BJT Circuit Analysis
Author:
Email:
Phone:
Notes Status:
Leigh Milner
lmilner@spri.levels.unisa.edu.au
(08) 8302 3805
Incomplete
Brief: This is an unauthorized personal set of notes on how to understand simple low
frequency analysis of the Bipolar Junction Transistor. The information contained in
these notes is correct to my knowledge and by no means copied from any textbook,
therefore, no references are required.
1
The Start
The Bipolar Junction Transistor (BJT) is an extremely common electronic device to all
forms of electronic circuits. It can be used for a number of useful applications such as
an amplifier, a switch, a buffer, an oscillator, a nonlinear circuit – so forth.
The BJT is made by P and N type semiconductor material, which should be familiar
from the study of diodes. The BJT is a three terminal device (Fig 1).
Collector
Base
Emitter
Figure 1 – The BJT
The three terminals are Base, Collector and Emitter. The emitter terminal always has an
arrow. The collector is always on the opposite side of the emitter and the base is the
other remaining terminal on the left. Note that this is the conventional schematic
diagram of a BJT transistor. Furthermore, there are two types of BJT transistors. They
are the NPN type, and the PNP type. Figure 2 illustrates this:
e
c
c
e
N
b
b
P
b
P
b
N
N
P
e
c
e
(a) NPN
(b) PNP
Figure 2 – The two types of BJT
1
c
The letters b, e, and c have been used for abbreviations for the base, emitter and
collector terminals respectively. An NPN transistor is always drawn with the arrow
pointing outwards whilst the PNP transistor always has the arrow pointing inwards.
And of course, remember that the arrow is always the emitter terminal. So which type is
the transistor in Fig 1? – It is NPN. The other diagrams shown in Figure 2 illustrate
why the transistors are called either NPN or PNP. It’s simply due to the semiconductor
material used for each terminal.
Now, lets take a more detailed look:
Ic
Ie
Ib
Ib
Ie
Ic
Figure 3.
The arrows show the direction of DC current flow for both the NPN and PNP cases. In
both cases the base current (Ib) is a very small current in the order of microamps whilst
the collector current (Ic) and emitter current (Ie) are larger and in the order of
milliamps. Note that for the NPN transistor, the base current flows into the transistor
but for the PNP transistor, the base current flows out the transistor. Also note Ic and Ie
always flow in the same direction and in the direction of the (black) arrow, the same
arrow that tells us whether the transistor is PNP or NPN.
Now for the voltages:
Vc
Vb
Ve
The voltage at the base is normally written as Vb.
The voltage at the collector is normally written as Vc.
The voltage at the emitter is normally written Ve.
That part was easy, but what about the voltage between the collector and the emitter? Is
it written as Vce or Vec? The convention is that the first subscript letter is the voltage
that you are measuring and the second subscript letter is the reference. That means, if:
Vc = 6V (The voltage at the collector is 6 volts)
Ve = 2V (The voltage at the emitter is 2 volts)
2
Then Vce is 4V because the voltage at the collector is 4V higher than the voltage at the
emitter. Also, Vec = -4V because the voltage at the emitter (measuring point) is 4V
lower than the voltage at the collector (reference point). This concept is important and
If you’re a bit lost read it again. The following diagram should summarize. This is the
convention used for measuring voltages between terminals of the NPN and PNP
transistors. The reason for this is that in these examples the first subscript letter is
usually of higher voltage than the second, hence all variables listed below will have
positive values.
Veb
Vcb
Vec
Vce
Vbc
Vbe
Figure 4.
2
Transistor DC Parameters
There are some important equations we need to look at first. Recall that Kirchoffs
Current Law (KCL) states that the sum of all currents entering a node (a point) must
equal the sum of all currents leaving the node. By taking a look at Fig 3 we can see then
that for both the NPN and PNP transistors:
Ie = Ic + Ib
(1)
i.e. Current flowing into the transistor (Ic and Ib) equals current
flowing out of the transistor (Ie) for the NPN, and Current in (Ie)
equals current out (Ic and Ib) for the PNP.
There is a parameter called  (Beta) for every transistor, which is a constant. The value
of  for transistors is normally between 50 – 500. Equation 2 states that the collector
current is  times bigger than the base current. Hence  is simply a ratio between
collector and base current. Recall that the base current is relatively small and the
collector current is relatively large.
Ic = .Ib
(2).
For a transistor with a  =100 and Ic=1mA, then from equation 2, Ib = 10uA. Run
through this in your head to make sure…
We can now substitute equation 2 into equation 1: The highlight shows the substitution
Ie = .Ib + Ib, which simplifies to:
Ie = (+1)Ib. (3)
We can now substitute equation (2) into equation (3) to obtain:
Ib = Ic/
(equation 2 rearranged)
3
Ie = (+1).Ic/ , and rearrange to obtain:
Ic 

(   1)
Ie
We now define a new parameter  (alpha) where


 1
(4)
Hence:
Ic = .Ie.
(5)
And that’s it. I highly recommend you go through the mathematics yourself and verify
every step that I have done. Only after you do this will you fully understand.
In summary you should definitely try to remember the first two following equations as
they crop up all the time. It’s also handy to remember the third one.
Ic = .Ib
Ie = (+1)Ib
 = /+1
E.g. For the transistor with  = 100,  = 100/(100+1) = 0.99
Hence from equation 5 you can see that Ic  Ie. This is true for all transistors with high
. Now take a look at the NPN transistor in Figure 3 again. The reason Ic is only
approximately equal to Ie is because of the small base current that adds in to make Ie
just a little bigger. You are now ready to do DC analysis of transistor circuits:
3
DC Analysis
The DC Analysis of transistor circuits involves solving for all (or most of) the currents
and voltages in the circuit. The most important DC parameters to solve are Ic and Vce.
There is a definition called the “Q-point” of a transistor, which is simply the Values of
Ic and Vce that are present in the transistor circuit. So if you are ever asked to find the
Q point of a transistor, solve for Ic and Vce. Lets look at the following circuit in Figure
5 overleaf. I’ve named the resistors Rb, Rc and Re to mark the base, collector and
emitter resistances respectively. Vcc is usually used for the power supply voltage and
the ground symbol is also shown. The first step in the DC analysis of any transistor
circuit is to solve for one of the unknown currents, i.e, Ic, Ie or Ib. If you solve for one
of these unknowns, the other two can be found just buy using equations 2 and 3 and the
given value of . The recommended way of solving for one of the currents is to write a
Kirchoffs Voltage Law (KVL) loop. Recall that KVL states the sum of all voltages
Va
V1
V2
4
around a closed loop equals zero. The figure on the previous page also shows that the
total voltage between a power supply and ground is equal to the sum of voltage drops
across components in a series connection to ground. If that was a bit too much to
swallow, look at the figure again:
Va = V1 + V2.
Applying this principal, and Ohms law (which states that the voltage drop across a
resistor is equal to the resistance multiplied by the current) we can write the loop as
follows for the path through the base of the transistor to ground:
Vcc = Ib.Rb + Vbe + Ie.Re
(6)
,note that each term has the units of volts.
Also note that if instead of being connected to ground, if Re was connected to another
voltage supply (Vx) the equation would be: Vcc = Ib.Rb + Vbe + Ie.Re + Vx
Vcc
Rc
Ic
Rb
Ib
Vc
Vbe
Ve
Ie
Re
ground (OV)
Figure 5.
It is now an appropriate time to mention that Vbe is approximated as a constant of 0.7V
for transistor circuits. This is because the B-E terminals form a forward biased PN
junction, and you should recall that a forward biased PN junction has approximately a
0.7V voltage drop from diode theory.
Hence with Vcc, Rb, Rc, and Re given and Vbe=0.7V assumed, the only unknowns are
Ie and Ib.
Remember that equation 3 shows a relationship between Ie and Ib?
Ie = (+1)Ib
(3)
We can substitute this into equation (6) for Ie to obtain:
Vcc = Ib.Rb + Vbe + (+1)Ib .Re
(7)
and hence we can solve for Ib:
5
Vcc  Vbe  Ib( Rb  (   1) Re)
Ib 
Vcc  Vbe
Rb  (   1) Re
For typical values of Vcc=12V, Vbe=0.7V, =100, Rb = 100k and Re = 1k
We obtain:
Ib 
12  0.7
 56.2 A
100  10  (100  1).1  10 3
3
and now:
Ic = .Ib = 100  56.2A = 5.62mA
also remember that for large ,   1, hence:
Ie = Ic = 5.62mA, and hence all the currents are solved. Next we can solve for the
voltages:
The voltage at the collector (Vc) is simply the supply voltage minus the voltage drop
across the collector resistance. See Fig 5. Hence:
Vc = Vcc - Ic.Rc.
(8)
If you cant remember whether the signs in the equations for a KVL loop should be
positive or negative, just work out the direction of current and remember that the
voltage arrow goes in the opposite direction:
I
Continuing our example, for Rc = 500 ohms,
Vc = 12 - 5.62mA 500 = 12 - 2.81 = 9.19V
and of course, the voltage at the emitter is simply the emitter resistance multiplied by
the emitter current:
Ve = Ie.Re = 5.62mA  1k = 5.62V
finally, Vce = Vc - Ve = 9.19 - 5.62 = 3.57V
Hence the Q-point of this circuit is Vce = 3.57V, Ic = 5.62mA.
6
Here's another example. The base of the transistor is connected to a voltage divider
circuit between Vcc and ground. Also note that the emitter is biased with a supply
voltage Vee. In this example we will use a negative supply of Vee = -5V.
Vcc
Rc
Ic
R1
Vb
R2
Ie
Re
Vee
Figure 6.
The simplest way of solving this circuit is to just calculate the voltage Vb using the
voltage divider rule and proceed with a KVL loop from Vb, down to through the
transistor to Vee. However, this method assumes that the small base current Ib flowing
into the transistor is actually zero, and hence introduces a small error in the results.
Such an error can be negligible in circuit design. I'll step you through the process:
From inspection of the circuit you can see that the voltage Vb is actually just the
voltage across resistor R2 since R2 is connected to ground. I remember the voltage
divider rule by a phase that goes 'The one you want, over the sum of the other two'. By
that I mean the resistors.
The equation for finding Vb using the voltage divider rule is:
Vb  Vcc
R2
R1  R 2
(9)
Hence the voltage Vb, which is the voltage across R2, is Vcc multiplied by the one you
want (R2) divided by the sum of the other two (R1 + R2). For Vcc = 12V as before ,
and R1=33k, R2 = 10k.
Vb  12
10k ,
10
 12.  2.79V
10k  33k
43
Next you can write the loop through the transistor:
Vb = Vbe + Ie.Re +Vee, and solving for Ie,
7
Vb  Vbe  Vee
Re
2.79  0.7  ( 5)
Ie 
 7.09mA
1k
Ie 
from here it is simple to solve for the remaining currents and voltages as shown before.
However, note that Ve = Ie.Re +Vee = 7.09m  1k + (-5) = 2.09V. This answer could
have been negative if the emitter resistance was a little smaller.
Now, the second method for solving this circuit is to model the base divider as a
Thevenins equivalent circuit.
Vcc
Vcc
Rc
Ic
Vb
Rc
Ic
R1
Rb
Vb
R2
Ie
Ie
Re
Vee
Re
Vee
Figure 7.
The thevenins voltage source is just the voltage at the point of interest, which is Vb in
series with the equivalent resistance of the circuit, which is denoted Rb.
Vb is found by using the voltage divider rule as before, and Rb is just R1 in parallel
with R2. To keep these notes short I won't go into this but you can derive these results
from Thevenins theorem from previous subjects. So, from before:
Vb  Vcc
R2
R1  R 2
and Rb = R1 || R2 = R1.R2/(R1+R2) or 1/( 1/R1 + 1/R2) = 7.67k for R2=10k and
R1=33k.
Writing the same loops as in the first example:
Vb = Ib.Rb + Vbe + Ie.Re + Vee
and substituting in equation 3:
Vb = Ib.Rb + Vbe + Ib(+1)Re + Vee
8
Ib 
Vb  Vbe  Vee
2.79  0.7  ( 5)

 65.2A ,
Rb  (   1) Re 7.67k  (100  1).1k
and hence Ie = (+1)Ib = 101  65.2 = 6.52mA.
Note our previous method produced a small error since the value of Ie obtained was
7.09mA. Again, the error was due to the fact that the base current was assumed to be
zero in the first method.
One last circuit worth mentioning is the 'Self Bias'. It's worth mentioning because It's a
little tricky if you haven't tried it before, and especially if it crops up in a test.
Vcc
Rc
Ic
Vc
Rb
Ie
Re
0V
Figure 8.
Without too much detail, all you have to do is use the Variable Vc in two equations:
Vc = Ib.Rb + Vbe + Ie.Re
Vc = Vcc - Ic.Rc.
solve equal to each other:
Ib.Rb +Vbe +Ie.Re = Vcc - Ic.Rc,
Substitute in for Ib:
Ib.Rb + Vbe + (+1)Ib.Re = Vcc - Ib..Rc
and rearrange for Ib:
Vcc  Vbe
Ib 
, and the circuit is then easily solved.
Rb  (   1) Re   .Rc
9
3
AC Circuit Analysis
AC analysis is pretty tricky to get a grasp on an first but with a good understanding of
the principals and a bit of practice it's a piece of cake.
I prefer to use the Hybrid  model for AC analysis but there are other models equally as
valid. The Hybrid  model for the transistor is shown below: r
ib
Collector
Collector
Base
r
Base
V
gm.V
Emitter
Emitter
Figure 9.
r is used to model the input resistance of the transistor. It's actually modelling the AC
resistance of the forward biased Base-Emitter PN junction. v is just the voltage across
r. Also note that the AC base current, ib, flows through r.
Lower case letters are always used for AC signals. For example, ib is the AC signal
base current whereas Ib is the DC base bias current. The exception to this is V, which I
always seem to write using a captial V, but that's just my personal choice.
The diamond shaped box with the arrow is a 'Voltage Controlled Current Source'. It's
important that you understand what this is. It is a current source, whose current is
controlled by a voltage which is somewhere else in the circuit. In this case, the current
is controlled by the voltage across r, which is V. gm is just a constant which
determines how much of a change in current is caused by a certain change of the
controlling voltage. The units of gm is amps per volt or, A/V.
An example,
for the voltage controlled current source shown above, a V of 5V peak to peak with a
gm of 0.5A/V would result in a current flow of:
0.5A/V x 5V = 2.5A peak to peak.
It's also worth mentioning that from previous subjects, you learned that the resistance of
an ideal current source is infinity. Hence whilst the input resistance to the circuit is r ,
the output resistance of the circuit is infinity. In practice this is not true and the output
resistance is just very large. This effect is modeled by another resistor ro connected
between collector and emitter at the output but is not shown on the above diagram. It
can be omitted as a simplification to the model.
So if gm is a constant for a given amplifier circuit, how do you work out it's value.
There is a relationship relating gm to the collector current, it is:
10
gm 
Ic
VT
VT is another constant which is given by:
kT 1.38  10 23.300K
VT 

 25mV , approximately
q
1.6  10 19
This equation should be familiar from diode study. Hence:
gm 
Ic
Ic

 40. I c (10)
VT 25mV
So all you have to do to find gm is to calculate the DC current Ic and multiply by 40,
remembering the units should be in A/V.
r is easy to calculate as well, it is given by:
r 

(11)
gm
Another useful thing to mention is that the Voltage controlled current source, "gm.V"
is sometimes expressed in another form:
g mV 

.V   .i b (12) , where equation 11 has been rearranged and substituted
r
in for gm and then V / r is of course just equal to ib --> You can see this from ohms
law looking at figure 9. The .ib model for the voltage controlled current source comes
in useful for the 'common collector' and 'common emitter with emitter resistor'
amplifiers.
So that's the Hybrid  model, all you have to do is replace the transistor in your circuit
with this model, and do some analysis.
3.1
Common Emitter Amplifier
The common emitter amplifier is the simplest circuit and therefore is presented first.
The first thing you will note different about this circuit is the capacitors that have been
introduced. The horizontal capacitors on the left and right are just DC blocking
capacitors. The capacitors are assumed to be large enough to be short circuits to AC
signals. The vertical capacitor joined to the emitter is what makes this circuit a
'Common Emitter Amplifier'. The capacitor shorts out the emitter to ground for AC
signals. Other noticeable features of the common emitter amplifier are that the input
signal is applied to the base of the transistor and the output is taken from the collector.
11
Vcc
short
for AC
Rc
Ic
R1
vo
C
c
b
C
e
R2
C
Re
Ie
Figure 10 - Common Emitter Circuit
The first step in drawing an AC Model is to short circuit the supply voltage Vcc, why?
Because a DC voltage supply is a short circuit to AC signals. This is something you just
have to know and learn - A DC supply is a short circuit to AC signals. My method for
drawing the AC equivalent circuit is to start from the left and work to the right.
ib
b
c
Vo
V
Vi
R1
R2
r
gm.V
Rc
e
Figure 11 - AC Model of Common Emitter Circuit
First we have the AC signal source, which is on the left, then we move rightwards. The
coupling capacitor is a short circuit to AC signals, hence it can be ignored. We then
move to the point between the two resistors in figure 9. Notice that both of these
resistors is connected to ground. R2 is connected directly to ground and R1 is
connected to ground via the power supply, hence both these resistors in Figure 9 appear
in parallel to ground. We are now at the base of the transistor b and are ready to insert
the Hybrid  model for the transistor - so that's what we do. Remember that the emitter
of the transistor in Figure 10 is connected to AC ground? - making it a common emitter
amplifier, this agrees with what is happening in Figure 11. The emitter terminal is on
the bottom line which is ground. Now we get to the collector c in figure 10. Notice that
Rc is connected to ground via the power supply on the top - this is the reason Rc is
connected to ground at point c in Figure 11. Finally, the output coupling capacitor in
Figure 10 is of course a short circuit, and can be omitted in Figure 11. And the output
voltage should be marked in Figure 11 also.
12
Now that we have drawn the AC equivalent circuit for the common emitter amplifier,
we can do some analysis to determine the voltage gain of the circuit.
Ri
ib
b
Ro
c
Vo
V
Vi
R2
R1
r
gm.V
Rc
e
Starting at the output we notice that the current gm.V is flowing in a loop through Rc.
Notice that the direction of current causes the voltage polarity across the resistor as
indicated. This means that the output voltage being produced is negative because
remember, the bottom line is ground. using ohms law, we can simply see that:
Vo = -(gm.V).
Rc
(13)
, current multiplied by resistance.
DONT FORGET THE MINUS SIGN! - which indicates that the voltage is negative.
Now we can rearrange equation 13 a little:
Vo
  g m Rc
V
Do you notice that Vi = V in the above equation? I hope so because things in parallel
always have the same voltage across them.
V = Vi ,
hence,
Vo
 Av   g m Rc
Vi
(14)
This is the expression for the voltage gain of the amplifier.
The input resistance of the amplifier Ri can be seen from inspection to be the parallel of
all the resistors in the left part of the circuit, hence:
Ri = R1||R2||r
(15)
The output resistance of the circuit can be seen from inspection to be just Rc, because
remember that the resistance of an ideal current source is infinity. And an infinite
resistance in parallel has no effect.
Ro = Rc
(16),
This completes the common emitter amplifier, it's easy.
13
3.2
Common Emitter with Emitter Resistance
The 'Common Emitter with Emitter Resistance' amplifier is slightly different from the
CE amplifier (3.1) in that there is no emitter bypass capacitor. The emitter is no longer
at AC ground.
Vcc
short
for AC
Rc
Ic
R1
C
c
b
C
e
R2
Re
Ie
Figure 12 - Common Emitter with Emitter resistance.
The solution of the bias currents and voltages is obviously the same as for the CE case.
The AC model is a little different though, and requires a different approach.
ib
b
c
Vo
Vi
R1
R2
r
gm.V
Rc
e
Re
Figure 13 - AC Model for CE amp with Emitter Resistance.
Again, as we approach from the left we have the AC source, and the horizontal DC
coupling capacitor is a short circuit to AC and is hence omitted. At the base of the
transistor we have the two resistors R1 and R2 connected to ground, remembering that
R1 is also connected to ground because the top DC supply Vcc is a short circuit to AC
signals, and remember of course that we are trying to draw and AC equivalent circuit.
At the base we can insert the Hybrid- model for the transistor. Remember that the
emitter is NOT directly connected to AC ground. The emitter is connected to ground
via the emitter resistance Re. The rest of the Hybrid- model includes the Voltage
controlled current source, and from the collector we have Rc going to ground because
Vcc is a ground for AC signals. Also, the output coupling capacitor is omitted, and I
needent repeat the reason for this again.
14
ib
b
c
Vo
Vi
r
R2
R1
gm.V
Rc
e
ie
Re
x
y
Figure 14
Starting at the output, we draw a current arrow through the Resistor Rc as we did before
for the common emitter amplifier. For the common emitter amplifier the current arrow
was just equal to gm.V , but in this case it isn't that obvious. Could the current arrow
instead be ie , which is flowing through the emitter resistor? This is a bit confusing but
it will soon be explained. Remember that the bottom line in the AC model is ground.
This means that points x and y are just ground. If the current arrow drawn was ie, this
would mean that ie would have to flow from points x to y in order to loop through and
travel through Rc. This is impossible because once a current reaches ground, it stays at
ground (disappears).
c
Vo
gm.V
Rc
e
Re
Figure 15
Another way of clearing up this confusion would be to draw the circuit shown above
(only output shown). In this cirucit, instead of all the grounds joined together with a
common line on the bottom, they are shown individually. (Note that this is exactly the
same). It is therefore visually easier to see that the current arrow at the output is indeed
just gm.V. The point I wanted to make after all that was: just remember that although
the bottom line is used to join up all the grounds at the bottom, NO current can flow
along this line.
Now, the equation for Vo is exactly the same as for the common emitter circuit,
Vo = - gmV.
Rc
(17).
(ohms law - current times resistance)
Remember that the minus sign shows the voltage is more negative at the top of the
resistor compared to the bottom of the resistor (which is at ground).
Now for Vi.
15
You can see that Vi is equal to the voltage across r plus the voltage across Re (Ve).
Look at Figure 14 and verify this for yourself. The two voltage arrows show that the
voltages just sum to equal Vi. - Remember that things connected in parallel have the
same voltage across them.
Hence we can write,
Vi = V + Ve. (18)
To work out Ve, we have to find the current flowing through the emitter resistance and
simply multiply by the resistance (ohms law). - Ve = Ie.Re.
Using Kirchoffs Current Law (KCL) at the emitter node, which states that the sum of
currents entering a node equals the sum of currents leaving the node, we can see that
the emitter current ie is equal to the current flowing through r plus the current flowing
from the 'Voltage controlled current source (gmV).
c
Vo
V
ib
r
gm.V
e
ie
Re
Figure 16
There are two ways to proceed from here. You can use the gm.V model for the current
source, which is shown in the above diagram, or you can use the ib (equation 12, page
11) model. I will show both ways, and that the gain of the amplifier turns out to be the
same either way.
The gm.V way first,
using KCL,
ie = V/ r + gm.V,
Notice that V/ r has been used for the base current (ohms law). We want to do this
because the current source has V in it, and this will allow easy simplification using
algebra.
We can now substitute this ie into equation (18), noting that the equation Ve = Ie.Re has
also been used:
Vi = V + (V/ r + gm.V).Re
(19)
Now we are ready to divide Vo by Vi and start simplifying the equation:
16
Vo
 Av 
Vi
 g mV Rc
V
V  (   g mV ) Re
r
You can see that V is present in every term in the above equation, hence it can be
cancelled,
Vo
 Av 
Vi
 g m .Rc
1
1  (  g m ) Re
r
Next we will use the rearranged version of equation 11 to substitute in for gm
gm 

r
Vo
 Av 
Vi


.Rc
r
1 
1  (  ) Re
r r
Next multiply top and bottom line by r ,
Vo
  .Rc
(20)
, and this is the answer. When you are asked this
 Av 
Vi
r  (   1) Re
question in a test or exam this is the result you should get. However, you may simplify
the equation further will some assumptions:
Recall that in the transistor circuits you have studied so far the collector current is in the
order of milliamps. Lets take the example of Ic = 1mA. Using the same method as
shown in equations 10 and 11, you can calculate the r for the circuit.


100
 2.5k , this is for a transistor with a typical beta of 100,
g m 40 Ic 40mA / v
which is close enough. Hence, for most transistor circuits the r is in the order of 1 to
10k. Recall that the value for Re in a typical circuit is in the order 1 to 10k also. Now
take a look at the bottom line of equation 20. You can see that because Re is multiplied
by the (+1) factor, it is much larger than r.
r 


(+1).Re >> r
Hence because the (+1).Re term on the bottom line of equation 20 dominates, r can
be ignored, Hence the equation can be simplified to:
Vo
  .Rc
 Av 
Vi
(   1) Re

Also note that
is  1 for large , hence the equation further simplifies to:
 1
17
Vo
 Rc
 Av 
Vi
Re
(21).
This result shows that the voltage gain of the 'Common Emitter with Emitter
Resistance' amplifier is approximately equal to the effective collector resistance divided
by the effective emitter resistance. If you designed an amplifier like this with Rc=10k
and Re=1k, the theoretical gain would be -10, but note the actual gain obtained in a real
circuit would be just a little less because of the assumptions made along the way to
obtain equation 21.
Now we'll solve the circuit again, but this time using the ib model for the transistor,
c
Vo
V
ib
r
ib
e
ie
Figure 17
Re
This time I'll go through the maths a lot faster, and let you work it out.
ie = ib + ib
hence,
Vi = ib.r + (ib + ib).Re
Vi = ibr
+ ib.(+1)Re
Vi = ib(r
+
(+1)Re) hence,
  .ib .Rc
Vo
 Av 
Vi
ib ( r  (   1) Re)
Vo
  .Rc
 Av 
Vi
r  (   1) Re
, and now ib just cancels off:
, and the result is the same as equation 20.
Taking a look at the AC model again, we can now work out the input and output
impedance's of the amplifier.
ib
b
c
Vo
18
Vi
R1
R2
r
gm.V
e
Rc
The input impedance is simply R1 in parallel with R2 in parallel with the resistance
combination in grey. You will notice that we have r and Re in a series combination in
the grey part. The challenge is to find the equivalent series resistance of the two, and it
isn't just a case of adding them together. The reason for this is that the same current
does not flow in each of the transistors. You can only sum two resistors in series if the
same current flows in each of the resistors, which is of course what happens when you
put two resistors in series. Technically then, r and Re are not in series in the grey
circuit above.
Ri
ib
r
Figure 19
Magic point
ib(+1)
Re
Recall that from equation 3:
Ie = (+1)Ib, The emitter current is (+1) times the base current.
hence,
Vi = ibr
+ ib.(+1)Re
(22)
Now, the input impedance, shown by the arrow is just the voltage in divided by the
current in. This is important to remember, it crops up all the time in these circuits. The
impedance looking into a stage is simply the voltage in divided by the current in. Since
Vi is the voltage in and ib is the current in then Ri = Vi/ib.
Hence by rearranging equation 22,
Ri = Vi/ib = r + (+1)Re
This is can be summarized by a rule called the beta reflection rule which states that:
- When you move a resistance from the emitter to the base you must multiply it by
(+1).
19
- When you move a resistance from the base to the emitter you must divide it by (+1).
Hence you can see from inspection that Ri is just r + (+1)Re , because the resistors
are in series (technically) and when you move the emitter resistance from where it is to
up with r you must multiply it by (+1). I have heard of some people calling this the
magic point, I have no idea why.
Finally, the input impedance to the AC circuit is:
Ri = R1||R1|| (r
+
(+1)Re)
The output impedance is the same as the common emitter amplifier, and for exactly the
same reasons (please refer).
Ro = Rc.
3.3
Common Collector Amplifier
This is also known as the emitter follower amplifier. It is loosely called this because the
AC voltage at the emitter approximately follows the AC voltage at the base.
Vcc
short
for AC
Ic
Figure 20
R1
C
c
b
C
e
R2
Vo
Re
Ie
In the common collector amplifier the input is again applied to the base but the output
is taken as the voltage across the emitter resistor. You will also notice that there is no
collector resistor Rc. The collector is at AC ground, hence why this circuit is called a
common collector circuit.
ib
b
c
20
Vi
R1
R2
r
gm.V
Vo
The analysis is very similar to the 'Common Emitter with Emitter Resistance Circuit',
and I won't go into as much detail for that reason.
Vi = ib.r + (ib + ib).Re
Vi = ibr
+ ib.(+1)Re
Vi = ib(r
+
(+1)Re)
The voltage out is just the voltage across Re, hence:
Vo = ib.(+1)Re.
Notice that unlike the other amplifier studied so far, the equation for Vo does not have a
negative sign. You can see this because the current flows through Re from top to
bottom, which makes the voltage at the top of the resistor higher than the voltage at the
bottom of the resistor, Hence Vo is positive.
The voltage gain is then:
ib (   1) Re
Vo
 Av 
Vi
ib ( r  (   1) Re)
, ib can again cancel from the top and bottom line:
Vo
(   1) Re
 Av 
Vi
r  (   1) Re
Recall the simplification before that
(+1).Re >> r
Hence, r can be ignored from the bottom line and the voltage gain becomes:
Vo
(   1) Re
 Av 
1
Vi
(   1) Re
Hence the voltage gain for the common collector amplifier is approximately 1.
21
The input impedance is exactly the same as for the case of the 'Common Emitter with
Emitter Resistance' amplifier and is given by:
Ri = R1||R1||(r
+
(+1)Re). Please refer to pages 19 and 20 for an explanation of this.
The output impedance is again found using the beta reflection rule.
The result is:
r
, I'll explain this futher in a later revision of these notes, but it's
 1
getting late and I want to go to sleep......
Ro = Re ||
22