HWN - Real "Touch Sensing Hardware Engineer" Questions

Hi fellow (and future) engineers!
If you ever wondered how tech giants and start-ups actually test your knowledge during interviews, this video is for you! In this episode, and thanks to your requests, we examine a real-life interview question from another community member we helped prepare. This video show the power of open ended questions and how to apply your knowledge to answer them. Enjoy!
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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.
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[@aloksahu4127]

Hi, as you mentioned its an Ideal opamp. Why have you considered that output voltage will be left saturated at -Vcc. Ideally it should go to infinity.

[@patrickcordero6673]

what would happen if Vin becomes 0 again? Will the Vx ramp down to 0 at exponential decay while the Vout ramps up to 0 at linear rate?

[@paransaran6718]

Please add Where you referred this topic, like application note

[@placementdas3997]

Hello can you please provide us with a list of interview questions we can practice ?

[@Dariusissocool]

Capacitive touch devices work by detecting the electrical charge from your finger iirc

[@coolwinder]

Amazing!

[@hariharakumar891]

Here Vout = gain x (Vx- 0), and due to Vout not being 0 Vx is changing?

[@CEA9234]

Hmm. Thankfully my interview was not like this lol. They asked me bode plot question, I did derive a transfer function for op Amp circuit too I believe. Power + Analog questions

[@stlo0309]

i reviewed opamps and gave this another try. from what i understand, Vout is (-t/RC) till it reaches -Vsaturation. Till this point Virtual ground is valid. After Vout attains -Vsaturation, virtual ground is not valid and thats the time when Vx at inverting terminal varies as per RC charging circuit. Is this whats happening?

[@HardwareNinja]

Thank you for contributing to the community! You're onto something, please look at the the response we sent on your most recent comment. Cheers!

[@pankajdhingra9985]

Hi, Yes you are right. Vx will try to maintain itself at 0v upto the situation the output voltage doesn't reach -15v. When the opamps output voltage reached -15v, the opamp is no more having a feedback mechanism and the Vx voltage will increase from 0v to Vin (input voltage of input).

[@Maaz_Khurram]

For the Vout curve, shouldn't it decrease from 0 to -Vcc with an RC discharge profile rather than a linear downward slope (since this is an active LPF which would slow down any sharp edges such as a step at the input because they contain high frequencies). Please let me know if my understanding is wrong or if I missed something that makes the Vout drop linear instead of an exponential decay/RC discharge profile.

[@JoaoVictor-rw9he]

I'm about to have an exam in roughly a week that is exactly about this, so I'll either mess it up or get a good grade haha. But the thing is that this is an Integrator Amplifier, so the output is proportional to the integral of the input signal. As the input is a constant line, the output should be a linear function that over time saturates at -Vcc, as said in the video.

[@six286]

Another way to think about this is that as long as the Vx=0, the resistor current (=capacitor’s) is constant, so the capacitor is basically being charged with a constant current. Constant current means voltage ramp. Once the output rails out, this does not hold anymore

[@stlo0309]

ummm..... This did NOT made much sense to me probably because I'm not too comfortable with modes of opamps. But anyways, thanks a lot! Very cool stuff & I hope to see more in future!

[@HardwareNinja]

Thank you for the support! These are very common questions asked on interviews. If you will be interviewing in an analog or hardware domain, we encourage you to get very comfortable with this material.

[@koosi987]

actually, if you use inverse laplace transform on the t.f., you will get : Vout=-1/RC*integral from 0 to t on(Vin(t))dt. this formula is correct until you reach the negative supply voltage, thus you have -Vcc in some place. your explanation is way more acceptable i think. thanks for sharing.

[@StandforAjdetray]

The sama as you 😊

[@chamis8002]

These graphs and explanation are not quite right in my opinion. The step function 0 part is correct. When Vcc step function applied, Vout goes to -Vsat. Then the capacitor starts charging via the voltage difference Vcc and Vout =-Vsat. Capacitor stop charging when it reached Vcc.. So the Vx is goes up to Vcc. opamp out put stays at -Vsat. Please correct me if I am wrong.

[@HardwareNinja]

Chami, Just as general feedback, you cannot start your comment with "this is not quite right" and finish it with "correct me if I'm wrong". The first part already implies that you are sure of your answer while the second one contradicts it :). In any case, we're not sure why you'd say "Vsat", this is an ideal opamp. Hope that helps. Cheers!

[@akhi9980]

Vx will reach Vin, not VCC

[@firstlast-go6vl]

Vin is Vcc at that point.