AC Load Lines

ELL 100 - Introduction to Electrical Engineering LECTURE 24: BIPOLAR JUNCTION TRANSISTOR (BJT) Outline

 Static (dc) characteristics of common-base circuits

 Static (dc) characteristics of common-emitter circuits

 Transistor as an amplifier: DC and AC Load lines

2 COMPUTERS

. Transistors present everywhere: CPU, memory, I/O units, display, mouse, keyboard, speaker, wi-fi, etc. DISPLAYS (TFT-LCD)

. Controlling each pixel & color dot of the flat-panel display . Adjusting display back-light brightness STORAGE DEVICES : SDD

. Every bit of (non-magnetic) memory storage: Flash, D-RAM, (E)P-ROM, volatile static RAM. . Every bit of cpu register data & addresses & every bit of math or logic or encryption/decryption. STORAGE DEVICES : PEN DRIVE MOBILE PHONES

Transistors are part of : . Modulating, transmitting, receiving, de-modulating . WiFi/Bluetooth/cellular WIFI MODULE BLUETOOTH CIRCUIT STATIC (DC) CHARACTERISTICS OF COMMON-BASE CIRCUITS

npn

. Vary input current IE using the potentiometer R1.

. For each value of IE, vary output voltage VCB using the

potentiometer R2 and note the corresponding output current IC

9 STATIC (DC) CHARACTERISTICS OF COMMON-BASE CIRCUITS

IE direction is taken to be flowing into the emitter terminal of BJT

10 STATIC (DC) CHARACTERISTICS OF COMMON-BASE CIRCUITS

IC   at a constant VCB IE

current amplification factor for a common-base circuit

0.95 < α < 0.995

11 STATIC (DC) CHARACTERISTICS OF COMMON-EMITTER CIRCUITS

npn

. Vary input current IB using the input potentiometer.

. For each value of IB, vary output voltage VCE using the output

potentiometer and note the corresponding output current IC 12 STATIC (DC) CHARACTERISTICS OF COMMON-EMITTER CIRCUITS

13 STATIC (DC) CHARACTERISTICS OF COMMON-EMITTER CIRCUITS

IC   at a constant VCE IB    1

current amplification factor for a common-emitter circuit 20 < β < 200

14 EFFECT OF REVERSE-BIASED LEAKAGE CURRENT

IIIECB

If the emitter is not connected, IE = 0. However, we find there is still a small collector current ICBO – this is a leakage current crossing the reverse-biased collector–base junction.

IIICECBO ICBO IICB    IIIBCCBO  1 

IIICBCBO1 IIB  (  1) CBO  I  II IICBO B CEO CB11   BJT MODES OF OPERATION • The two p-n junctions of the BJT can be independently biased, to result in four possible transistor operating modes as summarized in Table • A junction is forward-biased if the n material is at a lower potential than the p material, and reverse-biased if the n material is at a higher potential than the p material. BJT MODES OF OPERATION

• Saturation denotes operation (with VCE ~ 0.2 V and VBC ~ 0.5 V for Si devices) such that maximum collector current flows and the transistor acts much like a closed switch (i.e. short- circuit) from collector to emitter terminals.

• Cutoff denotes operation with near zero current, where the transistor acts much like an open switch. Only leakage current (of reverse-biased diode) flows in this mode of operation. Thus,

IC = ICEO  0 for Common-Base connection, and IC = ICBO  0 for Common-Emitter connection.

• Active or linear mode describes transistor operation in the region to the right of saturation

(VCE > 1 V) and above cutoff (IC >> ICEO). Here, near-linear relationships exist between the output and input terminal currents, thus used in amplifier circuits.

• Inverse mode is a little-used, inefficient active mode with the emitter and collector interchanged. TRANSISTOR AS AN AMPLIFIER

(DC supply)

(Small (Amplified a.c. signal) a.c. signal)

The transistor is required to operate in a unidirectional (active/linear) mode, otherwise the negative parts of the a.c. would cause, say, the emitter–base junction to be reverse biased and this would prevent normal transistor action occurring. As a result, it is necessary to introduce a d.c. bias. 18 TRANSISTOR AS AMPLIFIER: BIASING & LOAD LINE

IE IC

AC input source voltage

An npn transistor amplifier in common-base configuration with load resistance R

19 LOAD LINE FOR BJT IN IE IC COMMON-BASE CONFIG 1 V 6 V 100 Ω “Load-line” 1 kΩ IC = (B2 ‒VCB)/R = 6 ‒VCB

With AC source S off, bias point D is obtained

IE = -(B1 ‒VBE)/RS = -(1 – 0.7)/0.1 = -3 mA

With AC source S on,

input current IE varies between -1 & -5 mA (operating points E and C) 20 LOAD LINE FOR TRANSISTOR AMPLIFIER

. The function of capacitor C is to eliminate the d.c. component of the voltage across R from the output voltage.

. If Ic is the RMS value of the a.c. component of the collector current,

IcR gives the RMS value of the a.c. component of the output voltage and 2 Ic R gives the average output a.c. power.

. Note that a lower-case subscript (Ic) indicates an RMS a.c. value compared with

a capital subscript (IC) for the d.c. value.

21 TRANSISTOR AS AMPLIFIER: BIASING & LOAD LINE

C B

IB E

Simple common-emitter transistor amplifier 22 LOAD LINE FOR BJT IN COMMON-EMITTER CONFIG IC “Load-line” IB IC = (VS ‒VCE)/RC

Bias point Q is obtained

IB = (VS ‒VBE)/RB ~ VS/RB

With input AC source on, input base current varies as

IItBbm sin (operating points A and B) 23 TRANSISTOR AMPLIFIERS: RELEVANT QUANTITIES

III Current gain G oCACB i II2 ibm V o ΔI Voltage gain Gv  o Vi

ΔIi V P o o Vi Power gain GP  Pi

where Po is the output a.c. signal power in the load GGPvi G and Pi is the input a.c. signal power into the transistor. DYNAMIC CHARACTERISTICS & LINEARITY

Larger the value of load resistance Dynamic characteristics R, the greater the voltage and power gains, the maximum value of R for a given collector supply voltage being limited by the maximum permissible distortion of the output voltage (due to deviation from linearity of transistor characteristics).

25 DC BIAS (Q-POINT) OF TRANSISTOR AMPLIFIER • Supply voltages and resistors bias a transistor i.e. they establish a specific set of dc terminal voltages and currents, thus determining a point of active-mode operation (called the quiescent point or Q point). • Usually, the quiescent point is unperturbed by the application of a small a.c. signal to the circuit.

• e.g. Here D is the Q-point

26 BJT BIASING IN COMMON-EMITTER CONFIG

II  1 RR12 EQBQ   RB  RR12 R IEQ 1 VRVIR VVBBCC BBBBEQEQE RR12 1

27 BJT BIASING IN COMMON-EMITTER CONFIG

I VRVIREQ BBBBEQEQE1

VVBBBEQ IICQEQ RRBE/1  (α ~ 1)

If component values are such that RB/(β+1) << RE, then ICQ is nearly constant, regardless of changes in β (which can change due to temperature, voltage & current fluctuations) 28 BJT IN COMMON-EMITTER: DC LOAD LINE

Bias/quiescent point:

Q = (VCEQ, ICQ)

DC load line: . IBQ VVCCCE iC  Rdc iI BBQRdc = RC + RE

29 BJT IN COMMON-EMITTER: AC LOAD LINE

1. Coupling capacitors (CC) confine dc quantities to the transistor and its bias circuitry.

2. Bypass capacitor (CE) effectively removes the gain-reducing emitter resistor

RE for ac signals, while allowing RE to play its role in establishing -independent dc bias.

AC equivalent circuit

30 BJT IN COMMON-EMITTER: AC LOAD LINE The effective resistance seen by the d.c. bias collector current ICQ is Rdc = RC + RE.

The a.c. collector signal current ic sees a collector-circuit resistance

Rac = RCRL/(RC + RL).

Since Rac  Rdc in general, the concept of an a.c. load line arises. 31 BJT IN COMMON-EMITTER: AC LOAD LINE

viRcecac

Since ic = iC − ICQ and vce = VCEQ − vCE, the above equation can be written as

VvCEQCE iICCQ AC load line equation Rac

32 BJT IN COMMON-EMITTER: AC LOAD LINE

Bias/quiescent point: Q = (VCEQ, ICQ)

VVCCCE DC load line: IC  Rdc . IBQ Rdc = RC + RE

VvCEQ CE AC load line: iICCQ Rac

Rac = RC || RL 33 NUMERICAL

Q-1: In the given circuit, the silicon transistor has  = 75 and the collector voltage is VC = 9V.

Find the ratio RB / RC NUMERICAL

Solution: Given VC = 9 V,  = 75 IE Silicon transistor => VBE = 0.7 V 159 90.7  IE  IB RC RB IC

I IR6 6 RR B EB   1 BB   105.13 IRBC8.3 8.3 RRCC NUMERICAL Q-2: In the circuit shown, the transistor has β = 30.

If the input voltage Vi = +5 V, show that the transistor is operating in the active mode. NUMERICAL

Solution: In active mode, VBE = 0.7 V (base-emitter junction forward-biased)

=> IB = [(Vi – VBE)/15k] − [(VBE + 12)/100k]

 IB = [(5 – 0.7)/15] − [(0.7 + 12)/100] mA = 0.16 mA

 IC = βIB = 30×0.16 = 4.8 mA

 VCE = 12 – (2.2k)IC = 12 – (2.2×4.8) = 1.45 V

 VCB = VCE – VBE = 1.45 – 0.7 = 0.75 V  Thus the collector-base junction is reverse-biased NUMERICAL Q-3: Two perfectly matched silicon transistors are connected as shown below. Assuming the  of the transistors to be very high and the forward voltage drop in the diode to be 0.7 V, what is the value of the current I? NUMERICAL

00.70.75   ImA3.6 R 1k

Applying KCL at P, I 2 III2 I 2 C  I RCB C   C

Since  is very large, IImARC3.6 NUMERICAL Q-4: The transistor in the given circuit should be biased in the active region.

Take VCE (sat) = 0.2 V; VBE = 0.7 V.

What is the maximum value of RC that can be used? NUMERICAL

Solution: Applying KVL at the input, 4.3 50.72 kIImA 2.15 B B 2k Collector current,

IImACB  215 Applying KVL to the output loop,

50.215RVCCE

For transistor operation to be in active region, VVCE  0.2 4.8 0.215R  5  0.2 R = 22.32  C C 0.215 NUMERICAL

Q-5: In the given circuit,  = 100,

IBQ = 20 μA, VCC = 15 V, RC = 3 kΩ.

Find (a) IEQ and (b) VCEQ.

Solution: (a) IEQ = (β + 1)IBQ = (101)×(0.02) mA = 2.02 mA

(b) IEQ = βIBQ = 2 mA KVL around output loop,

VVICBQCCCQCEQ  RmA C kV 15 239 

VCBQCEQ  V CC  I CQ R C 15  2 mA  6 k  3 V NUMERICAL

Q-6: In the pnp Si transistor circuit shown,

RB = 500 k, RC = 2k, RE = 0, VCC = 15 V,

ICBO = 20 μA, and  = 70.

Find the Q-point collector current ICQ. NUMERICAL

Solution: IBQ = (-VEB – (-VCC))/RB = (-0.7 + 15)/500 mA = 0.0286 mA

ICQ = βIBQ + (β + 1)ICBO = (70×0.0286) + (71×0.02) mA = 3.42 mA

RB = 500 k, RC = 2k, RE = 0, VCC = 15 V, ICBO = 20 μA, and  = 70 NUMERICAL Q-7: The pnp Si transistor in the circuit has  = 0.99.

Also, VEE = 4 V and VCC = 12 V.

(a) If IEQ = 1.1mA, find RE.

(b) If VCEQ = -7 V, find RC. NUMERICAL Solution: (a) By KVL around the emitter-base loop,

VVEEBEQ 40.7  RE  3 ImAEQ 1.1

(b) VVVVCBQCBQBEQCEQ    70.76.3 

IImmACQBQ  0.99 1.11.089

By KVL around the collector-base loop,

VVCC CBQ RkC  5.234  ICQ NUMERICAL Q-8: The Si transistors in the differential amplifier circuit are identical with  = 60. Also,

RC = 6.8 k, RB = 10 k, and

VCC = VEE = 15 V.

Find the value of RE needed to bias the amplifier such that

VCEQ1 = VCEQ2 = 8V. NUMERICAL

Solution: IIEQEQ12

iIIIEEQEQEQ 121 2...... (1)

VIRVIREEBQBBEQEQE

 II CQEQ 1

1 VVIRVIRCC EE  EQ1 C C EQ 1  2 EQ 1 C .....(2) 1 1 NUMERICAL

111 1VVVCCBEQCEQ  60 1 15 8 0.7 ImAEQ1  1.18 1RRkkCB 60 6.810

1 10k VIRVEE EQ11 B BEQ 15 0.7 1.18m 1 1 60 1 RkE   5.97  2ImEQ1 2 1.15 NUMERICAL

Q-9: A bipolar transistor amplifier stage is shown in Figure (Next Slide) and the transistor has characteristics, shown in Figure (Next Slide) , which may be considered linear over the working range. A 2.2 kΩ resistive load is connected across the output terminals and a signal source of sinusoidal e.m.f. 0.6 V peak and internal resistance 10 kΩ is connected to the input terminals. The input resistance of the transistor is effectively constant at 2.7 kΩ. Determine the current, voltage and power gains of the stage. The reactances of the coupling

capacitors C1 and C2 may be considered negligible. NUMERICAL

Bipolar transistor amplifier stage

Bipolar transistor characteristics

52 NUMERICAL

Solution: First determine the extremities of the d.c. load line.

If ic=0 then Vs = 12V = vce

Vs 12 If vce=0 then ic  3  6.7 mA Rc 1.8 10

For the a.c load line, v 1.8 2.2 Rk  1  P I 1.8 2.2 NUMERICAL

Hence the slope of the a.c. load line is 1000   1/mAV 1 and the quiescent base current

12 IA60 B 200 103 The a.c. load line is therefore drawn with a slope of −1.0 mA/V through the quiescent point Q , which is given by the Equivalent input circuit intersection of the 60 μA characteristic and the d.c. load line. NUMERICAL

The input circuit consists of the base–emitter junction in parallel with the bias resistor, i.e. the signal passes through both in parallel. However, the transistor input resistance is 2.7 kΩ hence the shunting effect of 200 kΩ is negligible, and effectively the entire input circuit can be represented by Fig. 21.22.

0.6 The peak signal base current is  47 A 102.710 3 and hence the maximum signal base current is 60 47 107 A NUMERICAL and the minimum signal base current is 604713 A From the a.c. load line in Fig. 21.21,

imAc 5.10.94.22 This change in collector current is shared between the 1.8 kΩ collector resistor and the load resistor of 2.2 kΩ. Hence the change in output current is 1.8 103 i 4.2   1.9 mA o 1.8 2.2 103 NUMERICAL

The change in input current is iAi 1071394 

Therefore the current gain for the amplifier stage is

io Gi 20 ii Note that this is the current gain of the amplifier and should not be confused with the current gain of the transistor, which is given by

3 ic 4.2 10 Gi  6  45 ib 94 10 NUMERICAL

The change in output voltage Δvo is given by the change in collector– emitter voltage, i.e. vvVocepkpk 8.54.34.2 

The change in input voltage Δvi is given by the change in base–emitter voltage, i.e. 63 viiiipk   RV pk  39  102.7 10 0.25 

v0 4.2 Gv  17 vi 0.25

GGGP v i 17  20  340 UNSOLVED PROBLEMS-1

Q1. The transistor circuit shown uses a transistor with VBE = 0.7 V, IC = ≈ IE and a dc current gain of 100. The value of V0 is

Ans: 4.65V UNSOLVED PROBLEMS-2

Q2. In the following circuit, the transistor is in active mode and Vc = 2V. To get Vc = 4V, we replace Rc with Rc’. Then the ratio of Rc’/Rc is

Ans: 0.75 UNSOLVED PROBLEMS-3

Q3. The common emitter amplifier is biased using a 1 mA ideal current source. The approximate base current value is (in μA):

Ans: 10 UNSOLVED PROBLEMS-4

Q4. The common emitter forward current gain of the transistor is βF = 100. The transistor is operating in

Ans: Active Region UNSOLVED PROBLEMS-5

Q5. Two identical transistors made of silicon are connected as shown in Figure. The value of current I is

Ans: 4.3mA UNSOLVED PROBLEMS-6

Q6. Figure below shows a CE amplifier configuration. The quiescent collector voltage of the circuit is approximately

Ans: 14V UNSOLVED PROBLEMS-7

Q-7: What value of RB will result in saturation of the Si transistor. if VCC =20 V, RC = 5k, RE = 4k; = 50, and VCEsat = 0.2V?

Ans: RB ≤442.56k UNSOLVED PROBLEMS-8

Q-8: In the circuit, VCC = 20V;RC = 5k;RE = 4k, and RB = 500 k. The Si transistor has ICBO = 0 and = 50. Find ICQ and VCEQ. In what region the transistor is operating? If the transistor is replaced by a new transistor with ICBO = 0 and = 75. will it be in same region of operation?

Ans: 1.91 mA, 2.64V, active region, No, it will operate in Cut-off region UNSOLVED PROBLEMS-9

Q-9: The circuit of Fig. 3-29 illustrates a method for biasing a CB transistor using a single dc source. The transistor is a Si device VBEQ = 0:7V, =99, and IBQ = 30A.

Find (a) R2, and (b) VCEQ.

Ans: 3.36k, 6.06V REFERENCES

. Hughes, “Electrical and Electronic Technology”, Pearson Prentice Hall, Tenth Edition, 2008. . Robert L. Boylestad, Louis Nashelsky, “Electronic Devices and Circuit Theory”, Pearson Education Limited, Eleventh Edition, 2014.