Operational Amplifier basics

[krw]

In article <6v1206FgjlikU1@mid.individual.net>, notty@emailo.com
says...>

"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f4fa16439e97e2989944@news.individual.net...
Not sure how the op amp itself settles when V- = V+.

Negative feedback. If the output wanted to be too high, the
current in R1 would be too high forcing V- high, causing the output
to go lower. ...until the voltage between + and - is zero.

I just know that the amp acts like a variable PSU and what
voltage E1 needs to be to get V- to equal V+.

If it's not (the negative feedback can't stabilize it) the output
has to go to one rail or the other ("infinite gain").

The method to stabilize is feedack. Feedback through a resistor ensures
that V- = V+, but is it not an internal fuction of the OP Amp that Vout
stabilizes are the required voltage (Set by extermnal networks) when
V- = V+?

Right. If V- <> V+ the output is infinite, so they must be equal
or the feedback isn't doing its job.

I suppose you could make an Op Amp where
Vout stabilzes when there is 1V across V- and V+?

You wouldn't. To do so one would need a 1V reference, something
that is completely unnecessary and unwanted.


What I'm thinking about is this: If I had an audio amplifier speaking into
the microphone I might drive the ouput transistor into distortion.

Ok, if your drive exceeds the OpAmp's rails, sure. But that
doesn't have anything to do with the voltage across the OpAmp's
input.

But if I had some negative feedback I could ensure that the distortion was
less. But, the amplifier does not seek to drive the input to zero and
achieve stabilization there.

If the output voltage needed to drive the input voltage to zero
exceeds the power supply rails (minus some bit for most real
OpAmps), yes there will be distortion and, yes, the inputs will no
longer have 0V across them. It will *seek* (i.e. try) to drive the
input voltage to zero, but it can't.

 

[Rich]

"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f515449cb04d5f989946@news.individual.net...

In article <6v1206FgjlikU1@mid.individual.net>, notty@emailo.com
says...
"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f4fa16439e97e2989944@news.individual.net...
Not sure how the op amp itself settles when V- = V+.

Negative feedback. If the output wanted to be too high, the
current in R1 would be too high forcing V- high, causing the output
to go lower. ...until the voltage between + and - is zero.

I just know that the amp acts like a variable PSU and what
voltage E1 needs to be to get V- to equal V+.

If it's not (the negative feedback can't stabilize it) the output
has to go to one rail or the other ("infinite gain").

The method to stabilize is feedack. Feedback through a resistor
ensures
that V- = V+, but is it not an internal fuction of the OP Amp that
Vout
stabilizes are the required voltage (Set by extermnal networks) when
V- = V+?

Right. If V- <> V+ the output is infinite, so they must be equal
or the feedback isn't doing its job.

I suppose you could make an Op Amp where
Vout stabilzes when there is 1V across V- and V+?

You wouldn't. To do so one would need a 1V reference, something
that is completely unnecessary and unwanted.


What I'm thinking about is this: If I had an audio amplifier speaking
into
the microphone I might drive the ouput transistor into distortion.

Ok, if your drive exceeds the OpAmp's rails, sure. But that
doesn't have anything to do with the voltage across the OpAmp's
input.

But if I had some negative feedback I could ensure that the distortion
was
less. But, the amplifier does not seek to drive the input to zero and
achieve stabilization there.

If the output voltage needed to drive the input voltage to zero
exceeds the power supply rails (minus some bit for most real
OpAmps), yes there will be distortion and, yes, the inputs will no
longer have 0V across them. It will *seek* (i.e. try) to drive the
input voltage to zero, but it can't.

You are explaining what an Op Amp does when connected with negative
feedback. The Op Amp seeks to output the correct voltage so that V- = V+.

I'm just for a moment focusing *not* on what happens, on how we can get an
OP Amp to amplify by a determined gain, (which requires V- = V+) but why did
someone make an amplifier that worked that way? Did someone say - Mmm, we
can make a very useful amplifier if we create a circuit (an Op Amp) which
will give an output and stabilise the voltage output when V- = V+?

Correct me if I'm wrong, but I don't think the non inverting amplifier, with
negative feeback, where you can set the gain precisely, is entirely a matter
of feedback theory. Is it not the case that it works because someone
designed an Op Amp circuit and (kinda) said, When there is zero voltage
between pins V- and V+ our special amplifier will still give an output. And
when we apply standard negative feedback, that is, a voltage that
oppposes the input voltage, the output will settle down at a voltage
determined by an external network of resistors.

Normal amplifiers output zero when input is zero.

So, is it not the case that the issue that is making everything works is not
simply feedback theory? Although it is part of it.

 

[John Fields]

On Thu, 5 Feb 2009 21:10:44 -0000, "Rich" <notty@emailo.com> wrote:

"Rich" <notty@emailo.com> wrote in message
news:6v0mvfFh2i4fU1@mid.individual.net...

"John Fields" <jfields@austininstruments.com> wrote in message
news:tl4ko41kr1okpg22iu3mhjhsh6fvl9ieie@4ax.com...
On Wed, 4 Feb 2009 19:01:09 -0000, "Rich" <notty@emailo.com> wrote:
+V
|
E3--+-------------|+\
| | >----+--E1
[VOLTMETER] +--|-/ |
| | | |
+----------+ -V |
| |
E2--+---[R1]---+
|
[R2]
|
0V

The Invering amplifier which was drawn seeks to get V- and V+ to be equal.

Actually the circuit is a non inverting amplifier circuit.

---
Yup!

JF

 

[krw]

In article <6v178vFhk5h7U1@mid.individual.net>, notty@emailo.com
says...>

"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f515449cb04d5f989946@news.individual.net...
In article <6v1206FgjlikU1@mid.individual.net>, notty@emailo.com
says...
"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f4fa16439e97e2989944@news.individual.net...
Not sure how the op amp itself settles when V- = V+.

Negative feedback. If the output wanted to be too high, the
current in R1 would be too high forcing V- high, causing the output
to go lower. ...until the voltage between + and - is zero.

I just know that the amp acts like a variable PSU and what
voltage E1 needs to be to get V- to equal V+.

If it's not (the negative feedback can't stabilize it) the output
has to go to one rail or the other ("infinite gain").

The method to stabilize is feedack. Feedback through a resistor
ensures
that V- = V+, but is it not an internal fuction of the OP Amp that
Vout
stabilizes are the required voltage (Set by extermnal networks) when
V- = V+?

Right. If V- <> V+ the output is infinite, so they must be equal
or the feedback isn't doing its job.

I suppose you could make an Op Amp where
Vout stabilzes when there is 1V across V- and V+?

You wouldn't. To do so one would need a 1V reference, something
that is completely unnecessary and unwanted.


What I'm thinking about is this: If I had an audio amplifier speaking
into
the microphone I might drive the ouput transistor into distortion.

Ok, if your drive exceeds the OpAmp's rails, sure. But that
doesn't have anything to do with the voltage across the OpAmp's
input.

But if I had some negative feedback I could ensure that the distortion
was
less. But, the amplifier does not seek to drive the input to zero and
achieve stabilization there.

If the output voltage needed to drive the input voltage to zero
exceeds the power supply rails (minus some bit for most real
OpAmps), yes there will be distortion and, yes, the inputs will no
longer have 0V across them. It will *seek* (i.e. try) to drive the
input voltage to zero, but it can't.

You are explaining what an Op Amp does when connected with negative
feedback. The Op Amp seeks to output the correct voltage so that V- = V+.

Right. The direction of the feedback is such that it forces the
opposite of what the output does. If you kick the output positive,
the feedback forces a negative to compensate for it.

I'm just for a moment focusing *not* on what happens, on how we can get an
OP Amp to amplify by a determined gain, (which requires V- = V+) but why did
someone make an amplifier that worked that way? Did someone say - Mmm, we
can make a very useful amplifier if we create a circuit (an Op Amp) which
will give an output and stabilise the voltage output when V- = V+?

You have it somewhat backwards. The amplifier is designed to have
an "infinite" differential gain. That property is useful, using
negative feedback, to make an amplifier with a gain controlled
solely by external resistors. The three properties I've repeated
many times (because they're all important) is what allows the OpAmp
to be useful. It's not because someone said "hey, wouldn't it be
useful to have an amplifier that tries to balance its inputs".

Correct me if I'm wrong, but I don't think the non inverting amplifier, with
negative feeback, where you can set the gain precisely, is entirely a matter
of feedback theory. Is it not the case that it works because someone
designed an Op Amp circuit and (kinda) said, When there is zero voltage
between pins V- and V+ our special amplifier will still give an output. And
when we apply standard negative feedback, that is, a voltage that
oppposes the input voltage, the output will settle down at a voltage
determined by an external network of resistors.

Again, you have it backwards. The input voltage is zero *because*
of the negative feedback. Negative feedback is what makes the
OpAmp useful. By itself the (ideal) OpAmp has the three properties
I've stated before.

IOW, we analyze the circuit assuming the input voltage is zero
because the output, via negative feedback, forces it to be so.

Normal amplifiers output zero when input is zero.

No. There is often a DC offset in amplifiers. You'll often find
DC blocking caps (or AC coupling caps, your choice of terms)
between amplifiers that do not have a zero operating point. If you
run an OpAmp on a single rail you'll see the same thing because a
DC offset is added to the signal to keep the signal away from the
rails.

So, is it not the case that the issue that is making everything works is not
simply feedback theory? Although it is part of it.

No, it is *precisely* feedback that is allowing you to analyze the
circuit assuming the input voltage is zero. Without the negative
feedback it certainly won't be (zero).

 

[Andrew Holme]

"Rich" <notty@emailo.com> wrote in message
news:6v178vFhk5h7U1@mid.individual.net...

"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f515449cb04d5f989946@news.individual.net...
In article <6v1206FgjlikU1@mid.individual.net>, notty@emailo.com
says...
"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f4fa16439e97e2989944@news.individual.net...
Not sure how the op amp itself settles when V- = V+.

Negative feedback. If the output wanted to be too high, the
current in R1 would be too high forcing V- high, causing the output
to go lower. ...until the voltage between + and - is zero.

I just know that the amp acts like a variable PSU and what
voltage E1 needs to be to get V- to equal V+.

If it's not (the negative feedback can't stabilize it) the output
has to go to one rail or the other ("infinite gain").

The method to stabilize is feedack. Feedback through a resistor
ensures
that V- = V+, but is it not an internal fuction of the OP Amp that
Vout
stabilizes are the required voltage (Set by extermnal networks) when
V- = V+?

Right. If V- <> V+ the output is infinite, so they must be equal
or the feedback isn't doing its job.

I suppose you could make an Op Amp where
Vout stabilzes when there is 1V across V- and V+?

You wouldn't. To do so one would need a 1V reference, something
that is completely unnecessary and unwanted.


What I'm thinking about is this: If I had an audio amplifier speaking
into
the microphone I might drive the ouput transistor into distortion.

Ok, if your drive exceeds the OpAmp's rails, sure. But that
doesn't have anything to do with the voltage across the OpAmp's
input.

But if I had some negative feedback I could ensure that the distortion
was
less. But, the amplifier does not seek to drive the input to zero and
achieve stabilization there.

If the output voltage needed to drive the input voltage to zero
exceeds the power supply rails (minus some bit for most real
OpAmps), yes there will be distortion and, yes, the inputs will no
longer have 0V across them. It will *seek* (i.e. try) to drive the
input voltage to zero, but it can't.

You are explaining what an Op Amp does when connected with negative
feedback. The Op Amp seeks to output the correct voltage so that V- = V+.

I'm just for a moment focusing *not* on what happens, on how we can get an
OP Amp to amplify by a determined gain, (which requires V- = V+) but why
did
someone make an amplifier that worked that way? Did someone say - Mmm, we
can make a very useful amplifier if we create a circuit (an Op Amp) which
will give an output and stabilise the voltage output when V- = V+?

The concept of an "ideal" op-amp with infinite gain strikes me as a far from
ideal way to teach op-amp theory. It must be very confusing for many
students.

Real op-amps have a large but finite gain 'A'

Vout = A*(VP-VN)

Feedback makes the differential input voltage small, but not zero.

If I build a unity-gain non-inverting voltage follower, I have :

Vout = A*(Vin-Vout)

(A+1) * Vout = A*Vin

Vout ~= Vin

 

[krw]

In article <t%Jil.28009$jz5.7842@newsfe09.ams2>, ah@nospam.co.uk
says...>

"Rich" <notty@emailo.com> wrote in message
news:6v178vFhk5h7U1@mid.individual.net...

"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f515449cb04d5f989946@news.individual.net...
In article <6v1206FgjlikU1@mid.individual.net>, notty@emailo.com
says...
"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f4fa16439e97e2989944@news.individual.net...
Not sure how the op amp itself settles when V- = V+.

Negative feedback. If the output wanted to be too high, the
current in R1 would be too high forcing V- high, causing the output
to go lower. ...until the voltage between + and - is zero.

I just know that the amp acts like a variable PSU and what
voltage E1 needs to be to get V- to equal V+.

If it's not (the negative feedback can't stabilize it) the output
has to go to one rail or the other ("infinite gain").

The method to stabilize is feedack. Feedback through a resistor
ensures
that V- = V+, but is it not an internal fuction of the OP Amp that
Vout
stabilizes are the required voltage (Set by extermnal networks) when
V- = V+?

Right. If V- <> V+ the output is infinite, so they must be equal
or the feedback isn't doing its job.

I suppose you could make an Op Amp where
Vout stabilzes when there is 1V across V- and V+?

You wouldn't. To do so one would need a 1V reference, something
that is completely unnecessary and unwanted.


What I'm thinking about is this: If I had an audio amplifier speaking
into
the microphone I might drive the ouput transistor into distortion.

Ok, if your drive exceeds the OpAmp's rails, sure. But that
doesn't have anything to do with the voltage across the OpAmp's
input.

But if I had some negative feedback I could ensure that the distortion
was
less. But, the amplifier does not seek to drive the input to zero and
achieve stabilization there.

If the output voltage needed to drive the input voltage to zero
exceeds the power supply rails (minus some bit for most real
OpAmps), yes there will be distortion and, yes, the inputs will no
longer have 0V across them. It will *seek* (i.e. try) to drive the
input voltage to zero, but it can't.

You are explaining what an Op Amp does when connected with negative
feedback. The Op Amp seeks to output the correct voltage so that V- = V+.

I'm just for a moment focusing *not* on what happens, on how we can get an
OP Amp to amplify by a determined gain, (which requires V- = V+) but why
did
someone make an amplifier that worked that way? Did someone say - Mmm, we
can make a very useful amplifier if we create a circuit (an Op Amp) which
will give an output and stabilise the voltage output when V- = V+?

The concept of an "ideal" op-amp with infinite gain strikes me as a far from
ideal way to teach op-amp theory. It must be very confusing for many
students.

Real op-amps have a large but finite gain 'A'

Vout = A*(VP-VN)

Feedback makes the differential input voltage small, but not zero.

If I build a unity-gain non-inverting voltage follower, I have :

Vout = A*(Vin-Vout)

(A+1) * Vout = A*Vin

Vout ~= Vin

How is your "almost ideal" less confusing than going without the
almost part? The three characteristics of an ideal Op-Amp work
very well to analyze almost any OpAmp circuit, at least as a first
pass. Indeed the idealized OpAmp works better and is a whole lot
simpler than your "almost ideal" version.

 

[Andrew Holme]

"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f529de913830b598994b@news.individual.net...

In article <t%Jil.28009$jz5.7842@newsfe09.ams2>, ah@nospam.co.uk
says...
"Rich" <notty@emailo.com> wrote in message
news:6v178vFhk5h7U1@mid.individual.net...

"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f515449cb04d5f989946@news.individual.net...
In article <6v1206FgjlikU1@mid.individual.net>, notty@emailo.com
says...
"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f4fa16439e97e2989944@news.individual.net...
Not sure how the op amp itself settles when V- = V+.

Negative feedback. If the output wanted to be too high, the
current in R1 would be too high forcing V- high, causing the
output
to go lower. ...until the voltage between + and - is zero.

I just know that the amp acts like a variable PSU and what
voltage E1 needs to be to get V- to equal V+.

If it's not (the negative feedback can't stabilize it) the
output
has to go to one rail or the other ("infinite gain").

The method to stabilize is feedack. Feedback through a resistor
ensures
that V- = V+, but is it not an internal fuction of the OP Amp that
Vout
stabilizes are the required voltage (Set by extermnal networks)
when
V- = V+?

Right. If V- <> V+ the output is infinite, so they must be equal
or the feedback isn't doing its job.

I suppose you could make an Op Amp where
Vout stabilzes when there is 1V across V- and V+?

You wouldn't. To do so one would need a 1V reference, something
that is completely unnecessary and unwanted.


What I'm thinking about is this: If I had an audio amplifier speaking
into
the microphone I might drive the ouput transistor into distortion.

Ok, if your drive exceeds the OpAmp's rails, sure. But that
doesn't have anything to do with the voltage across the OpAmp's
input.

But if I had some negative feedback I could ensure that the
distortion
was
less. But, the amplifier does not seek to drive the input to zero and
achieve stabilization there.

If the output voltage needed to drive the input voltage to zero
exceeds the power supply rails (minus some bit for most real
OpAmps), yes there will be distortion and, yes, the inputs will no
longer have 0V across them. It will *seek* (i.e. try) to drive the
input voltage to zero, but it can't.

You are explaining what an Op Amp does when connected with negative
feedback. The Op Amp seeks to output the correct voltage so that V- =
V+.

I'm just for a moment focusing *not* on what happens, on how we can get
an
OP Amp to amplify by a determined gain, (which requires V- = V+) but
why
did
someone make an amplifier that worked that way? Did someone say - Mmm,
we
can make a very useful amplifier if we create a circuit (an Op Amp)
which
will give an output and stabilise the voltage output when V- = V+?

The concept of an "ideal" op-amp with infinite gain strikes me as a far
from
ideal way to teach op-amp theory. It must be very confusing for many
students.

Real op-amps have a large but finite gain 'A'

Vout = A*(VP-VN)

Feedback makes the differential input voltage small, but not zero.

If I build a unity-gain non-inverting voltage follower, I have :

Vout = A*(Vin-Vout)

(A+1) * Vout = A*Vin

Vout ~= Vin

How is your "almost ideal" less confusing than going without the
almost part? The three characteristics of an ideal Op-Amp work
very well to analyze almost any OpAmp circuit, at least as a first
pass. Indeed the idealized OpAmp works better and is a whole lot
simpler than your "almost ideal" version.

Yes, equating VP=VN is often the best/easiest way to analyze op-amp
circuits.

I just don't like infinity*zero = the correct answer.

 

[krw]

In article <BmKil.28010$jz5.18491@newsfe09.ams2>, ah@nospam.co.uk
says...>

"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f529de913830b598994b@news.individual.net...
In article <t%Jil.28009$jz5.7842@newsfe09.ams2>, ah@nospam.co.uk
says...
"Rich" <notty@emailo.com> wrote in message
news:6v178vFhk5h7U1@mid.individual.net...

"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f515449cb04d5f989946@news.individual.net...
In article <6v1206FgjlikU1@mid.individual.net>, notty@emailo.com
says...
"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f4fa16439e97e2989944@news.individual.net...
Not sure how the op amp itself settles when V- = V+.

Negative feedback. If the output wanted to be too high, the
current in R1 would be too high forcing V- high, causing the
output
to go lower. ...until the voltage between + and - is zero.

I just know that the amp acts like a variable PSU and what
voltage E1 needs to be to get V- to equal V+.

If it's not (the negative feedback can't stabilize it) the
output
has to go to one rail or the other ("infinite gain").

The method to stabilize is feedack. Feedback through a resistor
ensures
that V- = V+, but is it not an internal fuction of the OP Amp that
Vout
stabilizes are the required voltage (Set by extermnal networks)
when
V- = V+?

Right. If V- <> V+ the output is infinite, so they must be equal
or the feedback isn't doing its job.

I suppose you could make an Op Amp where
Vout stabilzes when there is 1V across V- and V+?

You wouldn't. To do so one would need a 1V reference, something
that is completely unnecessary and unwanted.


What I'm thinking about is this: If I had an audio amplifier speaking
into
the microphone I might drive the ouput transistor into distortion.

Ok, if your drive exceeds the OpAmp's rails, sure. But that
doesn't have anything to do with the voltage across the OpAmp's
input.

But if I had some negative feedback I could ensure that the
distortion
was
less. But, the amplifier does not seek to drive the input to zero and
achieve stabilization there.

If the output voltage needed to drive the input voltage to zero
exceeds the power supply rails (minus some bit for most real
OpAmps), yes there will be distortion and, yes, the inputs will no
longer have 0V across them. It will *seek* (i.e. try) to drive the
input voltage to zero, but it can't.

You are explaining what an Op Amp does when connected with negative
feedback. The Op Amp seeks to output the correct voltage so that V- =
V+.

I'm just for a moment focusing *not* on what happens, on how we can get
an
OP Amp to amplify by a determined gain, (which requires V- = V+) but
why
did
someone make an amplifier that worked that way? Did someone say - Mmm,
we
can make a very useful amplifier if we create a circuit (an Op Amp)
which
will give an output and stabilise the voltage output when V- = V+?

The concept of an "ideal" op-amp with infinite gain strikes me as a far
from
ideal way to teach op-amp theory. It must be very confusing for many
students.

Real op-amps have a large but finite gain 'A'

Vout = A*(VP-VN)

Feedback makes the differential input voltage small, but not zero.

If I build a unity-gain non-inverting voltage follower, I have :

Vout = A*(Vin-Vout)

(A+1) * Vout = A*Vin

Vout ~= Vin

How is your "almost ideal" less confusing than going without the
almost part? The three characteristics of an ideal Op-Amp work
very well to analyze almost any OpAmp circuit, at least as a first
pass. Indeed the idealized OpAmp works better and is a whole lot
simpler than your "almost ideal" version.


Yes, equating VP=VN is often the best/easiest way to analyze op-amp
circuits.

I just don't like infinity*zero = the correct answer.

Almost infinity is hardly correct for anything. Infinity is easy
because all your "almosts" drop out, greatly simplifying the math.
Since no one will measure the input voltage [*], infinity is close
enough.

[*] other non-idealized OpAmp issues come in *way* before this.

 

[John Fields]

On Thu, 5 Feb 2009 22:24:43 -0000, "Rich" <notty@emailo.com> wrote:

I'm just for a moment focusing *not* on what happens, on how we can get an
OP Amp to amplify by a determined gain, (which requires V- = V+) but why did
someone make an amplifier that worked that way? Did someone say - Mmm, we
can make a very useful amplifier if we create a circuit (an Op Amp) which
will give an output and stabilise the voltage output when V- = V+?

Correct me if I'm wrong, but I don't think the non inverting amplifier, with
negative feeback, where you can set the gain precisely, is entirely a matter
of feedback theory. Is it not the case that it works because someone
designed an Op Amp circuit and (kinda) said, When there is zero voltage
between pins V- and V+ our special amplifier will still give an output. And
when we apply standard negative feedback, that is, a voltage that
oppposes the input voltage, the output will settle down at a voltage
determined by an external network of resistors.

---
http://www.philbrickarchive.org/k3_series_components.htm
---

Normal amplifiers output zero when input is zero.

---
So does an opamp, either inverting or non-inverting.

For the non-inverting case:

Vin>-----|+\
| >----+--Vout
+--|-/ |
| |
+---[R1]---+
|
[R2]
|
0V

if Vin is 0V, what does Vout have to be to make the inverting input of
the opamp 0V?



For the inverting case:


+---[R1]---+
| |
Vin>--[R2]--+--|-\ |
| >----+--Vout
+--|+/
|
0V

Remember, for either case, the output voltage _must_ swing in the
direction which causes the inputs to have no voltage difference between
them, so if Vin is 0V, what _must_ Vout be?
---

So, is it not the case that the issue that is making everything works is not
simply feedback theory? Although it is part of it.

---
Feedback theory doesn't _make_ anything work, it explains why it does
when feedback is employed.

As a matter of fact, back in the early days, feedback was pooh-poohed as
being nonsensical because, and I paraphrase,: "How can something which
happens after an occurrence control that occurrence?"

JF

 

[Jasen Betts]

On 2009-02-05, krw <krw@att.zzzzzzzzz> wrote:

How is your "almost ideal" less confusing than going without the
almost part? The three characteristics of an ideal Op-Amp work
very well to analyze almost any OpAmp circuit, at least as a first
pass. Indeed the idealized OpAmp works better and is a whole lot
simpler than your "almost ideal" version.

there is no such thing as an ideal op-amp it's impossible for a number
of reasons. this can initially be a real distraction for those who
have more experience with real hardware than with mathematics,

 

[krw]

In article <gmgvg7$fv$2@reversiblemaps.ath.cx>, jasen@xnet.co.nz
says...>

On 2009-02-05, krw <krw@att.zzzzzzzzz> wrote:

How is your "almost ideal" less confusing than going without the
almost part? The three characteristics of an ideal Op-Amp work
very well to analyze almost any OpAmp circuit, at least as a first
pass. Indeed the idealized OpAmp works better and is a whole lot
simpler than your "almost ideal" version.

there is no such thing as an ideal op-amp it's impossible for a number
of reasons. this can initially be a real distraction for those who
have more experience with real hardware than with mathematics,

Complete hogwash. The "normal" operation of a real OpAmp is closer
to it's ideal than pretty much any other component. If you can't
deal with the simplicities of "zero" and "infinity" get out of the
business.

 

[Rich]

"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f5270dde1d5fab98994a@news.individual.net...

In article <6v178vFhk5h7U1@mid.individual.net>, notty@emailo.com
says...
"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f515449cb04d5f989946@news.individual.net...
In article <6v1206FgjlikU1@mid.individual.net>, notty@emailo.com
says...
"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f4fa16439e97e2989944@news.individual.net...
Not sure how the op amp itself settles when V- = V+.

Negative feedback. If the output wanted to be too high, the
current in R1 would be too high forcing V- high, causing the
output
to go lower. ...until the voltage between + and - is zero.

I just know that the amp acts like a variable PSU and what
voltage E1 needs to be to get V- to equal V+.

If it's not (the negative feedback can't stabilize it) the output
has to go to one rail or the other ("infinite gain").

The method to stabilize is feedack. Feedback through a resistor
ensures
that V- = V+, but is it not an internal fuction of the OP Amp that
Vout
stabilizes are the required voltage (Set by extermnal networks)
when
V- = V+?

Right. If V- <> V+ the output is infinite, so they must be equal
or the feedback isn't doing its job.

I suppose you could make an Op Amp where
Vout stabilzes when there is 1V across V- and V+?

You wouldn't. To do so one would need a 1V reference, something
that is completely unnecessary and unwanted.


What I'm thinking about is this: If I had an audio amplifier speaking
into
the microphone I might drive the ouput transistor into distortion.

Ok, if your drive exceeds the OpAmp's rails, sure. But that
doesn't have anything to do with the voltage across the OpAmp's
input.

But if I had some negative feedback I could ensure that the distortion
was
less. But, the amplifier does not seek to drive the input to zero and
achieve stabilization there.

If the output voltage needed to drive the input voltage to zero
exceeds the power supply rails (minus some bit for most real
OpAmps), yes there will be distortion and, yes, the inputs will no
longer have 0V across them. It will *seek* (i.e. try) to drive the
input voltage to zero, but it can't.

You are explaining what an Op Amp does when connected with negative
feedback. The Op Amp seeks to output the correct voltage so that V- = V+.

Right. The direction of the feedback is such that it forces the
opposite of what the output does. If you kick the output positive,
the feedback forces a negative to compensate for it.

I'm just for a moment focusing *not* on what happens, on how we can get
an
OP Amp to amplify by a determined gain, (which requires V- = V+) but why
did
someone make an amplifier that worked that way? Did someone say - Mmm, we
can make a very useful amplifier if we create a circuit (an Op Amp) which
will give an output and stabilise the voltage output when V- = V+?

You have it somewhat backwards. The amplifier is designed to have
an "infinite" differential gain. That property is useful, using
negative feedback, to make an amplifier with a gain controlled
solely by external resistors. The three properties I've repeated
many times (because they're all important) is what allows the OpAmp
to be useful. It's not because someone said "hey, wouldn't it be
useful to have an amplifier that tries to balance its inputs".

Okay I think I've got it. With feedack V- = V+ occurs. That happens with dc
amplifiers. Get a very high gain dc amplifier, have very high input
inpedance, low ouput impedance, great isolation between input and output
(internally), and when you connect it in a certain manner, principally
where the output is fed into the input to counteract changes in the input,
you get a situation where V- and V+ will become equal. That's it. Correct?

My electronics knowledge comes from studying ham radio. I've never studied
dc amplifiers. In my usual experience, when there is no input, there is no
output. So, when I think of this, it looks like something other than
feedback it at play in *addition to the things already mentioned*. I got to
thinking about setpoints in fact.

It's the bit about having an output when there is no input voltage that
throws me.

But of course if you do kind of have an output with "ac amplifiers" when
input is zero.. In the sense that there is a dc voltage on the other side
of the output capacitor. That dc voltage rises sinosoidally (say). But
when there is no input it goes to a quiesent value. Of course, it
it's sinusoildal voltages that you want mangnify, and dc voltage on
the other side of the output capacitor (or transformer, whatever)
is of no significance to the amplification of the input.

Is this getting there? Rich

 

[krw]

In article <6v2uurFhsl6aU1@mid.individual.net>, notty@emailo.com
says...>

"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f5270dde1d5fab98994a@news.individual.net...
In article <6v178vFhk5h7U1@mid.individual.net>, notty@emailo.com
says...
"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f515449cb04d5f989946@news.individual.net...
In article <6v1206FgjlikU1@mid.individual.net>, notty@emailo.com
says...
"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f4fa16439e97e2989944@news.individual.net...
Not sure how the op amp itself settles when V- = V+.

Negative feedback. If the output wanted to be too high, the
current in R1 would be too high forcing V- high, causing the
output
to go lower. ...until the voltage between + and - is zero.

I just know that the amp acts like a variable PSU and what
voltage E1 needs to be to get V- to equal V+.

If it's not (the negative feedback can't stabilize it) the output
has to go to one rail or the other ("infinite gain").

The method to stabilize is feedack. Feedback through a resistor
ensures
that V- = V+, but is it not an internal fuction of the OP Amp that
Vout
stabilizes are the required voltage (Set by extermnal networks)
when
V- = V+?

Right. If V- <> V+ the output is infinite, so they must be equal
or the feedback isn't doing its job.

I suppose you could make an Op Amp where
Vout stabilzes when there is 1V across V- and V+?

You wouldn't. To do so one would need a 1V reference, something
that is completely unnecessary and unwanted.


What I'm thinking about is this: If I had an audio amplifier speaking
into
the microphone I might drive the ouput transistor into distortion.

Ok, if your drive exceeds the OpAmp's rails, sure. But that
doesn't have anything to do with the voltage across the OpAmp's
input.

But if I had some negative feedback I could ensure that the distortion
was
less. But, the amplifier does not seek to drive the input to zero and
achieve stabilization there.

If the output voltage needed to drive the input voltage to zero
exceeds the power supply rails (minus some bit for most real
OpAmps), yes there will be distortion and, yes, the inputs will no
longer have 0V across them. It will *seek* (i.e. try) to drive the
input voltage to zero, but it can't.

You are explaining what an Op Amp does when connected with negative
feedback. The Op Amp seeks to output the correct voltage so that V- = V+.

Right. The direction of the feedback is such that it forces the
opposite of what the output does. If you kick the output positive,
the feedback forces a negative to compensate for it.

I'm just for a moment focusing *not* on what happens, on how we can get
an
OP Amp to amplify by a determined gain, (which requires V- = V+) but why
did
someone make an amplifier that worked that way? Did someone say - Mmm, we
can make a very useful amplifier if we create a circuit (an Op Amp) which
will give an output and stabilise the voltage output when V- = V+?

You have it somewhat backwards. The amplifier is designed to have
an "infinite" differential gain. That property is useful, using
negative feedback, to make an amplifier with a gain controlled
solely by external resistors. The three properties I've repeated
many times (because they're all important) is what allows the OpAmp
to be useful. It's not because someone said "hey, wouldn't it be
useful to have an amplifier that tries to balance its inputs".

Okay I think I've got it. With feedack V- = V+ occurs. That happens with dc
amplifiers. Get a very high gain dc amplifier, have very high input
inpedance, low ouput impedance, great isolation between input and output
(internally), and when you connect it in a certain manner, principally
where the output is fed into the input to counteract changes in the input,
you get a situation where V- and V+ will become equal. That's it. Correct?

Where the feedback is of the opposite polarity as the output, yes.
That is, a change in the output is fed back as the opposite signal
to the input, the inputs must balance or the output will "rail"
("infinite" voltage) trying.

My electronics knowledge comes from studying ham radio. I've never studied
dc amplifiers. In my usual experience, when there is no input, there is no
output. So, when I think of this, it looks like something other than
feedback it at play in *addition to the things already mentioned*. I got to
thinking about setpoints in fact.

Even your RF amplifiers have an output with no input. The (DC)
output is generally blocked with capacitors. Here we have no
capacitors, though for an AC amplifier one could and often does use
them.

It's the bit about having an output when there is no input voltage that
throws me.

You are just ignoring the output of your RF amplifier, with no
input. It's there.

But of course if you do kind of have an output with "ac amplifiers" when
input is zero.. In the sense that there is a dc voltage on the other side
of the output capacitor. That dc voltage rises sinosoidally (say). But
when there is no input it goes to a quiesent value. Of course, it
it's sinusoildal voltages that you want mangnify, and dc voltage on
the other side of the output capacitor (or transformer, whatever)
is of no significance to the amplification of the input.

Is this getting there? Rich

Yep! If you're concerned about DC you have to consider it. If you
aren't, block and ignore. ;-)

 

[John Fields]

On Fri, 6 Feb 2009 14:15:03 -0000, "Rich" <notty@emailo.com> wrote:

It's the bit about having an output when there is no input voltage that
throws me.

But of course if you do kind of have an output with "ac amplifiers" when
input is zero.. In the sense that there is a dc voltage on the other side
of the output capacitor. That dc voltage rises sinosoidally (say). But
when there is no input it goes to a quiesent value. Of course, it
it's sinusoildal voltages that you want mangnify, and dc voltage on
the other side of the output capacitor (or transformer, whatever)
is of no significance to the amplification of the input.

Is this getting there? Rich

---
Yes.

One thing you have to consider if you only have a single unipolar power
supply and you want to amplify a bipolar ac signal is biasing the opamp
so that the output voltage is quiescently between the rails.

For the inverting case what you'd do is connect a couple of resistors
across the rails and then connect their junction to the opamp's non
inverting input:

12V-----------+------------+
| |
[R1] +---|--[R4]--+
| | | |
Vin>--[C1]----|--[R3]--+--|-\ |
| | >------+-[C2]--->Vout
+-----------|+/
| |
[R2] |
| |
GND>----------+------------+


With a 12V supply you'd want the output to be at 6V with no AC signal
in, (so you could get a symmetrical +/-6V swing at Vout) so by making R1
and R2 equal, the + input would be at 6V and the output would have to
swing to 6V to make the to opamp input voltages equal.

Assuming Xc of C1 << R3 at any frequency of Vin, R3 and R4 are used to
set the gain of the stage:

R4
Av = ----
R3

and notice that since there's no DC path through R3 the gain and bias
settings are totally non-interactive.

JF

 

[Jasen Betts]

On 2009-02-06, krw <krw@att.zzzzzzzzz> wrote:

In article <gmgvg7$fv$2@reversiblemaps.ath.cx>, jasen@xnet.co.nz
says...
On 2009-02-05, krw <krw@att.zzzzzzzzz> wrote:

How is your "almost ideal" less confusing than going without the
almost part? The three characteristics of an ideal Op-Amp work
very well to analyze almost any OpAmp circuit, at least as a first
pass. Indeed the idealized OpAmp works better and is a whole lot
simpler than your "almost ideal" version.

there is no such thing as an ideal op-amp it's impossible for a number
of reasons. this can initially be a real distraction for those who
have more experience with real hardware than with mathematics,

Complete hogwash.

I made three points and I know you can't disprove any of them.

 

[krw]

On 8 Feb 2009 09:05:16 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2009-02-06, krw <krw@att.zzzzzzzzz> wrote:
In article <gmgvg7$fv$2@reversiblemaps.ath.cx>, jasen@xnet.co.nz
says...
On 2009-02-05, krw <krw@att.zzzzzzzzz> wrote:

How is your "almost ideal" less confusing than going without the
almost part? The three characteristics of an ideal Op-Amp work
very well to analyze almost any OpAmp circuit, at least as a first
pass. Indeed the idealized OpAmp works better and is a whole lot
simpler than your "almost ideal" version.

there is no such thing as an ideal op-amp it's impossible for a number
of reasons. this can initially be a real distraction for those who
have more experience with real hardware than with mathematics,

Complete hogwash.

I made three points and I know you can't disprove any of them.

Since you snipped my response, I'll assume you'd rather lie to cover
your inadequacy than discuss the issues. No, I don't approve of
liars, or incompetents for that matter

 

[John Fields]

On 8 Feb 2009 09:05:16 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2009-02-06, krw <krw@att.zzzzzzzzz> wrote:
In article <gmgvg7$fv$2@reversiblemaps.ath.cx>, jasen@xnet.co.nz
says...
On 2009-02-05, krw <krw@att.zzzzzzzzz> wrote:

How is your "almost ideal" less confusing than going without the
almost part? The three characteristics of an ideal Op-Amp work
very well to analyze almost any OpAmp circuit, at least as a first
pass. Indeed the idealized OpAmp works better and is a whole lot
simpler than your "almost ideal" version.

there is no such thing as an ideal op-amp it's impossible for a number
of reasons. this can initially be a real distraction for those who
have more experience with real hardware than with mathematics,

Complete hogwash.

I made three points and I know you can't disprove any of them.

---
Well, if:

"this can initially be a real distraction for those who
have more experience with real hardware than with mathematics,"

was one of the points, then that one is not only complete, it's complete
and _utter_ hogwash.

JF

 

[Jon Kirwan]

On Thu, 5 Feb 2009 22:24:43 -0000, "Rich" <notty@emailo.com> wrote:

snip
but why did
someone make an amplifier that worked that way? Did someone say - Mmm, we
can make a very useful amplifier if we create a circuit (an Op Amp) which
will give an output and stabilise the voltage output when V- = V+?

If you are interested in the history, you might see about finding a
copy of "Op Amp Applications," edited by Walt Jung, published by
Analog Devices, Inc. The history will help inform you a little more
about the "why did someone want to do that?" question.

You can get the PDFs for it at:
: http://www.analog.com/library/analogDialogue/archives/39-05/op_amp_applications_handbook.html

In particular, you might want to look at Chapter H and perhaps even
more particularly on the section called "Black's Feedback Amplifier."

Jon

 

[Rich]

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:506uo4ho64v6qlha655pvqt296koiqvk1f@4ax.com...

On Thu, 5 Feb 2009 22:24:43 -0000, "Rich" <notty@emailo.com> wrote:

snip
but why did
someone make an amplifier that worked that way? Did someone say - Mmm, we
can make a very useful amplifier if we create a circuit (an Op Amp) which
will give an output and stabilise the voltage output when V- = V+?

If you are interested in the history, you might see about finding a
copy of "Op Amp Applications," edited by Walt Jung, published by
Analog Devices, Inc. The history will help inform you a little more
about the "why did someone want to do that?" question.

You can get the PDFs for it at:
:
http://www.analog.com/library/analogDialogue/archives/39-05/op_amp_applications_handbook.html

In particular, you might want to look at Chapter H and perhaps even
more particularly on the section called "Black's Feedback Amplifier."

Jon

Brilliant. Thanks.

 

[Jasen Betts]

On 2009-02-08, krw <krw@att.bizzzzzzzzzzz> wrote:

On 8 Feb 2009 09:05:16 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2009-02-06, krw <krw@att.zzzzzzzzz> wrote:

Complete hogwash.

I made three points and I know you can't disprove any of them.

Since you snipped my response,

Again you go off on a tangent instead of dealing with what I wrote,

I'll assume

You do that too much.

you'd rather lie.

Nope.