Technically it is d.c., albeit with an a.c. ripple on top1) How can we justify that the output is DC, when the output is not actual DC?
well, arguably, yes, if a square wave is all on one side of 0v - i.e. the current always flows in the same direction - then it is d.c - even if it falls to 0 at times.2) In relation to the case above, can we even call any ossilating square waves as DC?
no. The coil should keep the current going via the diode when the switch is off. In a real buck convertor with capacitors the current through the load would be much smoother3) Wont the output voltage keep switching ON and OFF the electronic components connected to it?
There's often a misconception about back emf. It's the voltage on the input (driving side) of the inductor that reverses when the input source is opened (which causes current to flow through the free-wheel diode). The output polarity remains the same.................................................
If i'm right the flyback diode excites the load due to back emf. Doesn't it mean the polarity of voltage is reversed? We can't find that happening in the circuit.
So all these days, what i had in my mind was a misconception? People whom i know also think that, and so they did teach me that way. Back emf is not reverse voltage, but is the voltage with same polarity? i.e., current flows in the same direction as like before the switch is open. Yet, it is called as 'Back' emf.There's often a misconception about back emf. It's the voltage on the input (driving side) of the inductor that reverses when the input source is opened (which causes current to flow through the free-wheel diode). The output polarity remains the same.
If you follow the current through the inductor you will see how this works. The inductance acts rather like inertia to keep the current flowing once it is started. For example if you apply a plus voltage to the left side of the inductor then the current will flow through the inductor left to right and generate a plus voltage on the right (load) side also. Now if the switch on the left opens, the inductor will try to keep the current moving in the same direction. This means the voltage on the left side will go negative (back emf) until the diode is forward biased, keeping the current flowing (at least until all the inductive energy is spent). This means the voltage on the right remains positive.
Why is there any voltage even present across the inductor? We always accept a voltage across a resistor without argument because we know Ohm’s law (V = I × R) all too well. But an inductor has (almost) no resistance it is basically just a length of solid conducting copper wire (wound on a certain core). So how does it manage to “hold-off” any voltage across it?
In fact, we are comfortable about the fact that a capacitor can hold voltage across it. But for the inductor, we are not very clear!
A mysterious electric field somewhere inside the inductor! Where did that come from?
It turns out, that according to Lenz and/or Faraday, the current takes time to build up in an inductor only because of ‘induced voltage.’ This voltage, by definition, opposes any external effort to change the existing flux (or current) in an inductor. So if the current is fixed, yes, there is no voltage present across the inductor, it then behaves just as a piece of conducting wire. But the moment we try to change the current, we get an induced voltage across it. By definition, the voltage measured across an inductor at any moment (whether the switch is open or closed) is the ‘induced voltage.’
So let us now try to figure out exactly how the induced voltage behaves when the switch is closed. Looking at the inductor charging phase, the inductor current is initially zero. Thereafter, by closing the switch, we are attempting to cause a sudden change in the current. The induced voltage now steps in to try to keep the current down to its initial value(zero).
So we apply ‘Kirchhoff’s voltage law’ to the closed loop in question. Therefore, at the moment the switch closes, the induced voltage must be exactly equal to the applied voltage, since the voltage drop across the series resistance R is initially zero (by Ohm’s law).
As time progresses, we can think intuitively in terms of the applied voltage “winning.” This causes the current to rise up progressively. But that also causes the voltage drop across R to increase, and so the induced voltage must fall by the same amount (to remain faithful to Kirchhoff’s voltage law).
That tells us exactly what the induced voltage (voltage across inductor) is during the entire switch-closed phase.
Why does the applied voltage “win”? For a moment, let’s suppose it didn’t. That would mean the applied voltage and the induced voltage have managed to completely counter-balance each other — and the current would then remain at zero. However, that cannot be, because zero rate of change in current implies no induced voltage either! In other words, the very existence of induced voltage depends on the fact that current changes, and it must change.
We also observe rather thankfully, that all the laws of nature bear each other out. There is no contradiction whichever way we look at the situation. For example, even though the current in the inductor is subsequently higher, its rate of change is less, and therefore, so is the induced voltage (on the basis of Faraday’s/Lenz’s law). And this “allows” for the additional drop appearing across the resistor, as per Kirchhoff’s voltage law!
Jony130, I read the page 22 and following portion of that topic. (Img 1.3 inductor), i.e., Inductor circuit's corresponding waveforms has a cross mark. I didn't understand what that signifies. I also didn't understand how there would be a current when switch is open.And I highly recommend you to read this pdf, starting from page 22 "Understanding the Inductor".
https://www.elsevierdirect.com/samplechapters/9780750679701/9780750679701.PDF
And some quote from above pdf:
You have cross marks because this waveforms are wrong. Author show correct one on imag 1.6 page 31.Jony130, I read the page 22 and following portion of that topic. (Img 1.3 inductor), i.e., Inductor circuit's corresponding waveforms has a cross mark. I didn't understand what that signifies. I also didn't understand how there would be a current when switch is open.
A couple of years back I was looking for a good buck about switching power converters and I find this book on amazon an I both one.BTW, how did you find that book?
It is a reverse voltage but remember that voltages are relative. Voltages to ground from each end of the inductor are different than voltage across the inductor. As I noted the voltage (to ground) does reverse on the switch side of the inductor. That's the "back EMF". But the current direction is unchanged. The resulting inductor voltages are the result of the inductor's resisting any change in current flow. When the switch is ON the inductor current is increasing and the inductor is storing energy (the voltage across the inductor is plus-to-minus left-to-right). When the switch is open the current starts decreasing and the inductor is supplying energy (the voltage across the inductor is now minus-to-plus left to right). But in each case the inductor output voltage remains positive to ground because the current flow is always left to right.So all these days, what i had in my mind was a misconception? People whom i know also think that, and so they did teach me that way. Back emf is not reverse voltage, but is the voltage with same polarity? i.e., current flows in the same direction as like before the switch is open. Yet, it is called as 'Back' emf.
How can the direction of current be the same, if the voltage is reversed?
An inductor doesn't "block" AC, it just resists AC, and the amount of resistance (impedance) is proportional to the signal frequency and the inductance............................ Anyone please explain proper the working of 'Inductor'. It is supposed to block AC, but HowStuffsWork explaination seems to be wrong Link
Anyone please explain proper the working of 'Inductor'.
It is supposed to block AC
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?