OT: Is this question too challenging for a BSEE graduate?

[krw@att.bizzzzzzzzzzzz]

On Tue, 26 Oct 2010 10:53:32 -0700, "Joel Koltner"
<zapwireDASHgroups@yahoo.com> wrote:

"David Nebenzahl" <nobody@but.us.chickens> wrote in message
news:4cc67064$0$2444$822641b3@news.adtechcomputers.com...
What you might ought have said is that the *circuit*, including the feedback
loop, forces the inverting input to (virtually) the same voltage as the
noninverting input, right? The op amp, in and of itself, doesn't "do"
anything to (that is, out of) either input. It's only by virtue of the
feedback that this action occurs.

Yes... and of course you have to get the feedback "right" as well (negative
for the simple sorts of applications we're discussing here) -- the astute EE
101 student will point out that using the rules about infinite input
impedances and the inverting/non-inverting voltages being the same, you could
swap the inverting and non-inverting inputs and everything should still work,
yes?

Positive infinity is equal to negative infinity on your planet? ;-)

At least when I took the appropriate course, it was only about a week or so
between "here's the absolutely ideal op-amp model and use these rules to
figure out the gain" and "here's a real-world op-amp with finite gain" and
then a few more days to "...and finite frequency response, and offset
voltages, etc." -- so you didn't have to feel uneasy about the initial
hand-waving for too long.

I don't think I ever took a formal class using opamps. I learned about them
in a "special problems" classes, first with (tube/servo-multiplier) analog
computers then the IC versions.

 

[Joel Koltner]

<krw@att.bizzzzzzzzzzzz> wrote in message
news:16lec65ml3jnmcetseu7ffp3ustlmtlupm@4ax.com...

Positive infinity is equal to negative infinity on your planet? ;-)

Something like that...

Someone here once posted a link to a commercial op-amp data sheet (Maxim?)
where the example circuit had +/- swapped!

I don't think I ever took a formal class using opamps. I learned about them
in a "special problems" classes, first with (tube/servo-multiplier) analog
computers then the IC versions.

At the U. of Wisc where I was, it was covered in the first real EE class you
took. It went something like... DC analysis (Ohm, Norton, Thevenin), All
About Phasors, Op-Amps, general AC circuit analysis, and finally RC/RL/RLC
circuits (although the letter "Q" never came up once -- that was for later).

The teacher was a very good teacher, although unfortunately his knowledge
didn't extend much outside of the textbook. Strangely, he was a full tenured
guy, whereas the much more "practical" fellow teaching the "Circuits You Might
Actually Find Yourself Building In The Real World" classes missed obtaining
tenure several times and eventually retired early.

I occasionally wonder whatever became of some of my old professors and
teachers; one looks at them so differently when you're in school than once
you've been out in the working world for awhile.

---Joel

 

[krw@att.bizzzzzzzzzzzz]

On Tue, 26 Oct 2010 16:20:42 -0700, "Joel Koltner"
<zapwireDASHgroups@yahoo.com> wrote:

krw@att.bizzzzzzzzzzzz> wrote in message
news:16lec65ml3jnmcetseu7ffp3ustlmtlupm@4ax.com...
Positive infinity is equal to negative infinity on your planet? ;-)

Something like that...

Someone here once posted a link to a commercial op-amp data sheet (Maxim?)
where the example circuit had +/- swapped!

You trust the applications sections on datasheets?

I don't think I ever took a formal class using opamps. I learned about them
in a "special problems" classes, first with (tube/servo-multiplier) analog
computers then the IC versions.

At the U. of Wisc where I was, it was covered in the first real EE class you
took. It went something like... DC analysis (Ohm, Norton, Thevenin), All
About Phasors, Op-Amps, general AC circuit analysis, and finally RC/RL/RLC
circuits (although the letter "Q" never came up once -- that was for later).

I think I'm a little older than you. ;-) The 709 was available but very few
used them at the time. The first classes were just RLC circuits, no active
components at all.

The teacher was a very good teacher, although unfortunately his knowledge
didn't extend much outside of the textbook. Strangely, he was a full tenured
guy, whereas the much more "practical" fellow teaching the "Circuits You Might
Actually Find Yourself Building In The Real World" classes missed obtaining
tenure several times and eventually retired early.

I had a great prof for my more advanced circuits classes. He was a full
professor, a friend of the family (my father was an EE prof), and my
(academic) boss when I was a tech. I also had him for well over 10% of my
credits (which wasn't supposed to happen). I took the "special problems"
classes (8 semester hours, IIRC, another no-no) with him, too.

He was a great guy, though not all liked him. He gave miserable exams, but
then curved them so simple mistakes weren't a disaster. He planned the exams
so he could finish them in an hour, so the average was 50-60. One transfer
student got pissed because he had an 80 on the first exam. He studied his ass
off for the second and got everything right. The prof *never* had a perfect
exam, so took a point off for penmanship. The guy went ballistic. ...much
too serious.

I occasionally wonder whatever became of some of my old professors and
teachers; one looks at them so differently when you're in school than once
you've been out in the working world for awhile.

This prof retired and went became a lawyer for his second career. :-/

 

[brent]

On Oct 26, 7:20 pm, "Joel Koltner" <zapwireDASHgro...@yahoo.com>
wrote:

At the U. of Wisc where I was, it was covered in the first real EE class you
took.  It went something like... DC analysis (Ohm, Norton, Thevenin), All
About Phasors, Op-Amps, general AC circuit analysis, and finally RC/RL/RLC
circuits (although the letter "Q" never came up once -- that was for later).

Q is for Qan't Quite Qalculate the Quiirky parts of the Quantity
Qorrectly

 

[Michael A. Terrell]

PlainBill47@yawho.com wrote:

On Sun, 24 Oct 2010 21:14:00 -0400, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


PlainBill47@yawho.com wrote:

On Sat, 23 Oct 2010 13:05:23 -0700, "William Sommerwerck"
grizzledgeezer@comcast.net> wrote:

Any electronics technician should be able to solve this
by inspection; no calculator necessary. 3K/1k = x/40,
therefore x= 120 ohms. Come up with more difficult ones
next time.

What do you mean by "inspection"? Are you applying a formula you memorized?
Or do you /understand/ what's involved?
Probably better than you do. The voltage across R1 is 1/4 of +VDC. An op-amp tries to force both inputs to the same voltage. Since it was stipulated the op-amp is a 'classic, ideal' op amp, we can assume it has none of the defects found in the real world. As a result the voltage across R3 will also be 1/4 of VDC. The only way that can happen is if the effective resistance of Q1 is 3 times the resistance of R3, or 120 ohms.

NOW, what is less certain is the proper answer to the problem
"Calculate the equivalent resistance of this programmable load."
Given that R1, R2, and R3 are all part of the load, the proper answer
to the original diagram is 153.846 ohms. Except that circuit does not
show any evidence of being 'programmable'.


You 'program' it by changing reistors.
No kidding? You replace a simple variable power resistor, which only
requires a screw driver to change the resistance with three resistors,
an op-amp (which requires a separate power supply), and a mosfet. To
change the value of the virtual resistor you have to change a
resistor?

It would seem to me a potentiometer would improve usability greatly.

It would seem to me you aren't familiar with a customized BOM or
SIT. I worked with both in manufacturing.


--
Politicians should only get paid if the budget is balanced, and there is
enough left over to pay them.

 

[David Nebenzahl]

On 10/26/2010 12:05 PM John Fields spake thus:

On Tue, 26 Oct 2010 11:18:58 -0700, David Nebenzahl
nobody@but.us.chickens> wrote:

On 10/26/2010 10:53 AM Joel Koltner spake thus:

At least when I took the appropriate course, it was only about a
week or so between "here's the absolutely ideal op-amp model and
use these rules to figure out the gain" and "here's a real-world
op-amp with finite gain" and then a few more days to "...and
finite frequency response, and offset voltages, etc." -- so you
didn't have to feel uneasy about the initial hand-waving for too
long.

Yep. That stuff about the ideal op-amp--infinite gain, infinite
input impedance, zero output impedance, bandwidth to the far edges
of the electromagnetic spectrum--makes it sound like a
perpetual-motion machine ...

David,

Did you get the article I posted for you?

No; what article? (I take it this is the real John Fields, right?)


--
The fashion in killing has an insouciant, flirty style this spring,
with the flaunting of well-defined muscle, wrapped in flags.

- Comment from an article on Antiwar.com (http://antiwar.com)

 

[Michael A. Terrell]

tm wrote:

"David Nebenzahl" <nobody@but.us.chickens> wrote in message
news:4cc71c48$0$2445$822641b3@news.adtechcomputers.com...
On 10/26/2010 9:57 AM tm spake thus:

PlainBill47@yawho.com> wrote in message
news:nvudc65qpifjekd8bv7o0hr4grtc3slntk@4ax.com...

On Sun, 24 Oct 2010 21:14:00 -0400, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:

You 'program' it by changing reistors.

No kidding? You replace a simple variable power resistor, which
only requires a screw driver to change the resistance with three
resistors, an op-amp (which requires a separate power supply), and
a mosfet. To change the value of the virtual resistor you have to
change a resistor?

It would seem to me a potentiometer would improve usability
greatly.

Geese. It was just a quiz to see if an applicant understood how an opamp
works.

Yabbut, it says right there on the diagram "programmable load". So is it
or isn't it? To me, "programmable" means (or at least implies) changeable
by changing voltages or some other electronic parameter, not by physically
substituting components. Yes, a potentiometer would seem to be a better
choice--even if it is "just a quiz".

BTW, you didn't get the job.

I didn't want it anyway.


I was just pulling your chain. It was a shitty job anyway. Imagine working
for someone
that asked you that question on a job interview. I would more like to be
asked what
have I done that made someone some money. It's just business anyway.

Imagine the interview when a director of engineering asks "Can you
design an Op Amp" when applying for a job as a component engineer when
he meant, "Can you design a circuit using an op amp." A few days prior
to that he refused an ECO to correct a problem with an OpAmp that i had
submitted. Not because it was wrong, but because it was one of his
designs. I had to go to management to get it signed off.


How the same director of engineering giving a 'class' in the
lunchroom and using a microphone in his 'Power Point' presentation as a
video source for a telemetry system?


How about his proposed new inventory system that listed passive
components by their value. We had over a dozen 10K resistors in
inventory, but his system would have assigned a single part number to
all of them. The interview was over when I pointed out the flaws in his
system, nd his decision that there only be one stock number for an IC
available in different speeds, packages and multiple vendors.


--
Politicians should only get paid if the budget is balanced, and there is
enough left over to pay them.

 

[tm]

"Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message
news:3YSdndK57NrYGFrRnZ2dnUVZ_qWdnZ2d@earthlink.com...

PlainBill47@yawho.com wrote:

On Sun, 24 Oct 2010 21:14:00 -0400, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


PlainBill47@yawho.com wrote:

On Sat, 23 Oct 2010 13:05:23 -0700, "William Sommerwerck"
grizzledgeezer@comcast.net> wrote:

Any electronics technician should be able to solve this
by inspection; no calculator necessary. 3K/1k = x/40,
therefore x= 120 ohms. Come up with more difficult ones
next time.

What do you mean by "inspection"? Are you applying a formula you
memorized?
Or do you /understand/ what's involved?
Probably better than you do. The voltage across R1 is 1/4 of +VDC.
An op-amp tries to force both inputs to the same voltage. Since it
was stipulated the op-amp is a 'classic, ideal' op amp, we can assume
it has none of the defects found in the real world. As a result the
voltage across R3 will also be 1/4 of VDC. The only way that can
happen is if the effective resistance of Q1 is 3 times the resistance
of R3, or 120 ohms.

NOW, what is less certain is the proper answer to the problem
"Calculate the equivalent resistance of this programmable load."
Given that R1, R2, and R3 are all part of the load, the proper answer
to the original diagram is 153.846 ohms. Except that circuit does not
show any evidence of being 'programmable'.


You 'program' it by changing reistors.
No kidding? You replace a simple variable power resistor, which only
requires a screw driver to change the resistance with three resistors,
an op-amp (which requires a separate power supply), and a mosfet. To
change the value of the virtual resistor you have to change a
resistor?

It would seem to me a potentiometer would improve usability greatly.


It would seem to me you aren't familiar with a customized BOM or
SIT. I worked with both in manufacturing.

Ok, I get it now. It was for a cost plus government job.

tm

 

[ehsjr]

David Nebenzahl wrote:

On 10/25/2010 10:37 PM ehsjr spake thus:

I think you might be missing the fact that the circuit has feedback,
from the output to the inverting input. Take a look at the schematic
again.

The op amp "sees" a voltage on the + input and does whatever it can
to make the - input the same voltage. The output of the op amp is
connected back to the - input, so when the op amp raises or lowers
the voltage on the output pin, that voltage appears on the - input.
Thus, if you put X volts on the non-inverting (+) input, you'll get
X volts on the inverting (-) input.


Now that I understand things a little better, yes, I do get the feedback
here, and in other op amp circuits.

But just a small quibble with the way you and others have described
what's going on here. You say "the op amp ... does whatever it can to
make the - input the same voltage" (as the + input). In fact, it does no
such thing: the input is, after all, just an input.

In fact, _it most certainly does_. Without external connection - the
feedback - it will fail to make the inputs equal. But it will do all
it can do until those voltages are equal.

You seem to have missed the fact that my entire post was discussing
feedback. Sigh.

Ed

What you might ought have said is that the *circuit*, including the
feedback loop, forces the inverting input to (virtually) the same
voltage as the noninverting input, right? The op amp, in and of itself,
doesn't "do" anything to (that is, out of) either input. It's only by
virtue of the feedback that this action occurs.

Maybe just a semantic quibble. Other than that I'm with you here. Thanks
to all for explaining.

 

[David Nebenzahl]

On 10/26/2010 8:06 PM ehsjr spake thus:

David Nebenzahl wrote:

But just a small quibble with the way you and others have described
what's going on here. You say "the op amp ... does whatever it can to
make the - input the same voltage" (as the + input). In fact, it does no
such thing: the input is, after all, just an input.

In fact, _it most certainly does_. Without external connection - the
feedback - it will fail to make the inputs equal. But it will do all
it can do until those voltages are equal.

You mean the little op amp will huff and puff and turn blue? Not sure
what you're getting at here. Without feedback, the inputs will simply be
whatever they are.

You seem to have missed the fact that my entire post was discussing
feedback. Sigh.

See the remainder of my reply that you posted below but didn't reply to.

What you might ought have said is that the *circuit*, including the
feedback loop, forces the inverting input to (virtually) the same
voltage as the noninverting input, right? The op amp, in and of itself,
doesn't "do" anything to (that is, out of) either input. It's only by
virtue of the feedback that this action occurs.

--
The fashion in killing has an insouciant, flirty style this spring,
with the flaunting of well-defined muscle, wrapped in flags.

- Comment from an article on Antiwar.com (http://antiwar.com)

 

[Michael A. Terrell]

tm wrote:

"Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message
news:3YSdndK57NrYGFrRnZ2dnUVZ_qWdnZ2d@earthlink.com...

PlainBill47@yawho.com wrote:

On Sun, 24 Oct 2010 21:14:00 -0400, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


PlainBill47@yawho.com wrote:

On Sat, 23 Oct 2010 13:05:23 -0700, "William Sommerwerck"
grizzledgeezer@comcast.net> wrote:

Any electronics technician should be able to solve this
by inspection; no calculator necessary. 3K/1k = x/40,
therefore x= 120 ohms. Come up with more difficult ones
next time.

What do you mean by "inspection"? Are you applying a formula you
memorized?
Or do you /understand/ what's involved?
Probably better than you do. The voltage across R1 is 1/4 of +VDC.
An op-amp tries to force both inputs to the same voltage. Since it
was stipulated the op-amp is a 'classic, ideal' op amp, we can assume
it has none of the defects found in the real world. As a result the
voltage across R3 will also be 1/4 of VDC. The only way that can
happen is if the effective resistance of Q1 is 3 times the resistance
of R3, or 120 ohms.

NOW, what is less certain is the proper answer to the problem
"Calculate the equivalent resistance of this programmable load."
Given that R1, R2, and R3 are all part of the load, the proper answer
to the original diagram is 153.846 ohms. Except that circuit does not
show any evidence of being 'programmable'.


You 'program' it by changing reistors.
No kidding? You replace a simple variable power resistor, which only
requires a screw driver to change the resistance with three resistors,
an op-amp (which requires a separate power supply), and a mosfet. To
change the value of the virtual resistor you have to change a
resistor?

It would seem to me a potentiometer would improve usability greatly.


It would seem to me you aren't familiar with a customized BOM or
SIT. I worked with both in manufacturing.


Ok, I get it now. It was for a cost plus government job.

No, you don't get it. Customized BOM allows lower costs by reusing
most of a design for different customer needs. SIT (Select In Test)
allows equipment to be built that exceeds the normal specifications of
the current state of the art. HP and Tektronix used to do it in a lot
of their products, as well. There was no 'cost plus' jobs for anyone.
We would get a request for bid, or inquiry to see if we could supply
what a customer needed. We would bid it, based on a current base model
and all required customization to meet their needs. If they agreed to
the price we built it and shipped it on schedule and on budget. It
didn't matter who the customer was. NASA, NOAA, the ESA or numerous
other customers in the Aerospace industry. Some of that equipment was
in use 24/7 for over 30 years with no repairs.

If you think that you can build state of the art microwave receivers
with off the shelf parts, you need help.


--
Politicians should only get paid if the budget is balanced, and there is
enough left over to pay them.

 

[tm]

"Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message
news:3sqdnbd75qeRNVrRnZ2dnUVZ_qWdnZ2d@earthlink.com...

tm wrote:

"Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message
news:3YSdndK57NrYGFrRnZ2dnUVZ_qWdnZ2d@earthlink.com...

PlainBill47@yawho.com wrote:

On Sun, 24 Oct 2010 21:14:00 -0400, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


PlainBill47@yawho.com wrote:

On Sat, 23 Oct 2010 13:05:23 -0700, "William Sommerwerck"
grizzledgeezer@comcast.net> wrote:

Any electronics technician should be able to solve this
by inspection; no calculator necessary. 3K/1k = x/40,
therefore x= 120 ohms. Come up with more difficult ones
next time.

What do you mean by "inspection"? Are you applying a formula you
memorized?
Or do you /understand/ what's involved?
Probably better than you do. The voltage across R1 is 1/4 of
+VDC.
An op-amp tries to force both inputs to the same voltage. Since it
was stipulated the op-amp is a 'classic, ideal' op amp, we can
assume
it has none of the defects found in the real world. As a result
the
voltage across R3 will also be 1/4 of VDC. The only way that can
happen is if the effective resistance of Q1 is 3 times the
resistance
of R3, or 120 ohms.

NOW, what is less certain is the proper answer to the problem
"Calculate the equivalent resistance of this programmable load."
Given that R1, R2, and R3 are all part of the load, the proper
answer
to the original diagram is 153.846 ohms. Except that circuit does
not
show any evidence of being 'programmable'.


You 'program' it by changing reistors.
No kidding? You replace a simple variable power resistor, which only
requires a screw driver to change the resistance with three resistors,
an op-amp (which requires a separate power supply), and a mosfet. To
change the value of the virtual resistor you have to change a
resistor?

It would seem to me a potentiometer would improve usability greatly.


It would seem to me you aren't familiar with a customized BOM or
SIT. I worked with both in manufacturing.


Ok, I get it now. It was for a cost plus government job.


No, you don't get it. Customized BOM allows lower costs by reusing
most of a design for different customer needs. SIT (Select In Test)
allows equipment to be built that exceeds the normal specifications of
the current state of the art. HP and Tektronix used to do it in a lot
of their products, as well. There was no 'cost plus' jobs for anyone.
We would get a request for bid, or inquiry to see if we could supply
what a customer needed. We would bid it, based on a current base model
and all required customization to meet their needs. If they agreed to
the price we built it and shipped it on schedule and on budget. It
didn't matter who the customer was. NASA, NOAA, the ESA or numerous
other customers in the Aerospace industry. Some of that equipment was
in use 24/7 for over 30 years with no repairs.

If you think that you can build state of the art microwave receivers
with off the shelf parts, you need help.

I really do get it. I see you never worked on any "black" projects.

BOM and SIT are EE-101 so don't try to be so impressive with what you
think you know.

And by the way, I am not trying to piss you off. I was just trying to be a
bit sarcastic. By making a simple "resistor" using five parts and calling
it a "programmable" load sure sounds like someone was selling a design
at cost plus.

Regards (really),
tm

 

[John Fields]

On Tue, 26 Oct 2010 18:50:13 -0700, David Nebenzahl
<nobody@but.us.chickens> wrote:

On 10/26/2010 12:05 PM John Fields spake thus:



Did you get the article I posted for you?

No; what article? (I take it this is the real John Fields, right?)

---
Yeah.

Here's the message ID:

68oac6du5ec8alsq27p63a3hehpdcdvn8q@4ax.com

If you can't get it let me know and I'll email you a copy.

---
JF

 

[Michael A. Terrell]

tm wrote:

"Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message
news:3sqdnbd75qeRNVrRnZ2dnUVZ_qWdnZ2d@earthlink.com...

tm wrote:

"Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message
news:3YSdndK57NrYGFrRnZ2dnUVZ_qWdnZ2d@earthlink.com...

PlainBill47@yawho.com wrote:

On Sun, 24 Oct 2010 21:14:00 -0400, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


PlainBill47@yawho.com wrote:

On Sat, 23 Oct 2010 13:05:23 -0700, "William Sommerwerck"
grizzledgeezer@comcast.net> wrote:

Any electronics technician should be able to solve this
by inspection; no calculator necessary. 3K/1k = x/40,
therefore x= 120 ohms. Come up with more difficult ones
next time.

What do you mean by "inspection"? Are you applying a formula you
memorized?
Or do you /understand/ what's involved?
Probably better than you do. The voltage across R1 is 1/4 of
+VDC.
An op-amp tries to force both inputs to the same voltage. Since it
was stipulated the op-amp is a 'classic, ideal' op amp, we can
assume
it has none of the defects found in the real world. As a result
the
voltage across R3 will also be 1/4 of VDC. The only way that can
happen is if the effective resistance of Q1 is 3 times the
resistance
of R3, or 120 ohms.

NOW, what is less certain is the proper answer to the problem
"Calculate the equivalent resistance of this programmable load."
Given that R1, R2, and R3 are all part of the load, the proper
answer
to the original diagram is 153.846 ohms. Except that circuit does
not
show any evidence of being 'programmable'.


You 'program' it by changing reistors.
No kidding? You replace a simple variable power resistor, which only
requires a screw driver to change the resistance with three resistors,
an op-amp (which requires a separate power supply), and a mosfet. To
change the value of the virtual resistor you have to change a
resistor?

It would seem to me a potentiometer would improve usability greatly.


It would seem to me you aren't familiar with a customized BOM or
SIT. I worked with both in manufacturing.


Ok, I get it now. It was for a cost plus government job.


No, you don't get it. Customized BOM allows lower costs by reusing
most of a design for different customer needs. SIT (Select In Test)
allows equipment to be built that exceeds the normal specifications of
the current state of the art. HP and Tektronix used to do it in a lot
of their products, as well. There was no 'cost plus' jobs for anyone.
We would get a request for bid, or inquiry to see if we could supply
what a customer needed. We would bid it, based on a current base model
and all required customization to meet their needs. If they agreed to
the price we built it and shipped it on schedule and on budget. It
didn't matter who the customer was. NASA, NOAA, the ESA or numerous
other customers in the Aerospace industry. Some of that equipment was
in use 24/7 for over 30 years with no repairs.

If you think that you can build state of the art microwave receivers
with off the shelf parts, you need help.


I really do get it. I see you never worked on any "black" projects.

Sigh.

BOM and SIT are EE-101 so don't try to be so impressive with what you
think you know.

And by the way, I am not trying to piss you off. I was just trying to be a
bit sarcastic. By making a simple "resistor" using five parts and calling
it a "programmable" load sure sounds like someone was selling a design
at cost plus.

Regards (really),
tm

--
Politicians should only get paid if the budget is balanced, and there is
enough left over to pay them.

 

[David]

Regarding the subject line: When I first read this question and
looked at the schematic, it was not clear to me what the
programmable load that was asked about actually was. If it was
the FET, that should have been stated. There were many responses
that treated the entire circuit as the 'programmable load' hence
the 'in parallel with 4K ohms' type of answers. Furthermore, if
we assume the question was about the FET being the load to
calculate, that resistance has a DC and AC component and they are
very different. A question like this is very poorly worded and
should in no way be a judge of the person attempting to answer
it.
David

 

[ehsjr]

David Nebenzahl wrote:

On 10/26/2010 8:06 PM ehsjr spake thus:

David Nebenzahl wrote:

But just a small quibble with the way you and others have described
what's going on here. You say "the op amp ... does whatever it can to
make the - input the same voltage" (as the + input). In fact, it does
no such thing: the input is, after all, just an input.


In fact, _it most certainly does_. Without external connection - the
feedback - it will fail to make the inputs equal. But it will do all
it can do until those voltages are equal.


You mean the little op amp will huff and puff and turn blue? Not sure
what you're getting at here. Without feedback, the inputs will simply be
whatever they are.

Do you understand that the maximum output voltage that op amps
can produce differs between devices? Some go to the rail(s)
(or very close) and others go toward the rail(s) but can end up
well more than a volt away.

Assuming unequal input voltages:
If there is no feedback, the op amp responds to the unequal input
voltages and drives the output to whatever the maximum voltage is
for that specific op amp, in whichever direction the voltages on
the inputs dictate. It continues responding and driving forever,
or until the voltages are made equal by some external means.

If there is feedback, the output will go to whatever voltage
causes the input voltages to be equal, or will go to maximum
and keep on responding and driving the output to maximum if
the circuit feedback is insufficient to equalize the voltages.


In all cases, the op amp has done all that it can. In the first
case what it can do is limited by its maximum output voltage.
In the second case, where the feedback allows the voltages to become
equal, the op amp has done all that it can, because the equal
input voltages limit the Vout to the specific voltage that
creates equal inputs. In the third case, the op amp is once
again limited by whatever the maximum (or minimum) Vout it can
produce.

"An op amp does all that it can" does not mean or imply that it
somehow internally adjusts the input voltages. It can't do that.
It does what it _can_. It does everything that the inputs
allow and that the op amps own internal design allows.

I hope that clears it up for you. You're likely to encounter
the phrase again and again. It would seem that the smart course
is to try to understand it, rather than engage in what you termed
a semantic quibble.

Ed

You seem to have missed the fact that my entire post was discussing
feedback. Sigh.


See the remainder of my reply that you posted below but didn't reply to.

What you might ought have said is that the *circuit*, including the
feedback loop, forces the inverting input to (virtually) the same
voltage as the noninverting input, right? The op amp, in and of
itself, doesn't "do" anything to (that is, out of) either input. It's
only by virtue of the feedback that this action occurs.

 

[David Nebenzahl]

On 10/27/2010 2:45 AM John Fields spake thus:

On Tue, 26 Oct 2010 18:50:13 -0700, David Nebenzahl
nobody@but.us.chickens> wrote:

On 10/26/2010 12:05 PM John Fields spake thus:

Did you get the article I posted for you?

No; what article? (I take it this is the real John Fields, right?)

Yeah.

Here's the message ID:

68oac6du5ec8alsq27p63a3hehpdcdvn8q@4ax.com

Thunderbird knows not what to do with that address, just opens up a new
"compose" window.

If you can't get it let me know and I'll email you a copy.

Thanks, but as you can see, my email address is (intentionally) munged.
Post it to web-space and give us a link?

Anyhow, thanks for the effort.


--
The fashion in killing has an insouciant, flirty style this spring,
with the flaunting of well-defined muscle, wrapped in flags.

- Comment from an article on Antiwar.com (http://antiwar.com)

 

[Michael A. Terrell]

David Nebenzahl wrote:

On 10/27/2010 2:45 AM John Fields spake thus:

On Tue, 26 Oct 2010 18:50:13 -0700, David Nebenzahl
nobody@but.us.chickens> wrote:

On 10/26/2010 12:05 PM John Fields spake thus:

Did you get the article I posted for you?

No; what article? (I take it this is the real John Fields, right?)

Yeah.

Here's the message ID:

68oac6du5ec8alsq27p63a3hehpdcdvn8q@4ax.com

Thunderbird knows not what to do with that address, just opens up a new
"compose" window.

Like John said, it is a message ID, not a URL. It will open the
related message on most newsreaders, if you have access to the group it
was posted to.


--
Politicians should only get paid if the budget is balanced, and there is
enough left over to pay them.

 

[David Nebenzahl]

On 10/27/2010 4:52 PM Michael A. Terrell spake thus:

David Nebenzahl wrote:

On 10/27/2010 2:45 AM John Fields spake thus:

On Tue, 26 Oct 2010 18:50:13 -0700, David Nebenzahl
nobody@but.us.chickens> wrote:

On 10/26/2010 12:05 PM John Fields spake thus:

Did you get the article I posted for you?

No; what article? (I take it this is the real John Fields,
right?)

Yeah.

Here's the message ID:

68oac6du5ec8alsq27p63a3hehpdcdvn8q@4ax.com

Thunderbird knows not what to do with that address, just opens up a
new "compose" window.

Like John said, it is a message ID, not a URL.

I understand that. I clicked on it only half-expecting it to do anything
useful.

It will open the related message on most newsreaders, if you have
access to the group it was posted to.

Well, I certainly have access, as I'm reading s.e.r. using Thunderbird.
But I don't know how to search by message ID. Do you, using Tbird? I
assume it has this capability, but the online help is, as usual, useless.


--
The fashion in killing has an insouciant, flirty style this spring,
with the flaunting of well-defined muscle, wrapped in flags.

- Comment from an article on Antiwar.com (http://antiwar.com)

 

[Michael A. Terrell]

David Nebenzahl wrote:

On 10/27/2010 4:52 PM Michael A. Terrell spake thus:

David Nebenzahl wrote:

On 10/27/2010 2:45 AM John Fields spake thus:

On Tue, 26 Oct 2010 18:50:13 -0700, David Nebenzahl
nobody@but.us.chickens> wrote:

On 10/26/2010 12:05 PM John Fields spake thus:

Did you get the article I posted for you?

No; what article? (I take it this is the real John Fields,
right?)

Yeah.

Here's the message ID:

68oac6du5ec8alsq27p63a3hehpdcdvn8q@4ax.com

Thunderbird knows not what to do with that address, just opens up a
new "compose" window.

Like John said, it is a message ID, not a URL.

I understand that. I clicked on it only half-expecting it to do anything
useful.

It will open the related message on most newsreaders, if you have
access to the group it was posted to.

Well, I certainly have access, as I'm reading s.e.r. using Thunderbird.
But I don't know how to search by message ID. Do you, using Tbird? I
assume it has this capability, but the online help is, as usual, useless.

I still use Netscape 4.78 to access newsgroups so I'm no help with
Thunderbird. I tried it last year and didn't like it.


--
Politicians should only get paid if the budget is balanced, and there is
enough left over to pay them.