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

[RadioJ]

Maybe, those students have a lot of material thrown at them in a short
time. I've met both techs and engineers out of school who either "get"
op-amp theory or not... it's really just common sense on the basics. For
me, it was one practical app (and it was in a job interview that I
flunked) to cause me to get my a$$ in gear and figure it out for good.

Haven't done this stuff in while (used to do elect design but no BSEE..
just an "almost" AS in EET), and was sick/bored today, so here's some
bloated math... are you giving your potential candidates a couple of
hours to give the answer in terms of the resistors?


For the over-all load on VDC:

Req = (R1 + R2) / (1 + (R1 / R3))

For just the output network:

Req = ((R1 + R2) * R3) / R1

---------------------------------
Proof-

Since non-inverting input and VR3 will be at the same voltage:

VR3 = (R1 * VDC) / (R1 + R2)

Solving for VDC:

VDC = (VR3 * (R1 + R2)) / R1

......

IR3 = VR3 / R3

Resistance just for the output side, since current through JFET and R3
are same:

Rout = VDC / IR3

Substituting from above:

Rout = ((VR3 * (R1 + R2)) / R1) / (VR3 / R3)

Rout = ((R1 + R2) * R3) / R1

Check:

Rout = ((1k + 3k) * 40) / 1k = 160 ohms

Considering the input network, Rin:

Rin = R1 + R2

For overall load, using conductance, easier to calculate parallel:

Geq = (1 / Rout) + (1 / Rin)

Geq = (R1 / ((R1 + R2) * R3)) + (1 / (R1 + R2))

Geq = 1 / ((1 / (R1 + R2)) * (1 + (R1 / R3)))

Req = 1 / Geq

Req = (R1 + R2) / (1 + (R1 / R3))

Check:

Req = (1k + 3k) / (1 + (1k / 40))

Req = 153.85

----------------------


On 10/22/2010 10:59 PM, RosemontCrest wrote:

I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&current=programmableload.jpg

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.

Thanks,
Brian

 

[RosemontCrest]

On Oct 23, 2:31 pm, "William Sommerwerck" <grizzledgee...@comcast.net>
wrote:

So if this is so all-fired obvious, why are we getting
different answers?
A. 153.8 ohms
B. 120 ohms
Both can't be correct, obviously. Which one is?

I don't mean to be offensive, but it you /understood/ the principal of
anaysis -- which is actual quite trivial -- there would be no question.

The resistance of the JFET itself is 120 ohms, which is what we're looking
for, regardless of how the question is phrased.

Sorry. Q1 is a MOSFET in the test I give applicants. I replicated the
circuit at home and erroneously picked the first N-channel FET I
found.

 

[Robert Baer]

John Fields wrote:

On Fri, 22 Oct 2010 19:59:39 -0700 (PDT), RosemontCrest
rosemontcrest@yahoo.com> wrote:

I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&current=programmableload.jpg

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.


Since R2R1 is a voltage divider, the voltage on U1+ will be:

+VDC * R1
U1+ = -----------
R1 + R3

Now, since the voltage on U1- must be equal to the voltage on U1+, and
since R(Q1)R3 is another voltage divider,

(+VDC * R3) - ((U1+) * R3)
R(Q1) = ----------------------------
U1+

For the value, since the ratio of R2:R1 = 3 and the voltage across R1
and R3 are equal, the ratio of R(Q1) to R3 must also be 3, making the
FET's resistance 3 * R3 = 120 ohms.

---
JF
Aww...i wanted to solve it without your hints.

But a quick look at the schematic makes it obvious that Req (of the
fet) is 120 ohms; took me a long 5 seconds (should have been less than
one second, so my 70+ years seems to be slowing me down).
Work? What work? ...

 

[isw]

In article <i9vome$jd1$1@news.eternal-september.org>,
"William Sommerwerck" <grizzledgeezer@comcast.net> wrote:

So if this is so all-fired obvious, why are we getting
different answers?
A. 153.8 ohms
B. 120 ohms
Both can't be correct, obviously. Which one is?

I don't mean to be offensive, but if you /understood/ the principle of
anaysis -- which is actually quite trivial -- there would be no question.

Well, duh. Obviously, I don't. Perhaps someday I may.


The resistance of the JFET itself is 120 ohms, which is what
we're looking for, regardless of how the question is phrased.

So I ask, perhaps naively: is this basically the same as finding
the Thévenin equivalent of the circuit?

No. In fact, one of the things the testee has to recognize is that this
question has /nothing/ to do with the Norton or Thévenin equivalents, and
you're only going to confuse the hell out of yourself if you go in that
direction.

The principle is this... In a stable op-amp circuit, the feedback forces the
inverting and non-inverting inputs /to have the same voltage/. The rest is
trivial arithmetic.

That statement is so significant, and so rarely understood...

Isaac

 

[ehsjr]

RosemontCrest wrote:

On Oct 22, 8:04 pm, "Michael A. Terrell" <mike.terr...@earthlink.net
wrote:

RosemontCrest wrote:


I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&curr...

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.

Then make your challenge on the proper newsgroup:
news:sci.electronics.design which is where you'll find the EEs.

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


Thank you Michael. I was not sure to which sci.electronics.* group I
should post. Thanks for adding sci.electronics.design to this thread.

Maybe it's too easy. It is instantly obvious that Rfet has to be
3 times R3. So, an applicant may think it's a trick question, and
be wracking his brains looking for the trick. OTOH, it could
eliminate those who are not confident enough in their understanding
of it to say "160 ohms in parallel with 4000 ohms" or "about 153.8
ohms".

Ed

 

[Tim Williams]

"ehsjr" <ehsjr@nospamverizon.net> wrote in message
news:ia0bhf$3os$1@news.eternal-september.org...

Maybe it's too easy. It is instantly obvious that Rfet has to be
3 times R3. So, an applicant may think it's a trick question, and
be wracking his brains looking for the trick. OTOH, it could
eliminate those who are not confident enough in their understanding
of it to say "160 ohms in parallel with 4000 ohms" or "about 153.8
ohms".

Don't be so optimistic. Kids have a lot of trouble with controlled
sources.

Tim

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

 

[William Sommerwerck]

The principle is this... In a stable op-amp circuit, the
feedback forces the inverting and non-inverting inputs
/to have the same voltage/. The rest is trivial arithmetic.

That statement is so significant, and so rarely understood...

Indeed. National Semiconductor used to have an on-line course in op-amp
circuit design, and this principle -- which should be the very first words
out of the instructor's mouth -- is nowhere stated. Shame on you, Bob, shame
on you.

In case the reason isn't obvious -- an ideal op-amp has infinite gain. If
there were /any/ voltage difference between the inverting and non-inverting
inputs, the op-amp's output would slam up against the positive or negative
rail.

In practice, an op-amp has finite gain (usually between 100K and 1000K).
This means the actual voltage difference has to be something other than
zero. But it's is still so close to zero that it can be ignored for the
purposes of analysis.

By the way, I cut my op-amp teeth nearly 40 years ago on the wonderful
Philbrick brook. One of the greatest pieces of technical writing ever (I
keep a copy for inspiration), and still a classic.

 

[RadioJ]

On 10/24/2010 7:07 AM, William Sommerwerck wrote:

The principle is this... In a stable op-amp circuit, the
feedback forces the inverting and non-inverting inputs
/to have the same voltage/. The rest is trivial arithmetic.

That statement is so significant, and so rarely understood...

Indeed. National Semiconductor used to have an on-line course in op-amp
circuit design, and this principle -- which should be the very first words
out of the instructor's mouth -- is nowhere stated. Shame on you, Bob, shame
on you.

In case the reason isn't obvious -- an ideal op-amp has infinite gain. If
there were /any/ voltage difference between the inverting and non-inverting
inputs, the op-amp's output would slam up against the positive or negative
rail.

In practice, an op-amp has finite gain (usually between 100K and 1000K).
This means the actual voltage difference has to be something other than
zero. But it's is still so close to zero that it can be ignored for the
purposes of analysis.

By the way, I cut my op-amp teeth nearly 40 years ago on the wonderful
Philbrick brook. One of the greatest pieces of technical writing ever (I
keep a copy for inspiration), and still a classic.

That "inputs the same thing" is good to think about when you run into
integrators and differentiators too.... for the I the output is just
having to drive against the cap none too quickly to keep the inputs from
pulling away from each other while charging it. For D, the output gets
slammed trying to kept up with the quick input change on the cap going
from the input driving through the cap. My simple-minded way of thinking
of it anyhow.

 

[Proteus]

"Michael A. Terrell" <mike.terrell@earthlink.net> wrote in message
news:HaKdnQ3UkO1WzV_RnZ2dnUVZ_vednZ2d@earthlink.com...

RosemontCrest wrote:

I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&current=programmableload.jpg

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.


Then make your challenge on the proper newsgroup:
news:sci.electronics.design which is where you'll find the EEs.


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

FREAK! STOP POLICING ASSHOLE.

YOUR JOHN FOOL AND YOU ARE ABOUT THE SAME IN TECHNOLOGY, JOHN IS GOOD AT
COPY AND PASTE HIS FUCKING COPIED CAPACITOR FORMULA. KEEP FOLLOWING EACH
OTHER SKUNK!

I AM PROTEUS

 

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'.

PlainBill

 

[Robert Baer]

ehsjr wrote:

RosemontCrest wrote:
On Oct 22, 8:04 pm, "Michael A. Terrell" <mike.terr...@earthlink.net
wrote:

RosemontCrest wrote:


I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&curr...


or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.

Then make your challenge on the proper newsgroup:
news:sci.electronics.design which is where you'll find the EEs.

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


Thank you Michael. I was not sure to which sci.electronics.* group I
should post. Thanks for adding sci.electronics.design to this thread.

Maybe it's too easy. It is instantly obvious that Rfet has to be
3 times R3. So, an applicant may think it's a trick question, and
be wracking his brains looking for the trick. OTOH, it could
eliminate those who are not confident enough in their understanding
of it to say "160 ohms in parallel with 4000 ohms" or "about 153.8
ohms".

Ed
OOOhhhhh; you want a trick question based on real life?

Ages ago i went to an interview for a job.
They insisted that they had d'Arsonval meters that slowly decreased
their sensitivity...toward zero (!!).
The claim was that the Alnico magnets slowly lost their magnetism.
I told them that was completely impossible; maybe a magnetic loss of
one percent in a century, but not 90% in weeks.
Did not get the job, because they "knew" they were right.
I later found out those meters were in series with the plate of the
RF final for their plywood dryer product line.
Bypass on the low side of the meter was an electrolytic capacitor.
Like in Groucho Mark's Bet your Life, when i heard that, the duck
came down!

 

[Robert Baer]

Tim Williams wrote:

"ehsjr" <ehsjr@nospamverizon.net> wrote in message
news:ia0bhf$3os$1@news.eternal-september.org...
Maybe it's too easy. It is instantly obvious that Rfet has to be
3 times R3. So, an applicant may think it's a trick question, and
be wracking his brains looking for the trick. OTOH, it could
eliminate those who are not confident enough in their understanding
of it to say "160 ohms in parallel with 4000 ohms" or "about 153.8
ohms".

Don't be so optimistic. Kids have a lot of trouble with controlled
sources.

Tim

...like a few Curies of radium?

 

[TheGlimmerMan]

On Sat, 23 Oct 2010 13:28:58 -0700 (PDT), linnix <me@linnix.info-for.us>
wrote:

On Oct 23, 1:25 pm, hamilton <hamil...@nothere.com> wrote:
On 10/23/2010 1:35 PM, linnix wrote:







On Oct 23, 12:27 pm, "William Sommerwerck"
grizzledgee...@comcast.net>  wrote:
I don't think this is a very good "quiz" question,
It is a good quiz for fun, but not a good interview question.
What does it prove? The candidate is a good quiz solver?
It has no real engineering use. Are they hiring quiz solver
or engineer?

It's a great interview question. It shows whether the candidate can cut
through the clutter and see the basic principle involved. A good engineer
needs to be able to do that.

But do you base your hiring/firing decision on whether he can do it
during the interview?  Most engineers can solve it in more relaxed
environment, in the real world.  It only proved that many can't do it
under pressure, during the interview.

Maybe the OP is showing the new hire what to expect from the new manager.

Silly tests like this shows how a manager will behave after being hired.

These tests work both ways.

h

Yes, usually they are sweat shops hiring slaves.

Not if you have a title.

 

[brent]

On Oct 22, 11:04 pm, "Michael A. Terrell" <mike.terr...@earthlink.net>
wrote:

RosemontCrest wrote:

I routinely use the following question to test candidates for EE or TE
positions. For many years, it continues to stump all but one of many.
Is it really that difficult to solve?

http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&curr....

or

http://preview.tinyurl.com/2c8udf9

I see two ways to solve it. The preferred method is to use Ohm's law
and nodal analysis without regard to the value of +VDC. Another method
is to assign a value to +VDC and solve it that way; I find that method
lame. Is this question, or test, too challenging for a BSEE graduate?

For entertainment only, I invite any of you to provide the solution
using only nodal analysis without consideration of the value of +VDC
(showing or explaining your work). There are bonus points that have no
value for calculating the exact equivalent resistance.

   Then make your challenge on the proper newsgroup:
news:sci.electronics.design which is where you'll find the EEs.

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

It is pretty easy if you know about how the op-amp will do whatever it
can to make the voltage at pin 2 the same as pin 1.

 

[Michael A. Terrell]

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.


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

 

[hamilton]

On 10/24/2010 4:55 PM, TheGlimmerMan wrote:

On Sat, 23 Oct 2010 13:28:58 -0700 (PDT), linnix<me@linnix.info-for.us
wrote:

On Oct 23, 1:25 pm, hamilton<hamil...@nothere.com> wrote:
On 10/23/2010 1:35 PM, linnix wrote:







On Oct 23, 12:27 pm, "William Sommerwerck"
grizzledgee...@comcast.net> wrote:
I don't think this is a very good "quiz" question,
It is a good quiz for fun, but not a good interview question.
What does it prove? The candidate is a good quiz solver?
It has no real engineering use. Are they hiring quiz solver
or engineer?

It's a great interview question. It shows whether the candidate can cut
through the clutter and see the basic principle involved. A good engineer
needs to be able to do that.

But do you base your hiring/firing decision on whether he can do it
during the interview? Most engineers can solve it in more relaxed
environment, in the real world. It only proved that many can't do it
under pressure, during the interview.

Maybe the OP is showing the new hire what to expect from the new manager.

Silly tests like this shows how a manager will behave after being hired.

These tests work both ways.

h

Yes, usually they are sweat shops hiring slaves.


Not if you have a title.

"Slave" is a title.

;-)

 

[David Nebenzahl]

On 10/24/2010 3:19 PM brent spake thus:

[snip quiz]

It is pretty easy if you know about how the op-amp will do whatever it
can to make the voltage at pin 2 the same as pin 1.

But is this really true? This sounds like it might be either a gross
oversimplification or a possible falsehood.

DISCLAIMER: I'm just learning about this stuff. OK, I've been fooling
around with electronics for, lessee, about 40 years now, but have had no
formal training; I'm trying to rectify that by reading and studying.

So today I read up about op amps. Learned how a comparator works,
generally speaking. How they differ from op amps (open loop).

My understanding of a comparator is that the output will be forced to
one extreme or the other depending on the difference in voltage between
noninverting and inverting inputs. Any significant voltage difference
will drive the output to near the respective supply rail, positive or
negative.

But I don't see how an op amp can, to quote brent, "do whatever it can
to make the voltage[s the same]". After all, these are *inputs*, no? So
at least in the case of a comparator, let's say that there exists a
1-volt difference between noninverting and inverting inputs (which I
understand is a *huge* difference given the extremely high gain of the
amp). Let's say the difference is positive: this will drive the output
close to the + rail, correct? But the noninverting input will still be 1
volt positive w/respect to the inverting input, right? In other words,
the change in output doesn't affect the inputs.

Now, this may be different in other configurations (operational or
instrumentation amp), where there are connections between output and
input instead of open loop. So is it true that in these cases the inputs
will be forced to (near) equal? If so, how does that work?

This still sounds rather mysterious to me.


--
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)

 

[Dave Platt]

In article <4cc4e1a0$0$2450$822641b3@news.adtechcomputers.com>,
David Nebenzahl <nobody@but.us.chickens> wrote:

But I don't see how an op amp can, to quote brent, "do whatever it can
to make the voltage[s the same]". After all, these are *inputs*, no? So
at least in the case of a comparator, let's say that there exists a
1-volt difference between noninverting and inverting inputs (which I
understand is a *huge* difference given the extremely high gain of the
amp). Let's say the difference is positive: this will drive the output
close to the + rail, correct? But the noninverting input will still be 1
volt positive w/respect to the inverting input, right? In other words,
the change in output doesn't affect the inputs.

Correct, if the outputs and inputs are isolated from one another. In
this (open-loop) configuration, an op amp will behave pretty much like
a comparator... and not a terribly good one (it'll be slow to reverse
its output state after it has been driven into saturation).

Now, this may be different in other configurations (operational or
instrumentation amp), where there are connections between output and
input instead of open loop. So is it true that in these cases the inputs
will be forced to (near) equal? If so, how does that work?

Correct. What you do (in the usual closed-loop op amp configuration)
is to feed back a portion of the output, to the inverting input. This
may be a direct connection (for a unity-gain noninverting circuit), or
there may be other components between the output and inverting input,
and other connections to the inverting input as well.

In any case, the effect of this "feedback" is to create the sort of
"input forcing" you're referred to. Specifically, if the inverting
input voltage is below that of the noninverting input, the output
voltage will rise towards the positive rail... and a portion of this
voltage increase, going back through the feedback network, will raise
the voltage at the noninverting input, reducing the difference
between the two inputs. This process continues until (to a first
approximation) the two inputs are at equal voltage.

This still sounds rather mysterious to me.

Don't feel back about that. I understand that when the concept of the
feedback-looped high-forward-gain operation amplifier was first
presented to the U.S. Patent Office, the examiner rejected the
invention, claiming that it couldn't work and thus couldn't possibly
be useful.

--
Dave Platt <dplatt@radagast.org> AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

 

[tm]

"David Nebenzahl" <nobody@but.us.chickens> wrote in message
news:4cc4e1a0$0$2450$822641b3@news.adtechcomputers.com...

On 10/24/2010 3:19 PM brent spake thus:

[snip quiz]

It is pretty easy if you know about how the op-amp will do whatever it
can to make the voltage at pin 2 the same as pin 1.

But is this really true? This sounds like it might be either a gross
oversimplification or a possible falsehood.

snippers

But I don't see how an op amp can, to quote brent, "do whatever it can to
make the voltage[s the same]". After all, these are *inputs*, no? So at
least in the case of a comparator, let's say that there exists a 1-volt
difference between noninverting and inverting inputs (which I understand
is a *huge* difference given the extremely high gain of the amp). Let's
say the difference is positive: this will drive the output close to the +
rail, correct? But the noninverting input will still be 1 volt positive
w/respect to the inverting input, right? In other words, the change in
output doesn't affect the inputs.

Now, this may be different in other configurations (operational or
instrumentation amp), where there are connections between output and input
instead of open loop. So is it true that in these cases the inputs will be
forced to (near) equal? If so, how does that work?

This still sounds rather mysterious to me.

You are neglecting (negative) feedback. In its simplest form, the output
connects back
to the negative (inverting) input. Now it is the basic voltage follower.
Used as
a buffer in real life.

So, given your initial conditions, the positive (non-inverting) input is set
to 1 volt. The op-amp will drive the output in the positive direction until
both inputs are equal, or at 1 volt.


Regards,
tm

 

[brent]

On Oct 24, 9:48 pm, David Nebenzahl <nob...@but.us.chickens> wrote:

On 10/24/2010 3:19 PM brent spake thus:

[snip quiz]

It is pretty easy if you know about how the op-amp will do whatever it
can to make the voltage at pin 2 the same as pin 1.

But is this really true? This sounds like it might be either a gross
oversimplification or a possible falsehood.

It is a simplification, sort of. This simplification is assuming a

few things:

1. It assumes a stable feedback arrangement. Knowing if something is
stable is pretty deep into EE knowledge, but fortunately, using op-
amps allows stability to be achieved in most cases without going
through the stability analysis. This circuit feels stable by looking
at it, but there are many simple ways that a feedback circuit can go
unstable. But in a nutshell, anytime you have lots of gain (like in
an op-amp) and you try to implement a feedback circuit with the gain
(for real nice control and stability) you risk , or need to worry
about instability. Instability is when the circuit reacts so fast and
the delay is wrong and it can't make up its mind what to do so it
oscillates back and forth. You can also just screw up and put the
output on a rail too.


2. It is assuming an infinite gain opamp. This problem is a dc
(static) problem, so in reality there might be a dc gain of 100000.
This means that to achieve a voltage to control to FET there might be
like 1 uVolt of difference in voltage between the two pins. Close
enough to just say they are equal.

DISCLAIMER: I'm just learning about this stuff. OK, I've been fooling
around with electronics for, lessee, about 40 years now, but have had no
formal training; I'm trying to rectify that by reading and studying.

So today I read up about op amps. Learned how a comparator works,
generally speaking. How they differ from op amps (open loop).

My understanding of a comparator is that the output will be forced to
one extreme or the other depending on the difference in voltage between
noninverting and inverting inputs. Any significant voltage difference
will drive the output to near the respective supply rail, positive or
negative.

Correct.

But I don't see how an op amp can, to quote brent, "do whatever it can
to make the voltage[s the same]". After all, these are *inputs*, no? So
at least in the case of a comparator, let's say that there exists a
1-volt difference between noninverting and inverting inputs (which I
understand is a *huge* difference given the extremely high gain of the
amp). Let's say the difference is positive: this will drive the output
close to the + rail, correct? But the noninverting input will still be 1
volt positive w/respect to the inverting input, right? In other words,
the change in output doesn't affect the inputs.

Now, this may be different in other configurations (operational or
instrumentation amp), where there are connections between output and
input instead of open loop. So is it true that in these cases the inputs
will be forced to (near) equal? If so, how does that work?

This still sounds rather mysterious to me.

--
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)