Download small signal bjt amplifier

SMALL SIGNAL BJT
AMPLIFIER
SMALL SIGNAL OPERATION
INTRODUCTION
☺One of the primary uses of a transistor is to amplify ac signals.
☺This could be an audio signal or perhaps some high frequency
radio signal.
☺It has to be able to do this without distorting the original input.
☺Most transistors amplifiers are designed to operate in the linear
region
☺The common-emitter amplifier has high voltage and current
gain
Definition of Small Signal
• ac input signal voltages and currents are in
the order ±10% of Q-point voltages and
currents.
• eg
– If dc current is 10mA, the ac current (peak to
peak) is less than 0.1 mA.
HOW BJT AMPLIFIES?
•When a weak ac signal is given to the base, a small ac base current
start flowing.
•Due to BJT, a much larger ac current flow through RC
I C  I B
•Therefore a large voltage appear across the collector circuit.
• R1, R2, RE – form the biasing circuit & stabilization.
• C1(Coupling capacitor) – couple the signal to the
base.
• C2(Bypass capacitor) - use in parallel with RE to
provide low reactance path to amplified ac signal
• C3(Coupling capacitor)- couple the signal to the
next stage
Circuit Diagram
•
Common-Emitter Biasing(CE)
– input
– output
= VBE & IB
= VCE & IC
IC
VCE
IC
IB
VC
+
_
IB
C
IC
IB
= Common-emitter current gain =
I
= Common-base current gain = C
IE
The relationships between the two parameters are:


 1
Example
• Given the common-emitter circuit below
with IB = 25A,
VCC = 15V and  =
100. Find the ideal collector current.
•  = 100 = IC/IB
IC = 100 * IB = 100 * (25x10-6A) = 2.5 mA
Common-Emitter
Region
of
Operation
Active
Description
Small base current
controls a large
collector current
Collector-Current Curves
IC
Active
Region
IB VCE
Saturation VCE(sat) ~ 0.2V, VCE
increases with IC
Cutoff
Achieved by reducing
IB to 0, Ideally, IC will
also equal 0.
Saturation Region
Cutoff Region
IB = 0
DC Analysis of Transistor
Circuits Common-Emitter Circuit
Common-emitter circuit with an npn
transistor
• Assume B-E junction: forward biased  V drop is the cut-in /
turn-on V [VBE (on)]
•IC represented as a dependent I source (function of IB)
•Neglect reverse-biased junction leakage current & Early effect
VBB  VBE (on)
IB 
RB
I C  I B
VCE  VCC  I C RC
dc equivalent circuit, with piecewise
linear parameters
VCC  I C RC  VCE
Base-emitter Junction Characteristics And
The Input Load Line
Common- Emitter Transistor Characteristics
And The Collector-emitter Load Line
Common-emitter circuit
Load Line
Kirchoff’s voltage law equation (around B-E loop):
VBB VBE
IB 

RB
RB
• Both load line & quiescent IB change as either
or both VBB & RB change
Kirchoff’s voltage law equation (around C-E loop):
VCE  VCC  I C RC
IC 
VCC VCE
V

 5  CE (mA) • IC & VCE relationship represents DC
RC RC
2
load line
Amplifier Operation
•The boundary between cutoff and saturation is called the
linear region.
•A transistor which operates in the linear region is called a
linear amplifier.
•Only the ac component reaches the load because of the
capacitive coupling and that the output is 180º out of input
phase.
Example
• Assume Q-point value of iB=10A, ac signal peak
value is 5A and β=100.
• So at any instant, iC =100iB
iB
iC=100iB
VCE=VCC-iCRC
Output voltage
15A
1.5mA
4V
Positive peak
value
10A
1.0mA
6V
Signal current
zero
5A
0.5mA
8V
Negative peak
value
dc equivalent circuit of transistor
amplifier
• Only dc condition are to be considered
• No signal is applied (All ac source reduce
to zero)
• dc cannot flow through a capacitor (open
circuit)
• To calculate the dc currents & voltages
DC Equivalent for the BJT
Amplifier
DC Equivalent Circuit
• All capacitors in the original amplifier circuit are replaced by
open circuits, disconnecting vI, RI, and R3 from the circuit and
leaving RE intact. The the transistor Q will be replaced by its
DC model.
ac equivalent circuit of transistor
amplifier
• Only ac condition are to be considered
• dc voltage not important, considered zero
or ground. (VCC=0)
• the capacitors used to couple or bypass
the ac signal. (short circuit)
AC Equivalent for the BJT
Amplifier
• The coupling and bypass capacitors are replaced by short circuits.
• The DC voltage supplies are replaced with short circuits, which in
this case connect to ground.
(continued)
R  R R 10k 30k
B 1 2
R R R  4.3k100k
C 3
• By combining parallel resistorsinto equivalent RB and R, the equivalent AC
circuit above is constructed. Here, the transistor will be replaced by its
equivalent small-signal AC model (to be developed).
Graphical Analysis & AC
Equivalent Circuit
• Equations
RC
– Input loop
ic
vO
RB
vs
ib
+
vbe -
+
vce
-
vs  ib RB  vbe
 I BQ 
vbe
ib  
 VT 
0.026 V
– Output loop
ic RC  vce  0
AC equivalent circuit of C-E with npn
transistor
ic  ib
Small-signal hybrid-
equivalent circuit
gm=ICQ/VT
vbe = ibrπ
rπ
= diffusion resistance /
base-emitter input
resistance
1/rπ
r=VT/ICQ
= slope of iB – VBE
curve
vbe
VT
 FVT
 r 

,
ib
I BQ
I CQ
Using transconductance (gm) parameter
I CQ   F I BQ
Small-signal hybrid-
equivalent circuit
 ib
( I b )
Using common-emitter current gain (β) parameter
Small-signal hybrid-
equivalent circuit
Small-signal equivalent circuit
RB
B
Ic
C
Vo
+
Vs
Ib
Vbe
+
r
-
 r 
Vs
Vbe  
 r  RB 
gmVbe
E
RC
Vce
-
Vo  Vce  g mVbe RC
Output signal voltage
 r 
Vo

Small signal voltage gain, Av   g m RC 
Vs
 r  RB 
Input signal voltage
Example
Given :  = 100, VCC = 12V
VC
VBE = 0.7V, RC = 6k,
C
RB = 50k, and VBB = 1.2V
RC
Calculate the small-signal voltage
gain.
RB
vs
VBB
vO
Solutions
1.
I BQ 
VBB  VBE ( on)
RB

1.2  0.7
 10 A
50
2.
I CQ  I BQ  100(10A)  1 mA
3.
VCEQ  VCC  I CQ RC  12  (1)(6)  6V
4.
5.
6.
VT (100)(0.026)
r 

 2.6 k
I CQ
1
I CQ
1
gm 

 38.5 mA / V
VT
0.026
 r 
Vo
  11.4
Av   g m RC 
Vs
 r  RB 
Hybrid- Model and Early
Effect
transconductance
parameter
ro=VA/ICQ
current gain
parameter
ro = small-signal transistor
output resistance
VA = early voltage
Hybrid- Model and Early Effect
Early Voltage (pg 109)
Early Voltage
(VA)
Example
Given :  = 100, VCC = 12V
VCC
VBE(on) = 0.7V, RS = 0.5k,
RC = 6k, R1 = 93.7k, R2 = 6.3k
and VA = 100V.
Calculate the small-signal voltage
gain.
vs
R1
RS
CC
R2
RC
vO
Solution
Ri
Ro
RS
Vs
R1 \\ R2
B
C
r
gmV
rO
E
 R1 R2 r 

V
V  
 R1 R2 r  RS  s


Ri  R1 R2 r
Ro  ro RC
Vo  gmV ro RC 
 R1 R2 r
Vo

Av    g m 
 R1 R2 r  RS
Vs


r R 
 o C

Vo
RC
Exercises
Neamen book: Ch. 4
1. Question 4.1 (pg. 177)
2. Question 4.3 (pg. 179)