1. Recall Last Lecture Voltage Transfer Characteristic
A plot of Vo versus Vi
Use BE loop to obtain a current equation, IB in terms of Vi
Use CE loop to get IC in terms of Vo
Change IC in terms of IB
Equate the two equations to link Vi with Vo

2. Vo (V)
Vi (V) 5
0.7
Cutoff
Active
0.2 x
Saturation
= 4.8
Vi (V)
= 4.3

3. Bipolar Transistor Biasing

4. Bipolar Transistor Biasing
Biasing refers to the DC voltages applied to the transistor for it to turn on and operate in the forward active region, so that it can amplify the input AC signal

5. Proper Biasing Effect
Ref: Neamen

6. Effect of Improper Biasing on Amplified Signal Waveform
Ref: Neamen

7. Three types of biasing
Fixed Bias Biasing Circuit
Biasing using Collector to Base Feedback Resistor
Voltage Divider Biasing Circuit

8. Biasing Circuits – Fixed Bias Biasing Circuit
The circuit is one of the simplest transistor circuits is known as fixed-bias biasing circuit.
There is a single dc power supply, and the quiescent base current is established through the resistor RB.
The coupling capacitor C1 acts as an open circuit to dc, isolating the signal source from the base current.
Typical values of C1 are in the rage of 1 to 10 μF, although the actual value depends on the frequency range of interest.

9. Example – Fixed Bias Biasing Circuit
Determine the following: (a) IB and IC (b) VCE Assume VBE(on) = 0.7 V, and  = 50
(a)
VB – VE = 0.7
VB = 0.7V
IB = (12 – 0.7) / 240k = mA
IC = 50 (0.0471) = mA
VB = ??
VE = 0V
(b)
IC = mA = (12 – VC) / 2.2k
VC = V
VCE = – 0 = V
NOTE: Proposed to use branch current equations and node voltages

10. Biasing using Collector to Base Feedback Resistor
IC + IB = IE IB IC IE
Find RB and RC such that IE = 1mA , VCE = 2.3 V, VCC = 10 V and  =100.
NOTE: Proposed to use branch current equations and node voltages

11. Biasing using Collector to Base Feedback Resistor
IE = 1mA , VCE = 2.3 V, VCC = 10 V and  =100.
IB = (IE / (+1) = mA
(VC – VB ) / RB= IB
but VC = VCE
and VB = VBE = 0.7 V
(2.3 – 0.7) / RB = mA
RB = k
(VCC – VC ) / RC = IE
RC = 7.7 k VC VB
VE = 0V

12. Voltage Divider Biasing Circuit
This is a very stable bias circuit.
The currents and voltages are almost independent of variations in .

13. Analysis
Redrawing the input side of the network by changing it into Thevenin Equivalent
RTH
RTh: the voltage source is replaced by a short-circuit equivalent

14. Analysis VTH VTh: open-circuit Thevenin voltage is determined. VTH
Inserting the Thevenin equivalent circuit
Use voltage divider

15. Analysis
The Thevenin equivalent circuit

16. Example Find VCE ,IE, IC and IB given
β=100, VCC=10V, R1 = 56 k, R2 = 12.2 k,
RC = 2 k and RE = 0.4 k
VTH= R2 /(R1 + R2 )VCC
VTH = 12.2k/(56k+12.2k).(10)
VTH = 1.79V
RTH = R1 // R2
= 10 k

17. BJT Biasing in Amplifier Circuits
VTH = RTH IB + VBE + RE IE
1.79 = 10k IB k ( +1)IB
IB = 21.62mA
IC = bIB = 100(21.62m)=2.16mA
IE = IC + IB = 2.18mA
VCC = RC IC + VCE + RE IE
10 = 2k(2.16m)+VCE +0.4(2.18m)
VCE = 4.8 V

18. Example – A Design Question
VCC = 10 V
IE = 1 mA
=100
RE = k
RTH = 33 k
Determine the values of R1 and R2

19. R2 = 38.8 k BE Loop: IBRTH + 0.7 + IERE – VTH = 0
VTH = 1.5 V
We know:
1.5 (220 + R2) = 10R2
330 = 8.5 R2
R2 = 38.8 k