bjt transistor modelling
TRANSCRIPT
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