Operational Amplifier basics

[Rich]

I'm trying to understand Op Amps first time round. Are the following
statements correct? I've rushed it a bit, so most of it might look like
nonsense.

Please correct or amplify.

1 An Op Amp is basically a DIFFERENTAL amplifier.

2 Unlike a (for want of a better word) a "regular" amplifier, an Op Amp
can have two input voltages.

3 When the Op amp is run as a DIFFERENTIAL amplifier it amplifies the
difference in voltage between the two inputs.

4 Operational amplifiers can be run as "regular" amplifiers (i.e. not
differential).

5 The Op Amp can be run from either a single supply or a split supply

6 The Op Amp run from a split supply power pins will often be
labelled Vcc + and Vcc-


(From hereon I'm a bit shaky):

I'm now trying to say where the reference point is for input and output.
Obviously it takes two points in a circuit to establish a voltage, either an
input or ouput voltage, I mean, just connecting one side of a load or a
generator or battery to one of the input or output points shown on an Op Amp
circuit is no good. I'm going to call the corresponding point the REFERENCE
POINT. I'll probably mess up here, but please correct me.

OP AMPS *NOT* SPECIALLY DESIGNED FOR SINGLE RAIL SUPPLY


SPLIT SUPPLY OPERATION

REFERENCE POINTS

INPUT

7 V+ input: The REFERENCE POINT is VCC- a negative potential with
respect to ground.

8 V- input: The REFERENCE POINT is VCC- a negative potential with respect
to ground.

9 The reverse is true (REFERENCE POINT is VCC+) if transistors are
different type (We have NPN or PNP types).

10 The REFERENCE POINT point is NOT ground.

11 With a split supply, ground (SYSTEM GROUND) is between + and - supply
voltages = 0v+ SUPPLY MIDPOINT.

OUTPUT

12 Vout: The REFERENCE POINT is ground or SUPPLY MIDPOINT = 0v = SYSTEM
GROUND


SINGLE SUPPLY OPERATION

REFERENCE POINTS

INPUT

13 Single-supply operation requires a VIRTUAL GROUND in an OP Amp stage.

14 A VIRTUAL GROUND is at a dc level above SYSTEM GROUND.

15 SYSTEM GROUND is 0v.

16 V+ input: The REFERENCE POINT is ground.

17 V- input: The REFERENCE POINT is ground.

18 Input source and output load will often be referenced to system ground.

19 If the input source of a stage is referenced to ground then the solution
is to ac couple signals to the Op Amp stage. Unless certain conditions
apply.

OUTPUT

20 Vout: The REFERENCE POINT is ground.


OP AMPS SPECIALLY DESIGNED FOR SINGLE RAIL SUPPLY


21 V+ input: The REFERENCE POINT is Ground.

22 V- input: The REFERENCE POINT is Ground.

OUTPUT

23 Vout: The REFERENCE POINT is ground.

 

[Rich]

16 V+ input: The REFERENCE POINT is ground.

17 V- input: The REFERENCE POINT is ground.

Should that read?:

16: V+ input: The REFERENCE POINT is VIRTUAL GROUND.

17: V+ input: The REFERENCE POINT is VIRTUAL GROUND.

I can see that the input source relates to where the input signal is coming
from. Which is also an output! The reference point is often ground.

In order for the OP Amp to work correctly, the Op Amp input should be
referenced to ground. But it can have a virtual ground in the Op Amp stage,
if you put a capacitor in the input. Won't work for a dc input voltage that
needs to be amplified.

 

[krw]

In article <6uu6vpFhak3oU1@mid.individual.net>, notty@emailo.com
says...>

I'm trying to understand Op Amps first time round. Are the following
statements correct? I've rushed it a bit, so most of it might look like
nonsense.

Please correct or amplify.

There are only three "rules" to know about the ideal op amp

1) Infinite input impedance.
2) Infinite differential gain.
3) Zero output impedance.

All else can be derived.

1 An Op Amp is basically a DIFFERENTAL amplifier.
^ input

2 Unlike a (for want of a better word) a "regular" amplifier, an Op Amp
can have two input voltages.

It only has one. From rule #2 above, assuming a non infinite output
voltage, we conclude the difference between the inputs is 0V.

3 When the Op amp is run as a DIFFERENTIAL amplifier it amplifies the
difference in voltage between the two inputs.

No. It can have *NO* difference in the input voltage. A non-idel
amplifier will have some difference but it can be assumed to be
zero (for most purposes).

4 Operational amplifiers can be run as "regular" amplifiers (i.e. not
differential).

No. They *must* be run as a differential amp, with 0V across the
input. You may only USE them in a single-ended configuration,
though the Op-Amp is still a differential amplifier.

5 The Op Amp can be run from either a single supply or a split supply

"It" doesn't know the difference.

6 The Op Amp run from a split supply power pins will often be
labelled Vcc + and Vcc-

Whatever floats your boat. I usually label them "+15V" and "-15V"
(even though they're really +/-12V ;-).

(From hereon I'm a bit shaky):

I'm now trying to say where the reference point is for input and output.
Obviously it takes two points in a circuit to establish a voltage, either an
input or ouput voltage, I mean, just connecting one side of a load or a
generator or battery to one of the input or output points shown on an Op Amp
circuit is no good. I'm going to call the corresponding point the REFERENCE
POINT. I'll probably mess up here, but please correct me.

OP AMPS *NOT* SPECIALLY DESIGNED FOR SINGLE RAIL SUPPLY

No, they're not (so designed). As stated above, they don't know
the difference. They only see the two rails. They aren't
connected to the "half-way" voltage.

SPLIT SUPPLY OPERATION

REFERENCE POINTS

Use ground (or really, anything you like) as your reference points.
Abide by the three rules above, sum the currents in the inputs and
feedback (output to -In), and you're done.

 

[Rich]

"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f3b15061714b0398992f@news.individual.net...

In article <6uu6vpFhak3oU1@mid.individual.net>, notty@emailo.com
says...
I'm trying to understand Op Amps first time round. Are the following
statements correct? I've rushed it a bit, so most of it might look like
nonsense.

Please correct or amplify.

There are only three "rules" to know about the ideal op amp

1) Infinite input impedance.
2) Infinite differential gain.
3) Zero output impedance.

All else can be derived.

Yes, but I'm looking at things from a practical wiring point of view. Being
able to point out where the two points are for an input or ouput voltages.

Maybe it's just me, but I've got to be able to, see the two points where the
input and ouput voltages are meant to be measured from.

I'm looking at something like this:

http://www.electronics-tutorials.ws/opamp/opamp_1.html

And saying to myself for the differential amplifier there are two input
points, V1 and V2, whose reference points are -VEE. Yet below in the
equivalent circuit there is just V1 and V2. From a practical point of view
I'm thinking both V1 and V2 will have a potential with respect to -VEE. But
you would not see this in the equivalent circuit.

 

[krw]

In article <6uuc4pFgvbngU1@mid.individual.net>, notty@emailo.com
says...>

"krw" <krw@att.zzzzzzzzz> wrote in message
news:MPG.23f3b15061714b0398992f@news.individual.net...
In article <6uu6vpFhak3oU1@mid.individual.net>, notty@emailo.com
says...
I'm trying to understand Op Amps first time round. Are the following
statements correct? I've rushed it a bit, so most of it might look like
nonsense.

Please correct or amplify.

There are only three "rules" to know about the ideal op amp

1) Infinite input impedance.
2) Infinite differential gain.
3) Zero output impedance.

All else can be derived.

Yes, but I'm looking at things from a practical wiring point of view. Being
able to point out where the two points are for an input or ouput voltages.

Maybe it's just me, but I've got to be able to, see the two points where the
input and ouput voltages are meant to be measured from.

I'm looking at something like this:

http://www.electronics-tutorials.ws/opamp/opamp_1.html

And saying to myself for the differential amplifier there are two input
points, V1 and V2, whose reference points are -VEE. Yet below in the
equivalent circuit there is just V1 and V2. From a practical point of view
I'm thinking both V1 and V2 will have a potential with respect to -VEE. But
you would not see this in the equivalent circuit.

You can use V- as your reference. You can use V+, or anything else
you choose. If you look at the difference between '+' and '-' it
doesn't matter what '+' and '-' are referenced to. The
*difference* matters (and is zero for an ideal Op-Amp, given non-
infinite outputs).

 

[Rich]

http://www.electronics-tutorials.ws/opamp/opamp_5.html

Does the circuit really work because of R4?

We need two points to measure a voltage.

This circuit will amplify the voltage difference between V- and V+.

So, how can you do that with two inputs?

The input voltage comes fron the input voltage sources.

So, where are the two points for the voltages on V1 and V2?

From the input source side, (i.e. looking left as it were into the sources)
point V1 and Ground and V2 and Ground. There is a circuit in the input
source that goes from V1 and V2 and ground. (Like a battery, there is in
internal circuit in the battery).

But what about lookng into the OP Amp devic, where is that cicuit?

Looking to the right, into the Op Amp, the internal circuit starts at V1 and
ends at V2.

So, from a sources POV, one source sees it's termination on as R1, through
the internal resistor inside the Op amp (infinity) then through R4 to
ground.

The other source sees it's termination as R3, then through R4.

I'm probably wrong, but that's the way I'm thinking.

I think it only amplifies the difference between V1 and V2 because both
input sources are connected to a common reference point, Ground. True.

Yet the Op Amp itself just sees a voltage between V- and V+. Which is an
amplification of the voltage difference from the sources.

Is this getting near?

 

[Peter Bennett]

On Wed, 4 Feb 2009 19:01:09 -0000, "Rich" <notty@emailo.com> wrote:

I'm trying to understand Op Amps first time round. Are the following
statements correct? I've rushed it a bit, so most of it might look like
nonsense.

Please correct or amplify.

1 An Op Amp is basically a DIFFERENTAL amplifier.

2 Unlike a (for want of a better word) a "regular" amplifier, an Op Amp
can have two input voltages.

An op-amp _does_ have two inputs, normally called "inverting" and
"non-inverting".

3 When the Op amp is run as a DIFFERENTIAL amplifier it amplifies the
difference in voltage between the two inputs.

An op-amp _always_ amplifies the voltage difference between its two
inputs.

4 Operational amplifiers can be run as "regular" amplifiers (i.e. not
differential).

A circuit using an op-amp can be a "regular" (single input) amplifier,
but the op-amp still has differential inputs.

5 The Op Amp can be run from either a single supply or a split supply

6 The Op Amp run from a split supply power pins will often be
labelled Vcc + and Vcc-


(From hereon I'm a bit shaky):

I'm now trying to say where the reference point is for input and output.
Obviously it takes two points in a circuit to establish a voltage, either an
input or ouput voltage, I mean, just connecting one side of a load or a
generator or battery to one of the input or output points shown on an Op Amp
circuit is no good. I'm going to call the corresponding point the REFERENCE
POINT. I'll probably mess up here, but please correct me.

OP AMPS *NOT* SPECIALLY DESIGNED FOR SINGLE RAIL SUPPLY


SPLIT SUPPLY OPERATION

REFERENCE POINTS

INPUT

7 V+ input: The REFERENCE POINT is VCC- a negative potential with
respect to ground.

The convention in electronics is that "ground" is always the
"reference point" for voltage measurements.

8 V- input: The REFERENCE POINT is VCC- a negative potential with respect
to ground.

9 The reverse is true (REFERENCE POINT is VCC+) if transistors are
different type (We have NPN or PNP types).

No - "ground" is always the reference point. The type of transistors
used in the op-amp do not change this.

10 The REFERENCE POINT point is NOT ground.

wrong.

11 With a split supply, ground (SYSTEM GROUND) is between + and - supply
voltages = 0v+ SUPPLY MIDPOINT.

OUTPUT

12 Vout: The REFERENCE POINT is ground or SUPPLY MIDPOINT = 0v = SYSTEM
GROUND


SINGLE SUPPLY OPERATION

REFERENCE POINTS

INPUT

13 Single-supply operation requires a VIRTUAL GROUND in an OP Amp stage.

14 A VIRTUAL GROUND is at a dc level above SYSTEM GROUND.

15 SYSTEM GROUND is 0v.

16 V+ input: The REFERENCE POINT is ground.

17 V- input: The REFERENCE POINT is ground.

18 Input source and output load will often be referenced to system ground.

19 If the input source of a stage is referenced to ground then the solution
is to ac couple signals to the Op Amp stage. Unless certain conditions
apply.

It is generally necessary to bias both input pins to about mid-supply,
whether using a single supply or two supplies.

OUTPUT

20 Vout: The REFERENCE POINT is ground.


OP AMPS SPECIALLY DESIGNED FOR SINGLE RAIL SUPPLY


21 V+ input: The REFERENCE POINT is Ground.

22 V- input: The REFERENCE POINT is Ground.

OUTPUT

23 Vout: The REFERENCE POINT is ground.

--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

[krw]

In article <6uue29Fhddg2U1@mid.individual.net>, notty@emailo.com
says...>

http://www.electronics-tutorials.ws/opamp/opamp_5.html

Does the circuit really work because of R4?

Without R4 it works but doesn't do the same thing as it does with
R4.

We need two points to measure a voltage.

Certainly. However, the "reference" is chosen to make the math
easier. The circuit doesn't care what you name things.

This circuit will amplify the voltage difference between V- and V+.

No. The voltage between '+' and '-' is (almost) zero. The idea of
this circuit is to amplify the voltage between V1 and V2.

So, how can you do that with two inputs?

You just said it takes two points to measure (define) a voltage.
Choose V1 and V2. It may be easier for you to see what the output
voltage is because of V1 (to ground) and V2 (to ground),
independently, then add the two together.

The input voltage comes fron the input voltage sources.

That's why we call them the "input". ;-)

So, where are the two points for the voltages on V1 and V2?

Pick one, ut be consistent. To make the math easier pick the
ground ('0V'). REMEMBER, there is zero voltage from '+' to '-'.

From the input source side, (i.e. looking left as it were into the sources)
point V1 and Ground and V2 and Ground. There is a circuit in the input
source that goes from V1 and V2 and ground. (Like a battery, there is in
internal circuit in the battery).

Ok, I'm no really following you, but...

But what about lookng into the OP Amp devic, where is that cicuit?

It's a black box, that has the characteristics of infinite input
impedance, infinite gain ((V+ - V-)/Vout) = infinity, and zero
output impedance.

Looking to the right, into the Op Amp, the internal circuit starts at V1 and
ends at V2.

It's a black box. Don't care what's in there. If you want to
analyze the black box forget what's outside.

So, from a sources POV, one source sees it's termination on as R1, through
the internal resistor inside the Op amp (infinity) then through R4 to
ground.

Forget this side for now.

The other source sees it's termination as R3, then through R4.

Yep, the input impedance is infinite, so forget it. Thus:
V+ = V2(R4/(R3+R4)) You now have half the answer.

Now, since the gain is infinite, the voltage on the input must be
zero. Therefor V- = V+. Now you can solve for the current in R1.
Since the input impedance of the Op-Amp is infinite, that same
current must be in R2.

V1-V+(from above) V+ - Vout
----------------- = -----------
R1 R2

I'm probably wrong, but that's the way I'm thinking.

I think it only amplifies the difference between V1 and V2 because both
input sources are connected to a common reference point, Ground. True.

Your conclusion is correct, but your reasoning is flawed. The
"reference" point doesn't matter. It's a philament of your
imagination. ;-)

Yet the Op Amp itself just sees a voltage between V- and V+. Which is an
amplification of the voltage difference from the sources.

The Op-Amp (ouput) forces its input to be 0V. The rest falls out
in the math.

Is this getting near?

Read the last line again (and again, and...). This is the *key*
point. The input voltage *MUST* be zero.

 

[John Fields]

On Wed, 4 Feb 2009 19:01:09 -0000, "Rich" <notty@emailo.com> wrote:

<Big snip>

Never mind all that stuff; here, in a nutshell, is how an opamp works:

The opamp output does whatever it has to to make the voltages on the
inputs equal.

That's basically it, but what does it mean?

(View in Courier

Let's say that you've got a circuit set up that looks like this:

.. E1
.. /
.. +-------------+
.. | |
..+------+------+ [R1]
..| VARIABLE | | E2 E3
..|POWER SUPPLY | |/ / +--------+
..+------+------+ +---[VOLTMETER]---|+ |
.. | | | FIXED |
.. | [R2] | POWER |
.. | | | SUPPLY |
.. +-------------+-----------------|- |
.. +--------+

Let's also say that R1 is equal to zero ohms, that R2 is equal to 1000
ohms, that E1 is equal to 1 volt and that E3 is equal to 2 volts.

So how do you make the voltmeter read zero volts?

You crank up the variable supply until E1 equals E3, thereby making the
voltage _difference_ across the voltmeter zero.

Now lets change R1 to 1000 ohms while leaving everything else the same.

What will that do? It'll increase the amount of voltage the variable
supply has to put out to get 0V into the voltmeter.

How much? R1R2 is what's called a voltage divider, and the relationship
between the voltages and resistances is given by:


E1 * R2
E2 = ---------
R1 + R2


We want to find E1, so we rearrange like this:


E2(R1 + R2)
E1 = -------------
R2


and solve:


2V ( 1000R + 1000R) 2V * 2000R
E1 = --------------------- = ------------ = 4 volts.
1000R 1000R


If we were to leave E3 alone and change E1 from 1000 ohms to 10000 ohms
in 1000 ohm steps, and solve for E1 at each step, we'd have a table that
looks like this:


R1 E1
OHMS VOLTS
-------+-------
0 2
1000 4
2000 6
3000 8
4000 10
5000 12
6000 14
7000 16
8000 18
9000 20
10000 22


Let's go back to when we had 1000 ohms for R1 and see what would happen
if we changed E3.


4V (1000R + 1000R)
E1 = -------------------- = 8V
1000R

So now let's make a table of V1 VS V3 for V3 = 0V to 10V when E1 = 1000
ohms:

E3 E1
VOLTS VOLTS
-------+-------
0 0
2 4
3 6
4 8
5 10
6 12
7 14
8 16
9 18
10 20

If we say E3 is the input and E1 is the output, then the circuit has a
voltage gain of 2

But how does this apply to opamps?

OK let's redraw things a little:

+V
|
E3--+-------------|+\
| | >----+--E1
[VOLTMETER] +--|-/ |
| | | |
+----------+ -V |
| |
E2--+---[R1]---+
|
[R2]
|
0V

Now pretend that R1 and R2 are both 1000 ohms, that E3 = 1V, and that
all that matters is that the opamp makes the voltage at E1 be whatever
it needs to be to make E2 equal to E3, which will make the voltmeter
read zero.

What would that voltage be?

JF

 

[JeffM]

Rich wrote:

I'm trying to understand Op Amps first time round.
[. . .]
10 The REFERENCE POINT point is NOT ground.

Here are some synonyms for "ground":

Common
Common Return Path
Common Reference Point
Zero-Volts Point

 

[Rich]

"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

Now pretend that R1 and R2 are both 1000 ohms, that E3 = 1V, and that
all that matters is that the opamp makes the voltage at E1 be whatever
it needs to be to make E2 equal to E3, which will make the voltmeter
read zero.

What would that voltage be?

JF

Midnight here. I'll figure it all out tomorrow. :c)

 

[Rich]

"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:

E3 E1
VOLTS VOLTS
-------+-------
0 0
2 4
3 6
4 8
5 10
6 12
7 14
8 16
9 18
10 20

If we say E3 is the input and E1 is the output, then the circuit has a
voltage gain of 2

But how does this apply to opamps?

OK let's redraw things a little:

+V
|
E3--+-------------|+\
| | >----+--E1
[VOLTMETER] +--|-/ |
| | | |
+----------+ -V |
| |
E2--+---[R1]---+
|
[R2]
|
0V

Now pretend that R1 and R2 are both 1000 ohms, that E3 = 1V, and that
all that matters is that the opamp makes the voltage at E1 be whatever
it needs to be to make E2 equal to E3, which will make the voltmeter
read zero.

What would that voltage be?

JF

E3 is set to 1V, that means E2 needs to be at 1V.

E1 = 1V (1000R + 1000R)
-------------------------- = 2V
10000

The Invering amplifier which was drawn seeks to get V- and V+ to be equal.
When there is a tendency for E3 to go up, the voltage of the "Variable PSU"
goes down, thus stabalizing the system. The system reaches stability when E2
= E3. And at that point there is a certain voltage at E1. The value of that
particular voltage is a function of the make-up of the voltage divider
network. It cannot exceed supply. Of course the amplifier is acting as if
it's a "variable PSU", like a generator in fact.Correct?

Not sure how the op amp itself settles when V- = V+. I just know that the
amp acts like a variable PSU and what voltage E1 needs to be to get V- to
equal V+.

 

[Rich]

"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:

E3 E1
VOLTS VOLTS
-------+-------
0 0
2 4
3 6
4 8
5 10
6 12
7 14
8 16
9 18
10 20

If we say E3 is the input and E1 is the output, then the circuit has a
voltage gain of 2

But how does this apply to opamps?

OK let's redraw things a little:

+V
|
E3--+-------------|+\
| | >----+--E1
[VOLTMETER] +--|-/ |
| | | |
+----------+ -V |
| |
E2--+---[R1]---+
|
[R2]
|
0V

Now pretend that R1 and R2 are both 1000 ohms, that E3 = 1V, and that
all that matters is that the opamp makes the voltage at E1 be whatever
it needs to be to make E2 equal to E3, which will make the voltmeter
read zero.

What would that voltage be?

JF

E3 is set to 1V, that means E2 needs to be at 1V.

E1 = 1V (1000R + 1000R)
-------------------------- = 2V
10000

The Invering amplifier which was drawn seeks to get V- and V+ to be equal.
When there is a tendency for E3 to go up, the voltage of the "Variable
PSU"
goes down, thus stabalizing the system. The system reaches stability when
E2
= E3. And at that point there is a certain voltage at E1. The value of
that
particular voltage is a function of the make-up of the voltage divider
network. It cannot exceed supply. Of course the amplifier is acting as if
it's a "variable PSU", like a generator in fact.Correct?

Not sure how the op amp itself settles when V- = V+. I just know that the
amp acts like a variable PSU and what voltage E1 needs to be to get V- to
equal V+.

Actually, it's like the OP Amp has an internal setting. A- = A+ is like a
setpoint which stays put all of the time. You get your desired results by
messing about with the external networks.

 

[krw]

In article <6v0mvfFh2i4fU1@mid.individual.net>, notty@emailo.com
says...>

"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:

E3 E1
VOLTS VOLTS
-------+-------
0 0
2 4
3 6
4 8
5 10
6 12
7 14
8 16
9 18
10 20

If we say E3 is the input and E1 is the output, then the circuit has a
voltage gain of 2

But how does this apply to opamps?

OK let's redraw things a little:

+V
|
E3--+-------------|+\
| | >----+--E1
[VOLTMETER] +--|-/ |
| | | |
+----------+ -V |
| |
E2--+---[R1]---+
|
[R2]
|
0V

Now pretend that R1 and R2 are both 1000 ohms, that E3 = 1V, and that
all that matters is that the opamp makes the voltage at E1 be whatever
it needs to be to make E2 equal to E3, which will make the voltmeter
read zero.

What would that voltage be?

JF

E3 is set to 1V, that means E2 needs to be at 1V.

E1 = 1V (1000R + 1000R)
-------------------------- = 2V
10000

Correct.

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

That's all you need to know (and that the input impedance is
infinite and output zero).

When there is a tendency for E3 to go up, the voltage of the "Variable PSU"
goes down, thus stabalizing the system. The system reaches stability when E2
= E3. And at that point there is a certain voltage at E1. The value of that
particular voltage is a function of the make-up of the voltage divider
network. It cannot exceed supply. Of course the amplifier is acting as if
it's a "variable PSU", like a generator in fact.Correct?

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").

 

[krw]

In article <6v0n9dFhreqsU1@mid.individual.net>, notty@emailo.com
says...>

"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:

E3 E1
VOLTS VOLTS
-------+-------
0 0
2 4
3 6
4 8
5 10
6 12
7 14
8 16
9 18
10 20

If we say E3 is the input and E1 is the output, then the circuit has a
voltage gain of 2

But how does this apply to opamps?

OK let's redraw things a little:

+V
|
E3--+-------------|+\
| | >----+--E1
[VOLTMETER] +--|-/ |
| | | |
+----------+ -V |
| |
E2--+---[R1]---+
|
[R2]
|
0V

Now pretend that R1 and R2 are both 1000 ohms, that E3 = 1V, and that
all that matters is that the opamp makes the voltage at E1 be whatever
it needs to be to make E2 equal to E3, which will make the voltmeter
read zero.

What would that voltage be?

JF

E3 is set to 1V, that means E2 needs to be at 1V.

E1 = 1V (1000R + 1000R)
-------------------------- = 2V
10000

The Invering amplifier which was drawn seeks to get V- and V+ to be equal.
When there is a tendency for E3 to go up, the voltage of the "Variable
PSU"
goes down, thus stabalizing the system. The system reaches stability when
E2
= E3. And at that point there is a certain voltage at E1. The value of
that
particular voltage is a function of the make-up of the voltage divider
network. It cannot exceed supply. Of course the amplifier is acting as if
it's a "variable PSU", like a generator in fact.Correct?

Not sure how the op amp itself settles when V- = V+. I just know that the
amp acts like a variable PSU and what voltage E1 needs to be to get V- to
equal V+.

Actually, it's like the OP Amp has an internal setting. A- = A+ is like a
setpoint which stays put all of the time. You get your desired results by
messing about with the external networks.

No, the output, via the negative feedback (R1 feeds back from
output to the '-' input, thus "negative") forces the inputs to be
at the same voltage. If the inputs are forced to be different the
output voltage will try to go to infinity (either positive or
negative). In real life this means the output will limit at the
positive or negative rail (known as the output "railing").
Feedback is lost at this point so the OpAmp no longer functions as
an OpAmp.

 

[Rich]

"krw" <krw@att.zzzzzzzzz> wrote in message

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

That's all you need to know (and that the input impedance is
infinite and output zero).

When there is a tendency for E3 to go up, the voltage of the "Variable
PSU"
goes down, thus stabalizing the system. The system reaches stability when
E2
= E3. And at that point there is a certain voltage at E1. The value of
that
particular voltage is a function of the make-up of the voltage divider
network. It cannot exceed supply. Of course the amplifier is acting as if
it's a "variable PSU", like a generator in fact.Correct?

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

 

[krw]

In article <6v0r0lFhi15nU1@mid.individual.net>, notty@emailo.com
says...>

"krw" <krw@att.zzzzzzzzz> wrote in message

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

That's all you need to know (and that the input impedance is
infinite and output zero).

When there is a tendency for E3 to go up, the voltage of the "Variable
PSU"
goes down, thus stabalizing the system. The system reaches stability when
E2
= E3. And at that point there is a certain voltage at E1. The value of
that
particular voltage is a function of the make-up of the voltage divider
network. It cannot exceed supply. Of course the amplifier is acting as if
it's a "variable PSU", like a generator in fact.Correct?

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.

 

[Rich]

"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.

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.

 

[Rich]

"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.

 

[John Fields]

On Thu, 5 Feb 2009 17:46:35 -0000, "Rich" <notty@emailo.com> wrote:

"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:

E3 E1
VOLTS VOLTS
-------+-------
0 0
2 4
3 6
4 8
5 10
6 12
7 14
8 16
9 18
10 20

If we say E3 is the input and E1 is the output, then the circuit has a
voltage gain of 2

But how does this apply to opamps?

OK let's redraw things a little:

+V
|
E3--+-------------|+\
| | >----+--E1
[VOLTMETER] +--|-/ |
| | | |
+----------+ -V |
| |
E2--+---[R1]---+
|
[R2]
|
0V

Now pretend that R1 and R2 are both 1000 ohms, that E3 = 1V, and that
all that matters is that the opamp makes the voltage at E1 be whatever
it needs to be to make E2 equal to E3, which will make the voltmeter
read zero.

What would that voltage be?

JF

E3 is set to 1V, that means E2 needs to be at 1V.

E1 = 1V (1000R + 1000R)
-------------------------- = 2V
10000

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

---
Since the output goes in the same direction as the input, it's a
non-inverting amplifier.

Here, I'll redraw it without the power pins to make it clearer:



.. E3--+-------------|+\
.. | | >----+--E1
.. [VOLTMETER] +--|-/ |
.. | | |
.. +----------+ |
.. | |
.. E2--+---[R1]---+
.. |
.. [R2]
.. |
.. 0V

When there is a tendency for E3 to go up, the voltage of the "Variable PSU"
goes down, thus stabalizing the system.

---
No, it goes up.
---

The system reaches stability when E2
= E3. And at that point there is a certain voltage at E1. The value of that
particular voltage is a function of the make-up of the voltage divider
network. It cannot exceed supply. Of course the amplifier is acting as if
it's a "variable PSU", like a generator in fact.Correct?

---
Yes, exactly, except that when E3 rises, E1 must also rise to make E2 =
E3.

In this case, because the voltage at E3 (the non-inverting input) will
be half the voltage at E1 (the output) when the voltage at E2 (the
inverting input) equals the voltage at E3, the circuit is said to have a
"voltage gain" of 2.

The gain of a non-inverting opamp can be stated by:

R1
Av = 1 + ----
R2

So if you had a 1V signal at the + input and you wanted a 10V output
you'd have to make the ratio of R1 to R2 equal 9 to get the needed gain
of 10.

So how do you choose the resistors? First you decide how much current
you want running through the divider and then you choose the resistors
so their total resistance allows that much current to flow and then set
the ratio for the gain you need.

For example, let's say that at 10V out you want 1mA through the string.

So, then:

Et 10V
Rt = ---- = -------- = 10000 ohms
I 0.001A


Next, you know you'll need two resistors and that one will have a
resistance 9 times higher than the other, so we can say:


x + 9x = 10000 ohms

and then solve for the lower resistance:

10000R
x = -------- 1000 ohms
10

So, the other resistor must be 9000 ohms, and your circuit now looks
like this:

.. E3--+-------------|+\
.. | | >----+--E1
.. [VOLTMETER] +--|-/ |
.. | | |
.. +----------+ |
.. | |
.. E2--+--[9kR]---+
.. |
.. [1kR]
.. |
.. 0V

Now, if you set E3 to 1V and run the numbers you'll see what happens to
the output voltage.
---

Not sure how the op amp itself settles when V- = V+. I just know that the
amp acts like a variable PSU and what voltage E1 needs to be to get V- to
equal V+.

---
Right.

It works because the opamp itself has a huge amount of internal gain "in
reserve" and it uses that as brute force to keep the output just where
it needs to be to make the two input voltages equal.

JF