Transistor Small Signal Analysis

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

Channel: Mark Andrews

Views: 141,068

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Id: 18x_xFnq4-M

Length: 36min 26sec (2186 seconds)

Published: Sat Jun 09 2018