electronics fundamentals

electronics fundamentals circuits, devices, and applications THOMAS L. FLOYD DAVID M. BUCHLA

Lesson 2: Transistors and Applications Electronics Fundamentals 8th edition Floyd/Buchla

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

Lesson 2 Introduction A transistor is a semiconductor device that controls current between two terminals based on the current or voltage at a third terminal. It is used for amplification or switching of electrical signals. The basic structure of the bipolar junction transistor, BJT, determines its operating characteristics. DC bias is important to the operation of transistors in terms of setting up proper currents and voltages in a transistor circuit. Two important parameters are αDC and βDC

Electronics Fundamentals 8th edition Floyd/Buchla


Lesson 2 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.

B (Base)

C (Collector)

n p n

Base-Collector junction B

Base-Emitter junction

E (Emitter) Electronics Fundamentals 8th edition Floyd/Buchla

Separating the regions are two junctions. C

p n p

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

Lesson 2 Bipolar junction transistors (BJTs)

FIGURE 17–2 Transistor symbols. Electronics Fundamentals 8th edition Floyd/Buchla


Lesson 2 BJT biasing

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

For the pnp npn transistor, this condition requires that the base is more negative positive than thanthe theemitter and the and emitter collector the collector is more is more positive than negative thanthe thebase. base.


BC reversebiased

+ pnp npn

+ +

+

BE forwardbiased


Lesson 2 BJT currents IE  IC  IB

FIGURE 17–4 Transistor currents. Electronics Fundamentals 8th edition Floyd/Buchla


Lesson 2 BJT currents

A small base current (IB) is able to control a larger collector current (IC). Some important current relationships for a BJT are: IC IE  IC  IB I C  αDC I E Where αDC (dc alpha) = IC/ IE

I C  βDC I B

IB

I

I

IE I

Where βDC (dc beta) = IC/ IB Electronics Fundamentals 8th edition Floyd/Buchla


Lesson 2 Voltage-divider bias Because the base current is small, the approximation  R2  is useful for calculating the base voltage. VB    VCC  R1  R2 

After calculating VB, you can find VE by subtracting 0.7 V for VBE. VE = VB - VBE Next, calculate IE by applying Ohm’s law to RE: V IE  E RE Then apply the approximation I C  I E

R1

RC VB

R2

VC VE RE

Finally, you can find the collector voltage from VC  VCC  I C RC VCE  VC  VE Electronics Fundamentals 8th edition Floyd/Buchla


Lesson 2 Voltage-divider bias

Calculate VB, VE, and VC for the circuit.  R2  6.8 k   VB   V   CC  15 V = 3.02 V  27 k + 6.8 k   R1  R2 

VE = VB  0.7 V = 2.32 V V 2.32 V IE  E   2.32 mA RE 1.0 k

I C  I E  2.32 mA

+15 V

R1

RC

27 k

2.2 k 2N3904

R2

RE

6.8 k

1.0 k

VC  VCC  I C RC  15 V   2.32 mA  2.2 k   9.90 V Electronics Fundamentals 8th edition Floyd/Buchla


Lesson 2 Voltage-divider bias Determine VB, VE, VC, VCE, IB, IE, and IC in the given Figure, The 2N3904 is a general-purpose transistor with a typical βDC = 200.



Lesson 2 Collector characteristic curves The collector characteristic curves are a family of curves that show how collector current varies with the collector-emitter IC voltage for a given IB. IB6 The curves are divided into three regions: The breakdown active region region is after saturation region the is after saturation the active region. region occurs when the baseThis and isisisthe characterized region for by emitter and the baseoperation rapid increase of class-A in collector collector junctions are operation. current. Operation in this both forward biased. region may destroy the transistor.


IB5 IB4 IB3 IB2 IB1 IB = 0 0

VCE


Lesson 2 Collector characteristic curves



Lesson 2 Collector characteristic curves Draw the family of collector characteristics curves for the circuit in the given figure for IB = 5 µA to 25 µA in 5 µA increments. Assume βDC = 100



Lesson 2 Collector characteristic curves Draw the family of collector characteristics curves for the circuit in the given figure for IB = 5 µA to 25 µA in 5 µA increments. Assume βDC = 100

IB

IC

5 µA

0.5 mA

10 µA

1.0 mA

15 µA

1.5 mA

20 µA

2.0 mA

25 µA

2.5 mA


I C   DC I B I C  100  5A  0.5mA


Lesson 2 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. 2.0 k +12 V


INL = 6.0 0 mA mA SL = VNL = 012VV SL =

I (mA)

6

Load line 4 2 0

0

4

8

12

V (V)


Lesson 2 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. I (mA)

Read the load current and load voltage from the graph if a 3.0 k resistor is the load. 2.0 k +12 V

IV curve for 3.0 k resistor

6 4

RL =

Q-point

2

3.0 k 0

0

4

8

VL = 7.2 V Electronics Fundamentals 8th edition Floyd/Buchla

12

V (V)

IL = 2.4 mA


Lesson 2 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 I (mA) curve and the base current establish the Q-point. 6

2.0 k 4

+12 V 2 0


0

4

8

12

V (V)


Lesson 2 Load lines Load lines can illustrate the operating conditions for a transistor circuit. Assume the IV curves are as shown: 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. 2.0 k +12 V

I (mA)

6 4 2 0


For this base current, the Q-point is:

0

4

8

12

V (V)

The load voltage (VCE) and current (IC) can be read from the graph. © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.

Lesson 2 Load lines For the transistor, assume the base current is established at 10 A by the bias circuit. Show the Qpoint and read the value of VCE and IC. The Q-point is the intersection of the load line with the 10 A base current. IB = 25 A

IC (mA) 2.0 k 6

+12 V

IB = 20 A IB = 15 A

4

IB = 10 A

2

Bias circuit

0

IB = 5.0 A 0

4

8

12

VCE (V)

VCE = 7.0 V; IC = 2.4 mA Electronics Fundamentals 8th edition Floyd/Buchla


Lesson 2 Signal (ac) 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. This increase is called amplification The ratio of the ac collector current (Ic) to the ac base current (Ib) is designated by βac (the ac beta) of hfe

Ic  ac  Ib FIGURE 17–13 An amplifier with voltage-divider bias with capacitively coupled input signal. Vin and Vout are with respect to ground. Electronics Fundamentals 8th edition Floyd/Buchla


Lesson 2 Signal (ac) 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. 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?

IC (mA) 6 4 2

IB = 25 A IB = 20 A IB = 15 A IB = 10 A

IB = 5.0 A

The operation along 0 VCE (V) 0 4 8 12 the load line is shown in red. Reading the collector current, IC varies from 1.2 mA to 3.8 mA. Electronics Fundamentals 8th edition Floyd/Buchla


Lesson 2 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. VC C

R1

RC

Output coupling capacitor

C2

C1

Input coupling capacitor


R2

RE

C3 Bypass capacitor


Lesson 2

Summary

CE amplifier The bypass capacitor increases voltage gain. It shorts the signal around the emitter resistor, RE, in order to increase the voltage gain. To understand why let us consider the amplifier without the bypass capacitor as explained the preceding equations. VC C

R1

RC

Output coupling capacitor

C2

C1

Input coupling capacitor


R2

RE

C3 Bypass capacitor


Lesson 2 Formulas Lowercase italic subscript indicate signal (ac) voltages and signal (alternating currents)

Av (Voltage gain) = Vout / Vin Vout = IcRC The signal voltage at the base is approximately equal to

Vb ≡ Vin ≡ Ie (re + RE) where re is the internal emitter resistance of the transistor.



Lesson 2 Formulas without the bypass capacitor

Av now can be expressed as Av = Vout / Vin = IcRC / Ie (re + RE) Since Ic ≡ Ie, the currents cancel and the gain is the ratio of the resistance.

Av = RC / (re + RE) If RE is much greater than re, then

Av ≡ RC / RE



Lesson 2 Formulas with the bypass capacitor If the bypass capacitor is connected across RE, it effectively shorts the signal to ground leaving only re in the emitter. Thus the voltage gain of the CE amplifier with the bypass capacitor shorting RE is:

Av = RC / re The transistor parameter re is important because it determines the voltage gain of a CE amplifier in conjunction with RC. A formula for estimating re is given without derivation in the following equation:

re≡ 25mV / IE Electronics Fundamentals 8th edition Floyd/Buchla


Lesson 2

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. re 

VCC = +15 V

25 mV 25 mV   10.8  IE 2.32 mA

V R Av  out  C Vin re

2.2 k   204 10.8 

RC

R1 C1

27 k

2.2 k 2N3904

2.2 F

R2 6.8 k

C2

1.0 F

RE 1.0 k

C3

100 F

Sometimes the gain will be shown with a negative sign to indicate phase inversion. Electronics Fundamentals 8th edition Floyd/Buchla


Lesson 2 Phase Inversion The output voltage at the collector is 180 degrees out of phase with the input voltage at the base. Therefore, the CE amplifier is characterized by a phase inversion between the input and output. The inversion is sometimes indicated by a negative voltage gain.

AC Input Resistance

Rin ≡ βac Ib re / Ib The Ib terms cancel, leaving

Rin = Vb / Ib

Rin ≡ βac re

Total Input Resistance of a CE amplifier:

Vb = Iere Ie ≡ βac Ib Electronics Fundamentals 8th edition Floyd/Buchla

Rin(tot) = R1║R2║Rin Rin(tot) = R1║R2║βac re RC has no effect because of the reverse-biased, base-collector junction. © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.

Lesson 2 Current Gain: The signal current gain of a CE amplifier is

Ai= Ic / Is Where Is is the source current and is calculated by Vin /

Rin(tot)

Power Gain: The power gain of a CE amplifier is the product of the voltage gain and the current gain.

Ap≡ Av Ai



Lesson 2 Decibel (dB) Measurement The decibel (dB) is a logarithmic measurement of the ratio of one voltage to another or one power to another, which can be used to express the input-to-output relationship .

VOLTAGE RATIO: dB = 20 log (Vout / Vin)

POWER RATIO: dB = 10 log (Pout / Pin)



Lesson 2 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 IC(sat)/DC


Lesson 2 The BJT as a switch

FIGURE 17–30 Ideal switching action of a transistor.



Lesson 2 The BJT as a switch

FIGURE 17–30 Ideal switching action of a transistor.

Conditions in Cutoff A transistor is in cutoff when the base-emitter junction is not forward biased.

VCE( cutoff )  VCC Conditions in Saturation When the base-emitter junction is forward-biased and there is enough base current to produce a maximum collector current, the transistor is saturated. The minimum value of base current to produce saturation is:

I B (min) 

I C ( sat)

 DC


VBE )  0.7V

VRB  VIN  0.7V

I C ( sat)  RB (max) 

VCC RC VRB I B (min)


Lesson 2 The BJT as a switch (a) For the transistor switching circuit in the given figure, what is VCE when VIN = 0 V? (b) What minimum value of IB will saturate this transistor if the βDC is 200? (c) Calculate the maximum value of RB when VIN = 5 V.



Lesson 2 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.

G (Gate)

p

n p

p

The gate is always operated with reverse bias on the pn junction formed between the gate and the channel. As S (Source) the reverse bias is increased, the n-channel JFET channel current decreases. Electronics Fundamentals 8th edition Floyd/Buchla

D

D (Drain)

G

n n

S

p-channel JFET


Lesson 2 The FET

FIGURE 17–33 Effects of VGG on channel width and drain current (VGG = VGS).



Lesson 2 The FET

FIGURE 17–34 JFET schematic symbols.



Lesson 2 The FET: MOSFET The MOSFET (Metal Oxide Semiconductor FET) 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. D (Drain)

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


p

n G (Gate)

Channel

D

p

G

n

Substrate S (Source)

S

n-channel

p-channel

MOSFET

MOSFET


Lesson 2 The FET: MOSFET

FIGURE 17–35 Representation of the basic structure of D-MOSFETs. Electronics Fundamentals 8th edition Floyd/Buchla


Lesson 2 The FET: MOSFET 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). FIGURE 17–36 Operation of n-channel D-MOSFET.




FIGURE 17–37 D-MOSFET schematic symbols. Electronics Fundamentals 8th edition Floyd/Buchla


Lesson 2 The FET: MOSFET 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.

G

D

D

n

n

p

G

p

n

n

S

S

induced channel E-MOSFET E-MOSFET with bias FIGURE 17–38 E-MOSFET construction and operation. Electronics Fundamentals 8th edition Floyd/Buchla



FIGURE 17–39 E-MOSFET schematic symbols. Electronics Fundamentals 8th edition Floyd/Buchla


Lesson 2 JFET biasing JFETs are depletion mode devices – they must be operated such that the gate-source junction is reverse biased. +VDD VDD 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. VG = 0 V The voltage drop across the source resistor essentially reverse biases RG the gate-source junction. Because of the reverse-biased junction, there is almost no current in RG. Thus, VG = 0 V. Electronics Fundamentals 8th edition Floyd/Buchla

RD

RD VG = 0 V

V S

+VS RS

n-channel

RG

RS

p-channel


Lesson 2 JFET biasing For the n-channel JFET, the gate-tosource voltage is VGS  VG  VS  0V  I D RS VGS   I D RS

For the p-channel JFET, the gate-tosource voltage is VGS   I D RS The drain voltage with respect to ground is V  V  I R D

DD

D

D

Since VS=IDRS, the drain-to-source voltage is V  V  I R  R  DS

DD


D

D

FIGURE 17–40 Self-biased JFETs (IS = ID in all FETs).

S


Lesson 2 JFET biasing Find VDS and VGS in the JFET circuit below, given that ID = 5mA



Lesson 2 D-MOSFET biasing D-MOSFETs can be operated in either depletion mode or in enhancement-mode. For this reason, they can be +VDD biased with various bias circuits. The simplest bias method for a DMOSFET 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.

RD VG = 0 V RG

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

n-channel D-MOSFET with zero bias



Lesson 2 D-MOSFET biasing

FIGURE 17–42 A zero-biased D-MOSFET.

Recall that depletion/enhancement MOSFETs can be operated with either positive or negative values of VGS. A simple bias method, called zero bias, is to set VGS = 0 V so that an ac signal at the gate varies the gate-to-source voltage above and below this bias point. Since VGS = 0 V, ID = IDSS as indicated. IDSS is defined as the drain current when VGS = 0 V.

The drain-to-source voltage is expressed as V  V  I R DS

DD

DSS

D



Lesson 2 D-MOSFET biasing Determine the drain-to-source voltage in the circuit of the given figure. The MOSFET data sheet give VGS(off) = -8 V and IDSS = 12 mA



Lesson 2 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. +VDD +V

DD

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

RD R1

RG

RD

R2

Drain-feedback bias Electronics Fundamentals 8th edition Floyd/Buchla

Voltage-divider bias


Lesson 2 E-MOSFET biasing FIGURE 17–44 E-MOSFET biasing arrangements.

Recall that enhancement-only MOSFETs must have a VGS greater than the threshold value VGS(th) . In either bias arrangement, the purpose is to make the gate voltage more positive than the source by an amount exceeding VGS(th) . In the drain-feedback circuit, there is negligible gate current and, therefore, no voltage drop across RG . As a result, VGS = VDS.  R2  Equation for the voltage-divider bias is VDD VGS   given by R  R 2   1 Electronics Fundamentals 8th edition Floyd/Buchla

VDS  VDD  I D RD


Lesson 2 E-MOSFET biasing Determine the amount of drain current in the given figure. The MOSFET has a VGS(th) of 3 V.



Lesson 2

Selected Key Terms

Bipolar A transistor with three doped semiconductor junction regions separated by two pn junctions. transistor (BJT) An amplifier that conducts for the entire input cycle and produces an output signal that is a Class A replica of the input signal in terms of its amplifier waveshape. The state of a transistor in which the output current is maximum and further increases of Saturation the input variable have no effect on the output.



Lesson 2

Selected Key Terms

Cutoff The non-conducting state of a transistor. Q-point The dc operating (bias) point of an amplifier. Amplification The process of producing a larger voltage, current or power using a smaller input signal as a pattern.

Common- A BJT amplifier configuration in which the emitter (CE) emitter is the common terminal. Class B An amplifier that conducts for half the input amplifier cycle. Electronics Fundamentals 8th edition Floyd/Buchla


Lesson 2

Selected Key Terms

Junction field- A type of FET that operates with a reverseeffect transistor biased junction to control current in a channel. (JFET) MOSFET Metal-oxide semiconductor field-effect transistor.

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

Enhancement The condition in a FET when the channel has mode an abundance of majority carriers. Electronics Fundamentals 8th edition Floyd/Buchla


Lesson 2

Quiz

1. The Thevenin circuit shown has a load line that crosses the y-axis at a. +10 V. b. +5 V.

5.0 k +10 V

c. 2 mA. d. the origin.



Lesson 2

Quiz

1. The Thevenin circuit shown has a load line that crosses the y-axis at a. +10 V. b. +5 V.

5.0 k +10 V

c. 2 mA. d. the origin.



Lesson 2

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.



Lesson 2

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.



Lesson 2

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.



Lesson 2

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.



Lesson 2

Quiz

4. The type of amplifier shown is a a. common-collector. b. common-emitter.

VC C

c. common-drain. d. none of the above.

C1

R1

R2


RE


Lesson 2

Quiz

4. The type of amplifier shown is a a. common-collector. b. common-emitter.

VC C

c. common-drain. d. none of the above.

C1

R1

R2


RE


Lesson 2

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. d. all of the above.



Lesson 2

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. d. all of the above.



Lesson 2

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.



Lesson 2

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.



Lesson 2

Quiz

8. A transistor circuit shown is a a. D-MOSFET with voltage-divider bias. +VDD

b. E-MOSFET with voltage-divider bias. c. D-MOSFETwith self-bias. d. E-MOSFET with self bias.

R1

RD

R2



Lesson 2

Quiz

8. A transistor circuit shown is a a. D-MOSFET with voltage-divider bias. b. E-MOSFET with voltage-divider bias.

+VDD

c. D-MOSFETwith self-bias. d. E-MOSFET with self bias.

R1

RD

R2



Lesson 2

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.

+VDD

b. more positive than the source voltage. c. equal to zero volts.

R1

RD

d. equal to +VDD R2



Lesson 2

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.

+VDD

b. more positive than the source voltage. c. equal to zero volts.

R1

RD

d. equal to +VDD R2

