Small Signal Analysis of BJT Video Lecture - Electrical Engineering (EE)

we now like to use our small signal
model for the transistor to analyze some
actual circuits there's a few steps that
we have to follow the first step when
analyzing a circuit is to short-circuit
the capacitors and the DC power supplies
remember that coupling capacitors are
chosen so that they provide DC isolation
that is they do not allow DC to flow
from one circuit to another also those
capacitors are chosen so that at the
frequencies that were interested in
amplifying the impedance of those
capacitors is very small and we replace
them with short circuits because we're
assuming that the impedance of those
capacitors is very small compared to the
other components that they're connected
to and we also know that a DC power
supply looks like an infinite capacitor
so we short that out as well we then
redraw the circuit so that it's as
simple as possible right the simple of
the circuit that list chance will have
of screwing it up we then insert the
small signal model for all of the
transistors and then we analyze the
circuit whatever that means it usually
means we're interested in things like
the small signal voltage gain the input
resistance the output resistance there's
a few other things like the current gain
and the power gain but we're not really
interested in those this year so let's
see how we might put this into practice
here's a circuit this is an amplifier
that actually works you can build this
and it will amplify signals no problem
so we've got a potential divider bias
Network here we're going to collect a
resistor we've got an input coupling
capacitor that couples the AC input
voltage to the base we have
output coupling capacitor that subtracts
the DC voltage at the collector so that
we can have an AC voltage across the
load RL so the first step is to
short-circuit the capacitors so C 2 is
replaced with a piece of wire C 1 is
replaced with a piece of wire and VC C
is replaced with a piece of wire which
means is a piece of wire that goes from
VCC down to our reference rail here and
when we do that we end up with this
circuit here all right so we've got rid
of our DC power supplies that means that
our AC signal is now driving everything
right the energy for our whatever is
happening in the AC world is coming from
this circuit here oh sorry there's some
the source here alright so we now have
to redraw the circuit as so it's as
simple as possible well there's a few
simplifications we can make do you see
that RC is in parallel with RL so this
end of RC and RL are connected together
at the collector terminal but so are the
other ends they are also connected
together so RC and RL are in parallel
also r1 and r2 are in parallel they are
connected here but you can see that the
other ends are also connected so we can
replace these two resistors with a
single parallel combination we can
combine RC and RL in a single parallel
combination that's about all we can do
and when we do that we end up with a
circuit here and this one's quite a bit
simpler we've now just got three
resistors the transistor and the AC
source that's driving the circuit
so there are no more simplifications we
can make so our job now is to replace
the transistor with its small signal
model and sometimes people come unstuck
doing that so if you're in any doubt
about which terminal goes where just
right on the circuit what the relevant
terminal labels are and just make sure
that they go to the correct terminals of
the model so when we insert the model we
end up with the circuit here and you can
see that the emitter goes to the emitter
resistor the collector goes to the
parallel combination of RC and RL and so
on everything else looks fine and now
all we have to do is analyze the circuit
now which model do you put in here
GM vbe or beta IB if I don't tell you
which ones put in then you can choose
whichever you prefer more often than not
I'll give you a clue as to which one to
use all right so why don't we have a go
at analyzing a few actual circuits and
we'll see how that goes following these
four steps all right so here's a similar
looking circuit to the one we've just
been examining when discussing the
general principles again resistor
divider biasing Network for the
transistor we've got the collector
resistor and the emitter resistor but
now we have also what's called the
emitter bypass capacitor so when we
short-circuit all of the capacitors will
be short-circuiting this one as well
which effectively removes our e as far
as small signals are concerned re is
still there for DC because remember at
DC CE e has in
that compares to the sorry infinite
impedance therefore it doesn't really do
anything as far as DC is concerned but
it gets removed when we're when we're
doing the AC analysis so let's see what
we can do the first thing we have to do
is come up with the small signal
equivalent circuit so we have to redraw
this so let's have a go at doing that so
can I fit all this in first thing we had
to do was short-circuit the capacitors
so let me just try and draw this all
right so Ari has disappeared because we
have short-circuited the emitter bypass
capacitor which took like r1 and r2 and
I see this is our output voltage here
here is our input voltage V in so that's
the first step short circuit the power
supply short circuit the input coupling
capacitor the output coupling capacitor
and the emitter bypass capacitor and now
we've got to redraw this so that it's as
simple as possible well it's um it's
pretty clear again that our 1 and R 2
are in parallel C is all by itself but
we'll draw it sort of down so we don't
have this Y going all around here
there's no load resistor in this circuit
so there's nothing to go in parallel
with our C so let's redraw the circuit
that's it it's relatively
straightforward the output is still
taken from the collector so that hasn't
changed so this is still the out let's
see this is our one in parallel with our
two and this is V in alright the next
thing we have to do is replace the
transistor with its small signal model
so let's do that
and we end up with this so I want an R
to stay there
now between the base and the emitter
we've got our PI so there's our PI from
the collector to the emitter we've got
our voltage sorry our current source and
across the current source we've got our
collector resistor so we've got our CV R
V out I'll just put this as beta IB
there's ID this is our PI this is our
one in parallel with our two and this is
V in so that's about as simple as it can
get right two resistors on the input
side one resistor on the output side PI
in the current source so we're now
interested in doing some calculations
for example working out the ratio of the
output voltage to the input voltage and
that will give us the small signal
voltage gain or we might be interested
in calculating the input impedance of
the circuit or the output impedance so
let's move on and have a look at those
so here's a slightly neater version of
what I've just drawn and we're
interested in calculating say thee which
will surely do first the the small
signal voltage gain so we're interested
in say a V which is equal to V out
divided by the in well let's see what we
can do here can we come up with
expressions for V out and V in probably
do you agree that V out is equal to
minus and I'm going to use the current
controlled current source for this one I
think the out is equal to minus ICRC
minus because the current is flowing up
through RC therefore the bottom is going
to be more positive than the top or the
top is more negative than the bottom the
bottom is already at zero so we must
have a negative there so V out is equal
to minus IC times RC which is minus beta
IB see alright there wasn't so hard
what about V n well V n is equal to
whatever this current is times r1 and
parallel with our two but that won't
lead us anywhere useful because we will
have an undecided current i1 and in
order to make things work out we'd have
to somehow Express i1 in terms of either
I'd be or IC so that's not useful this
is much more useful VN is as you can see
here just equal to vbe and VB is just
equal to IB times R PI and you see
crucially we've got an IB in both of
these equations which will cancel out so
we take those and we now substitute them
into our expression for a V a V equals V
out which is minus beta IB times our c /
ib times our pi and so the result is
minus beta c / r pi that's it that's a
small signal voltage game now it's
always a good idea to pause and think
about whether this result makes sense
the voltage gain according to our
equation has to be you know it has to
obey the laws of physics we need to do a
few sort of ordered checks on it to make
sure it's okay the one I like to start
with is doesn't have the right units
well beta is dimensionless RC has units
of ohms and our PI has units of ohms so
overall this thing is dimensionless so
that's good AV is directly proportional
to RC that sounds reasonable
right because if we go back to our
original signal circuit rather when the
voltage here goes up the base current
goes up so the collector current goes up
but the output voltage is taken across
RC we can see that from here so when the
voltage here goes down the voltage
across RC goes up
so if RC is bigger then the product IC
times RC is going to be bigger so it
seems perfectly reasonable that AV
should be directly proportional to RC
the other thing is that AV is negative
does that seem reasonable well again it
it does seem reasonable because as I
just said a second ago when the voltage
here goes up the voltage here go
down so if we had a sine wave which was
increasing here we'd have a sine wave
which was decreasing here that
corresponds to essentially a 180 degrees
phase shift and that's the same as
having a minus sign so we feel pretty
happy with this result it looks okay to
me
and this is the small signal voltage
gain why don't we now have a go at
calculating the small signal input
resistance so the small signal input
resistance is by definition the input
voltage divided by the input current the
input current isn't shown on here but it
clearly has to be that current there so
let's see what we can do
RN is equal to V n divided by n now
again we'd like to build up an
expression where we can cancel out
things like voltages and currents
because we're always looking for an
expression which depends only on
component values not things like
currents and voltages so let's try and
Express I in in terms of VN and see what
happens well I in clearly splits at this
node into the current that flows down
this parallel resistor branch and r1 and
r2 will call that I 1 the other current
is clearly IB so we could write our in
equals V in over I 1 plus IB well I 1 is
just V in divided by the parallel
combination right so let's put that in
here the N divided by our 1
hello with our - now what's VB I sorry I
be well ib is just V in / R PI right so
that would be V in over V in over 1 in
parallel with our 2 + V in over our PI
now we can see that all of the viens
cancel and we're left with an expression
which is clearly the formula for the
parallel combination of our one in
parallel with our two and then in
parallel with our pi so the final result
is our one in parallel with our two in
parallel with our PI how do we feel
about this
well dimensionally it's clearly all ok
did we really have to go to this length
can't we just look at the circuit and
say well clearly this is in parallel
with this and in this case yes we could
have done that we could have seen by
inspection these two resistors are in
parallel and we would have gone straight
to this answer here the trouble is that
it's not always this simple and so we're
just trying to build up a little bit of
experience in doing this more methodical
approach later on we might have a
resistor connecting the base and the
collector or there might be some emitter
resistor which is going to foul up the
approach and we can no longer just stare
at the circuit and figure out what is in
parallel with what something may not be
in parallel with anything okay things
aren't always either series or parallel
it's possible to be neither but in this
case it was pretty straightforward and
we've got this result here okay it's all
going well so far
shall we have a go at the output
resistance so the output resistance if
you recall requires that we short
circuit the input and apply a notional
voltage to the output and that will
drive a current into the output
terminals and we need to develop an
expression for that so our out is equal
to this notional applied voltage the X
to this that the resulting current that
flows when V in is equal to 0 so the
idea is that we apply a voltage source
here and we will call this the X it's
going to drive a current I X into the
output terminals and at the same time we
short circuit the input okay so that's
short circuited r1 and r2 and also vbe
do you see that if the if if our PI is
short circuited then vbe must be 0 if
vbe is 0 then IB is zero therefore IC is
0 so this current source will not be
generating any current which means that
the only component that's left in the
circuit is RC so that means that the
circuit can be replaced with just this
and it's pretty clear that now our out
is VX divided by X with V in equal to
zero and that must be equal to RC all
right
again that looks okay to me there's I
don't think there's anything contentious
going on here
the key to this is recognizing that this
short-circuited input shorts out our PI
which causes IB to be 0 which causes IC
to be 0 basically eliminating everything
except the collector resistor we haven't
ended up with anything stupid like our
out equals 1 over RC or you know minus
RC or anything like that so we're pretty
happy with this so always order your
results always take a moment to think
about what you've written down does it
make sense does it obey the laws of
physics have I done anything stupid
let's have a look at another example
here's a more interesting circuit here
we've got what's called a common emitter
amplifier again but this time we've got
some feedback from the collector to the
base we've got our coupling capacitors C
1 and C 2 and we'd like to do what we
just did with the previous circuit and
that is calculate the small signal
voltage gain our in and our out so the
first thing we have to do is redraw the
circuit after short-circuiting the
capacitors and DC supplies so let's have
a go at that you can keep that previous
circuit in your mind while I draw this
now this is a particular
circuit but it's not all series of all
parallel this feedback resistor here is
now connecting the output loop to the
input loop we'll see what that does for
us so let's try and do a calculation
again here's a neat aversion for what
I've just drawn let's start with the
voltage gain all right so what do we do
let's just remind ourselves what the
voltage gain is it's the output voltage
over the input voltage can we just build
up expressions like we did last time
well last time we just worked out the
voltage across RC by multiplying the
current through it by the value of the
resistance well clearly V out is equal
to I 1 times RC but what is R 1 I 1 does
I see flow appear well part of it does
but it's complicated because now we
don't have a clear simple loop here some
of the current can disappear through our
if so we're going to have to try
something else the sensible thing in my
view would be to do a nodal analysis at
this node here and see what happens so
but it's tri nodal analysis and we'll
we'll try it at the node V out so here
this is all the node V out if we sum the
currents leaving this node
we've got I 1 plus IC plus I if so let's
see where that goes
so we know that I 1 plus I see plus bi
if equal 0
what's I 1
well it's V out over RC what's I see
well it's beta IB and once I if it's V
out - million over RF and that must be
equal to 0 now we could actually make an
interesting simplification here and
instead of having beta IB for IC we
could make your life easier by choosing
GM vbe why is that well because GM vbe
vbe is the same as V end so we would
replace this term here with GM VN let's
do that so we get V out of RC plus G in
the end plus V out over RF minus the end
over RF equal zero alright so let's just
try and collect some terms so let's
click the V out on the left hand side
and we'll put V in on the right hand
side so we end up with V out into 1 over
RC plus 1 over RF equals V in and we'll
put a minus 1 over RF oops
plus 1 over RF there cause it's going to
be on the right-hand side
- gene now do you see that this thing
here that's one over C and parallel with
our if so let's write it like this
we've got V out divided by C and
parallel with our F equals I'm just
going to reverse the order of these two
things I've got a V in x - I screwed
that up the end sorry - V in into G in -
whenever R if you'll see why I do that
in a minute
and now I want to get V out over the end
so I'll multiply both sides by the left
hand side denominator and then I'll
divide both sides by V in so AV which is
V out over the N is minus g-m - whatever
our F times C in parallel with our reef
alright well good Lord how do we audit
this does this make sense
well dimensionally everything looks okay
I've got Siemens Siemens hose so Siemens
times ohms is dimensionless the voltage
gain is dimensionless
so that looks okay RC in parallel with
our if well you know at least one end of
them is connected together the maths has
conspired to somehow make them give us
this expression this doesn't mean that
they really are in parallel right it's
the details of the mathematics I've
pulled the minus sign out the front here
because I know that the voltage gain
should be negative why is that well just
because we keep dealing with these
common emitter circuits so as the
voltage here goes up the current goes up
the collector current goes up the
voltage across our C goes up so the
collector emitter voltage goes down so
when this is going up this is going down
which corresponds to a negative voltage
gain so I'm okay with this I think we'll
stick with that but you can see there's
no guarantee that things are going to
work out pretty you're elegant
such as this example here all right what
about the input resistance let's crack
on the input resistance R in our in
equals V n divided by I in all right
well here's I in so R in because V in
over I in that's the end over IB - I if
you know do you see that we can't just
stare at this circuit and by inspection
write down what the input resistance is
it's not like our peers in parallel with
our air or in parallel with our if in
the series with our C it's just not that
complicates just too complicated to be
able to to figure it out we're going to
have to just use our mathematical tools
to to nut it out all right well let's
Ford you here then V in / ib what's IB
well IB is V in of
our pie what's - I if well I if is V out
- VN / R if so - I F is V in minus V out
R if so I'm gonna write that as V in
minus IV out over RF hmm well can't
quite cancel all the viens can we write
these guys can all cancel but it's not
going to cancel with that but we can
write V out in terms of V n can't we
because we know that V out is equal to
the small signal voltage gain times the
input so that's minus AV v out the inner
mother over RF and now when we divide
top and bottom by the end we end up with
1 over whatever our PI plus 1 minus AV
over RF which means that this thing is
equal to our PI in parallel with our if
over 1 minus AV well that's nice it's
does involve the voltage gain here but
if we substitute our expression for the
voltage gain we're getting end up with a
hideous miss I prefer to leave it like
this
is it possible for this thing to blow up
like if a V is equal to 1 we're gonna
end up with parry four over zero which
is infinity well no because remember AV
is negative so that's always going to
come out to be a positive number down
they're dimensionally it looks okay
we've got ohms and palate with ohms
because AV is dimensionless so again
this is looking okay good let's do one
more thing let's see if we can work out
the output resistance okay
reminding ourselves our out equals V X
over I X with me in equal to 0 here we
go there's the X here's I X and here is
the input shorted out okay
now the input is shorted out that means
that vbe is equal to 0 if vbe U is equal
to 0 then IB is equal to 0 therefore IC
is equal to 0 and therefore we if we
were to redraw the circuit we just have
this here's our if with I if flowing
down there
here is our C with i1 flowing down there
here is the X driving a current I X in
there it's pretty clear that our out is
equal to VX / IX with V in and put a 0
that must be equal to C in parallel with
our F quick AutoCheck dimensionally it's
fine
I don't see anything wrong here we've
shorted out vbe IB has to go to 0 IC has
to go to 0 that just leaves RF connected
to ground in parallel with C so we can
write that down by inspection okay
that's transistor small signal analysis