Download 3 Small signal and Mid Band frequency transistor circuits analysis

3 Small signal and Mid Band frequency transistor circuits analysis and design
When the signals levels which we have to deal with , are small , then as mentioned
earlier , the exactness of results in the analysis of the concerned circuits become
important . Under such conditions , the transistor can be represented by a linear device (
called the small equivalent circuit ) and providing that the working frequency is low to
medium , it’s effect ( the transistor response ) on the overall operation of the transistor
can be neglected .
3.1. The general design criteria of small signal transistor circuits .
In designing a transistor circuit which is suitable for small signals application the
following approach may be adopted :
(a) - The DC bias levels , including the operating point , are first determined , using
the techniques developed in part I of the Electronic course
(b) - The small signal techniques are employed to determine the gain , impedance's
and other dynamic variants , at signal frequency ( mid band )
3.2
The transistor small signal equivalent model -- Hybrid representation
Considering that a transistor amplifier is represented as a black box , having input and
output terminals as shown in fig 3.1 ;
i1
<
v1
v2
Black Box
Fig 3.1
Then applying the two port theory we can write six pairs of equations relating the input
and output quantities . The most common pair of equations uses what is called the
Hybrid representation ( called hybrid or H - parameters because it produces a
hybrid sets of measurements ) which can be written as follows :
v1 = h11 i1 + h12 v2
I2 = h21 i1 + h22 v2
..... (17)
..... (18)
the input current i1 and output voltage v2 are taken to be the independent variables
and are considered to vary lightly ( very small variations -- small signals ) about a
quiescent point ( the operating point ) .
The parameters h11 , h12, h21 and h22 can be replaced for transistor circuit by
another set of parameters which describe , more accurately , the physical nature of
the parameters themselves , as shown in the following equations :
v1 = hi i1 + hr v2
..... (19)
I2 = hf i1 + ho v2
.... (20)
where ,
hi = v1 / i1 | at v2=0
( ohms ) ;
v2 is made short circuit by placing a large capacitor -- which will act as a short
circuit for ac signal -- across the output terminals as shown in the following schematic
diagram , Fig 3.2 ;
i1
Z
v1
Black Box
<
v2 =0 for
AC sig
Fig 3.2
hr = v1/ v2 | at
i1 =0
( unitless ) ;
i.e. . this describes the open - circuit reversed voltage gain , as represented by the
following schematic diagram , Fig 3.3 ;
i1= 0
<
v1
v2
Black Box
Fig 3.3
hf = i2/i1 | at v2=0
( unitless )
;
i.e. depicting the short circuit forward current gain , as represented by the following
schematic diagram , 3.4 ;
i1
<
v1
v2
Black Box
Fig 3.4
and ho= i2 / v2 | at i1=0
( semen's) ;
depicting the open circuit output admittance as shown in the following diagram , Fig
3.5 ;
i1
<
v1
Black Box
v2
Fig 3.5
It is clear that the input circuit can be represented by a Thevinins equivalent (
voltage source ) and the output circuit by a Norton equivalent ( current source ) .
Other equivalent circuits can be used , namely the Z - parameters and the Yparameters equivalent circuits . These equivalent circuits uses either a voltage
source or a current source but not both in the same circuit as in the H - parameters
equivalent circuit . Each type of equivalent circuit may be transformed to the other
types using specified formulas ( refer to text book ) .
The use of the H-parameters offers many advantages over the other types of
parameters , namely :
[1] The H-parameters are easily measurable .
[2] The hybrid equivalent model isolate input from output circuits
[3] Using the Hybrid circuit , the effects of the input circuit and output load , can
be investigated at ease .
3.2.1
The common emitter small signal equivalent circuit .
Consider the following circuit , Fig 3.6 :
<
RC
iB
VCC
VBB
RB
RE
CE
iE
Fig 3.6
and since we are dealing with small signal conditions , linear operation can be
assumed and thus using the superposition concept , the DC and AC load components
can be considered separately , . i.e. , the small signal equivalent circuit for the above
circuit ( replacing capacitors and power supplies with short circuits ) can be redrawn
as follows Fig 3.7 :
ib
ii
vi
RC
RB
Fig 3.7
Since the circuit is a common emitter one , we will add the subscript e to the hybrid
parameters . Such parameters can be found ( practically ) as follows :
hoe = i2 / v2
| i1 =0
= ic / vce
| ib=0
referring to the output characteristics , Fig 3.8 , of the transistor used in the circuit ,
then ic / vce can be calculated , at the operating point Q ( zero AC input current )
. Such a value is expected to be very small ( 10 -4 Semens ) . i.e. hoe may be
represented by a very large resistor in parallel with the load resistance ( much
smaller ) . This follows that hoe can be neglected for most applications .
iC
v CE
Fig 3.8
i.e.
hoe =0
Considering hfe which is given by i2 / i1 | v2=0
= ic / ib | at Q point
then such a parameter can be found also from the output characteristics curves of the
transistor , which should yield the dynamic current gain of the amplifier .
Considering the parameter hie which may be written as ;
hie = v1 / i1 | v2 =0 = vbe / ib | at Q point
= VT / IBQ = hfe VT / ICQ = hfe VT/ IEQ
where VT = 25 mV , at room temperature , refer to the diode notes --- Electronics I )
And finally hre can be written as ;
hre = v1 / v2 | i1 =0
= vbe / vce | i1 =0
= very small
=0
i.e. the complete equivalent small signal circuit for the common emitter amplifier can
be drawn as follows , Fig 3.9 a,b :
hie
RB
vi
ic
+
hie *
vce
hfe *
ib
il
1/hoe
RC
Fig 3.9 a
** Note that , the shaded device is a current generator
ii
ic
ib
vi
-il
hfe *
ib
hie
RB
RC
Fig 3.9 b ( approximated a )
The Ac current gain for the amplifier can be calculated using the following equation
Ai = iL / ii = ( iL / ib ) * ( ib / iL )
and using the current divider principle for the input circuit and knowing that ;
iL = - ic = - hfe * ib ,
then a value for the current gain for the amplifier may be realized .
Similar procedure may be adopted for calculating the input and output impedance’s
for the amplifier as shown in the following example .
Example :
In the following amplifier circuit ( HFE = 30 ) , calculate the voltage gain Av ( = vo/vi
) , the current gain Ai ( = io /ii ) , the input and output impedances .
<
6.8K
1K
CC
iB
2.2K
20 V
0.1K
CE
0.5
K
iE
The small signal equivalent model will be represented as follows :
ii
ic
ib
vi
hie
RB
hfe *
ib
-il
1K
0.5K
To find hie , we need the value of IBQ . Such value can be obtained from DC analysis
of the circuit as follows :
RB = 6.8 x 2.2 / 9 =1.66 K
VB = 20 x 2.2 /9 = 4.89 V
Using the equivalent input circuit ;
IBQ = 4.89 - 0.7 / ( 1.66 K + 0.1 x 30 K ) = 0.9 mA
This follows that , hie = 25mv / 0.9 mA =27.8 Ohms
Z in = RB // hie
Zo = 1K // 0.5 K
Av = vo/vi
vi = ib hie
vo = io x 0.5 K
V
io = ib hfe [ 1/(1+0.5 ) ]
thus Av =ib hfe 0.5 /1.5 ib hie = hfe/3hie =30/83.3=0.36
Ai = io/ii
using current divider principle ;
ib = ii RB/(RB + hie )
Ai = ib hfe RB / [1.5 ib ( RB+hie )] = 30 * 1.66 / 0.885 =52.5
3.2.2
The common base small signal equivalent circuit
Referring to the common base circuit shown in Fig 3.10 , the AC representation of the
circuit will be as shown in Fig 3.11 , and if the concept of the two port theory ( hybrid
model ) is applied to the AC circuit then we will arrive to the small signal equivalent
circuit , as shown in Fig 3.12 , which can be descried in the following hybrid equations
:
ic
Ri ie
iE Ri
Rl iC
R2
R! VCC
vi
Fig 3.10
vi
Rl
il
Fig 3.11
Ri
hib
E
ic
+
ie
hfb *
ie
hrb *
vcb
vi
C
-il
Rl
1/hob
B
Fig 3.12
veb
= hib i1 + hrb vcb
and
ic
= hfb i1 + hob vcb
where ;
hib = veb / i1 = veb / -ie | vcb =0
= = VT / IEQ = = hie / ( hfe +1 )
and ,
hfb = ic / i1 | vcb =0 = ic / -ie | vcb =0
= -  = - hfe / ( hfe + 1 )
and ,
hob = ic / vcb | i1 =0 = very small value =0
= hoe / ( hfe +1 )
and finally ,
hrb = veb / vcb | i1=0 = very samll quantity = 0
Example :
Given the following circuit , determine Zi , Zo, Av and Ai
iE
iC
ii
1K
vo
vi
2V
ic
ie
5K
8V
vi
1
K
20
hfb *
ie
Small sig . cct
Using KVL for the input cct , we may write the following equation :
IEQ =( 2 - 0.7 ) / 1K = 1.3 mA
hib= VT / IEQ = 25mV/1.3 mA = 20 
hfb = -  = -1
thus Zi = 1K // 20 
and Zo = 5 K
Also Ai = il /ii = -ie/-ie = 1
and Av =vo/vi = ( -5 K * ie ) /( -ie * 20 ) = 250
Notice the following :
5
K
il
vo
The common base circuit has low input impedance , it’s current gain is
almost equals to “ ONE “ and it’s voltage gain is high .
3.2.3 The common collector small signal equivalent circuit
Given the common collector circuit shown in Fig 3.13 and it’s equivalent AC circuit
as shown in Fig 3.14 and substituting the hybrid equivalent model ( common emitter )
circuit into the network will yield the following circuit , shown in Fig 3.15 , we may
write following expressions describing the input and output impedances , voltage gain
and current gain :
VCC
RB
ic
ib
vi
vi
RE
ie
RB
vo
RE
Fig 3.12
Fig 3.11
ib
hfe *
ib
hie
vi
zi
RB
io
RE
zo
Fig 3.13
Zi = The input impedance = RB // ( hie + ( 1+ hfe ) * RE )
See Electronic I notes -- ( Reflecting RE into the input circuit )
also ,
Zo = The ouput impedance =RE // ( hie / ( 1+ hfe ) )  Source is short ccted
Again , see Electronic I notes ( Reflecting hie into the output circuit )
i.e
ib = Vi / ( hie + ( 1 + hfe ) * RE )
However ,
ie = ib ( 1+ hfe )
it follows that ie = Vi * ( 1 + hfe ) / ( hie + ( 1 + hfe ) * RE )
= Vi / [ ( hie / ( 1 + hfe ) ) + RE ]
The expression for Zo ( above ) can be used to redraw an alternative circuit for the
output circuit as follows , Fig 3.14 :
hie/ (1+hie)
RE
vi
vo
Input cct seen by output
Fig 3.14
cct
From Fig 3.14 we can deduce an expression for the voltage gain for the circuit as
follows :
Av = vo / vi
= RE / [ RE + ( hie / ( 1+ hfe ) ) ] , using the potential divider principle
Using Fig 3.13 , an expression for the current gain for the circuit may be deduced as
follows :
A i = io / i1
= io * ib / i1 * ib
now , ib / i1 = RB / [ RB + ( hie + (1+hfe ) RE ) ]
and ,
io / ib = 1+ hfe
It follows that Ai = ( 1+ hfe ) * RB / [ RB + ( hie + ( 1+ hfe ) RE ) ]
It is clear that , for the common emitter circuit ,
Z i = v . high
, Zo = v. Low
and
Av = close to 1
and because Av = 1 the common emitter circuit is also called
Emitter Follower
circuit ( the output follows the input ) . Such a circuit is ideal for using as buffer
stage in multi stage amplifiers .
3.2.4 General rules for the analysis of transistor circuits
There are many transistor circuits which do not conform with the common emitter ,
common collector or common base circuits configuration . Such circuits may be
analyzed using the following general rules :
[1] Draw the actual wiring of the circuit neatly
[2] Mark the points B ( base) , C ( collector ) and E ( emitter ) on the circuit
diagram .
[3] Replace each transistor with it’s h - parameter model
[4] Transfer all circuit elements from the actual circuit to the equivalent circuit of
the amplifier . Keeping the relative positions of these elements intact
[5] Replace each independent dc source with it’s internal resistance ( The ideal
voltage source is replaced with a short circuit and the ideal current source is
replaced with an open circuit .
[6] Solve the resultant linear circuit using mesh analysis methods .
3.3 The Re Transistor Model
The Re model relays on the fact that a transistor may be represented by a diode and a
controlled current device
3.3.1 The Common Emitter Case
Referring to the following circuit and its Diode/current source representation ;
ic
ic
C
C
ib

ib
B
B
ie
ie
E
E
Fig 3.15
It is clear that the current through the diode = ib + ib  = ib ( 1+ )
Now if we consider the input impedance of the circuit we may write the following
equation :
Zi = vi /ib
However vi = ib ( 1+ ) * the resistance of the diode ( RE )
Note that RE = VT / IEQ = 26 mV / IEQ
from the diode section
vi = ib  RE
for  >> 1
Zi = Vbe / ib =  RE
thus we may represent the Common Emitter circuit by the following diagram :
C
ib hfe
ib
Zo
B
RE
Zi
E
Fig 3.16
Zo = The resistance between the collector and emitter terminals = very large = 
Similar to the Hybrid case
Using the Zi, Zo and ic=  ib then it follows the RE model for the common Emitter case
may be represented as follows :
ii
vi
ic
ib
hfe *
ib
RE* hfe
Zo
Fig 3.17
3.3.2 The Common Base Case
Similar procedure ( to the common Emitter Case ) may be adopted to represent the
common base circuit as shown in the following diagram :
E
Fig 3.18
B
3.3.3. The Common Collector Case
C
ie
ie
ic
C
 ie
RE
B
There is no need to develop a separate model for the Common Collector case . Instead ,
the Common Emitter case is used .
Solved Examples :
[1]
+ 12V
ic
220
hie=1275
hfe =98
ib
0.22K
vi
3.3K
vo
ie
3.3K
vi
ib
0.22K
1.275K
98 ib
vi
zi
3.3 K
zo
The above Fig shows the actual cct , the AC circuit and the small signal equivalent cct .
respectively .
Zi = 0.22 K // [ 1.275K + (1+98 ) 3.3K ]
== 132 Ohms
Zo = 3.3K // [ 1.275K/( 98+1) ]
== 13 Ohms
Using KVL ( output cct ) ;
Av = vo /vi
vo = 3.3K * ie
vi = 3.3K + [ 1.275K / (98+1 )]
Thus Av = 0.996 == 1
Ai = io / ii
io = ie = ib+ 98 ib
ii = ib * .22K / [ .22K + 1.275K+ 99 * 3.3 K ]
Thus Ai= 40
[2] Given the following circuit , determine Zi , Zo , Av , Ai
ix
12 V
3K
120K
68K
iC
vo
iB
vi
hfe= 140
68K
vi
120 K
Help :
Perform DC analysis for the above circuit in order to find IBQ ( or ICQ )
Find the value of hie
Draw the small signal equivalent circuit and find the required results as in question 1
Ans. : IBQ = 18.5  A , hie = 1.35 K
[3] Given the following circuit , calculate Zi , Zo , Av , Ai :
15V
0.56 K
vi
560K
hfe =65
3.3K
2.2K
0V
vo
3K
vo
iC
0.56 K
ib
vo
560 K
vi
2.2 K
3.3 K
AC circuit
0.56 K
ib
hie
560K
ib hfe
vi
zi
2.2 K
3.3K
vo
Small Sig . cct
Help : DC analysis will yield the value of IBQ and thus hie
Zi = [( 2.2K // 3.3.K ) ( 1+ 65 ) + hie ] // 560 K + 0.56 K
Zo =  [ ( 0.56 K // 560 K + hie ) / ( 1+ 65 ) ] // 560 K  // 3.3 K
Ai = io / ii = ( io / ib ) ( ib / ii )
io / ib = Can be obtained using current divider principle as follows :
io = current through the 2.2 K
= ie * 3.3 K / ( 3.3 K + 2.2 K )
However ie = ib * hfe + ib
and thus an expression for io / ib can be deduced
Similarly an expression for ib / ii can found using the same procedure( current divider
principle
Finally Av can be obtained by using the following equivalent circuit ( input circuit
seen by the output ) :
Rth/(1+hfe)
hie/(1+hfe)
ie
vth
3.3.K / 2.2.K
vo
Where Rth and Vth are the thevenin’s parameters looking into the source as shown in
the following diagram :
0.56K
vi
560 K
vth , R th
R th = 560 K // 0.56 K
v th = vi [ 560 K /( 560 K + 0.56 K ) ]
vo in terms of vth can also be obtained using Fig , therefore vo / vi can be obtained