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Role of Schottky Diode in XL4015 module

eagle1109

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Hello,

I'm wondering the role of a Schottky diode on the output of XL4015 buck converter module. I opened the datasheet of this diode and found different features and could of course have different uses. Now on this buck converter, what would be its role ?

XL4015_2 - Copy.png
 

betwixt

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Explaining as simply as possible, it fills in the gaps between current pulses from the XL4015.

When the internal switch between VIN and SW is closed, current flows straight through , then through L1 to the output. In doing so, a magnetic field builds up around the core of L1. When the internal switch opens the SW pin becomes disconnected inside the XL4015 and with no current passing from it, the magnetic field in L1 collapses. The polarity of voltage across L1 is reversed from then the field was created, this makes D1 conduct and the energy from the magnetic field is added to VOUT. All this happens very fast, the XL4015 switches at 10s of KHz. The diode is a Schottky type because unlike normal diodes, they store very little charge across their PN junction and that allows then to start and stop conducting very rapidly. Obviously this is essential when the switch is operating so fast.

Brian.
 

Easy peasy

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The diode, that diode is an essential element in a buck converter, when the driving switch is OFF, the established current in the buck choke still has a path to flow, through this diode, thus keeping the diode current continuous & preventing large voltage stresses across Vin and the switch o/p if the diode is not there - due to the ability of the buck choke to raise its voltage ( to quite high levels ) to keep its current flowing if it is interrupted ...
 

eagle1109

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it fills in the gaps between current pulses from the XL4015.

so the XL4015 is like the switcher IC that drive the MOSFET at hot area in SMPS. It provides the chopping voltage for current/voltage regulation.
And this one is more compact than the ones in SMPS units.

So now it chops the Vin to the Vout in PWM, there are gaps, how it fills the gaps ? there have to be gaps in PWM signals.

When the internal switch between VIN and SW is closed, current flows straight through , then through L1 to the output. In doing so, a magnetic field builds up around the core of L1. When the internal switch opens the SW pin becomes disconnected inside the XL4015 and with no current passing from it, the magnetic field in L1 collapses.

Yep, I know this sequence of charging the inductor and that it charges up with current and the magnetic field collapses in a reverse direction.

In this point I have a concern of what's the difference between the discharge voltage of the inductor vs capacitor ?

I know capacitor hold the voltage between metallic plates and it can hold the voltage charge for some time up to minutes or hours.
But my understanding of the inductor that, it's of course different than the capacitor, the electromagnetic field builds up between the rounds of wires, when the voltage or current source is disappeared, this electromagnetic field collapses in reverse direction according to the polarity of charge across it same like the capacitor.

The polarity of voltage across L1 is reversed from then the field was created, this makes D1 conduct and the energy from the magnetic field is added to VOUT.

My question here, is where the current passed through D1 goes ?
Also I didn't understand "the energy from the magnetic field is added to VOUT". Since the charge across the inductor has gone through the diode, which energy is added to the output ? Is it the voltage on the Cout ?

All this happens very fast, the XL4015 switches at 10s of KHz. The diode is a Schottky type because unlike normal diodes, they store very little charge across their PN junction and that allows then to start and stop conducting very rapidly. Obviously this is essential when the switch is operating so fast.

Brian.

Yep, got this part, so normal diodes are used in SMPS bridge rectifiers, because they run at 60Hz, but when considering frequencies more than kHz the Schottky diodes are the one that can be counted for.

Also this means I can use certain components in a wide range of operating frequencies below its max capable frequency.
--- Updated ---

thus keeping the diode current continuous

Why to keep the diode current continuous ? where should the diode pass a current ? from the circuit diagram, the current should go to ground rail.

due to the ability of the buck choke to raise its voltage ( to quite high levels )
who would the choke raise the output voltage if it has a collapse current in reverse direction ?

to keep its current flowing if it is interrupted ...
what would interrupt the current ?
 
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Akanimo

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Wow! Lots of questions here. Will attempt to answer the much that I have time to do so.

About inductor current interruption:
During the ON time of the PWM, the main switch is turned ON. During the OFF time, the main switch is turned OFF. Turning OFF the main switch during this PWM OFF time interrupts the current that was flowing through the inductor during the ON time.

About "...the current should go to the ground rail":
Notice the polarity of the Schottky. It is conditioned to conduct only in the forward bias.

About "...is where the current passed through D1 goes ?":
For this also check the diode polarity and the polarity of the inductor voltage during the OFF time. For there to be current flow, there must be a closed loop. At any given instance, during the OFF time, the current flowing through the diode is the same current flowing through the inductor. For simplicity, we can say that the charges that constitute this current are dumped into the output capacitor (and load). This should, hopefully, answer the question on "the energy from the magnetic field is added to VOUT".

About "...what's the difference between the discharge voltage of the inductor vs capacitor ?":
At steady state, during the PWM OFF time, the voltage across the inductor would be approximately equal to the voltage across the output capacitor.

Overall, the Schottky is forced to conduct during the PWM OFF time so as to provide a complete path (or loop) for the inductor current (due to the collapsing magnetic field) to flow.
 

eagle1109

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About "...the current should go to the ground rail":
Notice the polarity of the Schottky. It is conditioned to conduct only in the forward bias.

I want to know the situation of the coil charge polarity during ON and OFF times.

My guess is that during ON time, the polarity is as follows:

XL4015_charge.png



I learned that from a controlled rectifier with inductive load that when the polarity of the input voltage turns to the negative side, the charge in the inductor goes back in the opposite path of the original charge when not using a freewheel diode.

Single-phase-half-wave-controlled-rectifier-with-resistive-inductive-load.ppm




About "...is where the current passed through D1 goes ?":
For this also check the diode polarity and the polarity of the inductor voltage during the OFF time.

Is it the same as in AC circuit where the +ve pulse ends and the -ve starts, the charge in the inductor changes direction and goes in the opposite direction.

Like in relay transistor circuit:
140a82ff5042175ac778ef4c8cb0177c.png



For there to be current flow, there must be a closed loop. At any given instance, during the OFF time, the current flowing through the diode is the same current flowing through the inductor. For simplicity, we can say that the charges that constitute this current are dumped into the output capacitor (and load). This should, hopefully, answer the question on "the energy from the magnetic field is added to VOUT".

Overall, the Schottky is forced to conduct during the PWM OFF time so as to provide a complete path (or loop) for the inductor current (due to the collapsing magnetic field) to flow.

I understand all that but my only concern is just in OFF time I think the polarity of inductor charge get changed in the opposite direction, so how would that work with the diode bias ? Otherwise I get the idea that in OFF time, the diode would complete the path of current to the load so it doesn't hit the XL4015 back just like the freewheel diode.

About "...what's the difference between the discharge voltage of the inductor vs capacitor ?":
At steady state, during the PWM OFF time, the voltage across the inductor would be approximately equal to the voltage across the output capacitor.

The voltages are the same, so does that mean the discharge is the same ?
 

Akanimo

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Okay, let's look at one misconception that I have noticed that you have that might clear up everything. Your diagrams of Post #7 showing ON and OFF times are almost correct except the current direction that you indicated in the bottom one.

Recall that if you charge an inductor, the magnetic field created is in such a direction that when it collapses it induces a current that is in the same direction that the charging current was. So in your bottom diagram, the inductor current is supposed to flow in a direction opposite to what you indicated.

Another way to look at it is that the MOSFET is turned OFF during the OFF time and the so the voltage source is disconnected and the inductor becomes the source as it discharges. Current flows from the positive polarity of the source into the load.
 
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Easy peasy

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Yup the inductor current is generally one way in a buck ckt ( can be reversed when a fet is used for the bottom device - but the control usually limits this effect )
 

eagle1109

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Recall that if you charge an inductor, the magnetic field created is in such a direction that when it collapses it induces a current that is in the same direction that the charging current was. So in your bottom diagram, the inductor current is supposed to flow in a direction opposite to what you indicated.

OK I by nature must agree with you, it's the same as in capacitor work.

But in AC SCR circuits where the load is inductive one, there showing negative spikes with the output signal ? so how to differentiate that with the case of DC chopping with inductive element ?

Another way to look at it is that the MOSFET is turned OFF during the OFF time and the so the voltage source is disconnected and the inductor becomes the source as it discharges. Current flows from the positive polarity of the source into the load.

Yep, got that understood, during OFF time, the stored current would pulsed directly to the load.
 

Akanimo

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...
But in AC SCR circuits where the load is inductive one, there showing negative spikes with the output signal ? so how to differentiate that with the case of DC chopping with inductive element ?
...
I'm sorry, I didn't get the question.
 

c_mitra

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Refer your diagram in post #7.

Inductors and capacitors are very similar if you replace the current in inductors with voltage in capacitors.

The diagram labeled on time is correct. Current in the inductor is increasing and voltage at the left end of the inductor is higher.

When the current in the inductor is constant, the voltage is zero across the inductor. Yes, right. Ideal inductor has zero resistance. And cosnt current produces zero voltage across the inductor.

The diagram labeled off time is not correct. The red arrow below the inductor is pointing in the wrong direction.

When the switch is off, the current cannot increase via the inductor. The voltage on the left side of the inductor is now lower (than the right side).

This causes the current in the inductor to decay at some rate (do not worry much; we are talking qualitatively). The current in the inductor is now decreasing and the voltage across the inductor is now reversed.

So the current in the inductor is now also going in the same direction but is decreasing at some rate. The current goes via the load and the diode. This diode is active only during this off time.

Without the diode, the current would be forced to decay at a very high rate (determined by the capacitor leakage) and that will produce a very high reverse voltage. We do not want that (obvious, right?)

Without the actual load, the R1 R2 pair acts as the load. But you know the rest of the story.

The key point here is that the rate of change of current determines the voltage across the inductor.
 

c_mitra

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It will be useful to remember the basic ideas about capacitors and inductors:

1. For an inductor, const current casues zero voltage drop across it.

2. For a capacitor, const voltage across it causes zero current to flow.

3. For an inductor, if the current is increasing, it acts like a voltage source (with a negative sense). Voltage sources have zero resistance.

4. For a capacitor, if the voltage is increasing, it is acting like a current source. Current sources have infinite resistance.

5. When the current is falling in an inductor, it is a voltage source with a positive sense (can supply positive voltage to a load).

6. If the voltage is falling across a capacitor, the capacitor acts like a current sink.

If the voltage across an inductor is const (non zero), the current increases to infinity.
If a capacitor is fed with a const current (non zero), the voltage across it increases to inifinity.

A circuit with a capacitor OR an inductor must have a resistance in series to limit current or voltage.

When you go into AC circuit analysis, these concepts will become clearer.
 
Z

zenerbjt

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Here is LTspice sim of buck converter...you can run it and see operation of buck and all of its components.
LTspice is free download, and is by far the best way to introduce yourself to an SMPS
--- Updated ---

Also here is a free SMPS course for you
 

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eagle1109

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It will be useful to remember the basic ideas about capacitors and inductors:

First of all, thank you for this reply with a more deep basics into principles of inductors & capacitors.

1. For an inductor, const current casues zero voltage drop across it.
2. For a capacitor, const voltage across it causes zero current to flow.

ok got these, but are these conditions ideal or practical ?


3. For an inductor, if the current is increasing, it acts like a voltage source (with a negative sense). Voltage sources have zero resistance.
Is it the lagging of voltage when connecting an inductor to AC circuit ?

4. For a capacitor, if the voltage is increasing, it is acting like a current source. Current sources have infinite resistance.
the current is sourced from negative to positive, right ?


5. When the current is falling in an inductor, it is a voltage source with a positive sense (can supply positive voltage to a load).
is the voltage supplied to the load al the same voltage that caused the inductor to be filled with current ? or it is a spike voltage that x100 the source voltage ?

6. If the voltage is falling across a capacitor, the capacitor acts like a current sink.

is it going to the ground of the negative side of the capacitor ?

If the voltage across an inductor is const (non zero), the current increases to infinity.

would it reach the same ampere as the voltage value ? e.g. if voltage is 10V supplied to an inductor then would the current be 10A ?

If a capacitor is fed with a const current (non zero), the voltage across it increases to inifinity.

to the point that the capacitor would explode, right ?

A circuit with a capacitor OR an inductor must have a resistance in series to limit current or voltage.

also using a diode and caps with inductor minimize the ripple in the buck converter. got this one from this channel.

How INDUCTOR's work & How to make your own
@3:55
when he disconnects the supplied voltage and the current collapses of inductor, the voltage spike went to 400V ! how this voltage didn't fry the oscilloscope ?
 

c_mitra

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1. For an inductor, const current casues zero voltage drop across it.
2. For a capacitor, const voltage across it causes zero current to flow.

ok got these, but are these conditions ideal or practical ?

Yes, for all practical purposes. Real inductors always have some resistance (unless they are part of the superconducting magnet coils) but that is usually small. Large inductors also have some capacitance (inter turn capacitance; important in some applications). Most capacitors hIs it the lagging of voltage when connecting an inductor to AC circuit ?ave parallel resistance that is so high that we can ignore it (electrolytic types have a small series resistance that can be important in some applications). Also real capacitors have some inductance (rolled foil capacitors in particular) that can be important in some applications.

But when you model them, you show them as independent quantities.

3. For an inductor, if the current is increasing, it acts like a v A circuit with a capacitor OR an inductor must have a resistance in series to limit current or voltage. oltage source (with a negative sense). Voltage sources have zero resistance.
Is it the lagging of voltage when connecting an inductor to AC circuit ?

I find it more useful to consider like this: you have a negative sign before dI/dt and that means when the current is increasing the voltage induced is in the opposite direction and when the current in the inductor is decreasing, the voltage induced is in the same direction. So your markings in the post #7 is correct. Just imagine that the inductor is acting like a battery (that is what I meant to be a voltage source).
If the voltage across an inductor is const (non zero), the current increases to infinity.
4. For a capacitor, if the voltage is increasing, it is acting like a current source. Current sources have infinite resistance.
the current is sourced from negative to positive, right ?

In this case there is no negative sign before dV/dt; the current sense is in the same direction; positive current when the voltage is increasing and the current is negative when the voltage is decreasing.

5. When the current is falling in an inductor, it is a voltage source with a positive sense (can supply positive voltage to a load).
is the voltage supplied to the load al the same voltage that caused the inductor to be filled with current ? or it is a spike voltage that x100 the source voltage ?

No; no.
There is no spike voltage because the current is not shut down abruptly. The voltage is shut down abruptly (the controller produces square pulses) but the current is allowed some path. Without the diode, there will be a big spike voltage applied to the load.

6. If the voltage is falling across a capacitor, the capacitor acts like a current sink.
is it going to the ground of the negative side of the capacitor ?

To fix ideas, consider a model case in which one terminal of a capacitor is grounded and the other terminal is connected to a voltage source. When the voltage is constant, tIf the voltage across an inwould it reach the same ampere as the voltage value ? e.g. if voltage is 10V supplied to an inductor then would the current be 10A ?ductor is const (non zero), the current increases to infinity. here will be no current through the capacitor. If the voltage is increasing, there will be a positive current going into the ground. If the voltage is decreasing, there will be a negative current (current flowing from the ground end).
A circuit with a capacitor OR an inductor must have a resistance in series to limit current or voltage.
If the voltage across an inductor is const (non zero), the current increases to infinity.
would it reach the same ampere as the voltage value ? e.g. if voltage is 10V supplied to an inductor then would the current be 10A ?

We need equation to be used. That depends on the amount of energy the inductor could store during the pulses. Voltage cannot be constant because both charging and discharging processes are exponential.

If a capacitor is fed with a const current (non zero), the voltage across it increases to inifinity.
to the point that the capacitor would explode, right ?

Possible; if you can find a current source with high enough compliance.also using a diode and caps with inductor minimize the ripple in the buck converter. got this one from this channel.

A circuit with a capacitor OR an inductor must have a resistance in series to limit current or voltage.
also using a diode and caps with inductor minimize the ripple in the buck converter. got this one from this channel.

The diode has nothing to do with the ripple; the capacitor acts like a filter. More accurately, a simple low pass filter. The inductor has only one role: to supply current during the off period. It is the controller that takes care of the output voltage by modifying the pulse train sent to the inductor.
 

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