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

2. 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.

3. 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.

4. 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–Collector junction is forward biased

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

6. 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 that it does not go into saturation or cutoff when an a ac signal is applied.
IB   IC and VCE 
IB IC  and VCE 

7. 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 mA (IB = 300 A) of collector current is best for maximum amplification, giving equal amount above and below the Q-point. Note that this is three different scenarios of collector current being viewed simultaneously.

8. 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.

9. 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

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

11. WAVEFORM DISTORTION

12. 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.

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

14. 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.

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

16. EXAMPLE
Given that VBE = 0.7 V and βDC = 100, determine the Q-point values

17. EXAMPLE
Given that VBE = 0.7 V, RB=22kΩ, RC=100 Ω and βDC = 90, determine the Q-point values

18. BJT bias circuit with emitter resistor.
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.
An npn transistor with emitter bias. Polarities are reversed for a pnp transistor. Single subscripts indicate voltages with respect to ground.
BJT bias circuit with emitter resistor.

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

20. EMITER BIAS –RE + DC Voltage Supply
Collector – Emitter loop
VCC – ICRC – VCE – IERE + VEE = 0
IE  IC
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

21. 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.

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

23. 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

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

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

26. EXAMPLE
Given that Vcc = +12V, VEE = -12V, RB=100kΩ, RC=750 Ω, RE =1.5kΩ, β=200. Find the value of IB,IE and VCE

27. EXAMPLE
Given that Vcc = 5V, VEE = -5V, RB=10kΩ, RC=1.0k Ω, RE =2.2kΩ, β=100. Find the voltage of terminal with respect to ground

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