P
Phil Hobbs
Guest
On 4/28/2014 9:39 PM, Ray Otwell wrote:
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
On Mon, 28 Apr 2014 18:44:31 -0400, Phil Hobbs
hobbs@electrooptical.net> wrote:
On 4/28/2014 5:29 PM, Gone Postal wrote:
On Mon, 28 Apr 2014 12:05:12 -0700, Jim Thompson
To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:
On Mon, 28 Apr 2014 12:54:20 -0400, Gone Postal
gone_postal@it.doesn't.exist> wrote:
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
Try and find a copy of Tom Frederiksen's booklet, "Intuitive OpAmps".
That should answer about any remaining questions you have.
...Jim Thompson
Thank you for your assistance!
GP
P.S. Is it really necessary to go by my real name in this group? I
prefer to try and keep Google blank when someone searches my real
name, though s.e.b. isn't a bad group to be associated with I guess. I
don't nym-shift without a really good reason, like the debacle in
a.b.e.t. last year.
Going by your real name is actually a big plus on Usenet, unless you
have self-control issues. Some not-insignificant amount of my business
comes from people looking at Usenet archives. Besides, as the old
saying goes, "tell the truth and shame the devil."
Cheers
Phil Hobbs
Okay, here goes nothing!
And a thanks to everyone that has helped today.
Ray
Welcome aboard, Ray!
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