Recall%20Last%20Lecture

1
Recall Last Lecture

• DC Analysis and 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
Vo (V)
5
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
• ßIB (5 VO) / 4
• IB (5 VO) / 480
• Equate the 2 equations
• (Vi 0.7) / 100 (5 VO) / 480

ß 120
A linear equation with negative slope
6
Vo (V)
5
Vi (V)
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
Voltage Transfer Characteristics - pnp
Vo (V)
Vo 4.8
ß 80

• 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

Vi (V)
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 start to
  become active is based on the equation which is
  derived from active mode operation

9

• BE Loop
• 200IB 0.7 Vi 5 0
• IB (4.3 Vi ) / 200
• CE Loop
• ICRC - VO 0
• IC VO / 8
• 80 IB VO / 8
• IB VO / 640
• Equate the 2 equations
• (4.3 - Vi) / 200 VO / 640

ß 80
A linear equation with negative slope
10
Vo (V)
Vo 4.8
ß 80
Vi (V)
x
2.8 V

• 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

when Vo 0, then, Vi 4.3 V
11
Bipolar Transistor Biasing
12
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

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Proper Biasing Effect
Ref Neamen
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Effect of Improper Biasing on Amplified Signal
Waveform
Ref Neamen
15

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

16
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.

17
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