High gain amplifier?

[powersupplies]

http://www.btinternet.com/~powersupplies/high_gain_amp.gif

Can someone help with the study of the above application,
what is the gain formula, why do we need constant current
flowing in the collector?

 

[Kevin Aylward]

powersupplies wrote:

http://www.btinternet.com/~powersupplies/high_gain_amp.gif

Can someone help with the study of the above application,
what is the gain formula, why do we need constant current
flowing in the collector?

A high impedance, i.e. a constant current load, is what gives the high
gain. The gain is nominally Rc/re, where Re is 40Ic_bias. However, for a
perfect currant source load, the turns out to be Va/25mV, where Va is
the Early voltage of the transistor.

See, for example,
http://www.anasoft.co.uk/EE/bipolardesign1/bipolardesign1.html

Note that the circuit you illustrated is a daft circuit made by apps
engineers who are usually clueless about real design, and would do
anything to flog such parts. No one in there right mind would use such a
pathetic circuit.

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.

"quotes with no meaning, are meaningless" - Kevin Aylward.

 

[powersupplies]

A high impedance, i.e. a constant current load, is what gives the high
gain. The gain is nominally Rc/re, where Re is 40Ic_bias. However, for a
perfect currant source load, the turns out to be Va/25mV, where Va is
the Early voltage of the transistor.

what's Rc, and why is a current source a high impedance?

See, for example,
http://www.anasoft.co.uk/EE/bipolardesign1/bipolardesign1.html

wow, great site, is it yours, it's great. I started reading, and about a
third
down you have

For the example above, this allows an easy calculation of the small signal
voltage across the diode as Vd = Vi (re/(R + re)) i.e. by the use of
conventional formulas.

but you do not explain how you got this, and which example exactly are
you refering too? what is Vd, R, Vi, which circuit is that?

 

[Animesh Maurya]

"Kevin Aylward" <kevindotaylwardEXTRACT@anasoft.co.uk> wrote in message news:<GOh6c.114$tp.80@newsfep3-gui.server.ntli.net>...

powersupplies wrote:
http://www.btinternet.com/~powersupplies/high_gain_amp.gif

Can someone help with the study of the above application,
what is the gain formula, why do we need constant current
flowing in the collector?

A high impedance, i.e. a constant current load, is what gives the high
gain. The gain is nominally Rc/re, where Re is 40Ic_bias. However, for a
perfect currant source load, the turns out to be Va/25mV, where Va is
the Early voltage of the transistor.

See, for example,
http://www.anasoft.co.uk/EE/bipolardesign1/bipolardesign1.html

Note that the circuit you illustrated is a daft circuit made by apps
engineers who are usually clueless about real design, and would do
anything to flog such parts. No one in there right mind would use such a
pathetic circuit.

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.

"quotes with no meaning, are meaningless" - Kevin Aylward.

Nice tutorial Kevin.

I am still confused to say BJT as voltage controlled current source.

Textbook I follow says BJTs as current controlled current source,
rather they call FETs as voltage controlled current source.

After interpreting the results of BJT output characteristics, one can
easily say that in the active region small base current controls a
large collector current, so it should be a current controlled device.

Can you please explain me what's the fact underlying it.

While going through it all around I found a little error, second last
line above the circuit diagram says emitter current as (1+hfe)Ic which
should be (1+hfe)Ib as mentioned in equation 2.

Thanks,

Animesh Maurya

 

[Kevin Aylward]

Animesh Maurya wrote:

"Kevin Aylward" <kevindotaylwardEXTRACT@anasoft.co.uk> wrote in
message news:<GOh6c.114$tp.80@newsfep3-gui.server.ntli.net>...
powersupplies wrote:
http://www.btinternet.com/~powersupplies/high_gain_amp.gif

Can someone help with the study of the above application,
what is the gain formula, why do we need constant current
flowing in the collector?

A high impedance, i.e. a constant current load, is what gives the
high gain. The gain is nominally Rc/re, where Re is 40Ic_bias.
However, for a perfect currant source load, the turns out to be
Va/25mV, where Va is the Early voltage of the transistor.

See, for example,
http://www.anasoft.co.uk/EE/bipolardesign1/bipolardesign1.html

Note that the circuit you illustrated is a daft circuit made by apps
engineers who are usually clueless about real design, and would do
anything to flog such parts. No one in there right mind would use
such a pathetic circuit.


Nice tutorial Kevin.

Thanks.

I am still confused to say BJT as voltage controlled current source.

Textbook I follow says BJTs as current controlled current source,
rather they call FETs as voltage controlled current source.

Those text books are wrong. Both devices are inherently voltage
controled devices.

After interpreting the results of BJT output characteristics, one can
easily say that in the active region small base current controls a
large collector current, so it should be a current controlled device.

The fact that there is some vague correlation between collector current
and base current does not imply that one is controlling the other. The
base current that exists is incidental. A super ideal transistor would
have zero base current, yet the emitter current would still follow the
exponential law. A bipolar should be thought of as a fet that has been
spoiled by a base emitter resistor.

The fundamental approximate equation is

Ie = Is.(exp(Vbe/Vt) - 1)

There is no base current term. Its a matter of basic physics. Applying a
potential to the base emitter injects carriers into the base region,
that get swept up into the collector. Its just a nuisance that some leak
out through the base. Such a leak, dose not effect the exponential
relation between base emitter voltage and emitter current.

Can you please explain me what's the fact underlying it.

While going through it all around I found a little error, second last
line above the circuit diagram says emitter current as (1+hfe)Ic which
should be (1+hfe)Ib as mentioned in equation 2.

Oh dear...

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.

"quotes with no meaning, are meaningless" - Kevin Aylward.

 

[Kevin Aylward]

powersupplies wrote:

A high impedance, i.e. a constant current load, is what gives the
high gain. The gain is nominally Rc/re, where Re is 40Ic_bias.
However, for a perfect currant source load, the turns out to be
Va/25mV, where Va is the Early voltage of the transistor.

what's Rc, and why is a current source a high impedance?


See, for example,
http://www.anasoft.co.uk/EE/bipolardesign1/bipolardesign1.html

wow, great site, is it yours,

Yes.

it's great.

Thanks.

I started reading, and
about a third
down you have

For the example above, this allows an easy calculation of the small
signal voltage across the diode as Vd = Vi (re/(R + re)) i.e. by the
use of conventional formulas.

but you do not explain how you got this, and which example exactly are
you refering too? what is Vd, R, Vi, which circuit is that?

Seems I missed a figure. It is fig.1 but with an emitter resistor. Vd is
the diode voltage of the base emitter diode. R is the emitter resistor.
The voltage at the base emitter (Vd) is just a simple potential divider
of R and the dynamic impedance re with the input voltage, Vi.

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.

"quotes with no meaning, are meaningless" - Kevin Aylward.

 

[John Popelish]

Kevin Aylward wrote:
(snip)

The fact that there is some vague correlation between collector current
and base current does not imply that one is controlling the other. The
base current that exists is incidental. A super ideal transistor would
have zero base current, yet the emitter current would still follow the
exponential law. A bipolar should be thought of as a fet that has been
spoiled by a base emitter resistor.
(snip)

Not so much a linear resistor leakage as an exponential leakage (a
nuisance diode in parallel). The distinction is especially important
at high frequency, where the base current is largely capacitive, but
that component of base current doesn't contribute to the collector
current (and may actually be counter productive). It is the voltage
that counts.

--
John Popelish

 

[powersupplies]

Seems I missed a figure. It is fig.1 but with an emitter resistor. Vd is
the diode voltage of the base emitter diode. R is the emitter resistor.
The voltage at the base emitter (Vd) is just a simple potential divider
of R and the dynamic impedance re with the input voltage, Vi.

in practice gm or 1/gm are not used for bipolars simply because the
datasheets give us hfe and so this is what we use, I think!

 

[Kevin Aylward]

powersupplies wrote:

Seems I missed a figure. It is fig.1 but with an emitter resistor.
Vd is the diode voltage of the base emitter diode. R is the emitter
resistor. The voltage at the base emitter (Vd) is just a simple
potential divider of R and the dynamic impedance re with the input
voltage, Vi.

in practice gm or 1/gm are not used for bipolars simply because the
datasheets give us hfe and so this is what we use, I think!

You think incorrectly. gm is always used by designers that know what
they are doing. Its indispensable, and is, essentially, the same for
*all* bipolar transistors.

gm = 40.Ic

at 25 deg C. (q/KT)

Of course hfe is required in a full design.

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.

"quotes with no meaning, are meaningless" - Kevin Aylward.

 

[powersupplies]

in practice gm or 1/gm are not used for bipolars simply because the
datasheets give us hfe and so this is what we use, I think!

You think incorrectly. gm is always used by designers that know what
they are doing. Its indispensable, and is, essentially, the same for
*all* bipolar transistors.

gm = 40.Ic

at 25 deg C. (q/KT)

Of course hfe is required in a full design.

I have looked it up in Millman and Grabel, and what you are talking
about is knwon as the hybrid-pi model.

beta = hfe = gm r_pi (page 119)
so given hfe you can get r_pi (as gm is known, 40Ic)
but normally the prefered model is the h-parameter model
i.e. hie/hfe/hre/hoe because this is what is given in datasheets
and can be "plugged" directly into the equivalent circuit.
In FETs the gm is given, so we use gm, but not for bipolars!!

 

[powersupplies]

"powersupplies" <steve@wwwusenet.com> wrote in message
news:c3c1ol$ahe$1@news.gngidc.net...

http://www.btinternet.com/~powersupplies/high_gain_amp.gif

Can someone help with the study of the above application,
what is the gain formula, why do we need constant current
flowing in the collector?

I think we can say

Ic=Vin/R1*hfe
and
Ic+Vout/RL = 2Amp (due to the constant current)
How do we obtain Av = Vout/Vin ? am I missing an equation?

 

[Kevin Aylward]

powersupplies wrote:

in practice gm or 1/gm are not used for bipolars simply because the
datasheets give us hfe and so this is what we use, I think!

You think incorrectly. gm is always used by designers that know what
they are doing. Its indispensable, and is, essentially, the same for
*all* bipolar transistors.

gm = 40.Ic

at 25 deg C. (q/KT)

Of course hfe is required in a full design.

I have looked it up in Millman and Grabel, and what you are talking
about is knwon as the hybrid-pi model.

No. That is, yes, the hybrid pi model uses gm, but gm stands on its own.
It doesn't rely of the hybrid-pi model.

beta = hfe = gm r_pi (page 119)
so given hfe you can get r_pi (as gm is known, 40Ic)
but normally the prefered model is the h-parameter model
i.e. hie/hfe/hre/hoe because this is what is given in datasheets

Hie is a *derived* parameter, it is hie = rbb' + (1+hfe)re

rbb' is the base spreading resistor, re=1/gm

and can be "plugged" directly into the equivalent circuit.
In FETs the gm is given, so we use gm, but not for bipolar!!

No!No!No!

You still misunderstand. It is a *fundamental* *physics* fact. The
emitter current is a direct function of the applied *voltage*. Whether
the diode is a standalone one, or the diode junction of a transistor, it
still follows the Id=Is.exp(Vd.q/KT) law. The base current only hase 2nd
order effects on the current, e.g. ib.rb voltage loss.

The gm is *the* basic parameter of a design. It sets the fundamentally
output current verses input voltage. The bipolar transistor is a voltage
controlled device. End of story.

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.

"quotes with no meaning, are meaningless" - Kevin Aylward.

 

[powersupplies]

"Kevin Aylward" <kevindotaylwardEXTRACT@anasoft.co.uk> wrote in message
news:3GS6c.21$Kt4.5@newsfep3-gui.server.ntli.net...

powersupplies wrote:
in practice gm or 1/gm are not used for bipolars simply because the
datasheets give us hfe and so this is what we use, I think!

You think incorrectly. gm is always used by designers that know what
they are doing. Its indispensable, and is, essentially, the same for
*all* bipolar transistors.

gm = 40.Ic

at 25 deg C. (q/KT)

Of course hfe is required in a full design.

I have looked it up in Millman and Grabel, and what you are talking
about is knwon as the hybrid-pi model.

No. That is, yes, the hybrid pi model uses gm, but gm stands on its own.
It doesn't rely of the hybrid-pi model.


beta = hfe = gm r_pi (page 119)
so given hfe you can get r_pi (as gm is known, 40Ic)
but normally the prefered model is the h-parameter model
i.e. hie/hfe/hre/hoe because this is what is given in datasheets

Hie is a *derived* parameter, it is hie = rbb' + (1+hfe)re

rbb' is the base spreading resistor, re=1/gm

I'll have to look this up.

and can be "plugged" directly into the equivalent circuit.
In FETs the gm is given, so we use gm, but not for bipolar!!

No!No!No!

You still misunderstand. It is a *fundamental* *physics* fact. The
emitter current is a direct function of the applied *voltage*. Whether
the diode is a standalone one, or the diode junction of a transistor, it
still follows the Id=Is.exp(Vd.q/KT) law. The base current only hase 2nd
order effects on the current, e.g. ib.rb voltage loss.

The gm is *the* basic parameter of a design. It sets the fundamentally
output current verses input voltage. The bipolar transistor is a voltage
controlled device. End of story.

the thing is, if you say a "voltage controlled device" you are implying
something
that has high input impedance, which only cares about voltage. Now with
bipolars, hie is not high impedance and therefore your voltage source
will be reduced by it's own internal impedance (Rs). All the formulas in
your
tutorial use the input voltage Vi = Vbe. This is not so good for bipolars
because Vbe will not be equal to you input due to hie and your source's
Rs, and when we calculate gain we use Vs and not Vbe.

Also, do you know what is hfe? hfe is *defined* as a current controlled
current source.

 

[Kevin Aylward]

powersupplies wrote:

"Kevin Aylward" <kevindotaylwardEXTRACT@anasoft.co.uk> wrote in
message news:3GS6c.21$Kt4.5@newsfep3-gui.server.ntli.net...
powersupplies wrote:
in practice gm or 1/gm are not used for bipolars simply because
the datasheets give us hfe and so this is what we use, I think!

You think incorrectly. gm is always used by designers that know
what they are doing. Its indispensable, and is, essentially, the
same for *all* bipolar transistors.

gm = 40.Ic

at 25 deg C. (q/KT)

Of course hfe is required in a full design.

I have looked it up in Millman and Grabel, and what you are talking
about is knwon as the hybrid-pi model.

No. That is, yes, the hybrid pi model uses gm, but gm stands on its
own. It doesn't rely of the hybrid-pi model.


beta = hfe = gm r_pi (page 119)
so given hfe you can get r_pi (as gm is known, 40Ic)
but normally the prefered model is the h-parameter model
i.e. hie/hfe/hre/hoe because this is what is given in datasheets

Hie is a *derived* parameter, it is hie = rbb' + (1+hfe)re

rbb' is the base spreading resistor, re=1/gm

I'll have to look this up.


and can be "plugged" directly into the equivalent circuit.
In FETs the gm is given, so we use gm, but not for bipolar!!

No!No!No!

You still misunderstand. It is a *fundamental* *physics* fact. The
emitter current is a direct function of the applied *voltage*.
Whether the diode is a standalone one, or the diode junction of a
transistor, it still follows the Id=Is.exp(Vd.q/KT) law. The base
current only hase 2nd order effects on the current, e.g. ib.rb
voltage loss.

The gm is *the* basic parameter of a design. It sets the
fundamentally output current verses input voltage. The bipolar
transistor is a voltage controlled device. End of story.

the thing is, if you say a "voltage controlled device" you are
implying something
that has high input impedance, which only cares about voltage.

It implies *nothing* of the sort. A voltage controlled device is any
device that its output is a *direct* function of its input voltage.
Period.

What part of Ie = Is.exp(Vbeq/KT) do you still fail to understand?

What part of, there is no base current term in this equation, do you
still fail to understand?

Now
with bipolars, hie is not high impedance and therefore your voltage
source will be reduced by it's own internal impedance (Rs).

Yes, but simply irrelevant. It is what ever voltage is at its input
terminals. The fact that this may not be the open circuit voltage of
some controlling source has no baring on the voltage controlled nature
of a transistor whatsoever.

All the
formulas in your
tutorial use the input voltage Vi = Vbe. This is not so good for
bipolars because

This is the correct way, its not debatable. You don't seem to understand
the basics of analysing circuits here. One calculates what output
current is produced by the input voltage. Of course, it may take some
more calculations to determine the input voltage if the source has
significant resistance, but this is besides the point.

Vbe will not be equal to you input due to hie and

hie has, little effect on the voltage across the base emitter junction.
hie is mostly a shunt resistance. I have already noted that they may be
a *small* drop due to rbb'.ib.

your source's Rs, and when we calculate gain we use Vs and not Vbe.

Sure, we include the source resistance to calculate Vbe. How do you
think this changes the fact that the transistor is fundamentally based
on the physics of a voltage controlled device.

You need to replace your unfounded preconceptions. I'm correcting the
old wifes tales that many hold.

Also, do you know what is hfe? hfe is *defined* as a current
controlled current source.

No its not. Look, dude. You aint goanna win on this. Your statement here
is completely nonsensical. hfe is simple the ratio of collector current
to base current. There is nothing implied about the base current
controlling the collector current. It doesn't.

Look, the flow of charge into the base is completely incidental to how a
transistor works. What part of the base emitter voltage injects carriers
into the base region where they are, now get this, collected, by the
collector, do you still fail to understand? Base current *leakage* is a
nuisance. That's it. Much effect is taken to design transisters so that
the base current is small, ideally, zero.

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.

"quotes with no meaning, are meaningless" - Kevin Aylward.

 

[powersupplies]

sorry, I am replying here because my server didn't pick up your
last post.

What part of Ie = Is.exp(Vbeq/KT) do you still fail to understand?

the problem with this formula is that Vbe is not something that can
be controlled, and therefore this formula is useless, unless you are
performing a computer simulation, in which case you may want to
use it (this is not my speciality so I don't know about writing cad
software). If you want a practical way to design a transistor circuit
you need to use the h-parameters equivalent circuit.

Also, do you know what is hfe? hfe is *defined* as a current
controlled current source.

No its not. Look, dude. You aint goanna win on this. Your statement here
is completely nonsensical. hfe is simple the ratio of collector current
to base current. There is nothing implied about the base current
controlling the collector current.

YES, it does imply that - you are wrong.

B.t.w. did you manage to find the gain for the original circuit, I am still
struggling with it.

 

[Winfield Hill]

powersupplies wrote...

Also, do you know what is hfe? hfe is *defined* as a
current controlled current source.

If so, the controlled "current source" is the base. If any
such "defining" is involved, it's the base current, defined
as the collector current (itself perfectly defined by the Vbe
voltage and g_m), divided by h_FE. Not the other way around.

h_FE, or beta, is in turn a very flaky parameter affected by
current, voltage, temperature, plus a substantial part-to-part
influence. Plus time and history elements as well. Sheesh!
Ic defined by Ib? No way. Sorry, transistor man! A useful
concept for beginners, but lasting only a few weeks at most.

Thanks,
- Win

whill_at_picovolt-dot-com

 

[Kevin Aylward]

powersupplies wrote:

sorry, I am replying here because my server didn't pick up your
last post.

What part of Ie = Is.exp(Vbeq/KT) do you still fail to understand?

the problem with this formula is that Vbe is not something that can
be controlled,

Of course it can. Its usually very easy to make a source that has a low
resistance relative to the input resistance of the effective emitter
junction. Its what we do when we design circuits correctly. If this
condition is not satisfied, the design is usually poor. It will often
result in excess noise, and low gain because the input signal is being
unduly attenuated. Typically, one aims to get at least Rs < 5.Rin.

For example, at 1ma re=25 ohms. Lets say hfe is > 100, then the input
resistance will be > 100*25 or > 2.5k. If the source resistance was say,
100 ohms, then the error in voltage would be around 0.1/2.5 = 4%.

and therefore this formula is useless,

Nonsense. You simply don't know the *first* thing about designing
circuits, yet you have the audacity to make grandiose claims.

unless you are
performing a computer simulation, in which case you may want to
use it (this is not my speciality so I don't know about writing cad
software).

Computer simulations usually account for all of the major effects. In
general, they can be very, very, accurate to real life. Writing cad
software has nothing to do with using the stuff as a tool.

If you want a practical way to design a transistor circuit
you need to use the h-parameters equivalent circuit.

Now I have to say it. You are simply clueless. Look sonny, there are
some us here who have actually been successfully, professionally,
designing analogue circuits for 25 years. You would do well to
appreciate the limits of your knowledge and the knowledge of your
betters.

Sure, one can do some rough sketch ideas with some simple models, but
the details take way more work then what can achieved by such a simple
model.

What we analogue i.c. design engineers actually do is use spice, pretty
much exclusively. Its indispensable. You try designing a 1000 transistor
circuit by writing down a h-parameter equivalent circuit. Of course,
such a model is useless for large signals. Now try solving the
non-linear differential equations for large signals. I doubt if you can
solve dy/dx=1/x.

Also, do you know what is hfe? hfe is *defined* as a current
controlled current source.

No its not. Look, dude. You aint goanna win on this. Your statement
here is completely nonsensical. hfe is simple the ratio of collector
current to base current. There is nothing implied about the base
current controlling the collector current.

YES, it does imply that - you are wrong.

Not a chance mate. Its obvious that you are a *complete* novice. Don't
get me wrong, we all have to learn sometime, but what you need to know
is just how limited you understanding of the most basic circuits is.
Tell me, where did you get *your* bachelors degree in Electronic
Engineering?

You need to go and get some text books on *physics*. All you knowledge
apears to be from bantam paperbacks, like "Electric Circuits For
Dummies". And I mean this seriously, not as a personal insult. You have
simply been reading books of limited technical quality.

B.t.w. did you manage to find the gain for the original circuit, I am
still struggling with it.

And you claim that I am wrong. Jesus wept, dude.

The gain of your circuit can not be calculated without more information.
You need to know the dynamic impedance of the regulator connected as a
current source.

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.

"quotes with no meaning, are meaningless" - Kevin Aylward.

 

[Winfield Hill]

powersupplies wrote...

If you want a practical way to design a transistor circuit
you need to use the h-parameters equivalent circuit.

No, the hybrid model was abandoned in the early 60s, just
about the time I started serious designing. I spent about
three months religiously using h-parameters, searching out
transistors with h-parameter curves, etc., and failing to
get accurate results. Then Carver Mead, who was a friend
of my boss, came by for a visit and I got a defining-moment
15-minute lecture on Ebers-Moll and its variants. Looking
in various textbooks, I realized that a few described this
method, which a small part of the community had been using
for years. Anyway, many difficult design issues suddenly
became easy, and my circuit results started matching my
calculations. Over the last 40 years, the awful h-parameter
approach has thankfully been relegated to the dustbin by the
engineering community, so that rarely do we see anyone now
espousing its use, let alone arguing it's the only true way.

Also, do you know what is hfe? hfe is *defined* as a current
controlled current source.

No its not. Look, dude. You aint goanna win on this. Your
statement here is completely nonsensical. hfe is simple
the ratio of collector current to base current. There is
nothing implied about the base current controlling the
collector current.

YES, it does imply that - you are wrong.

No, the collector current "controls" the base current is much
more accurate statement, albeit with limited usefulness.

B.t.w. did you manage to find the gain for the original circuit,
I am still struggling with it.

Kevin accurately told you what it'd be, using Va (the Early
voltage for the LM395 "transistor") and the transconductance
of the LM317 current source, but neither of these parameters
is available from the manufacturer, nor their manufacturing
spreads. But they are very easy to measure for a given part.

Thanks,
- Win

whill_at_picovolt-dot-com

 

[Kevin Aylward]

Winfield Hill wrote:

powersupplies wrote...

If you want a practical way to design a transistor circuit
you need to use the h-parameters equivalent circuit.

No, the hybrid model was abandoned in the early 60s, just
about the time I started serious designing. I spent about
three months religiously using h-parameters, searching out
transistors with h-parameter curves, etc., and failing to
get accurate results. Then Carver Mead, who was a friend
of my boss, came by for a visit and I got a defining-moment
15-minute lecture on Ebers-Moll and its variants. Looking
in various textbooks, I realized that a few described this
method, which a small part of the community had been using
for years. Anyway, many difficult design issues suddenly
became easy, and my circuit results started matching my
calculations. Over the last 40 years, the awful h-parameter
approach has thankfully been relegated to the dustbin by the
engineering community, so that rarely do we see anyone now
espousing its use, let alone arguing it's the only true way.

Also, do you know what is hfe? hfe is *defined* as a current
controlled current source.

No its not. Look, dude. You aint goanna win on this. Your
statement here is completely nonsensical. hfe is simple
the ratio of collector current to base current. There is
nothing implied about the base current controlling the
collector current.

YES, it does imply that - you are wrong.

No, the collector current "controls" the base current is much
more accurate statement, albeit with limited usefulness.

I like to use the phrase "the collector current is a *function* of the
base current". This doesn't imply any direct control, just a
correlation. We all know, that the fact that there is a relation between
lung cancer and smoking does not, by itself, mean that such a relation
is causal, i.e controlling. It takes more evidence to show a causal
relation. In the case of the transistor, Vbe causes both the base
current and collector current, hence the base current and collector
current will be correlated, but not due to each other.

Does having a pet reduce your stress, or is the nature of people that
have pets less likely to have stress?

B.t.w. did you manage to find the gain for the original circuit,
I am still struggling with it.

Kevin accurately told you what it'd be, using Va (the Early
voltage for the LM395 "transistor") and the transconductance
of the LM317 current source, but neither of these parameters
is available from the manufacturer, nor their manufacturing
spreads. But they are very easy to measure for a given part.

The key thing about a transistor, is that it is, essentially, a diode!

One puts a voltage across a diode, and this accelerating potential,
tries to accelerate carriers, ie create a current in it. This is pretty
basic physics. Its voltage that makes things do things. The base of the
transistor is simply a way of impressing a voltage across this basic
diode, without such (controlling) voltage actually taking much current,
so that the bulk of the current is supplied by the collector circuit.
All that matters is that the voltage gets there. The diode junction
current then must follow the diode equation, irespective of any current
that leaks out of the base.

This very common misnomer about about transisters being erroneosly
described as base current controlled really needs addressing in a full
page in a good book, so I'll leave that one to you Win for your next
edition.

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.

"quotes with no meaning, are meaningless" - Kevin Aylward.

 

[powersupplies]

What part of Ie = Is.exp(Vbeq/KT) do you still fail to understand?

the problem with this formula is that Vbe is not something that can
be controlled,

Of course it can. Its usually very easy to make a source that has a low
resistance relative to the input resistance of the effective emitter
junction. Its what we do when we design circuits correctly. If this
condition is not satisfied, the design is usually poor. It will often
result in excess noise, and low gain because the input signal is being
unduly attenuated. Typically, one aims to get at least Rs < 5.Rin.

For example, at 1ma re=25 ohms. Lets say hfe is > 100, then the input
resistance will be > 100*25 or > 2.5k. If the source resistance was say,
100 ohms, then the error in voltage would be around 0.1/2.5 = 4%.

and therefore this formula is useless,

Nonsense. You simply don't know the *first* thing about designing
circuits, yet you have the audacity to make grandiose claims.

unless you are
performing a computer simulation, in which case you may want to
use it (this is not my speciality so I don't know about writing cad
software).

Computer simulations usually account for all of the major effects. In
general, they can be very, very, accurate to real life. Writing cad
software has nothing to do with using the stuff as a tool.

If you want a practical way to design a transistor circuit
you need to use the h-parameters equivalent circuit.

Now I have to say it. You are simply clueless. Look sonny, there are
some us here who have actually been successfully, professionally,
designing analogue circuits for 25 years. You would do well to
appreciate the limits of your knowledge and the knowledge of your
betters.

Sure, one can do some rough sketch ideas with some simple models, but
the details take way more work then what can achieved by such a simple
model.

What we analogue i.c. design engineers actually do is use spice, pretty
much exclusively. Its indispensable. You try designing a 1000 transistor
circuit by writing down a h-parameter equivalent circuit. Of course,
such a model is useless for large signals. Now try solving the
non-linear differential equations for large signals. I doubt if you can
solve dy/dx=1/x.



Also, do you know what is hfe? hfe is *defined* as a current
controlled current source.

No its not. Look, dude. You aint goanna win on this. Your statement
here is completely nonsensical. hfe is simple the ratio of collector
current to base current. There is nothing implied about the base
current controlling the collector current.

YES, it does imply that - you are wrong.

Not a chance mate. Its obvious that you are a *complete* novice. Don't
get me wrong, we all have to learn sometime, but what you need to know
is just how limited you understanding of the most basic circuits is.
Tell me, where did you get *your* bachelors degree in Electronic
Engineering?

You need to go and get some text books on *physics*. All you knowledge
apears to be from bantam paperbacks, like "Electric Circuits For
Dummies". And I mean this seriously, not as a personal insult. You have
simply been reading books of limited technical quality.


B.t.w. did you manage to find the gain for the original circuit, I am
still struggling with it.

And you claim that I am wrong. Jesus wept, dude.

The gain of your circuit can not be calculated without more information.
You need to know the dynamic impedance of the regulator connected as a
current source.

I now realise, thanks to Winfield's post that there are two ways to
approach the transistor. I think what you and Winfield are
saying is that a transistor is a voltage controlled current
source because you are using the hybrid pi equivalent circuit,
whereas I am using the h-parameters equivalent circuit which
defines the transistor as a current controlled current source.
It seems that some books use hybrid pi, others use h parameters,
which is confusing. Apart from that, hfe is h21 of the h parameters
and implies a current controlled current source. Whether it is a good
way to describe a transistor, you may say - no, but hfe is a current
controlled current source.