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

Voltage Transfer Characteristic

VO versus Vi

Voltage Transfer Characteristics - npnVoltage 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

•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.

Vo (V)

Vi (V)

5

0.7

Cutoff

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 = - 480 Vi + 836 A linear equation with negative slope

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?

Vo (V)

Vi (V)

5

0.7

Cutoff

Active

x

0.2

Need to substitute in the linear equation Vi = 1.7 V

1.7 5

Saturation

and VO stays constant at 0.2V until Vi = 5V

To find point x, the coordinate is (x, 0.2)

Voltage Transfer Characteristics - pnpVoltage Transfer Characteristics - pnp

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

Vo (V)

Vi (V)

Vo = 4.8saturation

β = 80

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

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

VEC

VEB

β = 80

200

Vo = - 640 Vi + 2752 A linear equation with negative slope

Vo (V)

Vi (V)

Vo = 4.8saturation

Active

x

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: 200

Vo = - 640 Vi + 2752

54.3

cutoff

2.8 V

when Vo = 0, then, Vi = 4.3 V

VEC

VEB

β = 80

To find point x, the coordinate is (x, 4.8)

Bipolar Transistor Bipolar Transistor BiasingBiasing

• 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

Bipolar Transistor BiasingBipolar Transistor Biasing

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 Biasing Circuits – Fixed Bias Biasing CircuitCircuit

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.

Determine the following:(a) IB and IC

(b) VCE

(c) VB and VC

Example – Fixed Bias Biasing CircuitExample – Fixed Bias Biasing Circuit

NOTE: Proposed to use branch current equations and node voltages