Why is diode necessary in buck converter?

[alternatingcurrants]

Hello all,

If you look at Figure 1 here

https://www.maximintegrated.com/en/app-notes/index.mvp/id/2031

I am wondering why the diode to ground is necessary.

If the switch is closed, the inductor is charged, and then when the switch is opened, the inductor energy actually has nowhere else to go except to the capacitor (and load).

So why bother having the diode to ground?

When the switch is closed, the circuit already is grounded with the capacitor.

 

[(*steve*)]

What happens when there is current flowing in the inductor and the switch is opened?

Remember that you cant quickly change the current in an inductor.

 

[alternatingcurrants]

What happens when there is current flowing in the inductor and the switch is opened?

Remember that you cant quickly change the current in an inductor.

When the switch is opened, the current stops flowing into the conductor, the magnetic field begins to collapse, and voltage is generated, pushing current into the capacitor.

Back EMF is also generated, thus the field does not collapse instantaneously.

Whether or not a diode is there, I don't see how any back EMF or current can flow to ground. With the open switch, the wire is terminated, which is what a diode wants to attempt to be.

The only other thing I can think of is that the open ended wire may have floating voltage, because it's not ground referenced.

 

[(*steve*)]

Draw an arrow showing the way the current is flowing through the inductor. Where does it flow when the switch is open (hint, the diode is involved)

 

[alternatingcurrants]

Draw an arrow showing the way the current is flowing through the inductor. Where does it flow when the switch is open (hint, the diode is involved)

Ok I think I get it.

The current flows in the same direction as what the power supply was delivering, after the switch is opened.

But the circuit is not complete, without the diode-ground connection.

So, we can consider the ground to provide a sink for electrons, which are going into the ground, opposite the conventional flow, which is going into the capacitor.

Correct?

 

[(*steve*)]

No, the current flow is from ground, through the diode and then the inductor to the capacitor.

The diode is forward biased. It is reverse biased when the switch is closed.

 

[Arouse1973]

It also doesn't have to be a diode. MOSFETs are now used in some buck convertors, they are called synchronous convertors.
Adam

 

[alternatingcurrants]

Sorry I don't understand then. Why would the conventional current come from ground once the switch is opened?

The capacitor is grounded to the same reference point (same ground potential), which is lower, so ground current coming up through the diode doesn't make sense to me (it is the same lower potential as the capacitor ground).

 

[(*steve*)]

you need to review inductors. The reason is that there is energy stored in the inductor. The diode allows that energy to be dumped into the capacitor while preventing it from flowing from the cap back into the inductor.

 

[BobK]

Draw the circuit with switch open and no diode. Where is the path for the current? Remember, it has to be a closed circuit.

Or, think of it this way. What if the inductor is replaced by a battery. Does current flow into the capacitor?

Bob

 

[HellasTechn]

So is it there to protect the capacitor ?

 

[BobK]

No, it there to complete the circuit!

Look at the circuit without the diode. When the switch is off, the inductor has one side connected to nothing. How does current flow out of the inductor with no connection to one side?

Bob

 

[alternatingcurrants]

you need to review inductors. The reason is that there is energy stored in the inductor. The diode allows that energy to be dumped into the capacitor whine preventing it from flowing from the cap back into the inductor.

Yes I understand the energy is stored by a magnetic field, which later collapses. But why not just put a diode between the coil and the capacitor? That would block the current going back into the coil.

 

[alternatingcurrants]

No, it there to complete the circuit!

Look at the circuit without the diode. When the switch is off, the inductor has one side connected to nothing. How does current flow out of the inductor with no connection to one side?

Bob

Yes that's what I said before, but nobody commented on it.

I already said, is that grounded diode there to complete the circuit? BUT I also said that it doesn't make sense, because that grounded diode "completes" the circuit by making that side of the coil a LOWER voltage potential (since it is grounded), and the other side of the coil is ALSO grounded through the capacitor.

Both the diode and the capacitor are grounded at the same LOWER potential point. So, that doesn't make sense to me.

 

[Arouse1973]

If you were able to energise an inductor in free air and you remove the connections at exactly the same time. One side of the inductor goes negative and the other side positive. I dont know if that helps.
Adam

 

[(*steve*)]

That would block the current going back into the coil

But how would the stored energy get from the inductor to the capacitor without some path through ground?

Draw the inductor as a battery with the positive terminal nearest the capacitor. (its actually more like a current source, but I assume you're not familiar with those)

 

[NuLED]

Ok thanks for all the replies. I think I am starting to get it, but funnily enough not exactly the way you guys have been explaining it.

Here is how I see it now, please see if this makes sense.

I understand the circuit needing to be complete. So, ignoring the ground connections, let's just connect all the grounds together (that's what they are anyway). So now you have the complete circuit from the negative side of the capacitor, to the other negative side of the coil.

But when the switch is closed, the power supply needs to energize the inductor. If there is a direct short to ground without the diode, the coil would not get properly energized. That is why we need the diode there, blocking the power supply current, so that the current goes to the coil, and not to ground.

Once the switch is opened, the coil field collapses, and current goes into the capacitor. At the same time, the diode blocks current shorting to ground.

(However I don't understand the comment regarding the possibility of using a MOSFET instead of a diode.)

 

[davenn]

But when the switch is closed, the power supply needs to energize the inductor. If there is a direct short to ground without the diode, the coil would not get properly energized. That is why we need the diode there, blocking the power supply current, so that the current goes to the coil, and not to ground.

No, the diode is only in use when the switch gets opened ( when the inductor is de-energised - power supply removed from it)

have a read of the theory ....

https://en.wikipedia.org/wiki/Buck_converter