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Virtual ground 

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The idea of a "virtual ground" makes some simplifying assumptions and give us a really elegant and simple way to look at op-amp circuits.

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24 июл 2016

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Комментарии : 55   
@yukitanaka6854
@yukitanaka6854 2 года назад
Understood more watching 30 minutes of these videos than I did writing 4 hours worth of lecture notes. What is life without Khan Academy
@agstechnicalsupport
@agstechnicalsupport 6 лет назад
Video makes the concept of op-amp virtual ground easy to understand. Thank you !
@SSGSS24
@SSGSS24 2 года назад
U basically summed up 30% of my report in 5mins thx 🙏
@jaypandya8629
@jaypandya8629 3 года назад
after so many videos i found this one helpful !!!!
@vinaykumariitkgp8498
@vinaykumariitkgp8498 6 лет назад
Thank Khan academy for making life easy and simple ,and making better;
@boundlessgaming6545
@boundlessgaming6545 3 года назад
wait, so when an op amp is running how it should, both input voltages are equal? and when one of them is connected to ground that means both would be 0?
@whiteking80
@whiteking80 Год назад
How bout voltage follower, can it be used to create "virtual ground" at its output? Looks like it has same meaning.
@alexialu4224
@alexialu4224 3 месяца назад
Thank you so much from Milan, Italy 🙏🙏🙏
@brianm1916
@brianm1916 2 года назад
If both inputs are ground or virtually ground, what's the point of putting this into your circuit?
@diegoromero9433
@diegoromero9433 6 лет назад
Thanks, just thanks
@wtfchazpwnt
@wtfchazpwnt 4 года назад
Good stuff!
@anamikakapoor4745
@anamikakapoor4745 6 лет назад
but why would you take gain equal to 10^6 in a closed loop configuration ? since you have considered closed loop to explain virtual ground .
@anamikakapoor4745
@anamikakapoor4745 6 лет назад
Since gain is not even close to 10^6 in closed loop , how can virtual ground concept be applied to closed loop?
@harivigneshm9170
@harivigneshm9170 5 лет назад
That is a great question please do let us know if you got the answer
@fariahaslam9095
@fariahaslam9095 3 года назад
Did you get the answer to your question?
@aniket9806
@aniket9806 2 года назад
actually it is that much only in closed loop whereas it is infinity in the case of open loop
@sgtzio
@sgtzio 4 года назад
Thanks khan
@97m_it
@97m_it 3 года назад
What if i applied 1v input
@tomc642
@tomc642 5 лет назад
Why do you use the open loop gain “A” in your explanation when the circuits have clearly feedback loops, and the gain is way smaller by orders of magnitude. Maybe I miss something, but this is confusing.
@SadnanSanim00071
@SadnanSanim00071 4 года назад
this is called negative feedback lets say the the non inverting end of the opamp is V(in) and is equal to V(out) = 0 (initially) then when the V(in) from the (+) terminal is slightly increased the pd is multiplied by the infinite gain( ideal op amp property)(the open loop gain is usually very high in a practical opamp) this dramatically increases the V(out) A feedback loop is connected to the v(out) and (-) terminal the v(out) as it increases is fedbackj to the op amp which reduces the potential difference of the inputs which returns to zero the closed loop gain is found by dividing the V(out) to V(in) which returns a very very small value hence the gain is way smaller
@KK2k
@KK2k 6 лет назад
if the differential voltage (v+ = v-) is '0' then the output will be zero, right?
@olafdrake
@olafdrake 6 лет назад
well the thing is, it isn't zero. But for some calculations, like calculating the input impedance, this approximation comes in very handy. Obviously when you use the formula Vout = A*Vin you cannot make this approximation as it would result in an output voltage of zero.
@user-vi3pi9rf7w
@user-vi3pi9rf7w 5 лет назад
@@olafdrake then what do u do?? If Vo = A(Vin) Doesn't work???
@martinmartinmartin2996
@martinmartinmartin2996 5 лет назад
@@user-vi3pi9rf7w Vo ~A (Vin) where approximately is symbolized by ~ ! if A = -10^6 , V0 = -2v then Vin = -2v/-10^6 = +2uv ~ is NOT the same as = !
@117sandraantony3
@117sandraantony3 4 года назад
Great.....
@bidhankhirali
@bidhankhirali 7 лет назад
how does the power supply help getting that 6 volt from 0 volt???
@sohamsdays
@sohamsdays 7 лет назад
We are supplying +12 Volts and -12 Volts from the power supply. Its from that.
@blitzergstrikers1561
@blitzergstrikers1561 6 лет назад
If you're going to cross your T's, you better dot your I's.
@dudeawesome7096
@dudeawesome7096 5 лет назад
What's the application he use
@bradleysmith681
@bradleysmith681 5 лет назад
Virtual ground is used quite a bit in audio design, specifically in mixer applications...
@nachobalasch
@nachobalasch 2 года назад
1:34 Where does [ V(in) = V(0) / A ] come from? It makes no sense to me.
@andreaskern1982
@andreaskern1982 2 года назад
From v_in * A = v_out. So in words, the amplifier amplifies v_in by an amplification factor of A
@pcbekri340
@pcbekri340 3 месяца назад
most people in comments are mad not because of the video but because they couldn't understand the concept 😂although they watched and read alot like me 😅 what i concluded is that most of the people explaining this concept or the opamp in general don't really understand what they are talking about they just memorized the explanation because every single article every single video is saying the same thing word for word i mean come onnn people
@qzorn4440
@qzorn4440 3 года назад
Geee, why when learning algebra in school the books used semi meaningless examples... teach simple op-amps as real world algebra problems... this would also encourage students to learn a little of what engineering is about? most beginning algebra classes are vague on teaching the students on how to use algebra in earning a life-time income and not just to pass the class! thanks...:)
@pariwartanshrestha145
@pariwartanshrestha145 7 лет назад
what if opamp was saturated??
@sheetalswaroopburada
@sheetalswaroopburada 7 лет назад
Pariwartan Shrestha If +Vcc is 12v and Vo is 12v i.e saturation region. At Vo 12v the Vin=12ūV.≈0v. So Virtual Ground exists​ in Saturation Region.
@jianhaowu7368
@jianhaowu7368 2 года назад
Is this Leo's Bag of Tricks?
@JonDeth
@JonDeth 8 месяцев назад
For people still confused, it's a voltage divider. The node created between connecting the 2 resistors in series is half of the total supply; relative to when you first learn series circuits and are measuring voltage drops across a half dozens resistors or so when getting started in your education. If you check the positive of your supply with your meter to the center node, you will see 6v, and when you check the negative of your supply to the center node, you will see 6v. In relationship to the positive supply, the voltage drop across 1 resistor at that point is -, and more positive on it's opposite side, and in relationship to the negative side of your supply, the 6v is +, and more negative on the opposite side. Each resistor required 6 volts to "energize" the atoms into alignment so electrons can skip across their valiance rings. Some op amps and larger devices need dual source because of the bipolar semiconductor material in their integrated circuitry. *Relate it to when you bias the base current of a BJT, and connect a resistor from each side of your power supply to the single electrode of the base.* The op amp is shifting in polarity, and with each change of 1/2 cycle to the other, needs the opposite polarity reference to bias itself properly. It is only a reference of polarity, not actually quite the same as the base of a BJT requiring both at the same time to free up and activate charge carriers. *I've never understood why not a single video or website online bothers to explain it correctly as I just did when many of them are posted by college professors!* It's easily understood when the explanation relates it to a series circuit, voltage drops across devices in the series circuit and measuring each node to determine which side is more positive and which side is more negative.
@hafsamohammadi3489
@hafsamohammadi3489 3 месяца назад
Why does this sound like Nauman Ali Khan is teaching?😂
@paritoshpandey5103
@paritoshpandey5103 6 лет назад
You haven't explained virtual ground,you just explained the configuration which causes it,not satisfied
@tringuyen121
@tringuyen121 5 лет назад
Here's my asuumption, if v+ = v-, and v+ = 0, then v- = 0. This means that v- is "virtually grounded"
@saadanwar99
@saadanwar99 7 лет назад
so basically you can get 6V from 0V? wow
@atomskyjahid1533
@atomskyjahid1533 7 лет назад
Okay Bye You ignored the power supply.
@saadanwar99
@saadanwar99 7 лет назад
ahaan lol my bad
@sohamsdays
@sohamsdays 7 лет назад
hahah
@marcellobernabei4257
@marcellobernabei4257 7 лет назад
Va bene
@abdalgdooskhalil2704
@abdalgdooskhalil2704 6 лет назад
But u already use 12 v for Vcc Lol
@jamalel-omari5716
@jamalel-omari5716 5 лет назад
hek lool... thats an arabic word meaning like i want
@hxhdfjifzirstc894
@hxhdfjifzirstc894 4 года назад
Sounds like Alec Baldwin.
@martinmartinmartin2996
@martinmartinmartin2996 3 года назад
It is a TERRIBLE idea to use the term VIRTUAL GROUND to describe the summing junction of a inverting op amp ! How can one explain that the summing junction voltage v(+) minus V(-) increases >>6uv when Vin frequency > f cutoff ? ?
@nagasainagendra5949
@nagasainagendra5949 6 лет назад
I'm watching it at 1.25x rate
@kunalsingh1944
@kunalsingh1944 4 года назад
Man it's just a 5 min vid. Atleast need that much attention span
@VidarrKerr
@VidarrKerr 5 лет назад
Your algebra is still a sloppy mess. *Also*, it might help... If you are going to be using "v"s and "u"s and also writing your "v"s as "u"s, that you actually write them properly and consistently. Half your "v"s are "u"s and the other half are "v"s. The "u"s are turning into "v"s whenever there is a "u" sometimes, or something. There is almost no logic to it. Drawing your "v"s as "u"s is obviously contrived (maybe some person you want to emulate does it?), because when you are not thinking you draw them as "v"s. I saw you go back and draw the tail on them to make them "u"s after (they were "v"s). Confused?
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