HWN - You're guaranteed to be asked this question on your interviews - "Back to Basics"

Hi fellow (and future) engineers!
We're back after a week off. We're taking you back to basics with this video as a lot of community members have requested it. We'll post a few of these questions before we jump back to open ended questions.
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[@biprobhadra1102]

One baby doubt How you intuitively neglected 3R at the time of gain calculation i don't get it. Yeah! Gain will be -2 but how you approach to find the gain neglecting 3R that's can you please elaborate i just mess up at that point.

[@shimshonthehero]

so basically its like 2 voltage source one driving the 3R and one driving the others and you solve it with superposition ?

[@RitayanMitraggmu]

Don't know whether its a silly doubt. If so, sorry for that. What is the significance of 3R in this circuit? I mean without 3R also, we are getting a gain of -2. I want to understand what exactly this 3R has been used for? @Hardware Ninja

[@HardwareNinja]

Hi Ritayan, Thank you so much for being here and contributing to the discussion. It's not a silly doubt at all! If 3R was not there your input impedance would me higher. In fact, it'd be R instead of R/2. Would you ever need this in an actual circuit? It depends! Also, remember that these are interview questions to test your circuit analysis knowledge. Sometimes these questions won't have true applications but are more brain teasers :)

[@AyushAgarwal-YearBTechElectron]

how come I1 and I2 dont add up to 0 , where is the current draining in this circuit ??

[@mrc7153]

A faster way is to assume input impedance is high (so an open circuit) and output impedance is low (so basically a short). In that case we are looking at the resistance of 3R in parallel with R + 2R which equals R/2.

[@cpK054L]

This would make a bad example because the if the gain is 2, then 3R is technically "open circuit" But what do I know? Haven't seen an actual opening for Hardware Engineering in the US unless you're dealing with FPGAs

[@HardwareNinja]

Hi, Thank you so much for contributing to the discussion! Could you elaborate why you'd say that 3R is an "open circuit"? Maybe we're missing something and would love to let the community know about it.

[@cpK054L]

@@HardwareNinja the inverting OpAmp's gaim function is "-R feedback / R input" which just so happens to be -2R/R or a gain of -2 Which is basically what I believe I see here.... it could confuse some people who are not familiar with ideal opamp circuits, perhaps a different resistance value on the shunt to elaborate?

[@HardwareNinja]

​@@cpK054L Oh, we see what you mean. It's not that 3R is "open circuit". Instead, it should be taken as there's not feedback on 3R. If Vout is at a lower potential than Vin, then Vin will simply source current to the output of the opamp and the opamp, being ideal, will just sink that current. If Vout is at a higher potential than Vin, then the opposite happens. This is not a true implementation of a circuit but instead used to evaluate candidates in their knowledge. We hope we were able to give some more clarity here. Thank you so much for being here!

[@cpK054L]

Having a shunted resistor on the Vin and Vout is a very confusing implementation... The load you have with proportional resistance will let R3 and (R1 and R2) have the same current running across it. However, if you changed any of the values to say....R1 + R2 to have less resistance than R3, it will still allow R1 and R2 to have the same current, but draw more than R3. Sometimes I'm honestly confused if I actually saw implementation like this in real life. Even worse is when the designer doesn't write notes as to WHY they did it that way.

[@ahmadsadigh2439]

if the gain was +1 then yes, the R3 would be "open circuit" but in an "inverting amplifier" with "positive resistors" this wont happen.