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---It was the equivalent resistance of the *whole programmable load* that was
asked for.
That was my initial response, where the FET was the "programmable load",On Sat, 23 Oct 2010 06:16:57 -0700, Fred Abse
excretatauris@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.
Possibly. I took my formal electronics classes over 35 years ago.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
It's a great interview question. It shows whether the candidate can cutI 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?
OK, I freely admit that I do *not* understand how op-amps work and can'tI 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
It is a good quiz for fun, but not a good interview question. WhatSpehro Pefhany <speffS...@interlogDOTyou.knowwhat> wrote innews:Xns9E1A6190B2594speffinterlogcom@69.16.186.50:
RosemontCrest <rosemontcr...@yahoo.com> wrote in
news:5e68a1c4-aaf0-45d4-b186-53e1734a953f@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".
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
What do you mean by "inspection"? Are you applying a formula you memorized?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.
Maybe the OP is showing the new hire what to expect from the new manager.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.
Interviews can be unnerving. Well-qualified people can collapse underIt 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.
I don't mean to be offensive, but it you /understood/ the principal ofSo 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?
But do you base your hiring/firing decision on whether he can do itI 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.
153.85 Ohms is the equivalent resistance of the entire circuit. 120On 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?
http://smg.photobucket.com/albums/v18/RosemontCrest/?action=view&curr....
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)
Well, duh. Obviously, I don't. Perhaps someday I may.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.
So I ask, perhaps naively: is this basically the same as finding theThe resistance of the JFET itself is 120 ohms, which is what we're looking
for, regardless of how the question is phrased.
Yes, usually they are sweat shops hiring slaves.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
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).
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.
No. In fact, one of the things the testee has to recognize is that thisThe 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?
At least its technical.When I first graduated I had an interview with aSilly tests like this shows how a manager will behave after being
hired.
These tests work both ways.
h
Aha. Nice trick question, that.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.
Thank /you/ for thinking, and asking good questions.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.
Aha. Nice trick question, that.
So I learned something today. Thanks for your patience.
I had first hand experience. I got stomped-on by similar quiz by aIt 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.
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.