high freq cutoff beta

PRS

Joined Aug 24, 2008
989
It's not the same as in his original post. What he had in his original post is:

"is the expression Fb= 1/2\(\pi\)r(C\(_{}\pi\) + C\(_{}\mu\))...correct?"

Which may or may not be correct depending on what r he's referring to.

He posted in post #3 some pdf files (3.pdf, 4.pdf) that give this derivation:

\(f_\beta=\frac{f_\tau}{\beta_{mid}}=\frac{1}{2\pi r_\pi(C_\pi C\mu)}\)

so

\(f_\tau=\frac{1}{2\pi r_{e}(C_\pi C\mu)}\)

These expressions are equivalent to yours, but since the OP didn't specify which r he is referring to in his original post, we don't know which of your expressions might be equivalent to his. As his expression stands, none of yours are equivalent to his.

Also, he is somewhat confused over what Fb is. He said in post #1:

"the question is to find an expression for β cutoff frequency Fb which is simply a high freq current gain cutoff point of the transistor with collector & emitter terminals shorted as seen in the attached"

The file 4.pdf he attached says "The transit or cut-off frequency, Ft,...is defined as the frequency where the current gain falls to 1."

They've used the phrase "cutoff frequency" to mean "the frequency where the current gain falls to 1"

On the other hand, his reference material (3.pdf) does refer to a frequency \(f_\beta\), which is apparently the 3dB down frequency.

So, my best guess is that he wants:

\(f_\beta=\frac{1}{2\pi r_\pi(C_\pi C\mu)}\)

The schematic he posted has a resistor R∏ as a series element, which is not correct; R∏ is the shunt element. The series element is Rx in Sedra and Smith. Having R∏ as a series element has no effect on Ft when calculated for a transistor driven by a current source, but it will affect the response when driven from a more typical source.

Sedra and Smith don't include the shunt element RB, and the fact that the OP included it led me to think that perhaps his instructor intended it to be a part of the circuit for which Ft is to be calculated.
It would be nice if we could all gather at a table with pencil and paper. This tex formatting makes the discussion a little difficult. The derivation I gave is not my own; as I said I got it from my old textbook and therefore is most probably correct.
 
It would be nice if we could all gather at a table with pencil and paper. This tex formatting makes the discussion a little difficult. The derivation I gave is not my own; as I said I got it from my old textbook and therefore is most probably correct.
I realized, because you said so, that you got your derivation from Sedra/Smith. I also have that text. And, it's correct as far as I can see. It also matches the derivation in the OP's pdf files.

The issue I'm raising is that the OP's statement of his problem is a little ambiguous. There's no question what the correct expressions for Ft and Fβ (for a transistor alone) are. It's just not perfectly clear what the OP's problem is asking for.

Furthermore, I've been trying to lead the OP into doing the derivation himself, rather than just giving it to him.
 

Thread Starter

stupid

Joined Oct 18, 2009
81
it is an assignment given by the instructor.

btw how should i proceed from the last derivaive?

thanks
stupid
This is correct for the expression which doesn't include the effect of RB. Is the circuit you posted in post #1 something which is in your text book, or did your instructor give it to you?
 
it is an assignment given by the instructor.

btw how should i proceed from the last derivaive?

thanks
stupid
The result you got is:

r∏Gm / (1+jωCr∏)

Look at the denominator; when ωCr∏ equals 1, hfe will be down 3dB. That will give you Fb. So, you solve this equation for f: 2*∏*f*C*r∏ = 1
 

PRS

Joined Aug 24, 2008
989
I realized, because you said so, that you got your derivation from Sedra/Smith. I also have that text. And, it's correct as far as I can see. It also matches the derivation in the OP's pdf files.

The issue I'm raising is that the OP's statement of his problem is a little ambiguous. There's no question what the correct expressions for Ft and Fβ (for a transistor alone) are. It's just not perfectly clear what the OP's problem is asking for.

Furthermore, I've been trying to lead the OP into doing the derivation himself, rather than just giving it to him.
It's good to know you have a copy of Sedra and Smith. Mine is the 2nd edition c. 1987. If I ever run into a snag in understanding something in that text I'll know who to ask for help! And vice versa, if you want. Cheers! :)

By the way, concerning this problem, I think \(R_b\) is irrelevant as it is not a parameter of the device. Not even \(R_0\), is relevant in that there is a short at the collector. Only the device parameters \( C_\mu , C_\pi , and r_\pi \)have meaning when determining \(f_t\).
 
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By the way, concerning this problem, I think \(R_b\) is irrelevant as it is not a parameter of the device. Not even \(R_0\), is relevant in that there is a short at the collector. Only the device parameters \( C_\mu , C_\pi , and r_\pi \)have meaning when determining \(f_t\).
Yes, I know that this would be true for the conventional meaning of Ft as an intrinsic property of a transistor. But, as I said in post #8:

Upon thinking about this problem, it's not clear to me whether they want the cutoff frequency for the whole circuit, or just the part to the right of RB. I'm going to assume they want the whole thing. If they don't, it's easier to do the it without RB, and having done it with RB, you can go back and make the necessary deletions.
And in post #15:

The circuit he posted in post #1 has an extra resistor, Rb, which I assume is an equivalent to the biasing network. Ordinarily, this wouldn't be considered if one wanted the intrinsic Ft of the transistor. But the circuit he posted includes it.

If he finds Ft for the complete circuit he posted, then to delete the effect of Rb will be easy. He needs practice solving networks like this, and I decided to lead him through the steps for solving the complete circuit.

This is the homework help forum, after all.
And in post #19:

Sedra and Smith don't include the shunt element RB, and the fact that the OP included it led me to think that perhaps his instructor intended it to be a part of the circuit for which Ft is to be calculated.
And post #22:

The issue I'm raising is that the OP's statement of his problem is a little ambiguous. There's no question what the correct expressions for Ft and Fβ (for a transistor alone) are. It's just not perfectly clear what the OP's problem is asking for.
The instructor could have included Rb to see if the OP would realize that it has no effect of the intrinsic Ft of the transistor.

But, if you have a look at some of the OP's recent posts (http://forum.allaboutcircuits.com/showthread.php?t=35653 for example), you'll see that he has been calculating the h parameters for just plain old resistive networks.

Only the device parameters \( C_\mu , C_\pi , and r_\pi \)have meaning when determining \(f_t\).
Yes, this is true if you are asking about Ft for the device only, and that would be conventional, but an instructor may ask for more if he chooses.

It's certainly possible to define the unity gain cutoff frequency of a passive network (R's and C's only) as the frequency where the current transfer ratio, h21, becomes unity.

Since the OP included Rb in the circuit he posted, I thought it was a possibility the instructor wanted Ft for the complete circuit as posted, and calculating Ft for the complete network would be good practice.

As a real world example, there exist high frequency transistors where the biasing network is built-in, and also an impedance matching network. The actual transistor is not accessible from the external package terminals. The Ft one would measure from the package terminals will include the effect of those extra components.
 
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PRS

Joined Aug 24, 2008
989
Yes, I know that this would be true for the conventional meaning of Ft as an intrinsic property of a transistor. But, as I said in post #8:



And in post #15:



And in post #19:



And post #22:



The instructor could have included Rb to see if the OP would realize that it has no effect of the intrinsic Ft of the transistor.

But, if you have a look at some of the OP's recent posts (http://forum.allaboutcircuits.com/showthread.php?t=35653 for example), you'll see that he has been calculating the h parameters for just plain old resistive networks.



Yes, this is true if you are asking about Ft for the device only, and that would be conventional, but an instructor may ask for more if he chooses.

It's certainly possible to define the unity gain cutoff frequency of a passive network (R's and C's only) as the frequency where the current transfer ratio, h21, becomes unity.

Since the OP included Rb in the circuit he posted, I thought it was a possibility the instructor wanted Ft for the complete circuit as posted, and calculating Ft for the complete network would be good practice.

As a real world example, there exist high frequency transistors where the biasing network is built-in, and also an impedance matching network. The actual transistor is not accessible from the external package terminals. The Ft one would measure from the package terminals will include the effect of those extra components.
Electrician, you seem angry. I'm sorry if I said anything to upset you. I only hope to be your friend.
 
Electrician, you seem angry. I'm sorry if I said anything to upset you. I only hope to be your friend.
I'm not angry. I'm puzzled as to why you apparently ignored a point I raised several times, and then said:

By the way, concerning this problem, I think \(R_b\) is irrelevant as it is not a parameter of the device. Not even \(R_0\), is relevant in that there is a short at the collector. Only the device parameters \( C_\mu , C_\pi , and r_\pi \)have meaning when determining \(f_t\).
saying "by the way", as though I hadn't already mentioned that Ft is usually an intrinsic device property, but that the notion of an Ft, a unity gain frequency, could be applied to a larger circuit.

It occurred to me that perhaps you hadn't read all the posts in the thread, and I thought I would bring the relevant parts to your attention.

The OP, who, as you noticed, has chosen to call himself "stupid", when he is plainly not stupid, is trying to learn this skill, and he's working hard at it.

Solving the more difficult problem of determining Ft where Rb is part of a device in a package is something that would be good for the OP, and I can imagine an instructor intending just that.

In fact, the OP posed another problem in this forum (http://forum.allaboutcircuits.com/showthread.php?t=35745) which was going to be rather difficult. After I began helping him with it, the problem was downgraded (by his instructor, I suppose). At the end of the thread, I suggested to the OP that he had learned enough to tackle the original, more difficult, problem, and that perhaps he could solve it for extra credit.

I feel the same way about the circuit in this thread. The problem with Rb included is within his capacity to solve, and I hate to see it just totally ignored. While it is true that for an idealized single transistor, Ft is not affected by a shunt element like Rb, or by the transistor output admittance hoe, it isn't unreasonable to conceive of a real transistor in a package with built in equivalent to Rb. Then determining Ft for the package will have to include Rb in the calculation. This will be somewhat more difficult than without Rb, but a good instructor will add a little something out of the ordinary to challenge the student.

I'm prepared to help him solve it, but I won't just do it for him. I try to guide him, and it's definitely a lot more work to do that than to just provide the solution, and takes a lot more back and forth posts on the forum, but I'm willing to do that for somebody who will stay the course.
 

PRS

Joined Aug 24, 2008
989
I'm very sorry, Electrician. We have different approaches. I used to be a tutor of math, physics and chemistry. My way is not everyone's way. :) But how, for example, is one to know we can throw away the second term in a numerator because it is so small? That is not intuitive. We need to be shown. That's my way.
 
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I'm very sorry, Electrician. We have different approaches. I used to be a tutor of math, physics and chemistry. My way is not everyone's way. :) But how, for example, is one to know we can throw away the second term in a numerator because it is so small? That is not intuitive. We need to be shown. That's my way.
Why do you say that you're very sorry? What's to be sorry about?

I have no objection to your approach except that you did all the work for the OP. As I understand the philosophy of the homework help forum, the idea is to use a Socratic approach, and guide the student; not do it for them all at once.

Maybe when tutoring, immediate feedback from the student makes a different approach work better. You have that experience, but here it seems that doing the student's homework for them is not the favored approach.
 

PRS

Joined Aug 24, 2008
989
Why do you say that you're very sorry? What's to be sorry about?

I have no objection to your approach except that you did all the work for the OP. As I understand the philosophy of the homework help forum, the idea is to use a Socratic approach, and guide the student; not do it for them all at once.

Maybe when tutoring, immediate feedback from the student makes a different approach work better. You have that experience, but here it seems that doing the student's homework for them is not the favored approach.
I used to get books called Problem Solver. They're published for every learning subject. I found that by being shown how to do certain problems, it became a cinch to do other problems similar to them. I think we learn that way. Monkey see, monkey do. A calculus teacher took your approach, saying he was teaching the student to think. I beg to differ. We can't teach people how to think; we can only show them the material and how to solve particular problems.

And, by the way, if I were to include Rb I would do it just by defining a variable r' = \(r_\pi\) in parallel with Rb. The answer would then be the same, but with r' instead of \(r_\pi\)
 
I used to get books called Problem Solver. They're published for every learning subject. I found that by being shown how to do certain problems, it became a cinch to do other problems similar to them. I think we learn that way. Monkey see, monkey do. A calculus teacher took your approach, saying he was teaching the student to think. I beg to differ. We can't teach people how to think; we can only show them the material and how to solve particular problems.
I've been thinking about the apparent difference in our points of view here, and I think maybe it's the following:

When a teacher is delivering a lecture on a method of problem solution, he explains every step at the blackboard (whiteboard nowadays, I guess!); he solves the problem completely.

But, then he assigns a homework problem, and the student is expected to solve the problem on his own. He may ask the professor for help later, but the student is expected to work on it first.

When you're tutoring, you're like the teacher at the whiteboard.

A person who posts here is supposedly presenting a homework problem. Someone who helps is not in the same position as a teacher delivering a lecture at the whiteboard, and they shouldn't do the homework problem for the student from start to finish; they should guide the student through a solution, with the student doing most of the work.

And, by the way, if I were to include Rb I would do it just by defining a variable r' = \(r_\pi\) in parallel with Rb. The answer would then be the same, but with r' instead of \(r_\pi\)
That won't work, because the formula derived where Rb is absent assumes that all the input current contributes to output current, via gm.

If you just substitute Rb||R∏ for R∏ in that formula, that assumption still underlies the result.

But, when Rb is present, what actually happens is that some of the input current is shunted to ground through Rb, and doesn't contribute to the output current, and the correct formula will be different.
 

PRS

Joined Aug 24, 2008
989
That won't work, because the formula derived where Rb is absent assumes that all the input current contributes to output current, via gm.

If you just substitute Rb||R∏ for R∏ in that formula, that assumption still underlies the result.

But, when Rb is present, what actually happens is that some of the input current is shunted to ground through Rb, and doesn't contribute to the output current, and the correct formula will be different.
r' , being Rb in parallel with \(r_\pi\) works just fine. The current is Ib in either case. I(divider) is irrelevant.
 

JoeJester

Joined Apr 26, 2005
4,390
PRS,

As a tutor you have that give and take with your charges that we can only hope for in a worldwide forum. You have a grasp of the students baseline knowledge when you tutor them or you find out very quickly the baseline. That is a little more difficult here in the forums.

The Electrician is correct when stating this is more of the Socrates approach with guidance being the rule of thumb.

I grant you that some will produce the answer without any effort on the part of the OP. The tutor approach is illustrated in the eBooks. It certainly is within the capabilities of the various OPs to view the eBooks for guidance. If you don't like the eBooks here, there are others out in the web that will suffice.

My fear is some come here for instant gratification of their pending problem. I'm sure you will agree that is not the avenue we should pursue when helping our eventual replacements in the world of electronics. Mentorship comes to my mind when the rules for the homework forum was written.

Mentoring is what makes these forums different than the others. That esprit de corp permenates each technical section of these forums.

I'd be interested in hearing how the tutor model could work here when the intangible feedback is tough to discriminate with the written word. Especially when many are "english as a google translated language".

Our table is the white portions of these pages. Our pencils are the keys on the keyboard. Imposing these resitrictions on the OPs is tough enough. I'm sure those who answer the inquiries here would love to be in a position of face to face tutoring. We are limited by the resources at hand.
 

Thread Starter

stupid

Joined Oct 18, 2009
81
hi the electrician,
i actually assumed Rb was to be left out in order to beat the "extended date line.
i wanna include the Rb as part of Ft when time permits me to do so.


thanks
stupid
I'm not angry. I'm puzzled as to why you apparently ignored a point I raised several times, and then said:


In fact, the OP posed another problem in this forum (http://forum.allaboutcircuits.com/showthread.php?t=35745) which was going to be rather difficult. After I began helping him with it, the problem was downgraded (by his instructor, I suppose). At the end of the thread, I suggested to the OP that he had learned enough to tackle the original, more difficult, problem, and that perhaps he could solve it for extra credit.

I feel the same way about the circuit in this thread. The problem with Rb included is within his capacity to solve, and I hate to see it just totally ignored. While it is true that for an idealized single transistor, Ft is not affected by a shunt element like Rb, or by the transistor output admittance hoe, it isn't unreasonable to conceive of a real transistor in a package with built in equivalent to Rb. Then determining Ft for the package will have to include Rb in the calculation. This will be somewhat more difficult than without Rb, but a good instructor will add a little something out of the ordinary to challenge the student.

I'm prepared to help him solve it, but I won't just do it for him. I try to guide him, and it's definitely a lot more work to do that than to just provide the solution, and takes a lot more back and forth posts on the forum, but I'm willing to do that for somebody who will stay the course.
 
I think you're not talking about the same thing as I was when I said:

"As a real world example, there exist high frequency transistors where the biasing network is built-in, and also an impedance matching network. The actual transistor is not accessible from the external package terminals. The Ft one would measure from the package terminals will include the effect of those extra components."

To refer to the OP's circuit, I'm talking about a circuit where Ib is not the input current under consideration. I'm considering a current IB, which would be a current applied to the left of node B. That current will divide and part of it will pass to ground through RB, and part of it will pass to the right through the series connected R∏, becoming Ib.

So, if you put the circuit in a box, so that only the terminal B is accessible as an input, and calculate the ratio IL/IB, that ratio will have a different cutoff frequency when RB is present than if RB were not present.

Now, I want to be clear about what I think you're saying. I think we're both agreed that if the OP's circuit were in a black box with RB deleted, we would have a model of a transistor alone. And if we calculated the current gain from the B input terminal (not B', which is inaccessible) to the shorted output, it would be given by:

\(h_{fe} = \frac{g_m r_\pi}{1 + s(C_\pi + C_\mu)r_\pi]\)

which is straight out of Sedra/Smith. Their very next equation, (7.77) is:

\(h_{fe} = \frac{\beta}{1 + s(C_\pi + C_\mu)r_\pi]\)

From this we have \(\omega_\beta = \frac{1}{(C_\pi + C_\mu)r_\pi}\) which is equation (7.78) in Sedra/Smith

Finally, we have \(\omega_\tau = \frac{\beta}{(C_\pi + C_\mu)r_\pi}\) or \(F_\tau = \frac{\beta}{2 \pi(C_\pi + C_\mu)r_\pi}\)

Now, do I understand correctly that you are asserting that if we put RB back in the box as shown in the OP's original schematic, with input applied to the B node (not the B' node, which is inaccessible because of the box)), then we will just have to put r' = RB || R∏ in place of R∏ in this expression:

\(F_\tau = \frac{\beta}{2 \pi(C_\pi + C_\mu)r_\pi}\)

and that will be the correct Ft for the circuit in the black box, which now includes RB?

In other words, the expression:

\(F_\tau = \frac{\beta}{2 \pi(C_\pi + C_\mu)r'}\) which could also be written:

\(F_\tau = \frac{\beta}{2 \pi(C_\pi + C_\mu)(\frac{R_B*R_\pi}{R_B+R_\pi})}\)

is the correct Ft for the circuit in the box, including RB?
 

PRS

Joined Aug 24, 2008
989
Electrician, as you put the problem, you're right: the current going into the divider needs to be taken into account.
 

PRS

Joined Aug 24, 2008
989
PRS,

As a tutor you have that give and take with your charges that we can only hope for in a worldwide forum. You have a grasp of the students baseline knowledge when you tutor them or you find out very quickly the baseline. That is a little more difficult here in the forums.

The Electrician is correct when stating this is more of the Socrates approach with guidance being the rule of thumb.
To me showing the solution is the information a student needs. When he sees it he learns. That's been my experience.

I grant you that some will produce the answer without any effort on the part of the OP. The tutor approach is illustrated in the eBooks. It certainly is within the capabilities of the various OPs to view the eBooks for guidance. If you don't like the eBooks here, there are others out in the web that will suffice.

My fear is some come here for instant gratification of their pending problem. I'm sure you will agree that is not the avenue we should pursue when helping our eventual replacements in the world of electronics. Mentorship comes to my mind when the rules for the homework forum was written.
I'm not a rule-breaker, and I'm not sure I broke one. But if I have and an administrator reads this, let me know and I'll change my way.

Mentoring is what makes these forums different than the others. That esprit de corp permenates each technical section of these forums.
You could be right about that, but I don't see how this precludes giving the exact solution and the reason for it.

I'd be interested in hearing how the tutor model could work here when the intangible feedback is tough to discriminate with the written word. Especially when many are "english as a google translated language".
You do have a point. I have had many a student who just wanted me to do their homework. But I spotted these cases and I challenged them on it. I was and still am of the opinion that a student has to be willing to put in the work and the time. I stressed that once they were in upper mathematics they had to quit thinking of it as just a bunch of homework. They had to get really serious about it. Some students balked at this. Those who did seemed to want to think that you either were smart enough to get it done or you weren't. My position was this: It's just hard work. Your grade reflects the effort you put into it. I consider normal people to have the intelligence to get it done.

Our table is the white portions of these pages. Our pencils are the keys on the keyboard. Imposing these resitrictions on the OPs is tough enough. I'm sure those who answer the inquiries here would love to be in a position of face to face tutoring. We are limited by the resources at hand.
I understand. But we are still just arguing our personal approaches to teaching.
 
To me showing the solution is the information a student needs. When he sees it he learns. That's been my experience.
I see a dichotomy; the teacher lectures, showing how to solve certain kinds of problems. Then to find out if the student understood the lecture, and to help the student internalize the lesson, he assigns homework problems to be done by the student.

When you say "To me showing the solution is the information a student needs. When he sees it he learns.", certainly this is what happens during the lecture. But if this is done with homework it seems rather a sort of Schaum's Outline method; give the student a lot of worked problems.

Showing the solution to homework is ok if it's done after the student has attempted a solution on his own, but wouldn't it be better to show the student only as much as he needs to then complete the rest of the problem on his own? Why preempt his experience of working out as much of the solution as he can? It's doing the work himself that lodges his knowledge more permanently in his brain.

I found out to my dismay that when I attended lecture, but didn't do the homework, during exams I discovered that my understanding was deficient.

It's all too easy to think one understands when a problem solution is demonstrated; actually solving some problems on one's own is crucial.

When a tutor goes completely through a solution, this is more lecture, and that's ok, but there needs to be some problem solving by the student, where he has to do as much of the work without assistance as possible. It's when he finds he can't do it that he realizes his understanding is deficient and needs more work.

I understand. But we are still just arguing our personal approaches to teaching.
It's more specific than just differences in approaches to teaching; teaching (as I dichotomize it) encompasses lecturing, tutoring (which is just one-on-one lecturing) and assigning homework.

I think we all agree that the lecture ("...showing the solution...") is an integral part of teaching. I claim that tutoring is not homework; it's personalized lecturing.

Where we disagree is with respect to dealing with a student asking for help with homework. Doing as much of the work as possible by himself creates a stronger imprint on the brain than just watching someone else do it.

I think something vitally important is accomplished by a student who solves his homework with as little help as possible. It means more intellectual effort on his part to think about it long and hard until he sees how to do it. It's practice. It's just like learning to play the violin. Having somebody show you how to do it day after day can't substitute for doing it yourself, over and over. Work, work, work. That's why it's called homework.
 

JoeJester

Joined Apr 26, 2005
4,390
PRS,

No one said you broke any rules. You certainly may continue to assist those in need in whatever manner you wish. I was just giving you the background on the genesis of those "posted" guidelines.

You might consider the guidelines were written not with tutoring in mind, but with troubleshooting, specifically, where the OP went astray.

I applaud your efforts. It is a very difficult task to identify those who just want you to do their homework. Most times, that type just won't provide an attempt at the solution. How far you go in your assistance is certainly up to you. If you were going to use the "example" model, similiar to the OPs question, and one certainly used by tutors everywhere, that is commendable. Doing the OPs exact question is doing their homework, in my opinion.

We may disagree, but you would find there is more agreement than disagreement in the processes we use.
 
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