Christer said:
Yes, but in the emitter follower configuration, the base-emitter capacitance is connected in series with the load. Since the load is not being driven to a appreciable degree from the source via the base-emitter capacitance, the source sees a much lower effective input capacitance.
If the output transistor is configured as a common emitter amplifier (such as in a CFP), the base-emitter capacitance shunts the drive at higher frequencies to the respective supply rail, which is at AC ground, so the bootstrap effect does not occur.
Workhorse said:
Kanwar, no vertical MOSFET has as high a transconductance as a BJT. They are still way lower.
Cheers,
Glen
darkfenriz said:
Yes I have this one [swiss army knife], captured from a Russian KGB agent 😀 😀 😉
An externally hosted image should be here but it was not working when we last tested it.
G.Kleinschmidt said:
You mean the base (or gate), not the source, in the last sentence, I presume?
Maybe I am missing something, but it does seem to may that you are thinking in the wrong way, thereby fooling yourself.
Cbe is the base-emitter capacitance and does not depend on how the transistor is connected. The current flowing through it is
Icbe = Vbe/(1/sCbe) = sCVbe
and the effective capacitance is Vbe/Icbe = 1/sCbe, so the effective base-emitter capacitance is Cbe. It is independent on how you connect the transistor.
In the CFP case, I presume you are referring to the second transistor (the one supplying most of the current). Yes, there the emitter is usuallt connected to AC ground, so Cbe is connected from base to AC ground. However, Vb still differs from Ve (ground) only by the voltage Vbe, so there is no difference. The current through Cbe is sCVbe also in this case, and Vbe is directly related to Ic in both cases. Note that Vb of the second transistor does not follow Vb of the input transistor in the CFP.
Is it perhaps the case that you are rather talking about the effective base to ground capacitance than Cbe?
Christer said:
Yes, that's what Glen and I were discussing earlier - the total input capacitance. This would be something like:
Cin = Cob + Cib*(1 - Av)
where Cib = 1 / (2 * pi * re * fT)
and re = 26mV / Ie
Andy,
that clears things up, but the discussion was obviously a bit confused, since some people talked about "reducing the effective Cbe" and similar things, while obviously meaning the input capacitance. Since the discussion seened to start with the Cob and Cbe of various BJTs, I thought that was still the main discussion. Anyway, it seems it was all a terminology confusion (which is often the case when surprising disagreement arise).
Edit: Just to avoid further confusion. Andys and my formulae for Cbe/Cib are equivalent. They just differ in which parameters they use. I got rid of rb (hie) and used hfe instead, since that is the more basic parameter we get from datasheets. The connection is
rb = hie = hfe*vT/Ic
that clears things up, but the discussion was obviously a bit confused, since some people talked about "reducing the effective Cbe" and similar things, while obviously meaning the input capacitance. Since the discussion seened to start with the Cob and Cbe of various BJTs, I thought that was still the main discussion. Anyway, it seems it was all a terminology confusion (which is often the case when surprising disagreement arise).
Edit: Just to avoid further confusion. Andys and my formulae for Cbe/Cib are equivalent. They just differ in which parameters they use. I got rid of rb (hie) and used hfe instead, since that is the more basic parameter we get from datasheets. The connection is
rb = hie = hfe*vT/Ic
Christer said:
Yes. My intent wasn't to contradict what you were saying, but just to try to put all the relevant formulas into one post (which meant repeating an equivalent of a formula you had already posted).
No problem Andy. I was convinced we both knew the formulas were equivalent. I just wanted to point it out to other readers, in case somebody did not realize the equivalence.
BTW, in order to remove other potential confusions, as I have understood things, Cbe and Cib are the same thing and Cbc and Cob are the same thing. Do others share that view, or is there some subtle difference in definitions that I have failed to find?
BTW, in order to remove other potential confusions, as I have understood things, Cbe and Cib are the same thing and Cbc and Cob are the same thing. Do others share that view, or is there some subtle difference in definitions that I have failed to find?
jacco vermeulen said:
Clearing up potential confusion is good. If you don't agree with this, try reading the last few pages of the "output coils" thread
Not really, the last bit of this thread is very informative for types without the basic engineering training.
Often, you pro arguement lot skip the basic parts in a discussion.
If only all tech threads could be as complete.
I started reading this forum 2 weeks after NP joined in.
I've got a lot of respect for you guys, including Mr Kleinschmidt.
Often, you pro arguement lot skip the basic parts in a discussion.
If only all tech threads could be as complete.
I started reading this forum 2 weeks after NP joined in.
I've got a lot of respect for you guys, including Mr Kleinschmidt.
Christer said:
The "source" being that delivering the drive current to the base of the BJT. What else could it have meant in relation to a BJT?
Christer said:
But we talking about the effective input capacitance, as seen by the "source". That depends on how the transistor is connected.
Christer said:
Yes, but I am mostly talking about the contribution of Cbe to the effective base to ground capacitance.
In an emitter follower it is much less than when the emitter is tied to a supply rail, as with a common emitter connected output (not driver) transistor, as used in a CFP.
Cheers,
Glen
I think we should either close this thread or think on better ways of comparing mosfets with bipolars....Atleast this wouldn't be an off-topic😉
Just you wait, Kanwar. Someone will post some provocative statement and that will kick-start the whole thing again 🙂.
Workhorse said:
Well have I been out of line by attempting some kind of discussion on the performance potential of high fT BJT’s over low fT BJT’s?
I mean, there has been an obvious lack of interest, so perhaps I’m just a dummy talking from ignorance about something that everybody knows about already, but an appreciation of BJT performance does pertain to a comparison between BJT's & MOSFET’s, doesn’t it?.
The input capactiance of MOSFET's has be discussed here extensively. What about BJT's?
I think you're right on the money Glen. And what matters most (to me, anyway) is quality, not quantity of posts.