Quick quiz - Page 1
Forum section: electronics / general

author: allanw
author_rank: Frequent Contributor
timestamp: September 24, 2010, 06:05:27 am
content: Just did poorly on a technical phone interview for a certain graphics card company. The recruiter told me that a lot of people also have trouble with this seemingly simple problem:Qualitatively, what does the voltage at the resistor look like for this input assuming no initial charge on the cap?I was pretty caught off guard because I thought they'd ask me questions about embedded systems, which is what I was applying for... also the fact that input went back down got me a little bit confused and I couldn't do it. Of course, I know the answer now... I'd like to see what some other people's reasonings are for this circuit though.And no cheating with spice!

author: DJPhil
author_rank: Frequent Contributor
timestamp: September 24, 2010, 06:30:38 am
content: After all the book studying I've been doing the last six months, I feel like I should be able to nail this, but I'm not positive about my answer.After cheating with spice because I couldn't stand it: Ok, wow. I see what you mean.I was close, but for mostly the wrong reasons. I had the charge waveform right, but the discharge reversed in time because symmetry in my head became somehow more important than reality. Seeing it in spice makes what I was just guessing look like gibberish, which it was. I'm humbled.

author: EEVblog
author_rank: Unknown
timestamp: September 24, 2010, 06:49:35 am
content: Ah yes, an old trap for young players.Remember that the cap holds charge that doesn't go away instantly (unless you short it).It might help to think of it like this. Once the cap is charged, picture taking it out of the circuit (still with it's charge), and putting it into another circuit with the new conditions (i.e. the battery replaced by a link).Dave.

author: Zero999
author_rank: Super Contributor
timestamp: September 24, 2010, 07:36:05 am
content: I think the thing that will confuse most nubes is where the negative voltage comes from. Dave is right, here's another clue: when the capacitor is charged and its positive is connected to negative, what will be the voltage across the resistor?Of course it's a trick question anyway: the circuit never reaches steady state because the capacitor never charges to 100%.

author: EEVblog
author_rank: Unknown
timestamp: September 24, 2010, 07:41:24 am
content: Quote from: Hero999 on September 24, 2010, 07:36:05 amThat's the smart-arse theoretical answer, so = instant FAIL!Dave.

author: allanw
author_rank: Frequent Contributor
timestamp: September 24, 2010, 07:47:29 am
content: Quote from: EEVblog on September 24, 2010, 07:41:24 amI agree. It gets pretty close to 100% such that it doesn't matter for actually practically answering the question. The next part to the question is a bit more difficult but I won't share it yet...

author: Mechatrommer
author_rank: Super Contributor
timestamp: September 24, 2010, 12:22:15 pm
content: this is my guess (red line):top graph: the sourcemiddle: if after pulse, V = short.bottom: if after pulse = openedit: the max magnitude of red line should be the same as V, the picture showed smaller magnitude (mistake)

author: NiHaoMike
author_rank: Super Contributor
timestamp: September 24, 2010, 01:07:12 pm
content: To be fair, the ESR of the capacitor would mean the peak would be a little lower, but it's negligible in most signal circuits.

author: Strube09
author_rank: Regular Contributor
timestamp: September 24, 2010, 01:16:17 pm
content: Yep,Classic Differentiator... if the RC constant didn't reach stead state and the output taken off the cap it would be an integrator circuit.Strube

author: FreeThinker
author_rank: Frequent Contributor
timestamp: September 24, 2010, 02:19:31 pm
content: Well,Did you get the job?

author: KTP
author_rank: Frequent Contributor
timestamp: September 24, 2010, 03:25:43 pm
content: this is the kind of stuff I was trying to explain qualitatively to a fellow ee classmate several quarters back.Voltage cannot change instantly in a capacitor and current cannot change instantly in an inductor.So for this simple circuit, the capacitor with zero initial charge (0V) will have 0 volts at t = 0- and 0 volts at t = 0+. It charges up to V through the resistor R exponentially with a time constant of R*C. At t = x-, when the ideal capacitor is charged to V volts, it will remain at V volts at t = x+. Therefore the voltage across the resistor at t = x+ will be V - Vcap, or -V. The capacitor will then discharge through R exponentially with a time constant of R*C.Replace the capacitor with an inductor and you can do a similar analysis. Since the current cannot change in the inductor instantly, and it is in series with the resisitor, then the voltage across the resistor at t = 0- is the same as that at t=0+, or 0 volts. In fact if you look at the voltage across the inductor, it will look similar to the voltage across the resistor in the capacitor example (for certain values of L and R). The voltage across R will look similar to the voltage across the capacitor in the capacitor example also. This duality (is that the correct word?) is pretty dang cool.Btw, I failed at trying to help my fellow student realize this, and they just went back to trying to memorize formula to work all of the different circuit topologies that might be on the exam. It turned out the exam was simple enough that this worked.

author: JohnS_AZ
author_rank: Frequent Contributor
timestamp: September 24, 2010, 03:37:58 pm
content: This reminds me of my first job interview around 1978 to be a technician for Motorola.The guy asked me to diagram and explain the operation of a successive approximation AtoD converter.Got it mostly right, and did get the job.

author: allanw
author_rank: Frequent Contributor
timestamp: September 24, 2010, 06:00:57 pm
content: Ah, thinking of it as a differentiator would have helped.I have another phone interview today. These phone interviews are just screening candidates to fly to CA for a site interview, so it'll be a while before I know.

author: Zero999
author_rank: Super Contributor
timestamp: September 24, 2010, 07:47:53 pm
content: I think I would probably fail to answer this question if it was a telephone interview. In person I would've been fine as I would've been able to have drawn a diagram.Here's another simple but confusing question:What's the DC component of the waveforms attached?Draw the waveform if the signal is put through a differentiator with a very long time constant compared to the frequency and allowed to the steady state (yes I know it never gets there but assume it's 99.99% there)?

author: allanw
author_rank: Frequent Contributor
timestamp: October 27, 2010, 07:37:35 pm
content: update: got the internship despite missing this question. woot

author: DJPhil
author_rank: Frequent Contributor
timestamp: October 27, 2010, 07:45:40 pm
content: Congratulations!I hope you'll be allowed to share some neat stories, it doesn't seem like much makes it out from inside graphics card companies (for obvious reasons I guess).

author: David
author_rank: Frequent Contributor
timestamp: October 27, 2010, 08:46:58 pm
content: I had this exact question in an interview not long ago...now working for the company

author: allanw
author_rank: Frequent Contributor
timestamp: October 27, 2010, 08:48:20 pm
content: Haha cool. Do they do design work in the UK too?

author: David
author_rank: Frequent Contributor
timestamp: October 27, 2010, 09:01:43 pm
content: Quote from: allanw on October 27, 2010, 08:48:20 pmDo who? Do you mean is there much electronics design in the UK? If so, then there is still quite a lot of design going on. I work on engine control units for jet engines (it sounds a lot more exciting than it is...)

author: allanw
author_rank: Frequent Contributor
timestamp: October 27, 2010, 09:02:23 pm
content: Oops, thought you were talking about the same company as I was.

author: Zad
author_rank: Super Contributor
timestamp: October 27, 2010, 09:25:39 pm
content: Well doneMaybe, just maybe, you got it because you didn't write it off but went out of your way to find the right answer, andwhyit does it. That's what real world engineering is about. Most days are full of not knowing the full correct answer, that's what makes it interesting!