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Multistage amplifiers

  • Thread starter Drao92
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  • #1
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Greetings,
I want to tell me if i calculated well the following amplifications.
I took several cases when the first stage is common emitter/base/collector connected to second stage with common emiiter/base/collector.
Firsly i calculated the impedance of the stage 2 amplifier seen by the first amplifier and also i want to confirm me if my logic is good. When we connect the 2 amplifiers, each amplifier acts like an impedance for the other amplifier???
So, i calculated the impedance of the second stage seen by the first stage when the second stage is connected to the first stage with its emitter(RTA in photo), base(RTB in photo) and collector (RTC in photo).
Then i took all posibilities i could imagine of connections and i calculated the voltage gain of the first amplifier and i noted on circuit the second amplifier with RTA, RTB and RTC.
Im not sure on my work on calculating RTC, number 2 case when T1 is common base and the emitter is connected to the second stage.
RT(A,B,C) means that the impedance could be RTA, RTB or RTC, any of these 3.
Also i can have some sign mistakes on amplifications but ignore them.
Another question.
It seems like the impedance of the second amplifier seen by the first amplifier is always the input impedance of the second amplifier, its this somthing like a rule???
 

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  • #2
The Electrician
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Certainly the input impedance of the second transistor provides a load on the output of the first transistor, and the output impedance of the first transistor is the source impedance driving the second transistor.

However, some of the possible configurations just won't work in the real world. You can't use the grounded emitter configuration with the collector as input and the base as output, for example. The collector can't be an input in any configuration, and the base can't be an output.

Otherwise, you can cascade any allowable configurations, and calculate an overall gain. I can't say whether your results are correct because your images are mostly unreadable. You would need to do over your schematics, using a black ink and printing legibly.
 
  • #3
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When we connect the 2 amplifiers, each amplifier acts like an impedance for the other amplifier???
Correct.

It seems like the impedance of the second amplifier seen by the first amplifier is always the input impedance of the second amplifier, its this somthing like a rule???
Not always. Designers of multistage amplifiers make an effort to match the input impedance of the second stage to the output impedance of the first stage because this maximizes power gain. This can be done with resistor pads or transformers and at RF frequencies by capacitors and inductors.
 
  • #4
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Thanks very much for the informations. I think they are correct because i solved many problems using this and i got only correct answers.
 
  • #5
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Deleted
 
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  • #7
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Offhand it seems that the value of Vcc is very critical. What voltage are you planning to use for Vcc?
 
  • #8
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20 v
 
  • #9
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What are trying to do with this amplifier?
 
  • #10
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Its an exercise
Im trying to calculate its voltage gain and in the situation posted above im triying to calculate the voltage gain of the first stage.
 
  • #11
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I have another problem in which im not sure what im doing.
The T3 transistor doesnt amplify anything and it acts as an impedance and im not sure what`s the equivalent impedance.
If i use Y-A transform for Rbe,R5 and R2 the equivalent impedance is ((R2*R5)/(rbe+R2+R5)+((R2*rbe)/(rbe+R2+R5)+R3)||infinite impedance of current source)=(R2*R5)/(rbe+R2+R5)+((R2*rbe)/(rbe+R2+R5)+R3)
https://www.physicsforums.com/attachment.php?attachmentid=59024&stc=1&d=1369554518
 

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  • #12
NascentOxygen
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I have another problem in which im not sure what im doing.
I'm having trouble following what is going on. You have intentionally omitted all biasing arrangements, have you?

PNP T2 has its base grounded, so its emitter is +0.6V or so. This seems barely sufficient for T1 to operate. It also gives PNP T3 a collector voltage that is not only too small, but is also the wrong polarity (unless you are deliberately using it in the inverted mode?).

If i use Y-A transform for Rbe,R5 and R2 the equivalent impedance is ((R2*R5)/(rbe+R2+R5)+((R2*rbe)/(rbe+R2+R5)+R3)||infinite impedance of current source)=(R2*R5)/(rbe+R2+R5)+((R2*rbe)/(rbe+R2+R5)+R3)
I am not au fait with a Y-A transform, but it seems that, for typical circuit values, your expression evaluates to basically R3.

I evaluated the 'active load' arrangement from basics, and (assuming no errors) find the impedance presented is of the order of R3/β when operating.

Any idea what the answer should be, so we can check who is closest to being right?
 
  • #13
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No sorry, i dont have answers.
Now i think its R3/β when im looking at the second circuit from photo.
I dont know how to calculate the impedance in that circuit. Let I3=current in R3 resistor, then V=I3*R3+Ib*rbe+(β+1)*Ib*R5
The current in V source is I=β*ib+I3.
And to find the impedance i do V/I but here i have I3 which makes me troubles.
ALso, i realised its not correct as i did because if we calculate the equivalent impedance of common base from emmiter to ground the impedance is rbe/(β+1) but if we do as i did above it would be infinite impedance||rbe=rbe so i omitted 1/β

About the problem above, i dont think the circuit does anything, its just practice problem from homework so i dont know if it works in reality yet :D. The course is more about circuit analysis. Next semester we will do serious stuff.
 
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  • #14
NascentOxygen
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I dont know how to calculate the impedance in that circuit. Let I3=current in R3 resistor, then V=I3*R3+Ib*rbe+(β+1)*Ib*R5
The current in V source is I=β*ib+I3.
And to find the impedance i do V/I but here i have I3 which makes me troubles.
You can express Vb in terms of Ib for the transistor, then express I3 in terms of Vb and Ib for the potential divider.
 
  • #15
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This is so complicated.
I3=R3/(R2+R3)*Ib+Ib*rbe/R2+Ib*β*R5/R2is aproximative βIb.
Vb=R2*R3/(R2+R3)ib+ib*rbe+Ib*β*R5
V=β*ib*(R3+R2)
V/I=R3+R2 :confused:
R5=2.5k
R2=5k
R3=5k
rbe =1.25k
 
  • #16
NascentOxygen
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R5=2.5k
R2=5k
R3=5k
rbe =1.25k
Where have these values sprung from? On the face of it, those seem absurdly low values for R2 and R3.
 
  • #18
NascentOxygen
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I made aproximations when i said its 5k,2.5k etc.
You just guessed rbe did you??

The subcircuit has a resemblance to a conventional single transistor constant current source. The base bias of T2 sets the voltage across the current source so we can see it will be something like 10V.
 
  • #19
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Ic2 is about 2ma or 1ma, i dont remeber exactly.
And if the subcircuit is a constant current source its impedance is always infinite???
 
  • #20
NascentOxygen
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Ic2 is about 2ma or 1ma, i dont remeber exactly.
And if the subcircuit is a constant current source its impedance is always infinite???
If it is a constant current source, yes. But there is one crucial difference here, and that is that the bias resistors (as well as the collector) connect to the load. This changes things, as I think we are finding. (For your figures, I make Z roughly 2.R5 but as I said, I may have made a mistake.)

It would be good if you could run this in PSPICE, to compare with your analysis.
 
  • #21
The Electrician
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The impedance at the collector of T3 can be found in the usual way by writing the node equations for the circuit, with a test current applied to the collector node. It's convenient to arrange them in matrix form, Y*V=I, where Y is an admittance matrix, V is a vector of the unknown voltages and I is the stimulus vector of currents. For this problem it's possible to use only the admittance matrix. Taking its inverse is equivalent to applying a test current of 1 amp to the 3 nodes, one at a time.

Taking the emitter current to be 2 mA, we have re=13 ohms; β is 100 and the resistors are as indicated. Designate 3 nodes; V1 is the base, V2 is the collector and V3 is the emitter. Then we have an admittance matrix (Y matrix) for the circuit, and its inverse, an impedance matrix (a Z matrix). The diagonal elements of the Z matrix are the driving point impedances at the three nodes. The red expression is the impedance at the base; blue is the impedance at the collector and magenta is the impedance at the emitter.

attachment.php?attachmentid=59057&stc=1&d=1369687944.png


The numerical value of the impedance at the collector is 3127.57 ohms.
 

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  • #22
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You can express Vb in terms of Ib for the transistor, then express I3 in terms of Vb and Ib for the potential divider.
Well, after doing this for the 3rd time my impedance is: (R3+R2)*(rbe+β*R5)/(rbe+β*R5+R2*β) which is actualy 3097 ohms which is close to The Electrician solution :D. Thanks for your effort.
rbe is 1.25k becasue Ic3 is 2ma.
 
  • #23
The Electrician
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Since rbe/β = re, if you divide your numerator and denominator by β, you will get what I got except you are missing the term +(R2 R3)/(1+β) in your numerator.
 
  • #24
NascentOxygen
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Checking back, I found my blunder (I was determining Vin/Ib). With that fixed, I now find Zin exactly as The Electrician has.
 

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