FETs Vesus Bipolars, Why More Efficient?

[lemonjuice]

On Tue, 23 Nov 2004 07:54:52 GMT, "Kevin Aylward"
<salesEXTRACT@anasoft.co.uk> wrote:

Dr. Slick wrote:
"Kevin Aylward" <salesEXTRACT@anasoft.co.uk> wrote in message
news:<VYhod.19283$08.18967@fe2.news.blueyonder.co.uk>...

Well they correct your trivial misunderstandings. Like, the base
current is the relevant current in the diode equation. Yeah right
on. You put your foot right in it that time.


Never said the base current was as big as the
collector current.

You claimed (inferred) that one uses the base current as the current
to
be used in the base-emitter diode equation. This is incorrect.
No.

Why do you say it is incorrect to uses the base current as the current
to be used in the base-emitter diode equation? The base to emitter
forms a pn junction diode. An active bias causes minority carriers to
be injected into the base. The density of these are "first order"
(laughing at first order) proportional to the exponential of the base
to emitter voltage. The base current isn't exponentially dependent on
the bias (though it is on the linear dimensions) because you have a
recombining current of the majority carriers , also exponentially
dependent on the base emitter voltage(with an exponential linear
dimensional dependence) the 2 summed up give you an approx. (grin at
approx. ) constant base current.

Even your own "papers "snipped"
use beta or current gain!


Of course they do. Beta has a second order effect. However, this
doesn't
mean that base current controls the emitter current.

No to both of the last 2 assertions.

what do you mean second order effect.... secondary in importance ?
ic = beta * ib and delta ic = beta * delta ib ... beta does not
seem second order in these or the many other equations where it is
written.

Why doesn't the base current control the emitter current?
According to your hypothesis base emitter voltage controls the
emitter/collector current. Base emitter voltage ONLY reduces the
inbuilt potential of the semiconductor across the pn junction and
permits minority carriers to diffuse across the space charge area ...
but You seem to forget that majority carriers ENTER the semiconductor
through the ohmic contacts. These neutralize the diffused minority
carriers. BJTs aren't like FETS .... they use minority & majority
carriers for current conduction.
Part of this recombination current is the base current.
In other words ....No base current => no neutralization of base => no
neutralization of base => charge imbalance in base region=> no current
flowing through base from emitter to collector.
Therefore base current controls collector/emitter current.

 

[Ken Smith]

In article <pan.2004.11.23.16.49.27.353520@example.net>,
Rich Grise <rich@example.net> wrote:
[...]

Well, in all of my long experience, and through electronics tech school,
and according to the old military "physics of a transistor" book and
according to every circuit design I've ever had analyzed for me, base
current has always been the determining factor - Vbe is almost constant,
because it's a diode. This is why transistors need to be temperature
compensated.

I assume this is not large signal, RF experience. In large signal RF, you
can have buckets of base current with very little collector current.

--
--
kensmith@rahul.net forging knowledge

 

[Kevin Aylward]

lemonjuice wrote:

On Tue, 23 Nov 2004 07:54:52 GMT, "Kevin Aylward"
salesEXTRACT@anasoft.co.uk> wrote:

Dr. Slick wrote:
"Kevin Aylward" <salesEXTRACT@anasoft.co.uk> wrote in message
news:<VYhod.19283$08.18967@fe2.news.blueyonder.co.uk>...

Well they correct your trivial misunderstandings. Like, the base
current is the relevant current in the diode equation. Yeah right
on. You put your foot right in it that time.


Never said the base current was as big as the
collector current.

You claimed (inferred) that one uses the base current as the current
to
be used in the base-emitter diode equation. This is incorrect.
No.

No. Its yes.

Why do you say it is incorrect to uses the base current as the
current to be used in the base-emitter diode equation?

Because it is incorrect.

The base to
emitter forms a pn junction diode.

Yes

An active bias causes minority
carriers to be injected into the base.

Yes.

The density of these are
"first order" (laughing at first order) proportional to the
exponential of the base to emitter voltage. The base current isn't
exponentially dependent on the bias (though it is on the linear
dimensions) because you have a recombining current of the majority
carriers , also exponentially dependent on the base emitter
voltage(with an exponential linear dimensional dependence) the 2
summed up give you an approx. (grin at approx. ) constant base
current.

And what relevance does this have to the well known fact that the diode
equation is in reference to "emitter" current, not base current.

IE = is.(exp(Vbe/KT) - 1)

is is *not ib = is.(exp(Vbe/KT) - 1)

What matters is the *total* current going across the base emitter
junction. The base current is essentially, negligible.

Even your own "papers "snipped"
use beta or current gain!


Of course they do. Beta has a second order effect. However, this
doesn't mean that base current controls the emitter current.

No to both of the last 2 assertions.
what do you mean second order effect....

A 1st order calculation, ignoring beta, can give say, a gain accurate to
1%.

secondary in importance ?

Is 1% impotant? It depends on the situation.

ic = beta * ib and delta ic = beta * delta ib ... beta does not
seem second order in these or the many other equations where it is
written.

But one can use

ic = gm.vbe

thereby bypassing all mention of hfe.

Why doesn't the base current control the emitter current?

Because Electric field is what causes charges to move, not current.

F = q(E + vXB)

F is the force on a charge, q (carrier)

Assuming, of course, that the transistor is not magnetically operated.

According to your hypothesis

Its not my hypothesis. Its in all the standard text books on physics.
Non of what I am saying originated from me. I have simply covered the
material in formal physics classes. Clearly, you have not

.base emitter voltage controls the
emitter/collector current. Base emitter voltage ONLY reduces the
inbuilt potential of the semiconductor across the pn junction and
permits minority carriers to diffuse across the space charge area ...
but You seem to forget that majority carriers ENTER the semiconductor
through the ohmic contacts. These neutralize the diffused minority
carriers. BJTs aren't like FETS .... they use minority & majority
carriers for current conduction.

What has this got to do with the fact that it is Electric field that
makes carriers move. Hint where does the term "accelerating voltage"
come from?

Part of this recombination current is the base current.
In other words ....No base current => no neutralization of base => no
neutralization of base => charge imbalance in base region=> no
current flowing through base from emitter to collector.

Oh?

Therefore base current controls collector/emitter current.

Nope. Here we go again. A functional relation does not imply a causal
one. The electric field, Vbe, causes the base current. Mathematically
manipulating the equation to make vbe a function of the base current,
doesn't change the physics operation. The base current and the emitter
current are both *controlled* and instigated by the electric field, its
that simple.

F = q(E + vXB)

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Kevin Aylward]

Dr. Slick wrote:

Guess what? We don't disagree
as much as you would like!


Not at all. You are fundamentally wrong in claiming that the
collector current is controlled by base current. For the last time,
a functional relation is not proof of a casual relation.


I'm saying that the base current is controlled
by Vbe, so in essence, I'M AGREEING WITH YOU,

Well, it appeared that you were cliaming that base current controlled
Vbe.

BUTT-FUCKER. YOU CAN CONSIDER BJTs WITH
EITHER BETA OR Gm.

Sure, you can use a beta model, but that doesn't make a transistor
current controlled, which was the point being made.

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Kevin Aylward]

Dr. Slick wrote:

Active8 <reply2group@ndbbm.net> wrote in message
news:<rmm8ec59thly$.dlg@news.individual.net>...
On 22 Nov 2004 16:39:19 -0800, Dr. Slick wrote:

"Kevin Aylward" <salesEXTRACT@anasoft.co.uk> wrote in message
news:<VYhod.19283$08.18967@fe2.news.blueyonder.co.uk>...

Well they correct your trivial misunderstandings. Like, the base
current is the relevant current in the diode equation. Yeah right
on. You put your foot right in it that time.


snip

Point is,

No, your point was that Ib and not Vbe controls Ic.


Ok, so you need to apply a voltage
to get a current. No FUcking DUh!!

Then way are you disagreeing?

Or you can apply a current to
get a voltage too.

Lorentz force law:

F = q(E + vXB)

Are you suggesting that transisters work magnetically?

Otherwise, the only god damn way you will get a charge to move is to
apply an electric field.

you can take the derivative
of Ibe with respect to Vbe, and multiply
^^^^^^^^^^^^^^^^^^^^^^^
This may be true, but Ibe WRT Vbe is Vbe controlled, duh! That's
twice you've single handedly disproved yourself. Math aside, like
Kevin pointed out days ago, you need an electric field to get a
current flow.



So you can consider a BJT
as having current gain beta, OR
transconductance Gm. Period.

Sure, you can pretend in simple cases that "forcing" a current produces
a current. You can also pretend that current is water down a pipe. Whats
you point?

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Rich Grise]

On Tue, 23 Nov 2004 17:59:42 +0000, Ken Smith wrote:

In article <pan.2004.11.23.16.49.27.353520@example.net>,
Rich Grise <rich@example.net> wrote:
[...]
Well, in all of my long experience, and through electronics tech school,
and according to the old military "physics of a transistor" book and
according to every circuit design I've ever had analyzed for me, base
current has always been the determining factor - Vbe is almost constant,
because it's a diode. This is why transistors need to be temperature
compensated.

I assume this is not large signal, RF experience. In large signal RF, you
can have buckets of base current with very little collector current.

No. Actually, I've been studiously avoiding learning anything about any
semiconductor that operates any faster than about 20 KHz.

But, now that you mention it, in that application, would you consider the
part to be "voltage controlled" or "current controlled"?

Thanks!
Rich

[RTP wants to say, "It doesn't matter: it's part of a whole circuit with
an overall energy flow" or some such fluff.]

 

[Rich Grise]

On Tue, 23 Nov 2004 18:55:50 +0000, Kevin Aylward wrote:

Dr. Slick wrote:

I'm saying that the base current is controlled
by Vbe, so in essence, I'M AGREEING WITH YOU,

Well, it appeared that you were cliaming that base current controlled
Vbe.

The only distinction is which you consider the dependent variable.
And, of course, there's the temperature coeffecient.

How many times have you ever applied a rock-solid, exact voltage to the
base of a transistor relative to the emitter - and this is with no emitter
resistor, this is N volts directly from the base lead to the emitter lead
- and had the results predictable, or even repeatable?

Thanks,
Rich

 

[Paul Burridge]

Kevin pointed out days ago, you need an electric field to get a
current flow.

Oh dear. Is Mike still here posting snake oil? You can get a current
flow from a *magnetic* field. Back to school with you, Mike!
--

"What is now proved was once only imagin'd." - William Blake, 1793.

 

[Paul Burridge]

On Tue, 23 Nov 2004 21:44:01 GMT, Rich Grise <rich@example.net> wrote:

I don't know what his point is, but I just simply would like to present a
way that some people have been looking at this base voltage/base current
issue. Yes, the application of base voltage does ultimately result in an
effect on the collector current. But how many designs have you ever seen
where the designer relies on a particular Vbe drop to give the right
answer? I've _never_ seen a circuit like that, other than things like
current mirrors, and in that case, the Vbe is taken into consideration
either as an almost-constant, a la diode drop, or simply to be as equal as
is practicable to its partner in the circuit. But the easiest way I've
ever seen presented of doing the circuit analysis uses base current and
beta, with Vbe being a secondary (or less) consideration.

This is the only bone I have to pick here with you. If you want to call it
"voltage controlled", then fine, go ahead, but I know that every time I've
copied a circuit out of a book or whatever, the operative parameter has
been base current.

The Vbe/Ic way of looking at it works accurately over a much greater
range than the Ic/Ib relationship, Rich. Even *I* know that!
--

"What is now proved was once only imagin'd." - William Blake, 1793.

 

[Jamie Morken]

Rich The Philosophizer wrote:

On Sun, 14 Nov 2004 22:41:43 +0000, Paul Burridge wrote:


On Sun, 14 Nov 2004 15:40:52 +0100, "Fred Bartoli"
fred._canxxxel_this_bartoli@RemoveThatAlso_free.fr_AndThisToo> wrote:


"Jim Thompson" <thegreatone@example.com> a écrit dans le message de
news:cvgcp0hg2ud2ocgu21eovicehdnb5sgdqv@4ax.com...

On Sat, 13 Nov 2004 10:37:31 +0100, "Fred Bartoli"
fred._canxxxel_this_bartoli@RemoveThatAlso_free.fr_AndThisToo> wrote:


"Paul Burridge" <pb@notthisbit.osiris1.co.uk> a écrit dans le message de
news:16gap0138huajj8b3k0abf2h76opodmd39@4ax.com...

[snip]

FETs are certainly accurately modelled as VCCS in a certain regions of
operation. BJTs are *inaccurately* modelled as CCCS. Vbe's what you
really want to look at with BJTs.

Please, tell us more

Paul doesn't have a clue... he's just parroting ;-)


Yes indeed. And we also all know where his remark came from...
I just feel sometimes a bit irritated when, more often than not, he plays Mr
know it all and shouts things he picked here and there, and he have no clue
about.

I get criticised when I *don't* remember things I've been told and you
say you find it irritating when I *do* so I can't win, can I?


Nah. Can't break even either. Can't even quit the game. You're damned
if you do, and damned if you don't.

For these times, when you're at a crossroads of life, and the signs
all say, "STOP" "DO NOT ENTER" "NO RIGHT TURN" "NO LEFT TURN" "NO
U-TURN" "NO PARKING" and "Do Not Back Up, Sever Tire Damage!", I
always fall back on the short-form Serenity Prayer:

"Fuck it."

You do your name good justice!

;^j
Rich

 

[Rich Grise]

On Tue, 23 Nov 2004 22:02:22 +0000, Ken Smith wrote:

In article <pan.2004.11.23.20.45.07.542599@example.net>,
Rich Grise <rich@example.net> wrote:
On Tue, 23 Nov 2004 17:59:42 +0000, Ken Smith wrote:

In article <pan.2004.11.23.16.49.27.353520@example.net>,
Rich Grise <rich@example.net> wrote:
[...]
Well, in all of my long experience, and through electronics tech school,
and according to the old military "physics of a transistor" book and
according to every circuit design I've ever had analyzed for me, base
current has always been the determining factor - Vbe is almost constant,
because it's a diode. This is why transistors need to be temperature
compensated.

I assume this is not large signal, RF experience. In large signal RF, you
can have buckets of base current with very little collector current.

No. Actually, I've been studiously avoiding learning anything about any
semiconductor that operates any faster than about 20 KHz.

But, now that you mention it, in that application, would you consider the
part to be "voltage controlled" or "current controlled"?

It more closely resembles the "charge controlled" model. It is the
voltage gradient in the E-B depletion layer that really causes the
carriers to move from the emitter to the base. It easy for that to lag
what you see at the device terminals by quite a lot.

Thanks!

Rich

 

[Rich Grise]

On Wed, 24 Nov 2004 00:02:52 +0000, Paul Burridge wrote:

On Tue, 23 Nov 2004 21:44:01 GMT, Rich Grise <rich@example.net> wrote:

I don't know what his point is, but I just simply would like to present a
way that some people have been looking at this base voltage/base current
issue. Yes, the application of base voltage does ultimately result in an
effect on the collector current. But how many designs have you ever seen
where the designer relies on a particular Vbe drop to give the right
answer? I've _never_ seen a circuit like that, other than things like
current mirrors, and in that case, the Vbe is taken into consideration
either as an almost-constant, a la diode drop, or simply to be as equal as
is practicable to its partner in the circuit. But the easiest way I've
ever seen presented of doing the circuit analysis uses base current and
beta, with Vbe being a secondary (or less) consideration.

This is the only bone I have to pick here with you. If you want to call it
"voltage controlled", then fine, go ahead, but I know that every time I've
copied a circuit out of a book or whatever, the operative parameter has
been base current.

The Vbe/Ic way of looking at it works accurately over a much greater
range than the Ic/Ib relationship, Rich. Even *I* know that!

OK. I lose.

http://www.st-andrews.ac.uk/~www_pa/Scots_Guide/info/comp/active/BiPolar/bpcur.html

Kevin Aylward has been corroborated by The Internet. ;-)

Cheers!
Rich

 

[Rhett Auricle]

On Tue, 23 Nov 2004 23:57:30 +0000, Paul Burridge wrote:

Kevin pointed out days ago, you need an electric field to get a
current flow.

Oh dear. Is Mike still here posting snake oil? You can get a current
flow from a *magnetic* field. Back to school with you, Mike!

If you want to look at the big picture, magnetism came first.

;^j
R.

 

[Active8]

On Wed, 24 Nov 2004 03:43:02 GMT, Rich Grise wrote:

OK. I lose.
^^^^ "didn't win."

http://www.st-andrews.ac.uk/~www_pa/Scots_Guide/info/comp/active/BiPolar/bpcur.html

Kevin Aylward has been corroborated by The Internet. ;-)

Some months ago, Genome was posting on how loose measured beta was.

I don't think he got a tight family of curves over a number of
devices - beta/Ic curves, that is.

--
Best Regards,
Mike

 

[Active8]

On Tue, 23 Nov 2004 23:57:30 +0000, Paul Burridge wrote:

Kevin pointed out days ago, you need an electric field to get a
current flow.

Oh dear. Is Mike still here posting snake oil? You can get a current
flow from a *magnetic* field. Back to school with you, Mike!

Hey asshole. Faraday's law says that a changing magnetic field
gives an induced EMF, not a current flow, so fuck off.
--
Best Regards,
Mike

 

[Kevin Aylward]

Dr. Slick wrote:

"Kevin Aylward" <salesEXTRACT@anasoft.co.uk> wrote in message
news:<WILod.22907$08.22879@fe2.news.blueyonder.co.uk>...


I'm saying that the base current is controlled
by Vbe, so in essence, I'M AGREEING WITH YOU,

Well, it appeared that you were cliaming that base current controlled
Vbe.


My point, once again, was that the majority
of texts DO use beta and current gain to describe
BJTs, but never to describe FETs, which have to
high of a DC gate impedance to use current gain,
so they use transconductance instead.

I agree that the majority of popular elementary texts do this. However,
they are wrong if they claim that a bipolar transistor is current
controlled. It just introduces much confusion. The most appropriate
model is a voltage controlled device with a non-linear input resistance.

BUTT-FUCKER. YOU CAN CONSIDER BJTs WITH
EITHER BETA OR Gm.

Sure, you can use a beta model, but that doesn't make a transistor
current controlled, which was the point being made.


Hint: What's a current source? It's a voltage
source with a relatively high series resistance (as
compared to the resistance of the driven input
impedance.)

Yes. Which confirms my point.

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Active8]

On Wed, 24 Nov 2004 02:02:10 -0500, Active8 wrote:

On Tue, 23 Nov 2004 23:57:30 +0000, Paul Burridge wrote:

Kevin pointed out days ago, you need an electric field to get a
current flow.

Oh dear. Is Mike still here posting snake oil? You can get a current
flow from a *magnetic* field. Back to school with you, Mike!

Hey asshole. Faraday's law says that a changing magnetic field
gives an induced EMF, not a current flow, so fuck off.

Be nice, Mike.
--
Best Regards,
Mike

 

[lemonjuice]

On Tue, 23 Nov 2004 23:57:30 +0000, Paul Burridge
<pb@notthisbit.osiris1.co.uk> wrote:

Kevin pointed out days ago, you need an electric field to get a
current flow.

Oh dear. Is Mike still here posting snake oil? You can get a current
flow from a *magnetic* field. Back to school with you, Mike!
yes and you can get current from light using a photodiode

 

[Paul Burridge]

On Wed, 24 Nov 2004 03:43:02 GMT, Rich Grise <rich@example.net> wrote:

OK. I lose.

http://www.st-andrews.ac.uk/~www_pa/Scots_Guide/info/comp/active/BiPolar/bpcur.html

Kevin Aylward has been corroborated by The Internet. ;-)

Kevin know it, but by reading his "papers" you'd never guess. ;-)
--

"What is now proved was once only imagin'd." - William Blake, 1793.

 

[Kevin Aylward]

lemonjuice wrote:

On Tue, 23 Nov 2004 18:55:05 GMT, "Kevin Aylward"
salesEXTRACT@anasoft.co.uk> wrote:

lemonjuice wrote:
On Tue, 23 Nov 2004 07:54:52 GMT, "Kevin Aylward"
salesEXTRACT@anasoft.co.uk> wrote:

Dr. Slick wrote:
"Kevin Aylward" <salesEXTRACT@anasoft.co.uk> wrote in message
news:<VYhod.19283$08.18967@fe2.news.blueyonder.co.uk>...

Well they correct your trivial misunderstandings. Like, the base
current is the relevant current in the diode equation. Yeah right
on. You put your foot right in it that time.


Never said the base current was as big as the
collector current.

You claimed (inferred) that one uses the base current as the
current
to
be used in the base-emitter diode equation. This is incorrect.
No.


No. Its yes.

Sorry its a No.

Its a yes dude.

As I can easily prove to you below the current that
flows through the base emitter diode is related to the base emitter
voltage.

No problem with this. This isn't at issue. Its irrelevant that Vbe
controls ib, the point is determining the emitter current. The resulting
base current is not what determines Ie.

You seem to be arguing a different point.

But thats intuitive ... its a pn junction ... so what OTHER equation
do you think could be written for it?

The current that goes from emitter to base, is essentially, all
collector current. One dosnt use the "diode equation" with ib to get Ic.

The diode equaition is

Ie = is.exp(Vbe/Vt)

Sure one might write

Ib = is.exp(Vbe/Vt)/(1+hfe)

But this aint the diode equation as usually understood. We certainly
don't directly know what hfe is as a function of vbe.

The density of these are
"first order" (laughing at first order) proportional to the
exponential of the base to emitter voltage. The base current isn't
exponentially dependent on the bias (though it is on the linear
dimensions) because you have a recombining current of the majority
carriers , also exponentially dependent on the base emitter
voltage(with an exponential linear dimensional dependence) the 2
summed up give you an approx. (grin at approx. ) constant base
current.


And what relevance does this have to the well known fact that the
diode equation is in reference to "emitter" current, not base
current.

What I wrote above is proof that not only is Ic is related
exponentially to Vbe but so is Ib.

Of course there is a relation between vbe and ib, whats you point?

Why did I write it ? Because you
wrote above that it wasn't the case.

No I didn't. To the contrary, I explicitly stated that the base current
is a function of Vbe. I also said this is irrelevant.


{snip, as I already posted the full Spice equations}

IE = is.(exp(Vbe/KT) - 1)
you "probably" forgot the electron charge q
Ie = Is (exp qVbe/KT - 1)

is is *not ib = is.(exp(Vbe/KT) - 1)

yes I have proved it qualitatively and quantitatively above.
obviously the saturation currents are unequal.

Your blathering about a different issue, not in debate.

The "diode equation" as a statement in transistor theory, is the diode
equation relating Ie to Vbe. The base current is ignored.

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.