# Impedance for a BJT stage, with collector-base feedback resistor

Discussion in 'General Electronics Chat' started by patroclus, Feb 6, 2012.

1. ### RatchitWell-Known Member

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MrAl,

I can't figure what your point is.

Isn't that what I said at the end of post #55?

I said very specifically in post #40 that -R[T] was the output impedance to the left of all three resistors on the right. So -R[T] is the output impedance of the transistor without any resistors connected to the collector terminal. If you want to include R[c], it should be obvious that a parallel calculation need to be performed. I should not have to define that every time I reference it. I am glad you agree that my method got the right answer.

I gave the equations, described what they stood for, and showed how to derive them in the narrative of the attachment of post #40.

Agreed, I call that the Applied Source Method. It is certainly very direct and valid. Others will have to decide whether the method you used or the GIT is less work.

Isn't that a bit trivial? How about one I already did at http://www.electro-tech-online.com/...-thevenin-theorem-phasor-irwinext8-14.124126/ ?

To what end? We both know what R[in] and R[out] are now. The GIT method does not work for nonlinear circuits.

Ratch

Last edited: Feb 15, 2012
2. ### MrAlWell-Known MemberMost Helpful Member

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Hello,

I believe now that you believe that you wrote a 'clear' explanation and that you will never understand or accept anything else.

I didnt say the GIT method works for non linear circuits, i said a non linear method "to back up the linear analysis". You seem to have misinterpreted my post.

I'll try to get back later with some explicit equations for both linear and non linear.

3. ### RatchitWell-Known Member

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MrAl,

It appears I did. Looking forward to your examples.

Ratch

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5. ### MrAlWell-Known MemberMost Helpful Member

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Hello again,

Hey what is the problem with using a simple voltage divider as the illustration devce? I think this would make the workings of GIT much more transparent because it is quite easy to see what Zin and Zout are with such a simple circuit without any calculation, so comparison is immediate and open to very little interpretation differences.
I could post another transistor circuit but that would just complicate the issue.

The transfer function for the resistive divider with input Vin is:
Vout/Vin=R2/(R1+R2)
where
R1 is the upper resistor,
R2 is the lower resistor,
we take the output at the junction of R1 and R2 referenced to ground.
R1=10k, R2=40k, Vin=10v.
We want to calculate Rin and Rout.

If you want another transistor circuit, then using another 2N3904, biased with R1 from +Vcc to base, and R2 from base to input source voltage, no feedback resistor, collector resistor R3 to +Vcc, emitter resistor R4, output taken from collector referenced to ground as before.
We can take hfe=188.7, hie=812, hre=0, hoe=0, R1=150k, R2=20k, R3=1k, R4=1 ohms, Vcc=+15v.
We want to know Zin and Zout, which for our purposes will just be Rin and Rout.
If you want to call R4 zero that's ok i guess just to simplify this problem a little.

Last edited: Feb 16, 2012
6. ### RatchitWell-Known Member

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MrAl,

OK, here it is. I did it in a hurry, so I hope I did not make any mistakes. Will check this thread again tonight.

Ratch

7. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

I didnt check it super carefully yet, but it looks absolutely correct. I say this because THIS TIME you did it what i believe is the right way to do it. That is, add a load resistor for calculating the output impedance and adding a series input resistor for calculating the input impedance. You didnt do it that way with the previous transistor circuit, so questions naturally came up.

Adding those two resistors allows us to calculate either the input or output impedance, and also allows us to discover what the effect on each other have.

Now wasnt that a lot easier than before?

I believe that if you do the same with the previous transistor circuit you will end up with the explicit solutions without all the fuss.

The non linear equations are cool too. I'll probably have to get back here tomorrow for those though.

8. ### RatchitWell-Known Member

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MrAl,

Don't know about that. The second shorter way I did it was less work. But whatever way you like to do it is fine. Anyway, you seem to get the idea of how it works.

Ratch

9. ### MrAlWell-Known MemberMost Helpful Member

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Hello again,

Did I ? Sorry, i didnt mean to agree that fully. You've played with words through this whole thread and that's ok i guess if that's what you want to do, but now it's my turn to play a little

Sorry to say this but you've never solved for the true output impedance to date. The formulas we had been playing with only calculate the fictitious Rout, not the true Rout. And like all good jokes this one comes to an end now with an equally good punch line that i've saved for the very end. (ok we could do more later

The formula for the output impedance is in the attachment. It's the only correct result, and it comes out to about 307 ohms, not 356 ohms.

[Edit, defined values:]
hie=4000, hfe=180, R2=47000, R3=4700, R4=10000, Re=0

Last edited: Feb 18, 2012
10. ### RatchitWell-Known Member

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MrAl,

I never heard of "real" impedance and "ficticious" impedance. You will have to define what that is. Interesting formula with a square root you have, too. Please tell me how you got it. And, by the way, what value did you use for hie and hfe?

Ratch

11. ### MrAlWell-Known MemberMost Helpful Member

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Hi Ratch,

Let me provide a brief explanation, and i've included the values in my previous post, thanks for reminding me, i had forgotten to post the values i used in the calculation. Same values as before though:
R2=Rf=47000, R3=Rc=4700, R4=Rg=10000, hie=4000, hfe=180

Yes, the square root is interesting for sure, and one of the things that helps to ensure us that we have the right formula. Any formula without a square root can not be the true output impedance. I have to admit i forgot that over the years too as i dont really use this kind of transistor theory that much anymore, and we'll be hard pressed to find a reference to this on the web.

The square root equation comes from the fact that a problem arises when we try to actually use the amplifier in a true real world application and we try to use the formulas we previously posted in this thread. What happens is we calculate a value for Rout (which we will call Rout_1) with the previous formulas, and as you know this Rout_1 is the value that is 'supposed' to load the amplifier correctly, but once we connect that Rout_1 value the input impedance suddenly attains some real world value, we'll call this Rin_1. But then a very funny thing happens. Rout_1 is suddenly no longer the correct load impedance! We assumed Rin was zero to calculate Rout_1, but once we connect that Rout_1 to the output the input impedance changes to a value Rin_1. Now we try to connect an input circuit with a value of its Rout equal to Rin_1, but then we find that Rout changes to Rout_2!
The first thought is to remedy this situation is to start with Rin_1 as the input impedance instead of zero, then calculate Rout_2. So we do that, but then once we connect the real world value of Rout_2 to the circuit (remember we want something that works well in the real world, not just on paper) suddenly Rin_1 changes to Rin_2, another new value of Rin! And it doesnt end there. Connecting Rin_2 instead of Rin_1 leads to another output impedance Rout_3, and so on and so forth, until we vary the values so much that we may eventually land on the right two values.

The given formula with the square root is the solution to this little dilemma, so we dont have to go back and forth with trial and error values. We can calculate the one and only true value of Rout for the amplifier in one step.

I can post some detailed equations for this, but the basic procedure is not super difficult. It starts with two equations, one for Rout_Temporal and one for Rin_Temporal. As can be imagined, Rout_Temporal is an expression that takes as one parameter Rin_Temporal, and Rin_Temporal is an expression that takes as one parameter Rout_Temporal. Thus is is more correct to say that:
"The input and output impedances are interdependent"
rather than say that the output impedance depends on the input impedance, and/or the input impedance depends on the output impedance, although once we get the right value for Rout we're good to go with Rin_Temporal which we can change to simply Rin.
These two equations can be derived with either the perturbation methods or the GIT as we did before, and the equations we came up with for "Rout" are valid for this step, but for this step only, hence the name "Rout_Temporal".
The next step is to substitute Rin_Temporal into the equation for Rout_Temporal. This gives us an equation in Rout_Temporal only, without Rin_Temporal. We solve for Rout_Temporal, and the result is the one and only true value of the output impedance, Rout, and it includes a square root which is a tell tale sign that we might have the right equation.

If you like you can (and maybe should if you have the time and energy) go over this yourself and verify that the formula i posted is the correct derivation. I'll post my complete results a bit later today. BTW, thanks for the interesting chat.

12. ### RatchitWell-Known Member

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MrAl,

Whoa, Rout is the impedance of the circuit looking back from the load, not the load itself. In other words, it is the Thevinin impedance, not the load impedance.

That's right, it never was. Rout is the not the load impedance. Before we can go further, you have to define what the load impedance is. You calculated Rout in posts #32 and #34. The load impedance does not change due to the generator impedance and the generator impedance does not change due to the load impedance.

You are welcome. Yes, I look forward to your exposition.

Ratch

13. ### MrAlWell-Known MemberMost Helpful Member

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Yes that is correct, Rout is the impedance looking back from the load, not the load itself. That's why Rout_Temporal does not work, and that is what we calculated back in the thread with those other formulas where you added a parallel Rc at the end. Read on...

That is correct, the Rout is NOT the load impedance, and the load itself does not change, but the CORRECT load would have to change if it was the wrong value in the first place in order to use the amplifier correctly, which is the whole purpose of calculating Rout in the first place. When we calculate Rout we want to know what the best loading is for the amplifier, and we dont know that using Rout_Temporal.
Amazing as that sounds, it's a totally fictitious quantity. The true Rout is the equation that uses the square root.
I defined output impedance back in this thread.

You have to realize that we dont create a circuit and then hang it in deep space all by itself to do nothing. We create a circuit to actually use it in a real life application. When we do this we have to calculate the output impedance so that we know what kind of load we can connect to the output. We dont leave the output unconnected indefinitely. When we connect a load using our first formulas back in this thread we get a value that, if used, would change the input impedance of the circuit, unless the input impedance was matched with the value from Rin calculated using Rout from the square root equation. Any other value doesnt work.
That's why Rin (which is Rin_Temporal) using GIT wont work either unless we combine it with Rout_Temporal as i explained. Once combined however we get the correct value for Rsource after which when we connect RLoad the circuit is *perfectly* matched to the application. Interestingly, isnt it quite a coincidence that the units work out perfectly in the square root equation
If we choose a value for Rsource at random there is a slim chance it will be matched to the circuit's natural input impedance, and that means there is also a slim chance that the output load we think we need wont be right either, even if we use the previous formula without the square root to calculate Rout.

If you like we can go through this in more depth.

Last edited: Feb 18, 2012
14. ### RatchitWell-Known Member

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MrAl,

I calculated the output impedance looking back into the circuit from the three resistors located to the right in the attachment of post #40. Later I calculated the output impedance looking into the circuit from R[L] by putting R[c] in parallel. That is a perfectly legitimate operation. I did the same with the second solution of the potientiometer problem you gave me and got the correct answer.

The GIT method determines the output impedance looking back from the defined load. You can do anything you want with the output impedance. The load impedance value is not needed to determine the output impedance of the circuit. I have no idea what you mean by "temporal impedance".

As you stated, you mean. You did not explain what "temporal" or "fictitious" impedance means yet. Please do.

Why choose R[g] at random, unless for illustrative purposes? I would like to go into this more deeply. Keep your explanations simple and try to use examples.

Ratch

15. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

Yes i realize now that the term 'fictitious' isnt descriptive enough and just adds more mystery to this incredibly interesting problem of finding the output impedance and input impedance.

Since you are responding in a semi disbelief sort of way i take it you've never heard of this dilemma before this. That's not too much of a surprise to me because every book i can remember seeing describes output impedance in terms of either some measurement where there is no connection to the output, or in some other way that results in an impedance that is not accurate for use in the real world. This might be one of the best cases of where standard procedure by taking something for granted just because it is so common can lead to an unusable result.
We do have to keep in mind though that sometimes we get a value that is close, but theoretically it is still not the exact value. That's the way theory works anyway.

To clear a few things up, what we have been calculating prior to the equation with the square root is more commonly known as the "open circuit output impedance".
That is, after all, EXACTLY what it says it is, the output impedance with the output open circuited. But when do we ever use an amplifier with an open circuit? I guess that is possible, but the most common thing to do is to connect a load of some sort. That's how we really use an amplifier. Likewise, when do we use an amplifier with a circuit before it that has zero output impedance? Again that is possible, but that's not the most typical way to use an amplifier. This is why we want to calculate the input and output impedance, so we can use it most effectively in another circuit.

Now when we short the input and calculate the output impedance we get a value, 356 ohms (example from the original circuit with set parameters as described previously). What that is telling us is that we can connect a load to the output and see an excellent match between the output of the amp and the input of the next stage. But what about the previous stage? Well, we need to know what the input Z is also, so that we can connect a proper circuit to the input too. Then we calculate the input impedance, but the input impedance without a load will be different than with a load as it will be used in the actual circuit, so we have to connect the load to get a reliable measurement or calculation for the input impedance. So we calculate 11707 ohms. That means we need a circuit or generator with an output impedance of that value to match the input.
So now we go to check the output impedance with this new input resistance and we find that (after disconnecting the load as before) the output impedance changed to 307.72 ohms.

So what just happened here? We calculated the output impedance, supposedly 356 ohms, and we of course used that as a load because this is going into a circuit for later sale. Then using that load (as we must in order to get the right input impedance) we calculated the input impedance and then using a circuit that matches that input impedance we connected it to the input.
But then we go back and measure the output impedance again with this new source impedance and we find that the new measurement (or calculation) for the output resistance has changed. Granted it wasnt a huge amount (50 ohms out of 300 or 16.7 percent) but that means that the theory behind the original calculation could not have been exact. We can try using a different input impedance to start with, but the same thing happens, unless of course we get lucky and happen to choose the exactly correct value. It's only that one value that will mean that when we insert this circuit into another circuit that has the right value for the right side impedance and the right value for the left side impedance that none of the impedance values change, or to put it another way, no mismatch occurs.

It's interesting that using the square root formula i posted back a few posts calculates the exact output impedance so that we can then calculate the exact input impedance and nothing changes when we insert this circuit into another circuit (or use the required load and input drive).

Last edited: Feb 18, 2012
16. ### RatchitWell-Known Member

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MrAl,

Are you referring to cascaded amplifiers, which taken by themselves might be the same or different? And are you expounding on how the output impedance of one previous stage might affect the input impedance of the following one? No problem. I can string any number of individual small signal representations of amplifiers together, and find the transfer function of the first input of the first stage to the last output of the last stage. It might be a lengthy equation, with lots of parameters, but a good equation solver can handle it. Once you calculate the transfer equation, you can plot it and tune the parameters to whatever you want or whatever is possible.

The General Immittance Theorem (GIT) is really remarkable, especially its stark simplicity. As long as the circuit is linear, the denominator contains all the input/output impedance information between the source and load. It doesn't matter if the parameters are represented by phasors, Laplace transforms, D-operators, or just plain numbers. It doesn't matter if the circuit contains single or multiple dependent or independent sources. It doesn't matter if the dependent sources are VCVS, VCCS, CCVS, CCCS. It doesn't matter if the transfer function is Vout/Vin, Vout/Iin, Iout/Vin, Iout/Iin. As long as a linear transfer function can be found, the theorem can be applied. I know the theorem can be proved by Tellegen's Theorem, but I don't know how.

It is amazing that this theorem is so scarce in textbooks. It was first introduced in 1956. I first saw it when it was it published in ELECTRONIC DESIGN Feb 1, 1965, V. 13, Issue 3. Later the theorem was defended against detractors in the April 12, 1965, V. 13, Issue 8 of the same periodical. The title of the above article reads "A New Tool for Easier Network Synthesis", and the tag reads "Engineers may well discard Thevenin's and Norton's theorems. A new theorem, valid for both dependent and independent sources, gives the same result easier and faster."

I don't know about the above claim, but I sure like the GIT theorem.

Ratch

Last edited: Feb 18, 2012
17. ### The ElectricianActive Member

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These impedances at input and output are well known but not by the names "fictitious" or "true" impedance.

There are many references on the web; here's one:

http://en.wikipedia.org/wiki/Image_impedance

18. ### MrAlWell-Known MemberMost Helpful Member

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Hi Ratch and Electrician,

Electrician:
Yes i know the "open circuit output impedance" can be found on the web, but that's not what i am talking about here. We have a formula for the OCOI and we've discussed that. The problem comes in when it is not "open circuit" anymore. We want to find the best output resistive load for the amplifier and the best input source resistance.
What happens with the open circuit values is that the OC load causes a mismatch at the input, and the input causes a mismatch at the output.
The square root formula takes care of this problem because it considers BOTH input and output impedance values at the same time, not just one or the other. Thus, once these unique values are found the open circuit impedance works out to the exact value of RLoad, and the input works out to RSource, and RLoad and RSource are found using the square root formula.

Ratch:
It's not a question of whether or not GIT 'works' or not really, this is not for solving the open circuit output impedance until AFTER we've found the correct values for RLoad and RSource.

As to cascading, that is close to what this is about, but we dont want to have to do so much work.
Here's the problem:
Tell me what resistive load you would connect to this amplifier that would be matched to the output impedance of the amplifier, then tell me what input source resistance the previous stage (or generator) should have to be matched to the input of the amplifier.
So you have two tasks here, you have to find the load that best matches to the output of the amplifier, and the best source resistance for the input generator to match to the input of the amplifier.
So that means you should come up with two numerical values for resistances that best match to the amplifier:
R_Source=??
and the final circuit then will be an amplifier that has a resistive load R_Load and is driven by a generator with internal source resistant R_Source. You have to find these two values.
Again, assume R2=47000, R3=4700, R4=10000, hie=4000, hfe=180.

So to repeat the question, what is set of resistances for the load and source that best match to this amplifier:
R_Source=??

Remember now, you dont have to design another circuit for the output and you dont have to design another circuit for the input, just find those two values for this very amplifier with said parameter values.
In the past you've come up with 356 ohms for Rout using your formula using GIT, which means R_Load=356 ohms for your formula.

Last edited: Feb 19, 2012
19. ### The ElectricianActive Member

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I didn't suppose you were talking about open circuit output impedance. I assumed you were referring to the impedance you have calculated with a square root. Maybe if I include more of the quote that will be clear. You said:

"Yes, the square root is interesting for sure, and one of the things that helps to ensure us that we have the right formula. Any formula without a square root can not be the true output impedance. I have to admit i forgot that over the years too as i dont really use this kind of transistor theory that much anymore, and we'll be hard pressed to find a reference to this on the web."

These "unique values" are the image impedances. I already gave the Wikipedia reference:

http://en.wikipedia.org/wiki/Image_impedance

http://www.vias.org/albert_ecomm/aec05_electric_networks_012.html

http://www.transtutors.com/homework-help/Networks-Systems/two-port-network/image-impedance/

http://www.electro-tech-online.com/custompdfs/2012/02/400220.pdf

And, in another post, I explained all this as it applies to audio transformers in particular, but the theory applies to any circuit that can be reduced to a 2-port:

Another pair of unique input and output impedances of a 2-port are the "iterative impedances". The output iterative impedance is that impedance that when connected as a load to the output of the 2-port will cause the input impedance of the 2-port to be the same as the output iterative impedance. The iterative impedance arises in a long cascade of 2-ports. If many instances of a particular 2-port are cascaded, the input impedance of the first 2-port in the cascade, and the output impedance of the last 2-port in the cascade, approach the iterative impedances.

See: http://www.vias.org/albert_ecomm/aec05_electric_networks_013.html

The theories of both image impedances and iterative impedances were highly developed in the 1920s by Bell Telephone.

The transistor circuit you are discussing can be treated as a 2-port. Its image impedances are easily found using the open circuit/short circuit technique I explained in my long post about audio transformers:

11832.2 ohms
307.486 ohms

The iterative impedances are:

352.673 ohms
10316.2 ohms

To see if the iterative impedances work out properly, connect a load impedance of 10316.2 ohms to the output of the transistor circuit, and calculate the input impedance; it should also be 10316.2 ohms. Connect an generator impedance of 352.673 ohms to the input and calculate the output impedance; it should be 352.673 ohms.

I'm assuming in all this that the circuit under discussion in the most recent posts includes the 10000 ohm Rg in the input path.

Last edited: Feb 19, 2012
20. ### RatchitWell-Known Member

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MrAl,

Sorry, the GIT cannot help you do that. The GIT tells you what the imput/output impedance will be for a particular transfer equation. It does not tell you what the particular input/output values should be to match a particular set of conditions. There are other methods and formulas that will tell you what the input/output impedances should be for maximum power gain of the circuit. You already posted one of them. As you probably realize, the maximum power gain of a circuit is not the same as the maximum power that can be transferred. The latter occurs when the source impedance (Rg) is as small as possible.

Ratch

21. ### MrAlWell-Known MemberMost Helpful Member

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Electrician and Ratch,

Electrician:
Yes, that's the image impedance, sorry i didnt look at your original link i was in a hurry.
You've calculated the same values i found using help from GIT, which Ratch claims won help for this
One thing i dont think you explained (or i missed something) is what you mean by 'iterative impedance'. the 'iterative impedance' value i get is the same as the square root equation. Care to explain a little more about the method you mentioned?

Ratch:
You say that GIT wont help, but it does help, you just cant use it *once*, you have to use it *twice*. Alternately, you can use it iteratively until you find the right value, but im not sure yet if this is the same 'iteratively' that Electrician is talking about.
Notice the value for Rout that Electrician got is the same value i got previously. and that's not the same value that we get for the open circuit output impedance using GIT or current perturbation.

There's also a simpler equation using impedance parameters.

Quite interesting right?

Last edited: Feb 19, 2012