Op-amps and Transistors

[zalzon]

Without long explainations, is it fair to say :

1) A transistor uses change in voltage to control a large flow of
current.

2a) An op-amp amplifies small differences in voltages across its 2
inputs.

2b) An op-amp using negative feedback tries to make the amplified
output signal more like the input signal.

2c) The amplified voltage of an op-amp cannot be greater than the
supply voltage. e.g. if supply voltage is 5V, it cannot be amplified
to more than -2.5 to +2.5V.


Is all of the above right?

 

[Kevin Aylward]

zalzon wrote:

Without long explainations, is it fair to say :

1) A transistor uses change in voltage to control a large flow of
current.

Yes.

2a) An op-amp amplifies small differences in voltages across its 2
inputs.

Yes.

2b) An op-amp using negative feedback tries to make the amplified
output signal more like the input signal.

Yes.

2c) The amplified voltage of an op-amp cannot be greater than the
supply voltage. e.g. if supply voltage is 5V, it cannot be amplified
to more than -2.5 to +2.5V.

Yes.

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.

 

[John Popelish]

zalzon wrote:

Without long explainations, is it fair to say :

1) A transistor uses change in voltage to control a large flow of
current.

The word, 'large' adds no meaning. The rest is pretty iffy, also.

2a) An op-amp amplifies small differences in voltages across its 2
inputs.

This is good enough to be useful.

2b) An op-amp using negative feedback tries to make the amplified
output signal more like the input signal.

No. It uses negative feedback to make its two inputs more the same.

2c) The amplified voltage of an op-amp cannot be greater than the
supply voltage. e.g. if supply voltage is 5V, it cannot be amplified
to more than -2.5 to +2.5V.

Yes. The output voltage is the result of some sort of voltage divider
operation across the supply.

Is all of the above right?

I wouldn't go that far.
Besides, simple generalities are almost never 'right'.
But sometimes they are close enough to right to be useful.

--
John Popelish

 

[Kevin Aylward]

John Popelish wrote:

zalzon wrote:

Without long explainations, is it fair to say :

1) A transistor uses change in voltage to control a large flow of
current.

The word, 'large' adds no meaning.

A bit of a nit pick me thinks.

The rest is pretty iffy, also.

2a) An op-amp amplifies small differences in voltages across its 2
inputs.

This is good enough to be useful.

2b) An op-amp using negative feedback tries to make the amplified
output signal more like the input signal.

No. It uses negative feedback to make its two inputs more the same.

Depends on how one reads "more like the input signal". Your actual
statement here is obviously correct, but I read this as negative
feedback makes the "shape" of the output more like the shape of the
input, i.e. feedback reduces distortion errors, which is indeed what
feedback does. So, without further information on what the poster really
means, it seems that his description, although crude is ok.

2c) The amplified voltage of an op-amp cannot be greater than the
supply voltage. e.g. if supply voltage is 5V, it cannot be amplified
to more than -2.5 to +2.5V.

Yes. The output voltage is the result of some sort of voltage divider
operation across the supply.

Is all of the above right?

I wouldn't go that far.
Besides, simple generalities are almost never 'right'.
But sometimes they are close enough to right to be useful.

Well, I though he had a reasonable handle on it.

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.

 

[John Popelish]

Kevin Aylward wrote:

John Popelish wrote:
zalzon wrote:
(snip)
2b) An op-amp using negative feedback tries to make the amplified
output signal more like the input signal.

No. It uses negative feedback to make its two inputs more the same.

Depends on how one reads "more like the input signal". Your actual
statement here is obviously correct, but I read this as negative
feedback makes the "shape" of the output more like the shape of the
input, i.e. feedback reduces distortion errors, which is indeed what
feedback does. So, without further information on what the poster really
means, it seems that his description, although crude is ok.

What if the feedback includes diode junctions, zeners or other
nonlinear effects?

2c) The amplified voltage of an op-amp cannot be greater than the
supply voltage. e.g. if supply voltage is 5V, it cannot be amplified
to more than -2.5 to +2.5V.

Yes. The output voltage is the result of some sort of voltage divider
operation across the supply.

Is all of the above right?

I wouldn't go that far.
Besides, simple generalities are almost never 'right'.
But sometimes they are close enough to right to be useful.

Well, I though he had a reasonable handle on it.

With some unstated assumptions, yes.
I figured he would learn more if I nibbled at the edges of what he
knows.

--
John Popelish

 

[Bob Myers]

"zalzon" <zalzonishappy@zalll.com> wrote in message
news:5rrqj0lj4q5nvf9k269oh1pv1kmk7m0r2o@4ax.com...

Without long explainations, is it fair to say :

1) A transistor uses change in voltage to control a large flow of
current.

Yes.

2a) An op-amp amplifies small differences in voltages across its 2
inputs.

Yes, but that's not a sufficient definition for an "operational
amplifier" specifically. The above definition would apply
to any amplifier with differential inputs; to qualify as an
"operational amplifier," there are generally the additional
requirements of very high open-loop gain, very high input
impedance, and very low output impedance.

2b) An op-amp using negative feedback tries to make the amplified
output signal more like the input signal.

No. The operational amplifier itself does not do this.
Many useful circuits USING op-amps DO employ
negative feedback, but not all.

2c) The amplified voltage of an op-amp cannot be greater than the
supply voltage. e.g. if supply voltage is 5V, it cannot be amplified
to more than -2.5 to +2.5V.

This is true of any amplifier; more generically put,
the output signal swing cannot exceed the limits
of the "rails" (the positive and negative supplies, or
the supply and the reference or "ground" point).
In practice, it can only approach these limits, and the
output waveform will begin to distort as the limits
are closely approached.

Bob M.

 

[Bob Myers]

"Kevin Aylward" <salesEXTRACT@anasoft.co.uk> wrote in message
news:jHh%c.139101$a66.62764@fe2.news.blueyonder.co.uk...

John Popelish wrote:
zalzon wrote:

2b) An op-amp using negative feedback tries to make the amplified
output signal more like the input signal.

No. It uses negative feedback to make its two inputs more the same.

Depends on how one reads "more like the input signal".

Ahem. Methinks you both missed that fact that the
op-amp ITSELF does not necessarily employ negative
feedback in this manner AT ALL.

Bob M.

 

[Kevin Aylward]

Bob Myers wrote:

"Kevin Aylward" <salesEXTRACT@anasoft.co.uk> wrote in message
news:jHh%c.139101$a66.62764@fe2.news.blueyonder.co.uk...
John Popelish wrote:
zalzon wrote:

2b) An op-amp using negative feedback tries to make the amplified
output signal more like the input signal.

No. It uses negative feedback to make its two inputs more the same.

Depends on how one reads "more like the input signal".

Ahem. Methinks you both missed that fact that the
op-amp ITSELF does not necessarily employ negative
feedback in this manner AT ALL.

Maybe, but I major that's not what the poster cared about. One is trying
to answer the question as one believes the answer should be. One has to
make assumptions as to what the *main* point of the question is.

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]

John Popelish wrote:

Kevin Aylward wrote:

John Popelish wrote:
zalzon wrote:
(snip)
2b) An op-amp using negative feedback tries to make the amplified
output signal more like the input signal.

No. It uses negative feedback to make its two inputs more the same.

Depends on how one reads "more like the input signal". Your actual
statement here is obviously correct, but I read this as negative
feedback makes the "shape" of the output more like the shape of the
input, i.e. feedback reduces distortion errors, which is indeed what
feedback does. So, without further information on what the poster
really means, it seems that his description, although crude is ok.

What if the feedback includes diode junctions, zeners or other
nonlinear effects?

Ahmmm.... now your gettig a tad complicated....imo...

2c) The amplified voltage of an op-amp cannot be greater than the
supply voltage. e.g. if supply voltage is 5V, it cannot be
amplified to more than -2.5 to +2.5V.

Yes. The output voltage is the result of some sort of voltage
divider operation across the supply.

Is all of the above right?

I wouldn't go that far.
Besides, simple generalities are almost never 'right'.
But sometimes they are close enough to right to be useful.

Well, I though he had a reasonable handle on it.

With some unstated assumptions, yes.
I figured he would learn more if I nibbled at the edges of what he
knows.

I tried to take a somewhat different approach this time, you know, "Yes"
instead of my usual...

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.

 

[Bob Myers]

"Kevin Aylward" <salesEXTRACT@anasoft.co.uk> wrote in message
news:xGm%c.139410$a66.52695@fe2.news.blueyonder.co.uk...

Ahem. Methinks you both missed that fact that the
op-amp ITSELF does not necessarily employ negative
feedback in this manner AT ALL.


Maybe, but I major that's not what the poster cared about. One is trying
to answer the question as one believes the answer should be. One has to
make assumptions as to what the *main* point of the question is.

Maybe, but if the original poster leaves thinking that
this thing in the data book called an "op amp" behaves
as described, I would submit that they are in for a
rather rude awakening later on.

Bob M.

 

[Kevin Aylward]

Bob Myers wrote:

"Kevin Aylward" <salesEXTRACT@anasoft.co.uk> wrote in message
news:xGm%c.139410$a66.52695@fe2.news.blueyonder.co.uk...
Ahem. Methinks you both missed that fact that the
op-amp ITSELF does not necessarily employ negative
feedback in this manner AT ALL.


Maybe, but I major that's not what the poster cared about. One is
trying to answer the question as one believes the answer should be.
One has to make assumptions as to what the *main* point of the
question is.

Maybe, but if the original poster leaves thinking that
this thing in the data book called an "op amp" behaves
as described, I would submit that they are in for a
rather rude awakening later on.

Again, I disagree. Sure, an op-amp can be used for many things, but the
poster gave a pretty reasonable statement on what an op-amp *can* be
used for. The fact that the op-amp has other uses is not really
relevant. One has to start somewhere. What he stated was indeed,
essentially correct.

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.

 

[andy]

On Tue, 07 Sep 2004 08:26:46 +0000, zalzon wrote:

Without long explainations, is it fair to say :

1) A transistor uses change in voltage to control a large flow of
current.

2a) An op-amp amplifies small differences in voltages across its 2
inputs.

2b) An op-amp using negative feedback tries to make the amplified
output signal more like the input signal.

The point of op-amp circuits with negative feedback is that if the amp is
able to bring the difference between its inputs back to zero by changing
the output voltage, then it will do. Which means that the output voltage
depends (almost) only on the inputs and the feedback network you're using,
rather than the circuit that's been used to build the amp.


--
http://www.niftybits.ukfsn.org/

remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
bin unless notified with

 or [attachment] in the subject line.

 

[Bob Myers]

"Kevin Aylward" <salesEXTRACT@anasoft.co.uk> wrote in message
news:dkq%c.139629$a66.108850@fe2.news.blueyonder.co.uk...

Maybe, but if the original poster leaves thinking that
this thing in the data book called an "op amp" behaves
as described, I would submit that they are in for a
rather rude awakening later on.

Again, I disagree. Sure, an op-amp can be used for many things, but the
poster gave a pretty reasonable statement on what an op-amp *can* be
used for. The fact that the op-amp has other uses is not really
relevant. One has to start somewhere. What he stated was indeed,
essentially correct.

I think we're still disagreeing over the original poster's
confusion between "op-amp" and "op-amp-based circuits".
The statement that an op-amp, alone, with no further qualfication
or description, uses "negative feedback" to achieve such-and
-such a result is incorrect on the face of it. From the original
statements, I could easily see the original poster believing that
an op-amp, all by itself, can act as a low-gain linear amplifier,
and that's simply wrong.

Bob M.

 

[Kevin Aylward]

Bob Myers wrote:

"Kevin Aylward" <salesEXTRACT@anasoft.co.uk> wrote in message
news:dkq%c.139629$a66.108850@fe2.news.blueyonder.co.uk...
Maybe, but if the original poster leaves thinking that
this thing in the data book called an "op amp" behaves
as described, I would submit that they are in for a
rather rude awakening later on.

Again, I disagree. Sure, an op-amp can be used for many things, but
the poster gave a pretty reasonable statement on what an op-amp
*can* be used for. The fact that the op-amp has other uses is not
really relevant. One has to start somewhere. What he stated was
indeed, essentially correct.

I think we're still disagreeing over the original poster's
confusion between "op-amp" and "op-amp-based circuits".

Maybe, but I think its more of understanding what the poster thought he
wrote, rather then what he actually wrote. Most beginners can't
articulate what they mean correctly.

The statement that an op-amp, alone, with no further qualfication
or description, uses "negative feedback" to achieve such-and
-such a result is incorrect on the face of it.

I still disagree. By itself, with a reasonable assumption, the statement
is completly reasonable. There seems to be some implications here on the
difference between necessary and sufficient, and other like terms. An
op-amp circuit usually uses negative feedback to achieve such and such a
result. This is an indisputable fact, so I simply don't have a major
problem with that statement. The notion that there may be *additional*
factors as well to *complete* the description is not necessarily
relevant. Of course, the poster could believe how you interpret it.
Maybe he/she will renter the discussion.

From the original
statements, I could easily see the original poster believing that
an op-amp, all by itself, can act as a low-gain linear amplifier,
and that's simply wrong.

Maybe, if, so, the posters views would be incorrect, but I simply don't
read that as what the poster is meaning, although that may be the case.
I am filling in the blanks with the assumption that the poster has some
idea that one connects up other bits and bobs to get actual op-amp
circuit to work.

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.

 

[Ingvar Esk]

Depends how to interpret your questions. Litterally or trying to understand
what you "really mean".

"zalzon" <zalzonishappy@zalll.com> wrote in message
news:5rrqj0lj4q5nvf9k269oh1pv1kmk7m0r2o@4ax.com...

Without long explainations, is it fair to say :

1) A transistor uses change in voltage to control a large flow of
current.

No. A transistor amplifies current (hFE). But by "overstearing" the
transistor you can have it act as a current switch.

2a) An op-amp amplifies small differences in voltages across its 2
inputs.

Yes.

2b) An op-amp using negative feedback tries to make the amplified
output signal more like the input signal.

In "normal" circuits you provide negative feedback to limit the
amplification. Often a 100,000+ times amplification is to much.

2c) The amplified voltage of an op-amp cannot be greater than the
supply voltage. e.g. if supply voltage is 5V, it cannot be amplified
to more than -2.5 to +2.5V.

Yes, but your conclusion is wrong. If you have 5V supply, the output can
swing between 0V and 5V (allmost). If you have +/-5V supply the output can
swing between -5V and 5V.

Is all of the above right?

Sorry, no.

/Ingvar

 

[Kevin Aylward]

Ingvar Esk wrote:

Depends how to interpret your questions. Litterally or trying to
understand what you "really mean".

"zalzon" <zalzonishappy@zalll.com> wrote in message
news:5rrqj0lj4q5nvf9k269oh1pv1kmk7m0r2o@4ax.com...
Without long explainations, is it fair to say :

1) A transistor uses change in voltage to control a large flow of
current.

No.

No. You are wrong. The orginal claim was *100%* correct. Its not
debatable.

A transistor amplifies current (hFE).

Indeed it does as a side effect, however the transistor is absolutely
and fundamentally a *voltage* controlled and operated device
(http://www.anasoft.co.uk/EE/bipolardesign1/bipolardesign1.html).

Applying a *voltage* to the base emitter injects carriers into the base
region. These carriers are then swept up by the collectors accelerating
voltage. The fact that a few leak away through the base is just a
nuisance. Ideally, there would be no base current at all.

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.

 

[Ingvar Esk]

"Kevin Aylward" <salesEXTRACT@anasoft.co.uk> wrote in message
news:5cX%c.157$F73.138@fe2.news.blueyonder.co.uk...

Ingvar Esk wrote:
Depends how to interpret your questions. Litterally or trying to
understand what you "really mean".

"zalzon" <zalzonishappy@zalll.com> wrote in message
news:5rrqj0lj4q5nvf9k269oh1pv1kmk7m0r2o@4ax.com...
Without long explainations, is it fair to say :

1) A transistor uses change in voltage to control a large flow of
current.

No.

No. You are wrong. The orginal claim was *100%* correct. Its not
debatable.

Well, as you state in your paper "Despite much literature that implies other
wise..." I think it must be debatable, otherwise I can't see that "much
literature" would imply so

A transistor amplifies current (hFE).

Indeed it does as a side effect, however the transistor is absolutely
and fundamentally a *voltage* controlled and operated device
(http://www.anasoft.co.uk/EE/bipolardesign1/bipolardesign1.html).

Applying a *voltage* to the base emitter injects carriers into the base
region. These carriers are then swept up by the collectors accelerating
voltage. The fact that a few leak away through the base is just a
nuisance. Ideally, there would be no base current at all.

From my point of view. If I feed 1mA into the base the Ic would be some
hundreds (hFE) mA (if available), even though I know that hFE is not very
accurate. On the other hand if I supplied 0.1V to the base, nothing much
will happen at the collector. I can accept that it can be seen as a Voltage
amplifier, but only around its working point (Vb ~0.6-0.7V).

Ingvar Esk

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]

Ingvar Esk wrote:

"Kevin Aylward" <salesEXTRACT@anasoft.co.uk> wrote in message
news:5cX%c.157$F73.138@fe2.news.blueyonder.co.uk...
Ingvar Esk wrote:
Depends how to interpret your questions. Litterally or trying to
understand what you "really mean".

"zalzon" <zalzonishappy@zalll.com> wrote in message
news:5rrqj0lj4q5nvf9k269oh1pv1kmk7m0r2o@4ax.com...
Without long explainations, is it fair to say :

1) A transistor uses change in voltage to control a large flow of
current.

No.

No. You are wrong. The orginal claim was *100%* correct. Its not
debatable.

Well, as you state in your paper "Despite much literature that
implies other wise..." I think it must be debatable, otherwise I
can't see that "much literature" would imply so

Its not debatable in the sense that it is accepted by any physicists
that understands how transistors actually work. Its simply impossible to
derive and sensible device equations on the assumption that base current
controls emitter current. The most simple derived equation is:

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

Its called the diode equation. It is derived/shown in *any* standard
text book devoted to the device physics of the transistor. Its based on
applying a potential to the junction. There is no iffs or butts about
it.

Unfortunately there are rather a lot of non or semi technical books that
use explanations equivalent to water down a pipe. The sooner novices get
to grips with the fact that the ic=ib.hfe model, is a *grossly*
simplified model, with limited use, the better. It causes never ending
confusion.

A transistor amplifies current (hFE).

Indeed it does as a side effect, however the transistor is absolutely
and fundamentally a *voltage* controlled and operated device
(http://www.anasoft.co.uk/EE/bipolardesign1/bipolardesign1.html).

Applying a *voltage* to the base emitter injects carriers into the
base region. These carriers are then swept up by the collectors
accelerating voltage. The fact that a few leak away through the base
is just a nuisance. Ideally, there would be no base current at all.

From my point of view. If I feed 1mA into the base the Ic would be
some hundreds (hFE) mA (if available), even though I know that hFE is
not very accurate. On the other hand if I supplied 0.1V to the base,
nothing much will happen at the collector. I can accept that it can
be seen as a Voltage amplifier, but only around its working point (Vb
~0.6-0.7V).

You are looking at this from a way too naive point of view.

The transistor is a transconductance device. It outputs a current based
on its input voltage. It can be accurately exponential over 6 decades.
Sure, at low voltages, there is only a small current, but this doesn't
change how the transistor operates.

Its not a mater of what you want to accept. Its how it is. You can't
disagree with all the semiconductor physicista in the world, well not
unless your name is Einstein anyway-)

What I am explaining is *THE* *standard* *accepted* physics of the
situation. *Only* those *without* the academic background have
transistor operation mistaken. Go and have a read of *any* standard text
on semiconductor physics.

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.

 

[Bob Myers]

"Ingvar Esk" <ingvar.esk@ericsson.com> wrote in message
news:chpcsv$48b$1@newstree.wise.edt.ericsson.se...

1) A transistor uses change in voltage to control a large flow of
current.

No. A transistor amplifies current (hFE). But by "overstearing" the
transistor you can have it act as a current switch.

Sorry - that's one model you can use, and it's often a
useful one. But with respect to the actual physics of the
device, it is more accurate to say that slight VOLTAGE
variations across the base-emitter junction control the
current "into" the collector. The standard "hybrid pi"
model of the transistor is best for showing this in simple
form.


Bob M.

 

[Bob Myers]

"Ingvar Esk" <ingvar.esk@ericsson.com> wrote in message
news:chpgd7$64d$1@newstree.wise.edt.ericsson.se...

From my point of view. If I feed 1mA into the base the Ic would be some
hundreds (hFE) mA (if available), even though I know that hFE is not very
accurate. On the other hand if I supplied 0.1V to the base, nothing much
will happen at the collector. I can accept that it can be seen as a
Voltage
amplifier, but only around its working point (Vb ~0.6-0.7V).

But you can't "supply 0.1V to the base" - voltages do not
exist at isolated points, but only with respect to a
reference. In this case, it is the voltage across the base-
emitter junction (Vbe - the "v" should normally be written
lower-case in this context, but I'm afraid that would look
confusing here) which controls, through the transistor's
"transconductance" (Gm, again normally a lower-case G
with a subscript m), the collector current. I am afraid
you are confusing the DC bias conditions (and the
resulting DC forward drop across the B-E junction,
VBE) with the small-signal voltages. With the transistor
biased in the active mode, and that 0.6-0.7V drop you
mentioned established, yes, a relatively small *AC*
signal across this junction DOES control the collector
current.

Bob M.