Download Recall Lecture 11

Recall Last Lecture
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Load Line
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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:
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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
Voltage Transfer
Characteristic
VO versus Vi
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
Vo (V)
Cutoff
5
0.7
•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.
Vi (V)
Active Mode
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BE Loop
 100IB + VBE – Vi = 0
 IB = (Vi – 0.7) / 100
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CE Loop
 ICRC + VO – 5 = 0
 IC = (5 – VO) / 4
 βIB = (5 – VO) / 4
β = 120
 IB = (5 – VO) / 480
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Equate the 2 equations:
(Vi – 0.7) / 100 = (5 – VO) / 480
Vo = - 480 Vi + 836 A linear equation with
negative slope
100
Vo (V)
Cutoff
5
Active
Saturation
0.2
0.7
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?
1.7
5 V (V)
i
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
Voltage Transfer Characteristics - pnp
VCC – VEC (sat)
VEB
VEC
•Vo = VC and VE = VCC
0
x V –V
CC
EB (on)
•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

As Vi increases, VB will become
more positive than VC, the junction
C-B will become reverse-biased.
The transistor goes to active mode.
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The point (point x) where the
transistor becomes active is based
on the equation which is derived
from active mode operation
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BE Loop
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CE Loop
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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
VEB
VEC
 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:
(VCC - Vi – 0.7) / RB = VO / βRC

If Vo = 0, then, Vi = VCC – 0.7
Voltage Transfer Characteristic for
pnp Circuit - Example
β = 80
Vo = VCC – VEC sat
Vi = VCC – 0.7
Bipolar Transistor
Biasing
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
Proper Biasing Effect
Ref: Neamen
Effect of Improper Biasing on
Amplified Signal Waveform
Ref: Neamen
•
Three types of biasing
 Fixed Bias Biasing Circuit
 Biasing using Collector to Base Feedback Resistor
 Voltage Divider Biasing Circuit
Biasing Circuits – Fixed Bias Biasing
Circuit
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The circuit is one of the simplest transistor circuits is known as
fixed-bias biasing circuit.
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There is a single dc power supply, and the quiescent base current
is established through the resistor RB.
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The coupling capacitor C1
acts as an open circuit to dc,
isolating the signal source
from the base current.
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Typical values of C1 are in
the rage of 1 to 10 μF,
although the actual value
depends on the frequency
range of interest.
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