DMT 121 ELECTRONIC 1 Chapter 4 DC Biasing

DMT 121 – ELECTRONIC 1 Chapter 4 DC Biasing – Bipolar Junction Transistors (BJTs)

OBJECTIVES Discuss the concept of dc biasing of a transistor Analyze voltage-divider bias, base bias, emitter bias and collector-feedback bias circuits. Basic troubleshooting for transistor bias circuits.

INTRODUCTION For the transistor to properly operate it must be biased. There are several methods to establish the DC operating point. We will discuss some of the methods used for biasing transistors as well as troubleshooting methods used for transistor bias circuits.

BIASING & 3 STATES OF OPERATION Active or Linear Region Operation Base–Emitter junction is forward biased Base–Collector junction is reverse biased Cutoff Region Operation Base–Emitter junction is reverse biased Saturation Region Operation Base–Emitter junction is forward biased Base–Collector junction is forward biased

DC OPERATING POINT The goal of amplification in most cases is to increase the amplitude of an ac signal without altering it.

DC OPERATING POINT For a transistor circuit to amplify it must be properly biased with dc voltages. The dc operating point between saturation and cutoff is called the Q-point. The goal is to set the Q-point such that it does not go into saturation or cutoff when an a ac signal is applied. IB IC and VCE IB IC and VCE

DC OPERATING POINT Recall that the collector characteristic curves graphically show the relationship of collector current and VCE for different base currents. With the dc load line superimposed across the collector curves for this particular transistor we see that 30 m. A (IB = 300 A) of collector current is best for maximum amplification, giving equal amount above and below the Qpoint. Note that this is three different scenarios of collector current being viewed simultaneously.

DC OPERATING POINT With a good Q-point established, look at the effect of superimposed ac voltage has on the circuit. Note the collector current swings do not exceed the limits of operation (saturation and cutoff). However, as you might already know, applying too much ac voltage to the base would result in driving the collector current into saturation or cutoff resulting in a distorted or clipped waveform.

Key terms DC Load Line A straight line plot of IC and VCE for a transistor circuit. Q-point DC operating point along line between saturation and cutoff Linear Region A region of operation along the load line between saturation and cutoff

WAVEFORM DISTORTION Graphical load line illustration of a transistor being driven into saturation and/or cutoff

WAVEFORM DISTORTION

FIXED-BIAS (BASE-BIAS) CIRCUIT Simplest transistor bias configuration. Commonly used in relay driver circuits. Extremely beta-dependant and very unstable Fixed-bias circuit. DC equivalent circuit.

FIXED-BIAS (BASE-BIAS) CIRCUIT Base – Emitter loop Collector – Emitter loop Measuring VCE and VC. VCC – ICRC – VCE = 0 VCE = VCC – ICRC; then = VC – VE since VE = 0 VBE = VB – VE (since VE = 0) VBE = VB VCE = VC Since IC = IB, then Sensitive to Beta

Fixed-bias (base-bias) - Summary Circuit recognition : A single resistor (RB) between the base terminal and VCC. No emitter resistor. Q-point stability : Q-point is more dependent on βdc, so it becomes β dependent and unpredictable. Βdc varies with temperature and IC. Advantage: Circuit simplicity. Disadvantage: Q-point shift with temperature Applications: Switching circuits only. Rarely used in linear operation.

Cont’d Summary Load line equations: Q-point equations:

EXAMPLE Given that VBE = 0. 7 V and βDC = 100, determine the Q-point values

EXAMPLE Given that VBE = 0. 7 V, RB=22 kΩ, RC=100 Ω and βDC = 90, determine the Q-point values

EMITER BIAS Use both a positive and a negative supply voltage on emitter or it just contain an emitter resistor to improve stability level over fixed – bias configuration. BJT bias circuit with emitter resistor. An npn transistor with emitter bias. Polarities are reversed for a pnp transistor. Single subscripts indicate voltages with respect to ground.

EMITER BIAS – only RE Collector – Emitter loop VCC – ICRC – VCE – IERE = 0 IE I C VCC – ICRC – VCE –ICRE = 0 VCC – VCE = IC (RC + RE) BJT bias circuit with emitter resistor. Base – Emitter loop VCC – IBRB – VBE – IERE = 0 Since IC = IB, so IC also equivalent to IE = ( + 1) IB Then, VCC – IBRB – VBE – ( + 1)IBRE = 0. Less sensitivity to beta

EMITER BIAS –RE + DC Voltage Supply Collector – Emitter loop VCC – ICRC – VCE – IERE + VEE = 0 IE I C VCC – ICRC – VCE –ICRE + VEE = 0 VCC – VCE + VEE = IC (RC + RE) Base – Emitter loop VEE + IBRB + VBE + IERE = 0 IE = ( + 1) IB Then, VEE + IBRB + VBE + ( + 1)IBRE = 0. Less sensitivity to beta

EMITTER BIAS - Summary Circuit recognition: Dual-polarity power supply (+ve and -ve) and the base resistor is connected to ground. Stability : Adding RE to the emitter improves the stability of a transistor Stability refers to a bias circuit in which the currents and voltages will remain fairly constant for a wide range of temperatures and transistor Beta ( ) values. Advantage: The circuit Q-point values are stable against changes in β. Disadvantage: Requires the use of dual-polarity power supply. Applications: Used primarily to bias linear amplifiers.

EMITTER BIAS - Summary With DC Voltage supply + Resistor at Emitter Since IC = βIB , so With only Resistor at Emitter Since IC = βIB , so

EMITTER BIAS - Summary Previous analysis we use IE = ( + 1) IB; but if use IE IC IB, then from previous slide we can get. OR we also can use ( + 1) to get the same result. • If RE >>> RB/ then we can drop RB/ in equation Less sensitivity to beta or independent to beta • If VEE >>> VBE then Independent to VBE

EMITTER BIAS - Summary Load line equations: Q-point equations:

EMITTER BIAS - Summary Voltage with respect to ground : Emitter voltage; VE = VEE+IERE Base voltage; VB = VE + VBE Collector voltage; VC = VCC - ICRC

EXAMPLE Given that Vcc = +12 V, VEE = -12 V, RB=100 kΩ, RC=750 Ω, RE =1. 5 kΩ, β=200. Find the value of IB, IE and VCE

EXAMPLE Given that Vcc = 5 V, VEE = -5 V, RB=10 kΩ, RC=1. 0 k Ω, RE =2. 2 kΩ, β=100. Find the voltage of terminal with respect to ground

EXAMPLE From previous Figure, Find the voltage of terminal with respect to ground