TRANSISTOR BIAS CIRCUITS 1 09112020 CHAPTER 5 OBJECTIVES

TRANSISTOR BIAS CIRCUITS 1 09/11/2020 CHAPTER 5

OBJECTIVES Discuss the concept of dc biasing of a transistor for linear operation Analyze voltage-divider bias, base bias, emitter bias and collectorfeedback bias circuits. 09/11/2020 2

LECTURE OUTLINE 2. 3. 4. 5. 09/11/2020 1. Objectives Introduction DC operating point Voltage-divider bias Other bias methods Base bias Emitter bias Collector-feedback bias 6. Summary 3

INTRODUCTION biasing is used to establish a fixed level of current and voltage. Transistor must be properly biased with dc voltage to operate as a linear amplifier. If amplifier is not biased with correct dc voltages on input and output, it can go into saturation or cutoff when the input signal applied. 09/11/2020 4

09/11/2020 DC OPERATING POINT 5

DC OPERATING POINT The goal of amplification in most cases is to increase the amplitude of an ac signal without altering it. Improper biasing can cause distortion in the output signal. 09/11/2020 6

DC OPERATING POINT 09/11/2020 The purpose of biasing a circuit is to establish a proper stable dc operating point (Q-point). The goal of Q-point such that it does not go into saturation or cutoff when an ac signal is applied. 7

Q-point of a circuit: dc operating point of amplifier specified by VCE and IC. These values are called the coordinates of Q-point. Refer to figure a, given IB = 200μA and βDC=100. IC=βDCIB so IC=20 m. A and Figure b, VBB is increased to produce IB of 300μA and IC of 30 m. A. Figure c, VBB is increased to produce IB of 400μA and IC=40 m. A. So, VCE is: 8 09/11/2020

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DC OPERATING POINT-DC LOAD LINE Recall that the collector characteristic curves graphically show the relationship of IC and VCE for different IB. When IB and IC increases and VCE decreases or vice-versa. Each separate Q-point is connected through dc load line. At any point along line, values of IB, IC and VCE can be picked off the graph. Dc load line intersect VCE axis at 10 V, where VCE=VCC. This is cutoff point because IB and IC zero. Dc load line also intersect IC axis at 45. 5 m. A ideally. This is saturation point because IC is max and VCE=0. 09/11/2020 10

EXAMPLE 1 βDC=200. Q-point in figure below. Assume 09/11/2020 Determine 11

EXAMPLE 2 below shows the collector characteristic curves and dc load line. Determine the following a) Collector saturation current b) VCE at cutoff c) Q-point values of IB, IC and VCE 09/11/2020 Figure 12

09/11/2020 VOLTAGE-DIVIDER BIAS 13

VOLTAGE-DIVIDER BIAS Voltage-divider bias is the most widely used type of bias circuit. Only one power supply is needed and voltage-divider bias is more stable(independent) than other bias types. dc bias voltage at base of transistor is developed by a resistive voltagedivider consists of R 1 and R 2. Vcc is dc collector supply voltage. 2 current path between point A and ground: one through R 2 and the other through BE junction and RE. 09/11/2020 14

VOLTAGE-DIVIDER BIAS If IB is much smaller than I 2, bias circuit is viewed as voltage divider of R 1 and R 2 as shown in Figure a. If IB is not small enough to be neglected, dc input resistance RIN(base) must be considered. RIN(base) is in parallel with R 2 as shown in figure b. 09/11/2020 15

INPUT RESISTANCE AT TRANSISTOR BASE VIN is between base and ground and IIN is the current into base. By Ohm’s Law, 09/11/2020 RIN(base) = VIN / IIN Apply KVL, VIN=VBE+IERE Assume VBE<<IERE, so VIN≈IERE Since IE≈IC=βDCIB, VIN≈ βDCIBRE IIN=IB, so RIN(base)= βDCIBRE / IB RIN(base) = βDCRE 16

ANALYSIS OF VOLTAGE-DIVIDER BIAS CIRCUIT 09/11/2020 17

ANALYSIS OF VOLTAGE-DIVIDER BIAS CIRCUIT 09/11/2020 Total resistance from base to ground is: A voltage divider is formed by R 1 and resistance from base to ground in parallel with R 2. If βDCRE >>R 2, (at least ten times greater), then the formula simplifies to 18

ANALYSIS OF VOLTAGE-DIVIDER BIAS CIRCUIT 09/11/2020 Now, determine emitter voltage VE. VE=VB – VBE Using Ohm’s Law, find emitter current IE. IE = V E / R E All the other circuit values IC ≈ I E VC = VCC – ICRC To find VCE, apply KVL: VCC – ICRC – IERE – VCE =0 Since IC ≈ IE, VCE ≈ VCC – IC (RC + RE) 19

EXAMPLE 3 Determine VCE and IC in voltage-divider biased transistor circuit below if βDC=100. 09/11/2020 20

VOLTAGE-DIVIDER BIAS FOR PNP TRANSISTOR Pnp transistor has opposite polarities from npn. To obtain pnp, required negative collector supply voltage or with a positive emitter supply voltage. The analysis of pnp is basically the same as npn. 09/11/2020 21 a. Negative collector b. Positive emitter Supply supply voltage

ANALYSIS OF VOLTAGE BIAS FOR PNP TRANSISTOR Base voltage Emitter voltage By Ohm’s Law, And, 09/11/2020 22

EXAMPLE 4 Evaluate IC and VEC for pnp transistor circuit in Figure below. Given VEE = +15 V, R 1 = 63 kΩ, R 2 = 27 kΩ, RC = 1. 8 kΩ, RE = 2. 6 kΩ, βDC =120. 09/11/2020 23

EXAMPLE 5 09/11/2020 Figure below shown the schematic with a negative supply voltage, determine IC and VCE for a pnp transistor circuit with given values: R 1 = 25 kΩ, R 2 = 60 kΩ, RC = 6 kΩ, RE = 9 kΩ, VCC = -12 V, and βDC = 90 24

EXAMPLE 6 Construct a complete circuit required to replace the transistor in Figure below with a pnp transistor. Given VCC = 10 V, R 1 = 78 kΩ, R 2 = 100 kΩ, RC = 18 kΩ, RE = 8 kΩ. 09/11/2020 25

09/11/2020 OTHER BIAS METHODS 26 BASE BIAS EMITTER BIAS COLLECTOR-FEEDBACK BIAS

OTHER BIAS METHODS – BASE BIAS VCC – VRB – VBE = 0 or VCC – IBRB – VBE =0 09/11/2020 KVL apply on base circuit. Solving for IB, Then, apply KVL around collector circuit. VCC – ICRC – VCE = 0 We know that IC = βDCIB, 27

BASE BIAS This type of circuit is beta-dependent and very unstable. Recall that DC changes with temperature and collector current. Base biasing circuits are mainly limited to switching applications. 09/11/2020 From the equation of IC, note that IC is dependent on DC. When DC vary, VCE also vary, thus changing Q-point of transistor. 28

EXAMPLE 7 Determine IB, IC and VCE for a base-biased transistor circuit with the following values: βDC = 160, VCC = 18 V, RB = 43 kΩ and RC = 190Ω. 09/11/2020 29

EMITTER BIAS 09/11/2020 Npn transistor with emitter bias 30

EMITTER BASE 09/11/2020 This type of circuit is independent of βDC making it as stable as the voltage-divider type. The drawback is that it requires two power suppliers Apply KVL and Ohm’s Law, IBRB + IERE + VBE = VEE Since IC ≈ IE and IC = βDCIB, Solve for IE or IC, Voltage equations for emitter base circuit, VE = VEE - IERE VB = VE + VBE VC = VCC - ICRC 31

EMITTER BASE IBRB + IERE + VBE = VEE 09/11/2020 Apply KVL and Ohm’s Law, Since IC ≈ IE and IC = βDCIB, Solve for IE or IC, 32

EXAMPLE 8 Calculate IE and VCE for the circuit below using VE = -1 V and IC = IE, RB = 47 kΩ, RC = 4. 7 kΩ, RE = 10 kΩ, VCC = 15 V and VEE = -15 V 09/11/2020 33

COLLECTOR-FEEDBACK BIAS 09/11/2020 Collector-feedback bias is kept stable with negative feedback, although it is not as stable as voltagedivider or emitter. With increases of IC, VC decrease and causing decrease in voltage across RB, thus IB also decrease. With less IB, IC go down as well. 34

ANALYSIS OF COLLECTORFEEDBACK CIRCUIT By Ohm’s Law, 09/11/2020 Collector voltage with assumption IC>>IB. VC = VCC – ICRC And IB = IC / βDC So, collector current equation Since emitter is ground, VCE = VCC - ICRC 35

EXAMPLE 9 Determine IC, VB, VC and Power rating in figure below. 09/11/2020 36

09/11/2020 SUMMARY 37

6. SUMMARY 09/11/2020 The purpose of biasing is to establish a stable operating point (Q-point). The dc load line helps to establish the Qpoint for a given collector current. Q-point is used to avoid amplifier goes into saturation and cutoff The linear region of a transistor is the region of operation within saturation and cutoff. 38

6. SUMMARY 09/11/2020 Voltage-divider bias is most widely used because it is stable and uses only one voltage supply. Base bias is very unstable because it is β dependent. Emitter bias is stable but require two voltage supplies. Collector-back is relatively stable when compared to base bias, but not as stable as voltage-divider bias. 39