1. Recall Last Lecture Load Line
Input load line is based on the equation derived from BE loop.
Output load line is derived from 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) or VEC (for pnp) 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
0.2 ?
1.7
Vi (V)
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 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. Voltage Transfer Characteristics - pnp
VCC – VEC (sat)
VCC – VEB (on) x
VEC
VEB
Vo = VC and VE = VCC
Hence, VEC = VCC – VO  VO = VCC - VEC
As Vi starts from 0V, both diodes are forward biased. Hence, the transistor is in saturation. So, VEC = VECsat and Vo = VCC – VEC sat

8. As Vi increases, VB will become more positive than VC, the junction C-B will become reverse-biased. The transistor goes to active mode.
The point (point x) where the transistor becomes active is based on the equation which is derived from active mode operation

9. BE Loop CE Loop IB RB + VEB + Vi – VCC = 0 IB = (VCC - Vi – 0.7) / RB
ICRC - VO = 0
IC = VO / RC
βIB = VO / RC
IB = VO / βRC
Equate the 2 equations:
(VCC - Vi – 0.7) / RB = VO / βRC
VEC
VEB

10. Using the equation derived:
By increasing Vi even more, the potential difference between VEB becomes less than VEBON, causing junction E-B to become reversed biased as well. The diode will be in cut off mode. VO = 0V
Using the equation derived:
If Vo = 0, then, Vi = VCC – 0.7
(VCC - Vi – 0.7) / RB = VO / βRC

11. Voltage Transfer Characteristic for pnp Circuit - Example
β = 80
Vo = VCC – VEC sat
Vi = VCC – 0.7

12. Bipolar Transistor Biasing

13. 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

14. Proper Biasing Effect
Ref: Neamen

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

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

17. 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.

18. Example – Fixed Bias Biasing Circuit
Determine the following: (a) IB and IC (b) VCE (c) VB and VC
NOTE: Proposed to use branch current equations and node voltages