RG

+VDD

RD

RD

+VDD

R1

R2




Summary

FET amplifiers

FET amplifiers are voltage controlled and generally do not have as much gain or linearity as BJT amplifiers. The major advantage of FETs is high input resistance.The input resistance of a FET amplifier depends on the bias resistors. For the CS amplifier shown, Rin(tot) = RG because the gate-source resistance is a reverse biased pn junction. RG is made higher than the bias resistors in a BJT amplifier because of the negligible input current to the FET. Common-source amplifier

RG

RD

+VDD

RS

C1

C2

C3




Summary

Transconductance

An important parameter for FETs is the transconductance. Recall that conductance is the reciprocal of resistance, so from Ohm’s law:

The prefix “trans” is added to indicate the current and voltage are not measured in the same circuit (gate and drain).

dm

gs

Ig

V

For the common source amplifier, the drain current multiplied by the drain resistor is the output voltage. The voltage gain (ratio of output voltage to input voltage) can then be developed as

out d dv m d

in gs

V I RA g R

V V




Summary

FET amplifiers

If voltage gain is no required, the common-drain amplifier is a simple high input resistance amplifier.

Although the voltage gain is less than 1, the power gain is high because of the high input resistance. The circuit shown has the advantage of only two resistors (RS represents the load). The output is smaller and in phase with the input.

Common-drain amplifier

RGRS

+VDD

C




Summary

Feedback oscillators

An oscillator is a circuit that generates a repetitive waveform on its output. A feedback oscillator uses positive feedback from the output to sustain oscillations.

Av

Conditions for oscillations are

1. The phase shift around the loop must be 0o.

Feedback circuit

2. The closed loop gain must be 1 (unity gain).

B

Acl = AvB = 1




Summary

Feedback oscillators

The Colpitts and Hartley oscillators are examples of a feedback oscillator. The amplifier sections are nearly identical, but the LC feedback network is different.To obtain the required signal reinforcement (positive feedback), the amplifier inverts the signal (180o) and the feedback network shifts the phase another 180o.

Colpitts oscillator feedback network.Hartley oscillator feedback network

An additional capacitor is in the amplifier to block dc.




Summary

Troubleshooting

Assume a troubleshooter measures dc parameters for a circuit that has no signal output. For the amplifier shown, which of these dc readings would indicate a problem? If wrong, what problem is indicated?

RG

RD

+VDD= 12 V

RS

C1

C2

C3

3.9 k

560 10 M 100 F

0.1 F

a) VS = 1.5 V

c) VG = 0 V

b) VD = +12 V

Reading is expected.

Reading is too high; there may be a drain-source short.

Reading = VDD, indicating no drain current.




Bipolar junction transistor (BJT)

Class A amplifier

Saturation

Selected Key Terms

An amplifier that conducts for the entire input cycle and produces an output signal that is a replica of the input signal in terms of its waveshape.

A transistor with three doped semiconductor regions separated by two pn junctions.

The state of a transistor in which the output current is maximum and further increases of the input variable have no effect on the output.




Cutoff

Q-point

Amplification

Common-emitter (CE)

Class B amplifier

The dc operating (bias) point of an amplifier.

Selected Key Terms

A BJT amplifier configuration in which the emitter is the common terminal.

The non-conducting state of a transistor.

An amplifier that conducts for half the input cycle.

The process of producing a larger voltage, current or power using a smaller input signal as a pattern.




Junction field-effect transistor

(JFET)

MOSFET

Depletion mode

Enhancement mode

Metal-oxide semiconductor field-effect transistor.

A type of FET that operates with a reverse-biased junction to control current in a channel.

Selected Key Terms

The condition in a FET when the channel is depleted of majority carriers.

The condition in a FET when the channel has an abundance of majority carriers.




Common-source

Oscillator

Feedback

A circuit that produces a repetitive waveform on its output with only a dc supply voltage as an input.

An FET amplifier configuration in which the source is the common terminal.

Selected Key Terms

The process of returning a portion of a circuit’s output signal to the input in such a way as to create certain specified operating conditions.




Quiz

1. The Thevenin circuit shown has a load line that crosses the y-axis at

a. +10 V.

b. +5 V.

c. 2 mA.

d. the origin.

+10 V

5.0 k




Quiz

2. In a common-emitter amplifier, the output ac signal will normally

a. have greater voltage than the input.

b. have greater power than the input.

c. be inverted.

d. all of the above.




Quiz

3. In a common-collector amplifier, the output ac signal will normally

a. have greater voltage than the input.

b. have greater power than the input.

c. be inverted.

d. have all of the above.




Quiz

4. The type of amplifier shown is a

a. common-collector.

b. common-emitter.

c. common-drain.

d. none of the above.R1

R2RE

C1

VCC




Quiz

5. A major advantage of FET amplifiers over BJT amplifiers is that generally they have

a. higher gain.

b. greater linearity.

c. higher input resistance.





Quiz

6. A type of field effect transistor that can operate in either depletion or enhancement mode is an

a. D-MOSFET.

b. E-MOSFET.

c. JFET.

d. none of the above.




Quiz

7. For an FET, transconductance is the ratio of

a. drain voltage to drain current.

b. gate-source voltage to drain current.

c. gate-source current to drain voltage.

d. drain current to gate-source voltage.




Quiz

8. A transistor circuit shown is a

a. D-MOSFET with voltage-divider bias.

b. E-MOSFET with voltage-divider bias.

c. D-MOSFETwith self-bias.

d. E-MOSFET with self bias.RD

+VDD

R1

R2




Quiz

9. A Colpitts or Hartley oscillator both have

a. positive feedback.

b. amplification.

c. a closed loop gain of 1.





Quiz

10. If you were troubleshooting the circuit shown here, you would expect the gate voltage to be

a. more positive than the drain voltage.

b. more positive than the source voltage.

c. equal to zero volts.

d. equal to +VDD

RD

+VDD

R1

R2




Quiz

Answers:

1. c

2. d

3. b

4. a

5. c

6. a

7. d

8. b

9. d

10. b

chapter 17.ppt

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