Opamp resistor magnitude

[Gone Postal]

I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved? I've seen different circuits with resistors of, say
thousands, tens or hundered of thousands and even one that used
Megohm-range resistors.
Why the variation in magnitude of the resistors, assuming a fixed
ratio? Does it have anything to do with the overall impedance of the
opamp circuit block?
I thank you for whatever illumination that can be provided.

GP

 

[Phil Allison]

"Gone Postal" <

I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved? I've seen different circuits with resistors of, say
thousands, tens or hundered of thousands and even one that used
Megohm-range resistors.
Why the variation in magnitude of the resistors, assuming a fixed
ratio? Does it have anything to do with the overall impedance of the
opamp circuit block?
I thank you for whatever illumination that can be provided.

** Choosing a particular resistor value for a given spot in a circuit is a
major part of electronics design - so your Q is a mile wide and goes way
beyond just op-amps.

Try narrowing it down a bit.



..... Phil

 

[Jasen Betts]

On 2014-04-28, Gone Postal <gone_postal@it.doesn't.exist> wrote:

I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved?

mainly the impedance of the input.
the imput will have high resistance but not infinite.

--
umop apisdn


--- news://freenews.netfront.net/ - complaints: news@netfront.net ---

 

[Gone Postal]

On Mon, 28 Apr 2014 16:29:05 +1000, "Phil Allison" <phil_a@tpg.com.au>
wrote:

"Gone Postal"

I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved? I've seen different circuits with resistors of, say
thousands, tens or hundered of thousands and even one that used
Megohm-range resistors.
Why the variation in magnitude of the resistors, assuming a fixed
ratio? Does it have anything to do with the overall impedance of the
opamp circuit block?
I thank you for whatever illumination that can be provided.


** Choosing a particular resistor value for a given spot in a circuit is a
major part of electronics design - so your Q is a mile wide and goes way
beyond just op-amps.

Try narrowing it down a bit.



.... Phil

Thank you Phil.
Apparently I understand less than I thought (a common occurrence), I
had been under the impression my question was fairly narrow, but I can
(sort of) see how other circuit considerations would affect the choice
of specific resistors. I will spend some more time with LT Spice and
see what I come up with.

GP

 

[Gone Postal]

On 28 Apr 2014 06:13:27 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2014-04-28, Gone Postal <gone_postal@it.doesn't.exist> wrote:
I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved?

mainly the impedance of the input.
the imput will have high resistance but not infinite.

Thanks for the answer, that's what I was generally getting at.

GP

 

[George Herold]

On Monday, April 28, 2014 12:44:23 AM UTC-4, Gone Postal wrote:

I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved? I've seen different circuits with resistors of, say
thousands, tens or hundered of thousands and even one that used
Megohm-range resistors.
Why the variation in magnitude of the resistors, assuming a fixed
ratio? Does it have anything to do with the overall impedance of the
opamp circuit block?
I thank you for whatever illumination that can be provided.

I think 10k ohm resistors are my favorite. (then 1 k ohm)
In general I'd like to choose R's that are smaller.
First this tends to reduce the effects of stray capacitance.
(Some RC corner that move to higher frequency with smaller R)
And then if you care, smaller R has less noise.. though most of the time that doesn't matter much.
Now of course you can't make 'em too small.
First the opamp mya not have enough poop to drive it.
and second if the R's get very small (a few ohms) you then
start to worry about the resistance of the traces and contacts.

George H.

GP

 

[Phil Allison]

"Gone Postal"
"Phil Allison"

"Gone Postal"

I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved? I've seen different circuits with resistors of, say
thousands, tens or hundered of thousands and even one that used
Megohm-range resistors.
Why the variation in magnitude of the resistors, assuming a fixed
ratio? Does it have anything to do with the overall impedance of the
opamp circuit block?
I thank you for whatever illumination that can be provided.


** Choosing a particular resistor value for a given spot in a circuit is a
major part of electronics design - so your Q is a mile wide and goes way
beyond just op-amps.

Try narrowing it down a bit.




Thank you Phil.
Apparently I understand less than I thought (a common occurrence), I
had been under the impression my question was fairly narrow,

** Ignorance is bliss.

but I can
(sort of) see how other circuit considerations would affect the choice
of specific resistors. I will spend some more time with LT Spice and
see what I come up with.

** Go ahead - waste your time completely.


..... Phil

 

[Gone Postal]

On Mon, 28 Apr 2014 19:50:26 +1000, "Phil Allison" <phil_a@tpg.com.au>
wrote:

"Gone Postal"
"Phil Allison"

"Gone Postal"

I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved? I've seen different circuits with resistors of, say
thousands, tens or hundered of thousands and even one that used
Megohm-range resistors.
Why the variation in magnitude of the resistors, assuming a fixed
ratio? Does it have anything to do with the overall impedance of the
opamp circuit block?
I thank you for whatever illumination that can be provided.


** Choosing a particular resistor value for a given spot in a circuit is a
major part of electronics design - so your Q is a mile wide and goes way
beyond just op-amps.

Try narrowing it down a bit.




Thank you Phil.
Apparently I understand less than I thought (a common occurrence), I
had been under the impression my question was fairly narrow,

** Ignorance is bliss.


but I can
(sort of) see how other circuit considerations would affect the choice
of specific resistors. I will spend some more time with LT Spice and
see what I come up with.

** Go ahead - waste your time completely.


.... Phil

Do you think playing with Spice is the wrong approach? If so, what
would be a better approach to understanding this?

GP

 

[Phil Allison]

"Gone Postal "
"Phil Allison"

"Gone Postal"
"Phil Allison"

"Gone Postal"

I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved? I've seen different circuits with resistors of, say
thousands, tens or hundered of thousands and even one that used
Megohm-range resistors.
Why the variation in magnitude of the resistors, assuming a fixed
ratio? Does it have anything to do with the overall impedance of the
opamp circuit block?
I thank you for whatever illumination that can be provided.


** Choosing a particular resistor value for a given spot in a circuit is
a
major part of electronics design - so your Q is a mile wide and goes
way
beyond just op-amps.

Try narrowing it down a bit.




Thank you Phil.
Apparently I understand less than I thought (a common occurrence), I
had been under the impression my question was fairly narrow,

** Ignorance is bliss.


but I can
(sort of) see how other circuit considerations would affect the choice
of specific resistors. I will spend some more time with LT Spice and
see what I come up with.

** Go ahead - waste your time completely.



Do you think playing with Spice is the wrong approach?

** What was your first clue?

If so, what
would be a better approach to understanding this?

** Do what I asked you for a start - instead of using your own wrong
opinions.




.... Phil

 

[George Herold]

On Monday, April 28, 2014 12:39:54 PM UTC-4, Phil Hobbs wrote:

On 04/28/2014 11:06 AM, Gone Postal wrote:

On Mon, 28 Apr 2014 10:13:46 -0400, Phil Hobbs

snip previous stuff

There's also noise to worry about, of course. Your garden variety op
amp has an input noise voltage of very roughly 10 nV/sqrt(Hz), which is
about the same as the thermal (Johnson) noise of a 6k resistor. So
going too high will cost you voltage noise performance.

The other nice thing about 10 k ohm, is you can put a 1/4W 10k across
the +/- 15V rail and *not* let the magic smoke out.
And 10 k is very close the the "quantum" of conductance.
2*e**2/h =~ 1/12.9k ohm) Coincidence? I don't think so.
I think God also likes ~10k ohm :^)
(please excuse the extreme hubris in the above.)

George H.

Cheers



Phil Hobbs



--

Dr Philip C D Hobbs

Principal Consultant

ElectroOptical Innovations LLC

Optics, Electro-optics, Photonics, Analog Electronics



160 North State Road #203

Briarcliff Manor NY 10510



hobbs at electrooptical dot net

http://electrooptical.net

 

[Gone Postal]

On Mon, 28 Apr 2014 06:04:10 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

On Monday, April 28, 2014 12:44:23 AM UTC-4, Gone Postal wrote:
I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved? I've seen different circuits with resistors of, say
thousands, tens or hundered of thousands and even one that used
Megohm-range resistors.
Why the variation in magnitude of the resistors, assuming a fixed
ratio? Does it have anything to do with the overall impedance of the
opamp circuit block?
I thank you for whatever illumination that can be provided.

I think 10k ohm resistors are my favorite. (then 1 k ohm)
In general I'd like to choose R's that are smaller.
First this tends to reduce the effects of stray capacitance.
(Some RC corner that move to higher frequency with smaller R)
And then if you care, smaller R has less noise.. though most of the time that doesn't matter much.
Now of course you can't make 'em too small.
First the opamp mya not have enough poop to drive it.
and second if the R's get very small (a few ohms) you then
start to worry about the resistance of the traces and contacts.

George H.


GP

Thank you George. I've been following that as a general rule of thumb
(staying in the huuped singled K and the general 10's of Kohm range),
but I was seeing circuits that had other orders of magnatude and I was
just starting to wonder why so high, in particular with two of three
circuits I saw that just struck me as a wee bid odd..

GP

 

[Phil Allison]

"Gone Postal = Wanker "

Thank you George. I've been following that as a general rule of thumb
(staying in the huuped singled K and the general 10's of Kohm range),
but I was seeing circuits that had other orders of magnatude and I was
just starting to wonder why so high, in particular with two of three
circuits I saw that just struck me as a wee bid odd..

** FFS - why not post links to them ????

You pompous, boring, illiterate, anonymous fart.




..... Phil

 

[Jim Thompson]

On 28 Apr 2014 06:13:27 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2014-04-28, Gone Postal <gone_postal@it.doesn't.exist> wrote:
I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved?

mainly the impedance of the input.
the imput will have high resistance but not infinite.

At reasonable gains (well below the gain-bandwidth product of the
OpAmp), input impedance matters not at all. But...

(1) What can matter in bipolar OpAmps is input bias current, which can
cause an effective offset voltage if resistors are too large.

(2) Also, with very large feedback resistors, the stray capacitances
of the circuit can cause undesirable frequency roll-off.

(3) Too small valued resistors will tax the current output capability
of the OpAmp.

...Jim Thompson
--
| James E.Thompson | mens |
| Analog Innovations | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| San Tan Valley, AZ 85142 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[Phil Hobbs]

On 04/28/2014 03:06 AM, Gone Postal wrote:

On 28 Apr 2014 06:13:27 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2014-04-28, Gone Postal <gone_postal@it.doesn't.exist> wrote:
I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved?

mainly the impedance of the input.
the imput will have high resistance but not infinite.


Thanks for the answer, that's what I was generally getting at.

Capacitance is very important as well, both input and feedback.

Op amps always have at least 1.5 pF of input capacitance, and usually
more than that. Say you have a unity gain inverter with 1 meg input and
feedback resistors and Cin = 3pF.

Your feedback network will roll off starting at

f_c = 1/(2 pi (3pF) (500k)) = 106 kHz.

That means that the output swing will start to rise there, to keep the
input in balance. So you wind up with a gain peak. If you had picked
10k instead of 1 meg, the peak would start 100 times higher in
frequency, i.e. 10.6 MHz, which is liable to be outside the amplifier
bandwidth, so you wouldn't see it unless the op amp were reasonably fast.


The shunt capacitance of the feedback resistor is also important, though
it doesn't really matter for unity gain inverters, since the effect is
about the same on each one. It's typically about 0.12 pF for a 1/4 W
axial resistor and 0.05 pF for an 0603. (You also have to watch out for
the capacitance between pads--a nearby ground plane is your friend here.)

However, if you had an amp with a gain of -100, i.e. 1 meg on the input
and 100 meg feedback, that 0.05 pF starts to dominate the feedback at

f_RC = 1/(2 pi (50 fF) 100 meg) = 32 kHz.

Layout problems will make this worse.

Adjusting the feedback capacitance to cancel out the effect of the input
capacitance is a typical way of controlling these problems when you
can't just use lower impedances.

Cheers

Phil Hobbs




--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

 

[Gone Postal]

On Mon, 28 Apr 2014 10:13:46 -0400, Phil Hobbs
<pcdhSpamMeSenseless@electrooptical.net> wrote:

On 04/28/2014 03:06 AM, Gone Postal wrote:
On 28 Apr 2014 06:13:27 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2014-04-28, Gone Postal <gone_postal@it.doesn't.exist> wrote:
I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved?

mainly the impedance of the input.
the imput will have high resistance but not infinite.


Thanks for the answer, that's what I was generally getting at.

Capacitance is very important as well, both input and feedback.

Op amps always have at least 1.5 pF of input capacitance, and usually
more than that. Say you have a unity gain inverter with 1 meg input and
feedback resistors and Cin = 3pF.

Your feedback network will roll off starting at

f_c = 1/(2 pi (3pF) (500k)) = 106 kHz.

That means that the output swing will start to rise there, to keep the
input in balance. So you wind up with a gain peak. If you had picked
10k instead of 1 meg, the peak would start 100 times higher in
frequency, i.e. 10.6 MHz, which is liable to be outside the amplifier
bandwidth, so you wouldn't see it unless the op amp were reasonably fast.


The shunt capacitance of the feedback resistor is also important, though
it doesn't really matter for unity gain inverters, since the effect is
about the same on each one. It's typically about 0.12 pF for a 1/4 W
axial resistor and 0.05 pF for an 0603. (You also have to watch out for
the capacitance between pads--a nearby ground plane is your friend here.)

However, if you had an amp with a gain of -100, i.e. 1 meg on the input
and 100 meg feedback, that 0.05 pF starts to dominate the feedback at

f_RC = 1/(2 pi (50 fF) 100 meg) = 32 kHz.

Layout problems will make this worse.

Adjusting the feedback capacitance to cancel out the effect of the input
capacitance is a typical way of controlling these problems when you
can't just use lower impedances.

Cheers

Phil Hobbs

Thank you Phil Hobbs, precisely some of what I'm looking for! Thank
you!

GP

 

[Gone Postal]

On Mon, 28 Apr 2014 07:08:47 -0700, Jim Thompson
<To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:

On 28 Apr 2014 06:13:27 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2014-04-28, Gone Postal <gone_postal@it.doesn't.exist> wrote:
I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved?

mainly the impedance of the input.
the imput will have high resistance but not infinite.

At reasonable gains (well below the gain-bandwidth product of the
OpAmp), input impedance matters not at all. But...

(1) What can matter in bipolar OpAmps is input bias current, which can
cause an effective offset voltage if resistors are too large.

(2) Also, with very large feedback resistors, the stray capacitances
of the circuit can cause undesirable frequency roll-off.

(3) Too small valued resistors will tax the current output capability
of the OpAmp.

...Jim Thompson

Thank you Jim Thompson, the rest of what I'm looking for!

GP

 

[John Fields]

On Mon, 28 Apr 2014 00:44:23 -0400, Gone Postal
<gone_postal@it.doesn't.exist> wrote:

I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved? I've seen different circuits with resistors of, say
thousands, tens or hundered of thousands and even one that used
Megohm-range resistors.
Why the variation in magnitude of the resistors, assuming a fixed
ratio? Does it have anything to do with the overall impedance of the
opamp circuit block?
I thank you for whatever illumination that can be provided.

GP

---
Using a fixed-pitch font to display the basic configurations for
inverting and non-inverting opamps we have, for the supply
connections:

.. V+
.. |
.. --|-\
.. | >--
.. --|+/
.. |
.. V-

and for the inverting configuration:

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


Note that with the non-inverting (+) input at 0V, then Vout must
assume whatever value is required to force the - input to whatever
voltage is on the + input; 0V.

If Vin is at 1V, and if R1 = R2, then we have a voltage divider that
looks like this:

1V
|
[R1]
|
+---0V
|
[R2]
|
-1V

where -1V is the output of the opamp and 0V is the opamp's - input.

Accordingly, as the input voltage swings positive and negative, then
the output voltage will swing with the same magnitude as the input,
but 180 degrees out of phase.

The circuit, then, has a voltage gain of -1.


Assuming, now, that the signal generator driving R1 has an output
impedance of 10kohms and that R1 and R2 are each 10k, our circuit
now looks like this:

.. SIGNAL 10k
.. GENERATOR +--[R2]--+
.. +-----------+ 10k | |
.. | +--[Rg]--|--[R1]-+--|-\ |
.. | | 10k | | >--+-->Vout
.. | OSC | +--|+/
.. +-----------+ |
.. GND

If the signal generator has an open-circuit output of 1 volt, when
it's connected to R1 (since the end of R1 connected to the minus
input of the opamp will be at 0V) the current through RgR1 will be:

E 1V
I = --------- = ------ = 50 microamps
Rg + R1 20kR


Now, since Rg, R1, and R2 are in series and the current into the -
input of the opamp is miniscule, the current through Rg and R1 also
flows through R2.

Then, since R2 = 10kR and the current through it is 50 microamps,
the voltage across it must be:


E = IR = 50ľA * 10kR = 0.5V

Therefore, even though R1 = R2, the gain of the [entire] circuit
won't be -1 because of the effect of R1 loading the source.

Knowing that, and taking the source impedance into consideration,
the resistances of R1 and R2 can be selected so that the opamp's
output voltage will be whatever's desired.

The resistor selection criteria for the non-inverting amp are
arguably less stringent with, roughly, not starving the opamp -
input for bias current on one end and minimizing the noise into the
opamp on the other.

John Fields

 

[Phil Hobbs]

On 04/28/2014 11:06 AM, Gone Postal wrote:

On Mon, 28 Apr 2014 10:13:46 -0400, Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> wrote:

On 04/28/2014 03:06 AM, Gone Postal wrote:
On 28 Apr 2014 06:13:27 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2014-04-28, Gone Postal <gone_postal@it.doesn't.exist> wrote:
I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved?

mainly the impedance of the input.
the imput will have high resistance but not infinite.


Thanks for the answer, that's what I was generally getting at.

Capacitance is very important as well, both input and feedback.

Op amps always have at least 1.5 pF of input capacitance, and usually
more than that. Say you have a unity gain inverter with 1 meg input and
feedback resistors and Cin = 3pF.

Your feedback network will roll off starting at

f_c = 1/(2 pi (3pF) (500k)) = 106 kHz.

That means that the output swing will start to rise there, to keep the
input in balance. So you wind up with a gain peak. If you had picked
10k instead of 1 meg, the peak would start 100 times higher in
frequency, i.e. 10.6 MHz, which is liable to be outside the amplifier
bandwidth, so you wouldn't see it unless the op amp were reasonably fast.


The shunt capacitance of the feedback resistor is also important, though
it doesn't really matter for unity gain inverters, since the effect is
about the same on each one. It's typically about 0.12 pF for a 1/4 W
axial resistor and 0.05 pF for an 0603. (You also have to watch out for
the capacitance between pads--a nearby ground plane is your friend here.)

However, if you had an amp with a gain of -100, i.e. 1 meg on the input
and 100 meg feedback, that 0.05 pF starts to dominate the feedback at

f_RC = 1/(2 pi (50 fF) 100 meg) = 32 kHz.

Layout problems will make this worse.

Adjusting the feedback capacitance to cancel out the effect of the input
capacitance is a typical way of controlling these problems when you
can't just use lower impedances.

Cheers

Phil Hobbs

Thank you Phil Hobbs, precisely some of what I'm looking for! Thank
you!

GP

There's also noise to worry about, of course. Your garden variety op
amp has an input noise voltage of very roughly 10 nV/sqrt(Hz), which is
about the same as the thermal (Johnson) noise of a 6k resistor. So
going too high will cost you voltage noise performance.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

 

[Gone Postal]

On Mon, 28 Apr 2014 10:30:15 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Mon, 28 Apr 2014 00:44:23 -0400, Gone Postal
gone_postal@it.doesn't.exist> wrote:

I have a question about the magnitude of the two feedback resistors on
opamps. I am familiar with how to determine the ratio of the
resistors to get the desired gain, whether inverting or not. What I'm
trying to figure out is what determines the magnitude of the resistors
involved? I've seen different circuits with resistors of, say
thousands, tens or hundered of thousands and even one that used
Megohm-range resistors.
Why the variation in magnitude of the resistors, assuming a fixed
ratio? Does it have anything to do with the overall impedance of the
opamp circuit block?
I thank you for whatever illumination that can be provided.

GP

---
Using a fixed-pitch font to display the basic configurations for
inverting and non-inverting opamps we have, for the supply
connections:

. V+
. |
. --|-\
. | >--
. --|+/
. |
. V-

and for the inverting configuration:

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


Note that with the non-inverting (+) input at 0V, then Vout must
assume whatever value is required to force the - input to whatever
voltage is on the + input; 0V.

If Vin is at 1V, and if R1 = R2, then we have a voltage divider that
looks like this:

1V
|
[R1]
|
+---0V
|
[R2]
|
-1V

where -1V is the output of the opamp and 0V is the opamp's - input.

Accordingly, as the input voltage swings positive and negative, then
the output voltage will swing with the same magnitude as the input,
but 180 degrees out of phase.

The circuit, then, has a voltage gain of -1.


Assuming, now, that the signal generator driving R1 has an output
impedance of 10kohms and that R1 and R2 are each 10k, our circuit
now looks like this:

. SIGNAL 10k
. GENERATOR +--[R2]--+
. +-----------+ 10k | |
. | +--[Rg]--|--[R1]-+--|-\ |
. | | 10k | | >--+-->Vout
. | OSC | +--|+/
. +-----------+ |
. GND

If the signal generator has an open-circuit output of 1 volt, when
it's connected to R1 (since the end of R1 connected to the minus
input of the opamp will be at 0V) the current through RgR1 will be:

E 1V
I = --------- = ------ = 50 microamps
Rg + R1 20kR


Now, since Rg, R1, and R2 are in series and the current into the -
input of the opamp is miniscule, the current through Rg and R1 also
flows through R2.

Then, since R2 = 10kR and the current through it is 50 microamps,
the voltage across it must be:


E = IR = 50ľA * 10kR = 0.5V

Therefore, even though R1 = R2, the gain of the [entire] circuit
won't be -1 because of the effect of R1 loading the source.

Knowing that, and taking the source impedance into consideration,
the resistances of R1 and R2 can be selected so that the opamp's
output voltage will be whatever's desired.

The resistor selection criteria for the non-inverting amp are
arguably less stringent with, roughly, not starving the opamp -
input for bias current on one end and minimizing the noise into the
opamp on the other.

John Fields

John Fields, thank you very, very much for that thoughtful
explanation, it covers exactly what I was wanting to know but didn't
quite have the knowledge to ask correctly.

I had thought I understood op amps thoroughly, having used them in
some light-weight audio level circuitry in the past. I'm trying to
thoroughly understand them as I try to understand everything else,
before I embark on designing anything substantial. The most I've
managed to design and build from the ground up so far was a power
supply with current fold-back, from discrete parts, but it was a seat
of the pants design. I want to understand what's really happening in
a circuit so I can understand more or less what to expect and
understand why I expect it before I ever breadboard anything.

And thank you for treating a basic problem from a first time asker for
what it was.
Gone Postal

 

[Gone Postal]

On Mon, 28 Apr 2014 23:25:50 +1000, "Phil Allison" <phil_a@tpg.com.au>
wrote:

"Gone Postal = Wanker "


Thank you George. I've been following that as a general rule of thumb
(staying in the huuped singled K and the general 10's of Kohm range),
but I was seeing circuits that had other orders of magnatude and I was
just starting to wonder why so high, in particular with two of three
circuits I saw that just struck me as a wee bid odd..


** FFS - why not post links to them ????

You pompous, boring, illiterate, anonymous fart.




.... Phil








Because they're in Books and you can't link to a book.

GP