OT: Is this question too challenging for a BSEE graduate?
RosemontCrest
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¤t=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
Michael A. Terrell
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.
RosemontCrest
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?
>
>
> > 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.
tm
"RosemontCrest" <rosemo...@yahoo.com> wrote in message
news:14528147-a327-4a43...@l14g2000yqb.googlegroups.com...
4 * 40 ohms = 160 !! 4 kohms
RosemontCrest
> "RosemontCrest" <rosemontcr...@yahoo.com> wrote in message
>
> news:14528147-a327-4a43...@l14g2000yqb.googlegroups.com...
>
>
>
> >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?
>
>
> > 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
>
> 4 * 40 ohms = 160 !! 4 kohms
It really is that simple, isn't it? My frustration that prompted my
post is that recent BSEE graduates don't seem to be able to answer
this question.
isw
<14528147-a327-4a43...@l14g2000yqb.googlegroups.com>,
RosemontCrest <rosemo...@yahoo.com> wrote:
It's easier than either of the methods you propose; it's obvious by
inspection. Just notice that the two resistors in the op-amp's positive
input divider are in a 3:1 ratio, and so the same must be true for the
ones in the negative divider. So the FET must be acting as a 120 ohm
resistor.
Isaac
Rich Grise
Sheesh! Graduate EEs can't get this? I'm "just a tech," but it took me
about 13 seconds to get the answer, assuming I remember correctly that
I=E/R:
((Vdc/4)/40) A.
Cheers!
Rich
Michael
Graduates from where?
Nice carbine photos by the way.
(another) Michael
linnix
> On Fri, 22 Oct 2010 23:04:26 -0400, Michael A. Terrell 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?
>
>
> >> 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.
>
> Sheesh! Graduate EEs can't get this? I'm "just a tech," but it took me
> about 13 seconds to get the answer, assuming I remember correctly that
> I=E/R:
>
> ((Vdc/4)/40) A.
>
> Cheers!
> Rich
It's not asking for I, but Req.
I agree with last poster as (160 * 4000) / (160 + 4000) ohms
But I would not get the answer under pressure, in an interview. I
guess I am one of the bad EEs.
John Fields
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
William Sommerwerck
I'm always saying I'm smart because I look for the "fundamental principle"
underlying things. So... Let's see if I am...
This is a clever trick question. It assumes you understand /the/ basic rule
of op-amp circuit design -- if the circuit is stable, then the voltage
difference between the inverting and non-inverting inputs must be zero (or
in practice, vanishingly small).
Assuming Vdc is "stiff", then the voltage at the non-inverting input /must/
be Vdc/4. Right? The voltage at the inverting input /must/ be the same.
Ergo, the resistance of the JFET must be three times R3, or 120 ohms. Right,
too?
As the op amp draws no input current, the current through the JFET and R3
must be the same. Therefore, the load impedance must be 120 + 40.
QED?
Please note that only the most-trivial arithmetic is needed to solve the
problem. No fancy-shmancy algebra.
Bob Pease would be proud. I hope.
John Fields
R2 Rx
---- = ---- = 120 ohms
R1 R3
---
JF
Spehro Pefhany
news:5e68a1c4-aaf0-45d4...@26g2000yqv.googlegroups.com:
> 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?
No, it's trivial.
Since the voltage at the inputs of the opamp is 0.25 * Vdc, the
JFET drain/source current should be Vdc/160, or an equivalent
resistance of 160 ohms. The divider is in parallel.
So..
Req = 160 || 4000 ~= 153.8 ohms.
Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
sp...@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info : http://www.speff.com
Fredxx
"Michael A. Terrell" <mike.t...@earthlink.net> wrote in message
news:HaKdnQ3UkO1WzV_R...@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¤t=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.
>
>
To be honest I felt it wasn't obvious what the programmable load was. It's
also unusual to call a ground referenced load - a load, although I agree it
is, it does add to the confusion.
Also +VDC looks like a fixed voltage.
I would make it easier to interpret by saying:
VDC VDC
| |
circuit equiv to resistor
| |
GND GND
Apologies for non-fixed font if this gets messed up
David
news:14528147-a327-4a43...@l14g2000yqb.googlegroups.com...
>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¤t=programmableload.jpg
>
>or
>
>http://preview.tinyurl.com/2c8udf9
>
>Thanks,
>Brian
To me question seems ambiguous. From a DC viewpoint, the JFET
needs to conduct enough current to make the voltage across 40
ohms equal 1/4 of VDC so the JFET have to be 120 ohms. From an AC
viewpoint, the JFET looks like zero ohms.
David
William Sommerwerck
applicant understands "the basic principle of op-amp circuit design".
When I briefly attended CalTech, one of our physics tests had a question
about the Doppler shift of a satellite transmitter passing directly
overhead. You'd be amazed at how many students wasted time calculating it.
Spehro Pefhany
news:Xns9E1A6190B2594...@69.16.186.50:
> RosemontCrest <rosemo...@yahoo.com> wrote in
> news:5e68a1c4-aaf0-45d4...@26g2000yqv.googlegroups.com:
>
>> 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?
>
> No, it's trivial.
>
> Since the voltage at the inputs of the opamp is 0.25 * Vdc, the
> JFET drain/source current should be Vdc/160, or an equivalent
> resistance of 160 ohms. The divider is in parallel.
>
> So..
>
> Req = 160 || 4000 ~= 153.8 ohms.
>
>
One of the "issues" with this circuit lies in the interpretation of
"ideal" for the op-amp. Here, I (and others) have ASS-U-MEd that
it has infinite gain and zero offset voltage in zero input bias
current, but also that it will swing negative using the single
+12/0 supply (for example, it might have a built-in charge-pump
voltage converter). Most real op-amps won't do that, they'll swing down
to somewhere near the lower rail. In which case, with a JFET,
the op-amp will not be able to balance until the current
exceeds Idss for the JFET. It also won't work much above Idss
(regardless of the op-amp functionality) because the gate will begin to
conduct, so it would have only a narrow range of operation over which it
"looks" like a fixed resistor.
It will also behave differently if "+VDC" happens to be a negative
voltage.
I don't think this is a very good "quiz" question, it leaves too
many questions open and uses non-standard nomenclature. The proper
answer to this one is probably "what are you trying to do?", the
subtext being "whatever it is, this probably isn't going to do it".
William Sommerwerck
> "ideal" for the op-amp. Here, I (and others) have ASS-U-MEd that
> it has infinite gain and zero offset voltage in zero input bias
> current, but also that it will swing negative using the single
> +12/0 supply (for example, it might have a built-in charge-pump
> voltage converter). Most real op-amps won't do that, they'll swing down
> to somewhere near the lower rail. In which case, with a JFET,
> the op-amp will not be able to balance until the current
> exceeds Idss for the JFET. It also won't work much above Idss
> (regardless of the op-amp functionality) because the gate will begin to
> conduct, so it would have only a narrow range of operation over which it
> "looks" like a fixed resistor.
> It will also behave differently if "+VDC" happens to be a negative
> voltage.
> I don't think this is a very good "quiz" question, it leaves too
> many questions open and uses non-standard nomenclature. The proper
> answer to this one is probably "what are you trying to do?", the
> subtext being "whatever it is, this probably isn't going to do it".
I might be dead wrong, but it is a SUPERB question. It directly addresses
the question... "Does the applicant UNDERSTAND what goes on in a stable
op-amp circuit?"
The bottom line is that the "circuit" doesn't need the least bit of
"analysis" at all. If you understand that, in a stable circuit, the
inverting and non-inverting inputs have the same voltage on them, the
solution is utterly trivial.
John Fields
<excret...@invalid.invalid> wrote:
>It was the equivalent resistance of the *whole programmable load* that was
>asked for.
---
My take on it was that he was looking for the equivalent resistance of
the FET, which is 120 ohms.
---
JF
Fredxx
"John Fields" <jfi...@austininstruments.com> wrote in message
news:5vt5c69flvglk7ruf...@4ax.com...
That was my initial response, where the FET was the "programmable load",
hence my other post.
linnix
wrote:
> Spehro Pefhany <speffS...@interlogDOTyou.knowwhat> wrote innews:Xns9E1A6190B2594...@69.16.186.50:
>
>
>
> > RosemontCrest <rosemontcr...@yahoo.com> wrote in
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?
Plain...@yawho.com
<rosemo...@yahoo.com> wrote:
Possibly. I took my formal electronics classes over 35 years ago.
One of the instructors was an EE who couldn't understand the need for
bypass caps on the power supply lined of boards using TTL ICs.
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.
PlainBill
William Sommerwerck
> 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.
linnix
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.
David Nebenzahl
> 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¤t=programmableload.jpg
>
> or
>
> http://preview.tinyurl.com/2c8udf9
OK, I freely admit that I do *not* understand how op-amps work and can't
figure out this puzzle. (I hope to someday.) I'm just keeping score here
on this thread.
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?
--
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)
RosemontCrest
> On 10/22/2010 7:59 PM RosemontCrest spake thus:
>
> > 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?
>
>
> > or
>
> >http://preview.tinyurl.com/2c8udf9
>
> OK, I freely admit that I do *not* understand how op-amps work and can't
> figure out this puzzle. (I hope to someday.) I'm just keeping score here
> on this thread.
>
> 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?
>
> --
> 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)
153.85 Ohms is the equivalent resistance of the entire circuit. 120
Ohms is the equivalent resistance of Q1.
William Sommerwerck
> 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?
hamilton
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
linnix
Yes, usually they are sweat shops hiring slaves.
William Sommerwerck
>>> 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.
Interviews can be unnerving. Well-qualified people can collapse under
pressure. You would never use a single question, nor would you make the
answers to even a series of questions the basis for hiring or firing.
William Sommerwerck
> 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.
linnix
wrote:
I had first hand experience. I got stomped-on by similar quiz by a
junior engineer. After that, i had no intention of taking that job
and no interest in continuing the interviews. I should have just walk
out at the point. It was a total waste of my time and their
resources, but dinner was good and the company paid.
RosemontCrest
wrote:
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.
David Nebenzahl
>> 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.
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
Th�venin equivalent of the circuit?
Rich Grise
> On Oct 22, 9:53�pm, Rich Grise <richgr...@example.net> wrote:
>>
>> ((Vdc/4)/40) A.
>
> It's not asking for I, but Req.
>
Oh, OK: Vcd/(((Vdc/4)/40) A).
Hope This Helps!
Rich
William Sommerwerck
>>> 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 Th�venin 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 Th�venin 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.
Hammy
news:i9vgc0$l99$1...@news.eternal-september.org:
> Silly tests like this shows how a manager will behave after being
> hired.
>
> These tests work both ways.
>
> h
At least its technical.When I first graduated I had an interview with a
large comunications corp and not one question had anything to do with
electronics. I think the HR department gets their interview questions
from COSMO. I felt like I was answering one of those dumb womans
magazines quizes.
My experience was the smaller comapainies tended to be more intrested
in your technical understanding the larger ones seemed more intrested
in your psych profile.
I choked on one 6 guys asking me to reproduce the LNA I did in school I
think I screwed up all my symbols and couldnt explain the bias network
;brainfart.
Oh well it all worked out OK I make more money now then any of those
jobs so no compalints.
David Nebenzahl
[me wrote:]
>> So I ask, perhaps naively: is this basically the same as finding
>> the Th�venin 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 Th�venin 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.
Aha. Nice trick question, that.
So I learned something today. Thanks for your patience.
William Sommerwerck
>>> the Th�venin 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 Th�venin 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.
> Aha. Nice trick question, that.
> So I learned something today. Thanks for your patience.
Thank /you/ for thinking, and asking good questions.
RadioJ
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?
>
Robert Baer
> On Fri, 22 Oct 2010 19:59:39 -0700 (PDT), RosemontCrest
>
>> 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¤t=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.
>
>
>
> +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
"William Sommerwerck" <grizzle...@comcast.net> wrote:
That statement is so significant, and so rarely understood...
Isaac
ehsjr
> On Oct 22, 8:04 pm, "Michael A. Terrell" <mike.terr...@earthlink.net>
> 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?
>>
>>
>>>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.
>>
>>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
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
>> 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
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.t...@earthlink.net> wrote in message
news:HaKdnQ3UkO1WzV_R...@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?
>>
>>
>> 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
Plain...@yawho.com
>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
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
> "ehsjr" <eh...@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
>
brent
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.
TheGlimmerMan
wrote:
Not if you have a title.
Michael A. Terrell
You 'program' it by changing reistors.
hamilton
"Slave" is a title.
;-)
David Nebenzahl
[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.
brent
> 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.
>
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.
Dave Platt
David Nebenzahl <nob...@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 <dpl...@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" <nob...@but.us.chickens> wrote in message
news:4cc4e1a0$0$2450$8226...@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.
>
>
> 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
isw
"William Sommerwerck" <grizzle...@comcast.net> wrote:
Sounds like you predate me -- but not by much. I cut my "op-amp teeth"
on the earliest National Tech Notes (when they had that wonderful "NS"
logo where both glyphs were identical, with one rotated and flipped).
Isaac
William Sommerwerck
>> 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.
No, it's fact. It is, as I said, /the/ fundamental principle of op-amp
circuit design.
I can't think of a book that discusses this in a fairly simple way. Even the
Philbrick book -- which is hard to find these days -- doesn't address the
matter as directly as I'd like. But, trust me. Most op-amp circuits can be
analyzed by assuming the voltage is the same, then applying simple circuit
analysis. You might start with the basic op-amp inverting amplifier, and see
what happens.
William Sommerwerck
>> the wonderful Philbrick brook. One of the greatest pieces
>> of technical writing ever (I keep a copy for inspiration),
>> and still a classic.
> Sounds like you predate me -- but not by much. I cut my
> "op-amp teeth" on the earliest National Tech Notes (when
> they had that wonderful "NS" logo where both glyphs were
> identical, with one rotated and flipped).
Probably written by Bob Widlar. Unfortunately, NS is still using his
documents, which are now rather dated. Nothing wrong with continued respect
for his genius, but /someone/ needs to update them.
John Fields
<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.
>
>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.
---
If you think about a comparator as being nothing more than an opamp
with a huge amount of gain operating open-loop, then for its output to
swing to either rail requires an almost _insignificant_ difference in
voltage between its inputs.
Here: (View in Courier)
+10V
|
|U1
Vin+>-------|+\
| >--Vout
Vin->-------|-/
|
|
-10V
Let's say U1 is an ideal opamp with a gain of one million and that if
Vin+ goes more positive than Vin-, Vout will rise toward +10V, and if
it goes more negative than Vin-, Vout will fall toward -10V.
Now, if Vin+ = Vin-, Vout will sit between the rails at 0V, but since
U1 has a gain of one million and we have a +10V rail, all it'll take
for the output to rise to the positive rail is for Vin+ to be 10
microvolts (or more) more positive than Vin-.
Conversely, all it'll take for Vout to fall to the negative rail is
for Vin+ to be >=10�V more negative than Vin-.
Also, it's important to realize that it doesn't matter what voltage
Vin+ and Vin- are sitting at, what matters is the difference between
them.
>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.
---
>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?
---
Yes
---
>If so, how does that work?
---
Consider, first, the magical voltage divider: ;)
E1
|
[R1]
|
+---E2
|
[R2]
|
GND
Where, if GND is at 0V,
E1 * R2
E2 = ---------
R1 + R2
Just for grins, let's set this up:
10V E1V
|
[9k] R1
|
+---E2
|
[1k] R2
|
GND
and solve for E2:
10V * 1kR
E2 = ----------- = 1V
9kR + 1kR
Next, let's hook up the the voltage divider to an opamp in the
non-inverting configuration and apply 1V to the + input to see what'll
happen:
+V
|
1V>-------|+\
| >--+-- E1
+--|-/ |
| | [9k] R1
| -V |
+--------+-->E2
|
[1k] R2
|
gnd
Now, since the + input is sitting at 1V, the output will have to move
with sufficient magnitude and in the proper direction to make the
voltage on the - input equal to the voltage on the + input.
Since the + input is at +1V, the output will have to swing positive,
and for E2 to to get to +1V, we'll rearrange the voltage divider
equation to solve for E2, and we'll get:
E2R1 + E2R2 (1V * 9kR) + (1V * 1kR)
E1 = ------------- = ------------------------- = 10V
R2 1kR
So there you have it: a non-inverting opamp with its gain of one
million throttled down to 10. :-)
To adjust the gain, then, all that's necessary is to adjust the ratio
R1:R2 and the output will go wherever it has to in order to make the
inputs of the opamp equal to each other.
The gain of the circuit can be described as, simply,:
R1 + R2
Av = ---------
R2
---
>This still sounds rather mysterious to me.
---
Even now? ;)
---
JF
k...@att.bizzzzzzzzzzzz
<grizzle...@comcast.net> wrote:
>>> 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.
>
>No, it's fact. It is, as I said, /the/ fundamental principle of op-amp
>circuit design.
One of three principles:
1: Gain is infinite Vout/((V+) - (V-)) = infinity
2: Input impedance is infinite
3: Output impedance is zero
Of course none of these is really true, but close enough for most purposes.
k...@att.bizzzzzzzzzzzz
wrote:
>On 10/24/2010 3:19 PM brent spake thus:
The difference between an opamp and comparitor is that the comparitor is
designed to be operated with its output burried in the rails (some don't have
a current sourcing capability, either). Comparitors can be made out of most
opamps but they're generally pretty slow because they don't like their outputs
saturated.
Spehro Pefhany
news:mv9cc6pn3fclvj5k6...@4ax.com:
> On Mon, 25 Oct 2010 00:43:24 -0700, "William Sommerwerck"
> <grizzle...@comcast.net> wrote:
>
>>>> 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.
>>
>>No, it's fact. It is, as I said, /the/ fundamental principle of op-amp
>>circuit design.
>
> One of three principles:
>
> 1: Gain is infinite Vout/((V+) - (V-)) = infinity
> 2: Input impedance is infinite
> 3: Output impedance is zero
>
4: Phase shift is zero at all frequencies
Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
sp...@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info : http://www.speff.com
Spehro Pefhany
news:25acc616cisme9t9p...@4ax.com:
A comparator also designed to operate with a significant voltage between
the inputs. Many modern op-amps will allow significant current to flow
if there is more than a diode drop or so between the inputs. At least
one older op-amp would exhibit Vos drift over time in the presence of
continous voltage between the inputs.
isw
"William Sommerwerck" <grizzle...@comcast.net> wrote:
Actually, you can do it with bipolar transistor circuits too, to a first
approximation:
Base => inverting input
Emitter => non-inverting input
Collector => output
Of course, the two inputs have rather different impedances, but for
figuring out gains and so on, it works pretty well.
Isaac
ehsjr
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.
Ed
David Nebenzahl
> 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.
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.
Plain...@yawho.com
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.
PlainBill
tm
<Plain...@yawho.com> wrote in message
news:nvudc65qpifjekd8b...@4ax.com...
Geese. It was just a quiz to see if an applicant understood how an opamp
works.
BTW, you didn't get the job.
tm
Joel Koltner
> 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?
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. :-)
David Nebenzahl
> 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 Nebenzahl
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.
John Fields
<nob...@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?
---
JF
whit3rd
<grizzledgee...@comcast.net> wrote:
> >> 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.
>
> No, it's fact. It is, as I said, /the/ fundamental principle of op-amp
> circuit design.
I'd say the fundamental principle is high gain DC-and-up
amplification; the
balance of input potentials is derived from this, and is, rather, a
productive approximation. Useful, yes, but not fundamental.
It's this kind of 'principle' you have to drop when something
more important calls for your attention.
tm
"David Nebenzahl" <nob...@but.us.chickens> wrote in message
> On 10/26/2010 9:57 AM tm spake thus:
>
>> <Plain...@yawho.com> wrote in message
>> news:nvudc65qpifjekd8b...@4ax.com...
>>
>>> On Sun, 24 Oct 2010 21:14:00 -0400, "Michael A. Terrell"
>>> <mike.t...@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.
k...@att.bizzzzzzzzzzzz
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 ...
For many uses the idealized model works extremely well. It's kinda like the
ideal resistor; it doesn't really exist but for most problems reality is close
enough to practice that the difference doesn't matter.
k...@att.bizzzzzzzzzzzz
<zapwireD...@yahoo.com> wrote:
>"David Nebenzahl" <nob...@but.us.chickens> wrote in message
>news:4cc67064$0$2444$8226...@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
news:16lec65ml3jnmcets...@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
brent
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
k...@att.bizzzzzzzzzzzz
<zapwireD...@yahoo.com> wrote:
><k...@att.bizzzzzzzzzzzz> wrote in message
>news:16lec65ml3jnmcets...@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. :-/
David Nebenzahl
> David,
>
> Did you get the article I posted for you?
No; what article? (I take it this is the real John Fields, right?)
Michael A. Terrell
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.
Michael A. Terrell
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.
tm
"Michael A. Terrell" <mike.t...@earthlink.net> wrote in message
news:3YSdndK57NrYGFrR...@earthlink.com...
Ok, I get it now. It was for a cost plus government job.
tm
ehsjr
> 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
David Nebenzahl
> 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.
Michael A. Terrell
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.
tm
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
<nob...@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:
If you can't get it let me know and I'll email you a copy.
---
JF
Michael A. Terrell
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
David
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
> 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
David Nebenzahl
> On Tue, 26 Oct 2010 18:50:13 -0700, David Nebenzahl
> <nob...@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:
>
> 68oac6du5ec8alsq2...@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.
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
> > <nob...@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:
> >
> > 68oac6du5ec8alsq2...@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.
David Nebenzahl
> 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
>>> <nob...@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:
>>>
>>> 68oac6du5ec8alsq2...@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.
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
> >>> <nob...@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:
> >>>
> >>> 68oac6du5ec8alsq2...@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.