Common base BJT amplifier

You are correct of course, and a good designer needs all of this in the general case. Still, the question becomes how precise does the model have to be to understand this circuit and predict it's performance.

His analysis includes alpha=hfb and hib. It is clear that hib is critical and is highly dependent on temperature. Including alpha represents a 0.1%correction to the answer. Now including hob would be less than a 0.01% correction.
I wouldn't really expect the results from including all the h parameters to give a result any more useful in the real world than the simple approximations.

It is an exercise in circuit analysis.

If you take alpha to be 1, and use .025 mV to calculate re, then the gain at resonance is 8.000. If you include an alpha of .99 as hgmjr did, you get a gain of 7.92. This is a correction of 1%, not .1%, isn't it?

The value of 1/hob is about 13 megohms which is probably the source of your comment that including it would be less than an .01% correction. Actually, the output admittance of a transistor in CB configuration depends strongly on the impedance at the base. The parameter hob is the admittance when the base is open circuited, or driven from a very high impedance. In the case of the amplifier in this thread, at resonance the source impedance is zero, and the output resistance of the transistor is not 13 megohms, but rather about 300,000 ohms, which means including it will give about a .3% correction.

EDIT: The previous paragraph says "base" where I should have said "emitter". Thanks to Ron H for pointing this out. Here's the corrected paragraph.
-----------------------------
The value of 1/hob is about 13 megohms which is probably the source of your comment that including it would be less than a .01% correction. Actually, the output admittance of a transistor in CB configuration depends strongly on the impedance at the emitter. The parameter hob is the admittance when the emitter is open circuited, or driven from a very high impedance. In the case of the amplifier in this thread, at resonance the source impedance is zero, and the output resistance of the transistor is not 13 megohms, but rather about 300,000 ohms, which means including it will give about a .3% correction.
------------------------------

Including the reverse voltage transfer ratio, hrb, gives about a .1% correction.
 
Last edited:

steveb

Joined Jul 3, 2008
2,436
I wouldn't really expect the results from including all the h parameters to give a result any more useful in the real world than the simple approximations.

It is an exercise in circuit analysis.
Yes, i agree with all of your points. It's important to be able to estimate all of these parameter effects and include those that are relevant. Even when they are not significant, it is good to know what order of magnitude they are at (which I failed at that time ;)).

I didn't mean my comments to imply that i disagreed with you, but just wanted to add to your points by talking about that fine line we walk when deriving relations.
 

hgmjr

Joined Jan 28, 2005
9,029
It took me a while but I think I have address all of the issues raised by electrician. Thanks electrician for taking the time to read through my tome and catching the mis-steps. All of your comments were right on target.

Enjoy,
hgmjr
 

Attachments

Ron H

Joined Apr 14, 2005
7,014
I wouldn't really expect the results from including all the h parameters to give a result any more useful in the real world than the simple approximations.

It is an exercise in circuit analysis.

If you take alpha to be 1, and use .025 mV to calculate re, then the gain at resonance is 8.000. If you include an alpha of .99 as hgmjr did, you get a gain of 7.92. This is a correction of 1%, not .1%, isn't it?

The value of 1/hob is about 13 megohms which is probably the source of your comment that including it would be less than an .01% correction. Actually, the output admittance of a transistor in CB configuration depends strongly on the impedance at the base. The parameter hob is the admittance when the base is open circuited, or driven from a very high impedance. In the case of the amplifier in this thread, at resonance the source impedance is zero, and the output resistance of the transistor is not 13 megohms, but rather about 300,000 ohms, which means including it will give about a .3% correction.

Including the reverse voltage transfer ratio, hrb, gives about a .1% correction.
My understanding of hob is that it is the output admittance when the base is grounded, not open, and that hob is strongly dependent on the impedance at the emitter.
 
My understanding of hob is that it is the output admittance when the base is grounded, not open...[/B].
Aren't you making the same mistake I did? Since the base is already grounded, it's the condition of the emitter that's relevant. hob is the output admittance when the emitter is open

The h parameters are mixed parameters, unlike the Z and Y parameters. Two of them are measured with the output port shorted, and the other two are measured with the input port open.

A good reference for this sort of question is the venerable General Electric Transistor Manual. I've attached the relevant page from the 7th edition.

... and that hob is strongly dependent on the impedance at the emitter.
I said: "Actually, the output admittance of a transistor in CB configuration depends strongly on the impedance at the base."

Of course, what I meant to say is "...impedance at the emitter.", since the base is already grounded. :-(

And, of course, I also misspoke when saying "The parameter hob is the admittance when the base is open circuited, or driven from a very high impedance." It should be "...when the emitter is open circuited..."

I'm just not used to grounded base amplifiers, so I had grounded emitter on the brain.

The actual numbers I gave are for the emitter open (~13 Meg) and emitter grounded (~300k).
 

Attachments

Ron H

Joined Apr 14, 2005
7,014
Aren't you making the same mistake I did? Since the base is already grounded, it's the condition of the emitter that's relevant. hob is the output admittance when the emitter is open
You're the one who said the base was open. I was just correcting that.

The h parameters are mixed parameters, unlike the Z and Y parameters. Two of them are measured with the output port shorted, and the other two are measured with the input port open.

A good reference for this sort of question is the venerable General Electric Transistor Manual. I've attached the relevant page from the 7th edition.



I said: "Actually, the output admittance of a transistor in CB configuration depends strongly on the impedance at the base."

Of course, what I meant to say is "...impedance at the emitter.", since the base is already grounded. :-(

And, of course, I also misspoke when saying "The parameter hob is the admittance when the base is open circuited, or driven from a very high impedance." It should be "...when the emitter is open circuited..."

I'm just not used to grounded base amplifiers, so I had grounded emitter on the brain.

The actual numbers I gave are for the emitter open (~13 Meg) and emitter grounded (~300k).
I had sorta figured that you had common emitter on the brain. I just didn't want to leave your errors uncorrected for future readers. Thanks for straightening it all out.
 
You're the one who said the base was open. I was just correcting that.
What were you intending to correct; the "base" part or the "open" part?

The first half of your sentence seemed a little odd to me because you were "correcting" the "open" vs "shorted" question (even though you were wrong on that point), but you were not correcting the "base" vs "emitter" issue. I couldn't understand why you were "correcting" one error but not the other. Then in the second half of your sentence it is clear that you knew that "emitter" was the right word. So, I thought you had inadvertently said "base" in the first half when you really meant "emitter".

When you are exhibiting an error somebody else made, the way to make sure your readers don't think you are also making the same error is to put "(sic)" right after the error.

Had you said "My understanding of hob is that it is the output admittance when the base (sic) is grounded, not open, and that hob is strongly dependent on the impedance at the emitter.", it would have been clear that you knew that "base" was the wrong word to use there, that it was my error you were displaying.
 

Ron H

Joined Apr 14, 2005
7,014
What were you intending to correct; the "base" part or the "open" part?

The first half of your sentence seemed a little odd to me because you were "correcting" the "open" vs "shorted" question (even though you were wrong on that point), but you were not correcting the "base" vs "emitter" issue. I couldn't understand why you were "correcting" one error but not the other. Then in the second half of your sentence it is clear that you knew that "emitter" was the right word. So, I thought you had inadvertently said "base" in the first half when you really meant "emitter".

When you are exhibiting an error somebody else made, the way to make sure your readers don't think you are also making the same error is to put "(sic)" right after the error.

Had you said "My understanding of hob is that it is the output admittance when the base (sic) is grounded, not open, and that hob is strongly dependent on the impedance at the emitter.", it would have been clear that you knew that "base" was the wrong word to use there, that it was my error you were displaying.
You said,
Actually, the output admittance of a transistor in CB configuration depends strongly on the impedance at the base. The parameter hob is the admittance when the base is open circuited...
I said,
My understanding of hob is that it is the output admittance when the base is grounded, not open, and that hob is strongly dependent on the impedance at the emitter.
If you compare what you said with what I said, where is the ambiguity in my corrections? And you said I was wrong on the open vs shorted question. What is wrong with the first half of my sentence? And I did correct the base vs emitter issue, in the second half of my sentence. Are you still mixing CE and CB in your mind?:confused:
EDIT: perhaps you were thinking that since the circuit is common base, the fact that the base is grounded goes without saying, but that is the issue i was addressing in the first part of my sentence.
 
EDIT: perhaps you were thinking that since the circuit is common base, the fact that the base is grounded goes without saying, but that is the issue i was addressing in the first part of my sentence.
I think that's probably it. When you said, "...output admittance when the base is grounded, not open...", the "not open" part would seem to suggest that you considered the possibility that the base might not be grounded. But this wouldn't be possible if the circuit is a grounded base circuit, so I figured you must be referring to the input port, and inadvertently said "base" instead of "emitter".
 

Ron H

Joined Apr 14, 2005
7,014
I think that's probably it. When you said, "...output admittance when the base is grounded, not open...", the "not open" part would seem to suggest that you considered the possibility that the base might not be grounded. But this wouldn't be possible if the circuit is a grounded base circuit, so I figured you must be referring to the input port, and inadvertently said "base" instead of "emitter".
Whew! I'm glad we got that settled! I thought I was going nuts. :)
 

Thread Starter

anik321

Joined Oct 25, 2008
22
The response from you guys is certainly impressive. It is crazy how in depth you can analyze a circuit if wished to.

Thank you hgmjr, your solution has been most helpful in helping me understand. I do have a few more questions regarding this particular problem but I will save it for the time being.

Thank you.
 
Top