This Will Change How YOU Think of Digital Design 🤔| BCD Multiplier | Brain Teaser#1

[@anjisrikanth5535]

please make more such videos

[@saiprasadszhayi]

One more observation... here MSB BCD digit (b3, b2, b1, b0) is right shifted version of i/p. i. e {a3, a2, a1, a0)} >>1 = {0, a3, a2, a1}=={b3, b2, b1, b0}. and LSB BCD digit toggles b/w 0 and 5 ( a T-f/f with i/p 1 connected to c2 and c0 and rest bits being grounded). Of course you had a much better approach. Thank you!

[@KarthikVippala]

Namaskaram sai Prasad 🙏, thanks for explaining another approach , good luck & great health 👍😊

[@petepeterson5337]

I stopped the video at 0:25 and spent a bit over 5 minutes (shoulda been faster) and came up with: Zero Gate Solution: BCD input bits: in3, in2, in1, in0 BCD output bits most sig nibble, least sig nibble: msn_2, msn1, msn0, lsn3, lsn2, lsn1, lsn0 Solution: msn2 = in3, msn1 = in2, msn0 = in1 lsn3, lsn1 = 0. lsn2, lsn0 = in0 To change the multiplier that gets applied to the multiplicand, a soldering iron and possibly some gates will be needed.

[@KarthikVippala]

Well done boi😎

[@sun759]

Can you please tell what tools are you using to animate and edit videos? Also how long it took you too create this video?

[@KarthikVippala]

Namaskaram gameRoyal🙏, I use Canva , windows movie maker for editing thanks for asking , good luck & great health 👍😊

[@saintenlewis7861]

Hi Karthik Could you please make a video on lockups latches. Why they are used and its advantages and disadvantages

[@jogeshsingh854]

But sir, it doesn't guarantee the system to be stable unless we use any control signal to remove the unknown or high impedance state , would this design be valid?

[@arjunb8263]

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