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.
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.
ht
eer
Yes, usually they are sweat shops hiring slaves.
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.
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 in news: snipped-for-privacy@t8g2000yqk.googlegroups.com:
action=view&cur
That's much better.
Now, I have a question or two for _you_. If the 100 ohm resistor is still there, does it make the circuit more stable or less stable? Can it oscillate with an ideal op-amp? How about a real op-amp?
Best regards, Spehro Pefhany
-- "it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info : http://www.speff.com
Oh, OK: Vcd/(((Vdc/4)/40) A).
Hope This Helps! Rich
snipped-for-privacy@t8g2000yqk.googlegroups.com:
=A0
The 100 Ohm resistor can indeed make the circuit unstable; that's an extra-credit response.
hamilton wrote in news:i9vgc0$l99$ snipped-for-privacy@news.eternal-september.org:
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.
;-) ...Jim Thompson
-- | James E.Thompson, CTO | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona 85048 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | I can see November from my house :-)
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? ...
...with or without the !
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
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
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!
..like a few Curies of radium?
..
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.
Not if you have a title.
"Slave" is a title.
;-)
On 10/24/2010 3:19 PM brent spake thus:
[snip quiz]
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)
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