HWN - Apple Hardware Interview Question!

Hi fellow (and future) engineers!
This week we tackle yet another a real interview question using the foundations we learned last week and on one of our first videos. Let us know what you think!
Check this video before answering the question: • HWN - Hardware Intervi...
Welcome back to another episode of Hardware Ninja! Ever wondered how to get a job as a Hardware Designer in some of the top tech companies like Microsoft, Google, Tesla, Apple, Facebook, etc?
The interview process employs many application based questions whether you're a recent college grad or an experienced applicant. It doesn't matter if you want to work for Elon Musk, Tim Cook, or the up and coming start-up you heard about. We're here with curated material of real life technical interview questions.
Please consider subscribing to the channel and supporting our movement. Prospective job seekers (including yourself some day) will benefit from this resource.
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[@belaliqbal8858]

if anyone wondering how -1.5v and -0.9v come, just balance the charges across capacitor initially and finally. initial charge = final charge charge when Vin switches from 2.4V to 0V = charge when voltage Vx becomes stable. -1.8*C1-0.6*C2 = Vx*C1+(Vx+0.6)C2 Vx = (-1.8*C1-1.2*C2)/(C1+C2) if you assume C1=C2=C Vx= -1.5V Vx+0.6V = -0.9V (this is across C2) for the first few cycle there would be glitch at the Vx voltage , it won't go to -1.2V, -1.5v smoothly final volatage on the Vx= -1.8V and across C2 is -1.2V.

[@PPSRHD]

It's amazing how fewer channels are there for electrical engineers as compared to cs engineers. This is the first channel exclusively for electrical engineers i have found .

[@case_sensitive]

Ikr? It's so skewed

[@ajeysingh8024]

The question is simply amazing. But for those who were unable to solve it, it's a voltage doubler circuit taught in Analog Circuits course. Only the input signal has been converted to a Digital signal here. When I learnt first about this circuit, I personally had faced trouble in solving this circuit and even after I got the answer, I still had trouble in understanding how the voltage eventually doubled. I tried to search everywhere and only after quite a search, I found the right material to understand this circuit. Please refer Razavi "Fundamentals of Analog Circuits" book and search for voltage doubler circuits. Trust me you will find not a better resource than this to understand the above.

[@Adgjmptw12able]

"Fundamentals of Analog Circuits" is not available for purchase. Where can I buy one?

[@mrlapras024]

I think he is referring to Fundamentals of Microelectronics by Behzad Razavi. Razavi also has a Electronics I lecture series uploaded here on KZbin and it has an lecture on voltage doubler circuits. He is a great prof@@Adgjmptw12able​

[@six286]

I had this exact question in my interview with Apple a few months back…Landed the job!

[@tarunfirestoned]

Hi hardware ninja. A truly wonderful question. I am actually working as analog design engineer with 4 years of experience. I'm currently looking for opportunities in Europe. Can you tell me in Europe in general do they expect designers with my experience to just have good knowledge on the layout guidelines and practice (along with design knowledge) or do they expect the designers to have good amount of hands on experience drawing actual layouts?

[@jeraldvicente4339]

Correct me if I am wrong. But I think this solution only works if it is assumed that C1 and C2 are equal, which was not mentioned. C1 loses an amount of voltage that C2 gains, that is why on the beginning of the second cycle, Vout settled on -0.6V after gaining -0.6V which was lost by Vx that started on -1.8V and settled on -1.2V, giving D2 0.6V across it. The third cycle employs the same principle, but Vout now settled on -0.9V after gaining -0.3V which was lost by Vx that started on -1.8V (again) and settled on -1.5V, giving D2 0.6V across it (again). This is only guaranteed when C1 and C2 are equal as the same current flows thru them. To add, the words lost and gained are in the perspective of the capacitors.

[@justqassim]

Please explain the 2nd cycle. What does c1 lose what does c2 gain? VD1= -1.8 and VC1=1.8 when vin=0 is this correct? How does Vout =-0.6?

[@skywalkerlxy5213]

@@justqassim VC1 loses 0.6V from 1.8 to 1.2 and VC2 gets 0.6V from 0 to 0.6V. That's why Vx changes from -1.8V to -1.2V and Vout changes from 0 to -0.6V.

[@anujagrawal156]

hello ninja , could make playlist on your channel dividing it into different category.

[@LOLWUT281]

It should be a taylor series, where they keep getting half way closer to -1.8V... I think. I'm pretty sure it doen't go below -1.8V though. I think an application for this would be a boost inverter...

[@nirmalshah1166]

At the end of 1st cycle We got the values at Vx=-1.2V and Vout = -0.6V This would only be true if we are assuming C1 = C2 else we would get different answer??

[@ahmadsadigh2439]

Yes

[@kharactertheguru5104]

I’m currently a junior computer engineering student. Could you tell me what course I should have or will learn this in?

[@yuanruichen2564]

electronic circuit; analog circuit design. Those are hardware courses though, unless you really like circuits, sticking to computer engineering has better job prospects🙃

[@kharactertheguru5104]

@@yuanruichen2564 thanks! The curriculum at my school requires me to take hardware courses anyway so I think I’m in good shape. I take electronics next semester

[@ahmadsadigh2439]

At the end of 5th or 6th cycle the Vx is about -1.8 and Vout is about -1.2 Actually neither Vx nor Vout will not be -1.8 and -1.2 respectfully but the voltage at nodes tend to the aforementioned values gradually.

[@koosi987]

Why the delta v across c1 should be maintained?

[@yashwanthmokkapati2102]

capacitor resists the change in voltage...as they mentioned its ideal capacitor it means it wouldnt tolerate any change in voltage across the capacitor

[@senseilisalisa9804]

I can't understand the -1.5V and -0.9V part. Can you explane with equasition?

[@ahmadsadigh2439]

At this point, Vx is -1.8 again and Vc2 is -0.6, apparently the D2 diode lets C1 shares its electric charge with C2 until the voltage across the diode becomes 0.6, so after a very short period of time (ideal components) Vx stops at -1.5 because voltage across the D2 has became 0.6, hence the Vout is -0.9

[@steveocho]

@@ahmadsadigh2439 but why does Vx stop at -1.5V? because Vout is -0.9V? well then why is Vout -0.9V? because VX stops at -1.5V? that doesn't make sense.

[@AdamWright88]

You just said...at the end of the cycle Vout becomes -0.6V and Vx becomes -1.2V, but didn't explain how. You obviously think that part is important because of the graphic "Do you understand what happened here?" So why not explain it?

[@ahmadsadigh2439]

The D2 diode lets both capacitors to share charge, but also stops the sharing at a point which voltage across D2 is 0.6.

[@AdamWright88]

@@ahmadsadigh2439 I don't think it was explained like you just did. It was simply stated....not explained.. Anyway, thank you for the explanation

[@HardwareNinja]

Hi Adam, it's extremely important yet we decided not to explain it so as to let future interviewees work through it themselves so they understand it deeply. Also, in a way we explained it by mentioning everything happens "instantaneously' due to ideal components. If we assume limited resistance (i.e. non ideal devices) on the diodes then it'll be clearly seen what we meant. Hope that helps it clarify our reasoning as to not giving every detail away. Thanks for being here!!

[@AdamWright88]

@@HardwareNinja I love the videos, been studying a bit and they are helpful to practice! Thanks for the great content. I simulated the circuit in LTSPICE and I have a better understanding on how this functions

[@LOLWUT281]

Isn't the onerous on you to figure out how? Esp. if you're doing interview prep?

[@volodymyrsotnikov5121]

Good question, but not the explanation… For those who find it hard - google and read a couple of charge-pump papers. Try to understand how the positive and negative charge pumps are operating and how to get the waveforms. Check if your understanding is correct by calculating output voltage from 0 to 10 cycles in excel and compare it vs paper or vs spice simulation.

[@HardwareNinja]

Hi Volodymyr, thank you so much for the feedback. This time around we decided to switch our strategy to present the answer to the question. Instead of providing the entire solution step by step, we decided to skip steps in key places to let the ninjas work through them themselves. We believe that if they can work through the process, they'll get a deeper understanding of the structure and the question. Hope that helps clarify why we opted for this direction and why the explanation seemed "lacking" details