bjt transistor modelling

BJT Transistor Modelling

Prepared by:Engr. Mark S. Cañete

BJT Transistor Modelling

AC Equivalent Models AC equivalent models can be obtained by:

Setting all DC sources to zero and replacing them by a short- circuit equivalent

Replacing all capacitors by short-circuit equivalent

Removing all elements bypassed by the short –circuit equivalents introduced by step 1 and 2.

Redraw the network in a more convenient and logical form

AC equivalent Models Example:

AC equivalent Models

AC Equivalent Models

Note: R12 is the parallel combination of R1 and R2

Important Parameters in AC analysis

•Two port System

•Ii

•Io

•Vi•Zi •Zo

•Vo

Input Impedance

For the input Side, the input impedance is define as:

i

ii I

VZ

Note: For small signal analysis, once the input impedance has been

determined the same numerical value can be used for changing levels of applied signal. It is purely resistive in nature and depending on the manner in which the transistor is employed. An ohmmeter cannot be used to measure the small signal ac input impedance since it is in the DC mode.

Input ImpedanceConsidering the source resistance

Rs

Vs

Two port SystemViZi

Ii

Input Impedance

source

isi R

VVI

Input Impedance Example1:

Using two port system, solve for the input impedance given the following values:

mVV

mVV

kR

i

s

s

2.1

2

1

Output Impedance

The output impedance of a BJT transistor amplifier is resistive in nature and depending on the configuration and the placement of the resistive element Zo can varyfrom a few ohms that can be exceed 2Mohm.

o

oo I

VZ

Voltage Gain One of the most important

characteristic of an amplifier is the small signal ac voltage gain.

For no load voltage gain: iV V

VA 0

)( topencircuiRi

oV

L

NL V

VA

No load Voltage Gain

•Ii

•Vo

Rs

Vs•AVNL•Vi

•Zi

Voltage Gain

NLs Vsi

i

iV A

RZ

Z

V

VA

0

Voltage Gain Example2

• For the BJT Amplifier, Rs = 1.2K and Vs = 40mV, AVNL=320 determine:a. Vib. Iic. Zid. Avs

Ii

Vo=7.68 V

Rs

VsAVNLVi

Zi

Current Gain

ii I

IA 0

L

oo R

VI

L

ivi R

ZAA

Re Transistor Model

• Re model employs a diode and controlled current source to duplicate the behavior of a transistor in the region of interest

Re Transistor model• Common Base NPN Configuration

Figure 1. Common Base Configuration Figure 2. re Equivalent model

Figure 3. Common Base re Equivalent circuit

Re Transistor Model• Common Base PNP Configuration

Figure 1. Common Base Configuration Figure 2. re Equivalent model

Figure 3. Common Base re Equivalent circuit

Re Transistor Model• Useful Equation in common based

configuration

Ee I

mVr

26

re is the Ac resistance of the emitter diode. It is important because it determines the voltage gain.

CBei rZ For common base configuration, typical values of Zi ranges from a few ohms to a maximum of about 50

CBZo For common base configuration,

typical values of Zo are in the megaohm range

For CB typical input impedance is relatively low and output Z quite high

Re Transistor Model

e

L

e

LV r

R

r

RA

Common Base Voltage gain

Common Base Current gain

1 iA

Example3 For a common base configuration of figure below

with IE = 4 mA, = 0.98, and an AC signal of 2 mV applied between the base and emitter terminals:

a. Determine the input impedance b. Calculate the voltage gain if a load of 0.56 k is

connected to the output terminalsc. Find the output impedance and current gain

Re Transistor Model• Common emitter Configuration

Ic = Ib

Figure 1. Common Emitter Configuration Figure 2. re Configuration

Figure 3. re equivalent

IE Ib

Figure 4. re model for CE configuration

Re Transistor Model

be

bc

II

II

Collector Current and Emitter current

Input Impedance

ei rZ Output Impedance

00 rZ For CE configuration, typical values of Zo are in the range of 40 to 50 k

Voltage Gain

0,rCEe

Lv r

RA

Current Gain

iA

Given = 120 and IE = 3.2 mA for a common emitter configuration with ro =

a. Zib. Av if the load of 2 K is appliedc. Ai with the 2 K load.

example4