# 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,
Yes, I will have to look that over when I get more time.

Ratch

2. ### The ElectricianActive Member

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Confusion is arising here. The impedances known as "iterative impedances" have been known as such for about 100 years. You are describing the result of your square root equation as an "iterative impedance value" and it is not the same as the classical "iterative impedance". When I am discussing these things, I will use the word "iterative" only in connection with the classical concepts. I will use the phrase "repeated impedance value" to denote the number(s) you have calculated, and which are the same as the classical "image impedances". You can see the possibilities for confusion here, when the impedances that you are calculating, the "repeated impedances", are the same as the classical "image impedances", but not the same as the classical "iterative impedances" (explained further below).

The image impedances (Zi1 and Zi2) are these: when Zi2 is connected as a load to a 2-port, the impedance looking into the input of the 2-port will be Zi1; when Zi1 is connected to the input of a 2-port, the impedance looking into the output of the 2-port will be Zi2. The image impedances are the impedances which simultaneously match the input and output of the 2-port.

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

The classical "iterative impedance" (not the same as the "image impedance") is the result of cascading a lot of identical 2-ports and calculating the input and output impedances of the cascade; this is why they called it "iterative impedance". Imagine the last 2-port in the cascade; its input impedance becomes the load for the next to last 2-port. The input impedance of the next to last 2-port is affected by the fact that its load is the input impedance of the last 2-port. As you work your way back the cascade, eventually the input impedances of each 2-port approaches a particular value. That value is the impedance (it is the output iterative impedance), which, if it is the load of the 2-port, will cause the input impedance of the 2-port to be the same as that load. For the transistor circuit under discussion, its value is 10316.2 ohms.

And, of course, it works the same in the other direction. The input iterative impedance is that impedance, which, when connected as the driving impedance for the 2-port, will cause the output impedance to have the same value as that driving impedance (the input iterative impedance).

"Given a two port, measure the input impedance with the output port open-circuited; call that impedance Zoc. Then measure the input impedance with the output port short-circuited; call that impedance Zsc. Then Zi1 = SQRT(Zsc*Zoc); this value is the geometric mean of Zsc and Zoc. The same procedure can be followed to determine the output image impedance, Zi2.

This same procedure is well known as a method to determine the characteristic impedance of a transmission line. A length of transmission line is essentially a two-port; it has an input and output port.
"

I would use the word "repeated" where the red "iteratively" is above. It is being used to mean something other than what the classical "iterative impedance" means.

I used the word "iterative" in "iterative impedance"; I didn't use "iteratively".

3. ### MrAlWell-Known MemberMost Helpful Member

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I used the word 'iteratively' because that's what you do to get the result with the equations if you dont want to use the square root equation. To iterate to a result is a widely used procedure in other numerical calculations in electronics simulations, not limited to impedances (ie Gear's). The results of the iteration i mentioned leads to the 'image' impedance we talked about. It only takes about 3 or 4 iterations, but that could be with this particular circuit and so i would not count on that as a limit of some sort.

You should realize that the short circuit and open circuit techniques require measurements and that is not what Ratch and I were after. Of course if you use a measurement technique it's going to look simpler than a bunch of equations. What we would have to do if we wanted to use the shorts is we would have to replace the load with a zero impedance (for example) and so we'd have to calculate that too. For example, using impedance parameters we get the following two equations for the exact load and exact source resistances:
zL=sqrt(z22^2-(z12*z21*z22)/z11)
zS=sqrt(z11^2-(z11*z12*z21)/z22)

Looks simple right? But each of those impedance parameters has to be measured, and that's not what we really wanted to do, so that means each of those parameters would have to be calculated and used in the equation for Rout so we end up with an equation we can use without measuring anything except maybe hie.

BTW i did not say that the square root equation was the result of an iteration, the iteration procedure is used with the equations for Rin and Rout alone (no square root equations) if you dont want to use the square root equation. In other words, that's two different ways of calculating the 'image' impedances. The square root equation does not need iteration, it's a one time calculation that provides the exact value right then and there.

You *really* felt that you had to clear up the fact that you used the word "iterative" rather that "iteratively"? Isnt that a bit much? I think we all here can read English good enough to tell the difference without someone pointing it out

Last edited: Feb 19, 2012

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5. ### RatchitWell-Known Member

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MrAl,

I got 11832.23 for the input image impedance and 307.49 for the output image impedance.

Ratch

6. ### The ElectricianActive Member

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No they don't require measurements; they can also be used with calculated values. It's not hard to calculate the open circuit input impedance of the transistor circuit under discussion, and also to calculate the short circuit input impedance. The input image impedance is then the square root of the product of those two. Similarly for the output image impedance.

They don't have to be measured; they can be calculated. For the circuit under discussion, I calculate:

z11 = 10229.34457
z12 = 20.849506
z21 = -44075.8567
z22 = 265.831208

You said "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."

You were describing the possible use of the GIT in an iterative mode, hence the use of the adverb form of "iterative", describing a repetitive procedure. You then used that same adverb form, "iteratively", to refer to what I was talking about. This suggested to me that perhaps you thought I was using a repetitive procedure, which I was not. I wanted to be clear that I was not using a calculation "iteratively", but that the word "iterative" as I used it was just the name of the impedance, the "iterative impedance".

7. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

Ratch:
Great You didnt specify what method(s) you used, but if you dont want to that's ok i guess, your input is still appreciated.

Electrician:
Ok you cleared up some things nicely, but i didnt say that you 'had' to measure, i said you can either 'measure' or 'calculate'. You did a static 'calculation', which is fine, but that's still not what Ratch and I were after here. We were after a *single* formula, without any sub calculations, that would calculate the (open circuit) output impedance, and then later the 'image' impedances...again a formula not a static value. The static value comes at the end after the formula is developed.
In other words, we do want this:
356 or 307 ohms
but that's not all we were after. We also wanted this:
Rout_Image=function(R2,R3,R4,hie,hfe)
In other words, a formula with the resistors in it (not the static values but the variable names R2, R3, etc.) which usually comes out to some algebraic function.
This means if we want to 'calculate' the values with say the impedance parameters, we can not just say:
z11, z12, etc.,
we have to say:
z11(R2,R3,R4,hie,hfe), z12(R2,R3,R4,hie,hfe), etc.,
so each parameter becomes a function rather than a static value like 1.2345 or 56.321 or what have you.

You did however clear some things up very nicely so you did help a lot here with or without that kind of formula, so your input was greatly appreciated.

So you were using the word 'iteratively' as a placeholder for a more elaborate idea, that's completely understandable.

Here is the result of my iterated (or repeated) calculation:

Code (text):

iPass    Rout       Rin

307.49    11832.23  (predicted with Sqrt formula)

1      355.67    11707.97
2      307.72    11831.57
3      307.49    11832.22
4      307.49    11832.23
5      307.49    11832.23
6      307.49    11832.23
7      307.49    11832.23
8      307.49    11832.23
9      307.49    11832.23
10      307.49    11832.23

As can be seen above, it only takes 4 iterations (repetitions) to converge. This speed of convergence may not be typical however.

Last edited: Feb 20, 2012
8. ### Jony130Active Member

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Hi, I use Shekel's method to find this "real" Rout.
The Shekel's method is described here by Electrician

The circuit diagram that I use

The Y matrix and the Voltage gain ; Zin and Zout_open_circuit

Zoc = (Rc (re+Rf+re β))/(Rc+re+Rf+(Rc+re) β)

I also find Zout_short_circuit

And Rout_Image = √( Zsc * Zoc)

Rout_Image = 307.4855304

For
Rf = 47000; Rc = 4700; Rg = 10000; hie = 4000; Hfe = β = 180; and re = hie/(hfe +1)

I am sure that we can use this method also to find Zin_ Image.
I hop that Electrician tell me how to do it.

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9. ### RatchitWell-Known Member

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MrAl,

I used the GIT to get the "ordinary" input/output impedances in terms of R[L] and R[g]. Then I substituted R[g] into the output impedance expression, and equated R[L] with with that expression. Isolating R[L] in that expression gave me output image impedance. Then substituting that value into the ordinary input impedance expression gave me the input image impedance. It was easier than it sounds.

Ratch

Last edited: Feb 20, 2012
10. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

Nice work Jony and Ratch.

Jony:
That's very nice work there, but how did you get such a nice graphic posted? Did you link to another site or something?

Ratch:
Good, that's how i did it too, then followed up with a quick check using the iterative method (or if you prefer to call it the 'repeat open circuit to input/output image impedance' method).

So in conclusion we find that we sometimes have to think about impedance in more than one way to be able to use a circuit effectively. The open circuit impedance isnt always enough, as i tried to point out many moons ago. But i think we're all on the same page now so that's good

11. ### The ElectricianActive Member

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"You should realize that the short circuit and open circuit techniques require measurements"

And:

"Looks simple right? But each of those impedance parameters has to be measured, and that's not what we really wanted to do, so that means each of those parameters would have to be calculated"

So I guess you're saying, simultaneously, that they have to be measured AND they have to be calculated?

In the classical literature, the images impedances are usually denoted Zi1 and Zi2, and I use that same notation. If you want a formula involving symbolic variables, all you have to do is follow this procedure (the procedure works equally well with numerical or symbolic variables):

"Calculate the input impedance (using symbolic variables) with the output open-circuited; call that impedance Zoc. Then calculate the input impedance (using symbolic variables) with the output short-circuited; call that impedance Zsc. Then Zi1 = SQRT(Zsc*Zoc); this value is the geometric mean of Zsc and Zoc. The same procedure can be followed to determine the output image impedance, Zi2."

Jony130 shows one way to do it in post #87.

I used the word "iterative" when I was discussing the classical "iterative impedance". It's part of the name, and not descriptive of a procedure for calculating the values. I only used the word "iteratively" when responding to you.

Since you want a formula involving symbolic variables, how will this procedure get you such a formula?

12. ### MrAlWell-Known MemberMost Helpful Member

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Hello,

Geeze i cant believe you dont understand this yet.
How many times do i have to say that you can use EITHER the square root formula OR the iterative method. You dont seem to read my entire post before you comment. I also said that the iterative method was used for a 'backup' to the square root formula i posted. I didnt say you could use the iterative method for solving algebraically either, i dont know where you got that from and i also dont care.
I wont continue this silly jibber jabber it's a waste of time and space. Use what method you want, use one, use them all, use none. Take your pick. Use those theories, use other theories, make up new theories, dont make up new theories, take your pick.

This conversation is a little thing i like to call, "over"... time to move on.

Last edited: Feb 20, 2012