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Switching Mode Supply Wave-forms

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Electroenthusiast

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Well, a thread by me after many months.
I was in discussion about his topic with some fellow members of the forum, and we couldn't conclude with one answer.

Here's a circuit of a DC Chopper.
DC-Chopper.jpg


The output waveforms of a DC Chopper (Buck Converter) is as below.
350px-Buck_chronogram.png


We can see from the images that the output is astable, it keeps changing from 0 to 1. But, yet people call the output of dc chopper as Direct Voltage, as depicted in the image below.
Chopper-Circuit.jpg


My questions are: 1) How can we justify that the output is DC, when the output is not actual DC?
2) In relation to the case above, can we even call any ossilating square waves as DC?
3) Wont the output voltage keep switching ON and OFF the electronic components connected to it?

Chopper-Circuit.jpg

350px-Buck_chronogram.png

DC-Chopper.jpg
 
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Hi

Sticking my neck out here... in a nutshell this is what happens:
Choppers start with AC Input, rectify to smooth DC...and then switch the Chopper transformer at many KHZ (Primary side).
The mains isolated and regulated output (Secondary side) then has to deal with smoothing and rectifying a high frequency AC Voltage.

So, the Secondary has Diodes and Smoothing Caps. To make clean DC on it's side.

The Secondary caps are the ones that wear out, burst, pop. go dry etc. They have high frequency DC to smooth and deal with. They run hot.

Simpler than that I cannot explain it.

Regards,
tvtech
 
The circuit is missing 2 important things, the capacitor across the load and the feedback circuit.
The coil can not increase current instantaneously, so when the switch conducts, the voltage is at the input of the coil while it is still zero at the output. The current increases gradually till the capacitor charges, then the control circuit starts to play its role it it is a PWM or else.
Now the square wave is at the coil input, and during this time the current increases through it gradually building magnetic field, when the switch stops current, the magnetic field decays keeping the current to the load and decaying, there where the diode becomes a must as it provides the current pass to close the loop.
On the other hand, the capacitor will not allow sudden voltage change, so the coil actually compensates for what the load consumes making the change rather softer.
no one says it is clean DC but it includes a ripple of few milli-volts and can be easily be filtered more and more with LC low pass filter
 
1) How can we justify that the output is DC, when the output is not actual DC?
Technically it is d.c., albeit with an a.c. ripple on top

2) In relation to the case above, can we even call any ossilating square waves as 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.

3) Wont the output voltage keep switching ON and OFF the electronic components connected to it?
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 smoother
 
Glad to see you Guys here. Welcome :)

Fixing SMPS is what I do daily. TV's of course. SMPS all work the same way basically ...so I can maybe help in a small way :eek:

Maybe I can share a little practical experience and all. Like why theory is not always the answer in real situations...

Regards,
tvtech
 
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tvtech, that was a simple answer. But i wonder, why none of the circuit on internet (search term 'dc voltage chopper circuit') show those cap and diode. Well, even i felt the same while analyzing the chopper. One correction, DC choppers have DC input and DC output. Can you get me a practical circuit from anywhere?

Maged mab2
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.
 
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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.
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.
 
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.
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.
 
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:
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!
 
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:
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.

BTW, how did you find that book?
 
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.
You have cross marks because this waveforms are wrong. Author show correct one on imag 1.6 page 31.
Start reading from page 22 and read at least to 48.
BTW, how did you find that book?
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.
https://books.google.pl/books?id=zM...S7Aal-IFI&ved=0CDoQ6AEwAQ#v=onepage&q&f=false
 
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.
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.
 
crutschow, How can the direction of current be the same, if the voltage is reversed? I'm not getting it. You see.

+ -
---- [Inductor]---- Current Direction: ------->

- +
---- [Inductor]---- Current Direction: ------->

How?
 
You have all the answer in pdf
An intuitive (but not necessarily rigorous) way to visualize it is shown in Figure 1-8. Here
we note that when the switch is closed (upper schematic), the current is shown leaving the
positive terminal of the applied dc voltage source — that being the normal convention for
describing the direction of current flow. During this on-time, the upper end of the inductor
gets set to a higher voltage than its lower end. Subsequently, when the switch opens, the
input dc source gets disconnected from the inductor. But we have just learnt that the current
demands to keep flowing (at least for a while) — in the same direction as previously flowing.
So during the switch off-time, we can mentally visualize the inductor as becoming a sort of
“voltage source”, forcing the current to keep flowing. For that reason, we have placed an
imaginary (gray) voltage source (battery symbol) across the inductor in the lower half of the
figure — its polarity in accordance with the convention that current must leave by the
positive terminal of any voltage source. Thus we can see that this causes the lower end of
the inductor to now be at a higher voltage than its upper end. Clearly, voltage reversal has
occurred — simply by the need to maintain current continuity.

?temp_hash=8e32c722f26f72a9371bde5e714ef00c


?temp_hash=8e32c722f26f72a9371bde5e714ef00c

This example show that current through inductor flow in the same direction. From upper end to the lower end of the inductor.
 

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How can the direction of current be the same, if the voltage is reversed?

When the switch turns on, it puts voltage on the left side of the inductor. This voltage is higher than the voltage on the right side. Because of the voltage difference, the current through inductor strats to increase, but it doesn't increase immediately as it would with resistor, but rather does it slowly. Eventually it would go up as if the inductor was a resistor, but well before this, the switch turns off.

When the switch turns off, the diode puts low voltage on the left side. The voltage is lower than the voltage on the right side. Because of the voltage difference, the current through the inductor starts to decrease. If it was a resistor, the current would drop immediately (and would reverse if not for the diode on its path). However, with inductor, it doesn't happen immediately. It rather starts decreasing slowly. Eventually it would totally disappear, but well before this, the switch turns on.
 
Years of break from electronics, seems like i need to learn everything from first. 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, but HowStuffsWork explaination seems to be wrong Link
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.

Why do you think the HowStuffWorks explanation is wrong?
 
As nige says.
Expanding a little further an inductor tries to maintain current flow, if you break the supply to an inductor it will try to maintain current flow by increasing the voltage across it inducing a spark if necessary (or another component to fail).
They can also be used to drop voltage.
These are principles of a switching supply.
I hope you get back into electronics.
 
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