numerical challenge

Oct 23, 2012 144 Replies

You didn't want to work there, so it was a great interview.

This is my first-pass qualifier question.

+10V | | | | C +5V--------------B E | | 1K | | gnd

(drawn like a real schematic, on a whiteboard)

What is the base voltage? (one answer: 0.6!) What is the emitter voltage? What is the collector voltage?

What is the base current? What is the emitter current? What is the collector current?

Any other comments?

The best interview is to work on a real design problem, together on a whiteboard, for a couple of hours.

I also like to ask people what they designed last, and to describe some circuits.

John Larkin Highland Technology, Inc jlarkin at highlandtechnology dot com http://www.highlandtechnology.com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom laser drivers and controllers Photonics and fiberoptic TTL data links VME thermocouple, LVDT, synchro acquisition and simulation

171^4

=641 mod 1000

Ugh- and you think that is better than modular arithmetic? The method I dem onstrated actually has a name, the right-to-left binary method.

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I thought there might be some esoteric theorem leading to a short cut based on factorization of something or another, but there isn't that I can find.

Consider yourself lucky. Who would want to work for a worthless trouble-mak ing asshole with an ego to boot? People like that make me applaud when engi neering jobs are shipped to China, actually I don't care if they're shipped to Mars- whatever it takes to kick trash like that to the curb works.

!WRONG! Base voltage is 5V .

Okay, now that is 5V-0.6V .

Ummm, gimme a minute, that's a tough one.

Sorry, you stay in the 47%. Go get your food stamps.

--- Since the base voltage is as surely +5V as the collector voltage is

10V, don't you mean "base-to-emitter voltage"?

Version 4 SHEET 1 948 680 WIRE 256 48 -48 48 WIRE 256 144 256 48 WIRE 192 192 64 192 WIRE 256 288 256 240 WIRE -48 304 -48 48 WIRE 64 304 64 192 WIRE -48 448 -48 384 WIRE 64 448 64 384 WIRE 64 448 -48 448 WIRE 256 448 256 368 WIRE 256 448 64 448 WIRE -48 512 -48 448 FLAG -48 512 0 SYMBOL npn 192 144 R0 SYMATTR InstName Q1 SYMATTR Value 2N2222 SYMBOL res 240 272 R0 SYMATTR InstName R1 SYMATTR Value 1000 SYMATTR SpiceLine "" SYMBOL voltage 64 288 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V1 SYMATTR Value 5 SYMATTR Value2 "" SYMATTR SpiceLine "" SYMBOL voltage -48 288 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V2 SYMATTR Value 10 SYMATTR Value2 "" SYMATTR SpiceLine "" TEXT -24 480 Left 2 !.op TEXT 392 104 Left 2 ;--- Operating Point ---\n \nV(n001): 10 voltage\nV(n002): 5 voltage\nV(n003): 4.30708 voltage\nIc(Q1): 0.00428639 device_current\nIb(Q1):

2.06893e-005 device_current\nIe(Q1): -0.00430708 device_current\nI(R1): 0.00430708 device_current\nI(V2):

-0.00428639 device_current\nI(V1): -2.06893e-005 device_current

-- JF

I can't believe you SPICEd that. Did SPICE show the horrible oscillation?

(JL's '0.6' was one of the incorrect answers given by unsuccessful candidates.)

Cheers

Phil Hobbs

Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net

Really, I've had the strangest answers from presumed circuit designers:

The base voltage is 0.6

The transistor is saturated, so the collector voltage is 0.2

The transistor is saturated, so the emitter voltage is 10.

If anybody gets them right, and then notes the oscillation hazard, hire them.

Why does your newsreader double-space everything?

John Larkin Highland Technology, Inc jlarkin at highlandtechnology dot com http://www.highlandtechnology.com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom laser drivers and controllers Photonics and fiberoptic TTL data links VME thermocouple, LVDT, synchro acquisition and simulation

Those are the questions that I ask.

Applicants are supposed to answer immediately, not go off and run Spice.

John Larkin Highland Technology, Inc jlarkin at highlandtechnology dot com http://www.highlandtechnology.com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom laser drivers and controllers Photonics and fiberoptic TTL data links VME thermocouple, LVDT, synchro acquisition and simulation

Damn! I sure hope somebody read Sylvia's post. Very nice!

That way you filter out otherwise bright people who just can't concentrate when an interviewer is looking over their shoulder, while passing the people who just had the chance to know the method. Never mind that the former would solve the problem at home, while the latter might turn out to be unable to do so unless being shown how to do it.

I second the opinion, that this would be a stupid interview question, like all other a-ha!-type questions. Nice and inspiring puzzle though.

ae

1000

demonstrated actually has a name, the right-to-left binary method.

factorization of something or another, but there isn't that I can find.

I never made such a claim. Rather it being another way, which is perfectly valid.

Given the constrains of only pencil, paper and limited time, it immediately came to mind and I had the answer with very low efforts in less than 5 minutes, which is a good answer for an interview. And a bit of astuteness to save some operations don't hurt.

Guess I'd hire me :-)

Thanks, Fred.

Okay, well, if you think that's the answer then you need to see your doctor about necrosis of the brain. The circuit I see shows the base terminal pinned at

5V and the collector terminal pinned at 10V. How could it possibly saturate????

It would be a valid question if you apply for a job as a stock broker. Interestingly quite a few people know how to break it down. I wouldn't but that probably indicates there is little practical use for solving these kind of algebra problems :-)

Failure does not prove something is impossible, failure simply indicates you are not using the right tools... nico@nctdevpuntnl (punt=.) --------------------------------------------------------------

Well, OK, but since one usually refers to voltage with respect to 0V, (GND) as you have when referring to the collector voltage, your statement: "(one answer: 0.6!)" is flawed unless you're referring to the base-to-emitter voltage. That is, of course, unless the wire connecting the 5V supply to the base exhibits an extraordinarily high resistance. My point is that if someone with a critical eye was looking for employment and took your test, their proper response to the base voltage being 5V would be cause enough for you to leave them on the street since you're dead set on the answer being 0.6V.

Of course the answer was flawed. The guy who said that was an idiot, and I didn't hire him.

You would have obviously flunked the test.

John Larkin Highland Technology, Inc jlarkin at highlandtechnology dot com http://www.highlandtechnology.com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom laser drivers and controllers Photonics and fiberoptic TTL data links VME thermocouple, LVDT, synchro acquisition and simulation

That interview worked *just fine*. You didn't wanna work for him anyway...

Les Cargill

The last three digits of 171 ^ 172 are 172

The question didn't ask for the last three digits of the result.

Ed

This appears to originate from

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Solutions similar to those that I and others have posted in this thread are given there, but the original post implies that there's a single line solution. If there is, I can't see it.

I have noted that 171 = 900 - 9^3, which looks a bit interesting, but I haven't been able to make anything of it in the sense of achieving a one-line answer.

Sylvia.

[snip]

Maybe he just used you as an expert consultant without having to pay you. If he did the same thing with many knowledgeable applicants he might have gotten all the information he needed to redesign the product properly, without paying for an employee or contractor.

Paul

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