Пікірлер

• 
  @ThePowerofInspiration-ym7vrКүн бұрын

  Thank You beta, very helpful

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  @poojashah6183Күн бұрын

  Waiting since long, amazingly explained 🎉

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  @milestoneguy_212Күн бұрын

  Amazing

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  @mayurshah9131Күн бұрын

  Always happy with your video ❤🎉

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  @ALLABOUTELECTRONICSКүн бұрын

  _The link for the other useful videos:_ *1) Types of Resistors:* kzread.info/dash/bejne/mH6qkqZveMfRf8Y.htmlsi=E4Qv7r4bLziJEqOF *2) Types of Capacitor:* kzread.info/dash/bejne/h4F-3Khxj8LgqKQ.htmlsi=DXPLdfPhr-ygsu6W *3) Heatsink Explained:* kzread.info/dash/bejne/fnVqqsGrd9Kwo6w.htmlsi=CRLSxJDpgHNjHJZw

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  @alexisdauli8537Күн бұрын

  Thanks for this video. It helped a lot.

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  @4hdpraju896Күн бұрын

  Bro earned respect from my side 🙌

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  @Lina-025Күн бұрын

  Life saving, thank you

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  @inteluhdgaming1524Күн бұрын

  at the ending, you called Vo/Vin the attenuation, I though it was the gain and attenuation was the inverse?

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  @inteluhdgaming1524Күн бұрын

  Your derivation was so much better, my lecturer Laplace transformed the circuit and ended up getting 3x3 matrix and the derivation was worth over 5 pages

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  @ompurohit65802 күн бұрын

  mast madarchod lecture tha 😍😍😍😍😍😍😍😍

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  @marvin57292 күн бұрын

  Hello from Germany. Very nice explenation! I only have one question: What would the numer at minute 5:29 be in Decimal (its the -0.00011 * 2 ^-126). Could you give me a Calculation method for that pls <3 ?

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  @ALLABOUTELECTRONICSКүн бұрын

  Yes, sure. First, you need to separate the given HEX number in 32 bit binary number. Then write it in a different segments. (like sign, exponent and mantissa). Here, since exponent is zero and mantissa is non-zero, it means the given number represents de-normalized number. So, it will be in the form ± 0.000 x 2 ^-126. That means here, exponent is 2^-126. Now, here since the mantissa is 00011. So, overall number will be 0. 00011 x 2^-126. More over since sign bit is 1, so number is negative number. Therefore, the equivalent decimal number is - 0.00011 x 2^-126. I hope, it will clear your doubt.

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  @nidhinpgdi43762 күн бұрын

  bro i have exam in 2 weeks idk anything about binary thanks to you atleast i know half adder and full adder

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  @athoiba992 күн бұрын

  The point is to convert dc to Ac then why is there a sine wave input ?

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  @poojashah61832 күн бұрын

  Awesome explanation 🎉

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  @benahmad58672 күн бұрын

  Just excellent !!!

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  @Dytalks2 күн бұрын

  Everything clear ❤

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  @bhagyanagarapu27582 күн бұрын

  Please provide notes in pdf sir

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  @user-dn1ig6ug1f2 күн бұрын

  2:12 why the output of this xor gate is equal to 0? Maybe previous stage is 0. I didn't undarstate that.

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  @ALLABOUTELECTRONICS2 күн бұрын

  Here, just for explaining, the initial state of the XOR gate is assumed as 0.

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  @Maldiniiiiiii3 күн бұрын

  cihana selam bu videoyu izliyorsan bana ulaş .ALL ABOUT ELECTRONİC ADAMSIN

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  @chetanpatidar15623 күн бұрын

  If my exam question ask me to "explain dual slope dvm" so can I explain "dual slope ADC"? Means are both the terms synonym of each other or if there is any difference,then tell me the difference. 🙏🙏

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  @ALLABOUTELECTRONICS2 күн бұрын

  Yes you can write, as it works on the same principle.

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  @TukwasibweOsbertBoss-il8ec3 күн бұрын

  Better ee

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  @raavimounika56663 күн бұрын

  Thank you sir

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  @karanpalsingh60654 күн бұрын

  This single video covered all my doubts about sop , pos and k map which I was searching every where bit by bit from whole day, Thanks alot

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  @jeeva17634 күн бұрын

  Thank you so much

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  @Idk00374 күн бұрын

  thank you soooo much 🥹🥹🥹

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  @user-cy1rc2dd2r4 күн бұрын

  in active low pass filter at 9:07, if frequency is high gain is 1. then it is still giving amplitude equal to input then how did it become low pass. If should have reduced the output. please correct me

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  @ALLABOUTELECTRONICS4 күн бұрын

  Here, we are amplifying the input signal. So, at lower frequencies the amplification will be more, and at higher frequencies, the amplification will be less. And very high frequencies, the output is the same as input. That means higher frequencies are not getting amplified at all. For example, if your input is coming from a microphone (where the signal level is mV or even less than that, then higher frequencies will not get amplified and will not be able to be heard in the speaker). So, still, this circuit works as a low-pass filter with amplification. I hope, it will clear your doubt.

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  @ogarblessing77344 күн бұрын

  I am having exam in two days and I was having trouble with binary. Thanks alot

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  @chibuzordesmond39374 күн бұрын

  You mixed up which half is foward bias and which is reverse bias, in both the pnp and npn transistor diagrams. 2:55 and 2:32

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  @ALLABOUTELECTRONICS4 күн бұрын

  In PNP transistor, EB junction is forward-biased. (Emitter is P, and base is N), while CB junction is reverse-biased. (Collector is P and base is N). In the NPN transistor, BE junction is forward biased (Base is P and Emiiter is N), while BC junction is reverse biased. (Collector is N and base is P, so higher voltage is applied at Collector side). I hope, it will clear your doubt.

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  @chibuzordesmond39373 күн бұрын

  @@ALLABOUTELECTRONICS but if base-emitter in NPN is foward biased, that means current flows into emittor(N) and out of the base(P), because that is how a [NP] diode would behave in foward bias. But current is flowing out of the emitter in your NPN diagram, which makes the emitter-base reverse bias.

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  @ALLABOUTELECTRONICS3 күн бұрын

  @@chibuzordesmond3937 I would recommend you to watch the basic video on BJT. If you will watch it, then your doubts will get clear. Here is the link: kzread.info/dash/bejne/X4qrsrV9gdDKmrA.htmlsi=B9kH68XWb2-uVTNo

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  @chibuzordesmond39373 күн бұрын

  @@ALLABOUTELECTRONICS will do

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  @chibuzordesmond3937Күн бұрын

  @@ALLABOUTELECTRONICSI've watched it and my understanding now is that current always flows from + to - in the circuit, I've been taking the convention for granted but let me know if that is correct.

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  @shakibahmed35115 күн бұрын

  at 3.09 how the voltage at the base of Q1 is zero???

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  @ALLABOUTELECTRONICS3 күн бұрын

  Since Q2 is ON and operating in saturation, it can be assumed as short circuit. So, one end of R1 is connected to ground terminal. And other end is connected to Q1. That means the base of Q1 is connected to ground via R1. So, we can say that, the voltage at the base of Q1 is zero. (Diode D is reversed biased and will act as a open circuit) I hope, it will clear your doubt.

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  @AlberTesla10245 күн бұрын

  I think there is a mistake, in case of normalized numbers, the biased exponent can be between 0x01 to 0xFE, but in case of denormal number the exponent is 0, which makes the actual exponent equal to 2^-127 not 2^126. To differentiate whether the number has preceding one of not is identified by exponent value which is 1 for all normalized numbers but 0 for denormal numbers. Correct me if I am wrong.

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  @ALLABOUTELECTRONICS2 күн бұрын

  The de normalized numbers are used to represent numbers which are smaller than smallest possible normalized numbers. In single precision format, the smallest positive normalized number is 1. 0 x 2^-126. So, to represent numbers between 0 and this smallest number this denorms are used. The thing of the biased exponent holds true only for normalized numbers. For denorms, the exponent is always -126. And the numbers are represented as 0. BBB x 2^-126. So, with this representation, the numbers will be less than the smallest possible normalized numbers. For example, 0.11 x 2^-126. I hope, it will clear your doubt.

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  @ShaikhNazmusSadatSourav5 күн бұрын

  in my book i see a capacitor parallel with a resistor for entering set reset pulse.. why is it there?

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  @HetvikumariGamit5 күн бұрын

  Is it the part of computer engineering

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  @ALLABOUTELECTRONICS5 күн бұрын

  yes, you can refer it.

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  @md.shahriarabidswapnil6046 күн бұрын

  it was very very good. understood easily and properly. firstly it seemed hard then everything settled

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  @subjrajyotimandal89676 күн бұрын

  sir at 6:18 why are u using the opposite convention of assigning sign to potential differences

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  @ALLABOUTELECTRONICS6 күн бұрын

  Yes, that's right, but It will not make any difference in the final result. Because, if you take the negative sign outside, then we will get the same result. So, I have directly used the second step. I hope you got it.

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  @ric82486 күн бұрын

  How can you obtain Vout when the input signal's frequency is not constant (like in an audio signal)? In other words, is there a way to calculate Vout(t) in terms of Vin(t), R, and C, for any arbitrary instant t?

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  @RakeshNarthu7 күн бұрын

  Thank you Sir for explaining beautifully. can you please share me analog and digital communication ppt

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  @MampiKakoti7 күн бұрын

  Thanks sir for good explanation

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  @noob.48597 күн бұрын

  I have a dought sir, what is the difference between Vbb, Vee and Veb when the base is common. And the volatege is always apllied between base and emitter then why three different notations are used

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  @akashtr99727 күн бұрын

  very very wory🦍

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  @shaileshsonu57397 күн бұрын

  Best ❤❤❤❤❤❤

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  @muktipaul9807 күн бұрын

  After 3 hours 😭 examssssss 🗿🗿

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  @nikhilchaudhari75937 күн бұрын

  Just go slow ur too fast

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  @DebjaniRoy-rq9iw8 күн бұрын

  so helpful..........THANK YOU SO MUCH

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  @gabrielrosso15778 күн бұрын

  obrigado❤

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  @ashvikadoss61538 күн бұрын

  I was stuck on this topic for more than an hour, it simply took me 20 mins to understand this. This is the simplest yet the most efficient video i had seen on this topic, thank you so much sir! This was very very helpful for me

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  @TheHonouredOne078 күн бұрын

  5:33 Virtual Short

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  @singhvijay13248 күн бұрын

  Thanks a lot Sir❤❤❤❤

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  @user-ye7hw7qr5c8 күн бұрын

  I hope you can explain the last two-level carry-lookahead.

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  @vishal-vc8cl8 күн бұрын

  Arigatho