buck converter

Discussion in 'Mathematics and Physics' started by PG1995, Apr 4, 2014.

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Hi,

What voltage are you trying to find?

Also, the cap would not be 10pf as that is way too small, especially with a 1kHz switching frequency.

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2. PG1995Active Member

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Hi

The voltage on the left of inductor titled "Voltage?" in Figure 3 just before the switch is opened.

Then, you might make it 50pF. I just made up those numbers.

Thank you.

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Hi,

We already said what that voltage was didnt we? It's the same as the power supply voltage.

50pf is still far too small. More like 1uf or 10uf, but at 1kHz much much larger. Consider at 50kHz a typical value is 1000uf.

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5. PG1995Active Member

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Hi

I'm afraid that no one said anything about that voltage, or I might be missing it.

Let's step through Figure 2 and Figure 3 here again.

In Figure 2, the switch has just been closed. Before the switch was just closed, both inductor and capacitor were supplying enough current that voltage drop across 20 ohm load was 9.8V and this goes without saying that note before the switch was closed the capacitor was being discharged and inductor's current was decreasing. But the capacitor starts getting charged and inductor's current also starts increasing from its previously descending value as the switch is closed. Basically two things are happening. First, the voltage on the left of inductor starts decreasing from applied voltage value which in our case is 20V and meanwhile magnetic field of inductor also starts expanding. Second, the capacitor is getting charged up and at the same time voltage is being dropped across the load and this means that current is being supplied to both capacitor and load. These "two things" will continue until voltage drop across the load (or, capacitor's voltage) has reached the value of 10.2V and at that instant switched is opened again.

And you can make it 50uF if 50pF is too small. Thank you.

Regards
PG

6. NorthGuyWell-Known Member

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When the switch is conducting, you get the source voltage on the left of the inductor, load voltage on the right of the inductor. The inductor is capable of keeping this voltage difference only if the current through the inductor increases. If the rate of current increase slows down, the ability of the inductor to develop voltage diminishes and the voltage accross the inductor will start dropping. As a result of this voltage decrease accross the inductor, either voltage on the left has to decrease or voltage on the right has to increase (or a combination of both).

If you have industrial power supply on the left and resistor on the right, then power supply will wins. Voltage on the left keeps steady, voltage on the right increases.

If you have a solar panel on the left and battery on the right, then battery wins. Voltage on the battery changes little, but panel voltage drops.

If you have an industrial power supply on the left and battery on the right, they'll fight hard and you better be at a safe distance when inductor saturates.

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Hi,

In figure 2 when the switch is open the inductor is supplying
current to the cap and load (assuming continuous operation)
and the diode is conducting, so the voltage on the left of the
inductor is about -0.5 volts. The voltage across the inductor
is 9.8+0.5 volts with the right side more positive, and the
current through the inductor is near some minimum value but
is not zero, and is flowing into the cap too.

When the switch is first closed, the battery current flows
through the diode in the reverse direction and within a very
short time the diode recovers and the voltage on the left of
the inductor ramps up very quickly. As this voltage rises
the battery begins supplying current though the inductor at
a level equal to the level just before the switch was closed
with the inductor current now decreasing a little more.
At some point the ramping voltage becomes greater than 9.8 volts
and so the inductor current starts to increase and capacitor
voltage starts to rise more.
As the voltage across the diode increases more, eventually the
current through the series resistance of the battery limits the
voltage at the left side of the inductor to the source voltage
(20v) minus any drop in the series resistance. Normally this
drop would be small so almost the entire 20v appears at the
left of the inductor, and current through the inductor increases.
As the current continues the cap charges up to 10.2 volts.

Assuming the switch then opens, the voltage on the left of the
inductor ramps down very quickly as the inductor tries to maintain
the current through itself. Eventually this voltage reaches about
-0.5v and the diode again begins to conduct and we are back where
we started from.

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8. PG1995Active Member

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Thank you, NG.

Could you please check through the text below to see if my understanding isn't that much off the track. In the text below, redundancy and verbosity is intentional!

The voltage across the inductor is directly proportional to the time rate of change of the current: v = L*di/dt. Let's call it eq.1.

We can also write i = (1/L)∫v(t)dt +i(t0) where limits for integral are from "t0" to "t" and v(t) is voltage across the inductor. Let's call it eq.2.

So, if eq.1 tells that there has to voltage of 2V across an inductor then it means that left side of inductor is going to 2V more positive than the right assuming current is flowing from left to right.

These figures might be useful. Don't forget that in Figure 22, i(t) and v(t) are current thru the inductor and voltage across inductor respectively.

This is my interpretation of what you say above. If we have a supply on left of an inductor and on the right of inductor there is a load connected. Assuming that the current increases linearly thru the inductor. Slowly the voltage on the left of inductor starts decreasing but the current thru it is still increasing linearly which means that the voltage across inductor should remain constant. To make the voltage across inductor constant, the voltage across the load increases.

What do we really mean by the voltage across inductor? Above I had thought that it's calculated in relation to the current flow but now it seems that it's more of an absolute value of potential difference across an inductor.

Please have a look here. Before the switch is closed, the panel's voltage is Voc. When switch is closed, the voltage starts decreasing from Voc and current starts increasing. The battery's voltage won't change but panel's voltage keeps dropping. Note that as the voltage starts dropping, the current's rate of increase will also get lowered which means less voltage across the inductor. The reason for reduced rate of increase of current is that the battery will accept less current as the panel's voltage get lowered because the panel isn't pushing the 'electrons' with enough pressure.

In other words, when switch is closed, Voc starts dropping and at that moment current's rate of change is maximum and hence maximum voltage across the inductor. But when rate of change of current gets very low then voltage across the inductor also gets very low. Suppose, if battery is 6V then the panel's voltage might be somewhere around 6.5V when rate of change of current is very low and hence voltage across inductor is just 0.5V.

You have been specifically mentioning "industrial power supply" in this discussion and from the context I infer that it's a type of power supply which can maintain constant voltage at its terminals and can output a lot of current.

Note that a battery can take in a lot of current during bulk stage of charging.

Let's assume that the battery is 6V one and the supply is a 12V one. When the switch is closed, the voltage across inductor is 6V and current starts increasing. As the supply can supply a lot of current without any trouble therefore the current keeps on increasing in linear fashion which means that the voltage across inductor should remain constant. But there will come a time when inductor becomes saturated and it won't be able to maintain constant voltage across it and becomes a short circuit, and at this point the current increases infinitely rather than in linear fashion. This link might be useful here. Thank you.

Regards
PG

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9. PG1995Active Member

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10. NorthGuyWell-Known Member

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I would suggest di = 1/L*V*dt It is easier to interpret. Voltage put accross the inductor causes increasing current. As soon as there's a voltage, the current goes up. If the system manages to keep the voltage (as in case with battery and industrial power supply), the current will keep increasing until the inductor saturates and no longer works as an inductor. Otherwise, something must give up and voltage will decrease. The decrease in voltage will cause the decrease in current rise rate.

In case of solar panel, the voltage is maintained by the input capacitor, not by the panel.

Q1. I don't know
Q2/Q3: correct
Q4: The only thing connected is the load, wich draws relatively constant current from the capacitor, so it's relatively flat.

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Hi,

Q1:
That's the diode current where the diode has very very little forward voltage drop (for a synchronous rectifier the 'second' transistor switch is a mosfet).

Q2
The cap value would have been chosen to provide a smooth output voltage with the required load, therefore the voltage across the load will stay fairly constant so the cap current will stay fairly constant. In real life it will droop slightly depending on the output ripple spec for the design.

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12. PG1995Active Member

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Thank you, NG, MrAl.

Yes, it is. If a voltage V is put across the inductor for an infinitesimal amount of time "dt" then there is a corresponding infinitesimal increase "di" in current for the given infinitesimal time duration "dt". I hope I have it right.

Yes. But I wasn't thinking of any capacitor at that time. I had only a panel, switch, inductor and battery in mind.

It's still confusing. A load connected to a capacitor makes it an RC circuit where capacitor is discharging. For example, this figure shows discharging current for a capacitor which isn't constant in any way but we can assume it to be linearly decreasing. The diagram for Q4 represents a discontinuous mode of operation where inductor's current becomes zero at some point during off and then its only capacitor which is supplying current. Could you please help me with it? Thanks.

Regards
PG

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Hi,

What you are talking about in #51 is the ripple, and that diagram is not a good reference because that's not the way we view every circuit that has a capacitor and resistor discharging it.

The correct view is to think about the discharge current vs the time vs the capacitor value. If the value is high and the current low and the time short, the cap voltage drops by only a very small amount which would look like a small amount of ripple.
Remember the time constant RC and the discharge of the cap, it only discharges so much in little time and on the scale of the diagram it may not show up as even a one pixel difference.

14. PG1995Active Member

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Hi

I was going through Wikipedia article on buck converter and I found the highlighted portion confusion. In my humble opinion, it should have been "the amount of energy required by the load is too large" instead. Only if the load is too heavy (which translates to 'large amount of energy') then it will 'deplete' the inductor of all its 'stored current' during off period. Do I make any sense? Thank you.

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The article is correct. Discontinuous mode occurs for light loads when the average current in the inductor (in steady state) is less than the peak value of ripple current.

16. PG1995Active Member

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Thank you, Steve.

Let's see if I'm picturing it correctly. Note that light load equates to high resistance which does take a lot of power. In a buck converter if the load resistance is too high then not enough current will be able to pass thru the inductor which will result in less magnetic field around the inductor and hence it results in less ability of the inductor to sustain current when the switch is off. In other words, if it takes 1s to rise the current from 0A to 0.5A, the rise rate of current will be less than the case when the current only takes 0.5s to rise from 0A to 0.5A.

Does this mean that discontinuous mode cannot happen for a heavy load (i.e. low resistance)? Thanks.

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I dont think that is the right way to think of it.

Instead, think of the rate of current change being controlled by inductance and voltage because dI/dt =V/L. Now, the loading does not change the voltages, but only the current. In particular the load changes the average current. But current ripple is affected by voltages and inductance (and PWM frequency, of course), and not loading (as much). Hence, think of light loading as a case where current ripple is greater than average current, so to speak. If the ripple wants to be greater than the average, then the low side of the ripple will hit zero. That's all discontinuous mode is.