1. Recall Lecture 11 DC Analysis and Load Line
Input load line is based on the equation derived from KVL at BE loop.
Output load line is derived from KVL at CE loop.
To complete your load line parameters:
Obtained the values of IB from the BE loop
Get the values of x and y intercepts from the derived IC versus VCE.
Draw the curve of IB and obtained the intercept points IC and VCE (for npn) which is also known as the Q points

2. Voltage Transfer Characteristic
VO versus Vi

3. Voltage Transfer Characteristics - npn
A plot of the transfer characteristics (output voltage versus input voltage) can also be used to visualize the operation of a circuit or the state of a transistor.
Given VBEon = 0.7V,  = 120, VCEsat = 0.2V, Develop the voltage transfer curve

4. In this circuit, Vo = VC = VCE
Cutoff 5
0.7
Vi (V)
In this circuit, Vo = VC = VCE
Initially, the transistor is in cutoff mode because Vi is too small to turn on the diodes. In cut off mode, there is no current flow.
Then as Vi starts to be bigger than VBEon the transistor operates in forward-active mode.

5. Active Mode BE Loop 100IB + VBE – Vi = 0 IB = (Vi – 0.7) / 100 CE Loop
ICRC + VO – 5 = 0
IC = (5 – VO) / 4
βIB = (5 – VO) / 4
IB = (5 – VO) / 480
Equate the 2 equations:
(Vi – 0.7) / 100 = (5 – VO) / 480
β = 120
100
Vo = Vi + 836
A linear equation with negative slope

6. Vo (V)
Cutoff 5
0.7
Active 5
Saturation
To find point x, the coordinate is (x, 0.2)
0.2
Vi (V) x
1.7
However, as you increase Vi even further, it reaches a point where both diodes start to become forward biased – transistor is now in saturation mode.
In saturation mode, VO = VCEsat = 0.2V. So, what is the starting point, x, of the input voltage, Vi when this occurs?
Need to substitute in the linear equation  Vi = 1.7 V
and VO stays constant at 0.2V until Vi = 5V

7. Bipolar Transistor Biasing

8. 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

9. Proper Biasing Effect
Ref: Neamen

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

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

12. 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.

13. 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

14. 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

15. 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