Why can't an N-Channel FET be used instead a Buck Freewheeling Diode?

Hopefully, an electronic guru will provide some input here.

I've been curious about this ever since I accidentally swapped a FET for the freewheeling diode in a HX-1175b and it seemed to work fine. The driver seemed quite happy and had no heat generation other than the normal heat at the Sense Resisors. It was driving an MTG2 at 11.3 amps with 3S cells. BLF member wight had noticed my mistake and I immediately put back in the stock diode. The driver (with the stock diode) was significantly hotter then with the accidental FET. So I dialed back the current and heat sinked the driver.

Given the limitations of Diodes , a FET would seem much more attractive.

After thinking about it some, it seems like a FET could be used instead of a diode. Here is a simplified buck circuit schematic (Pretend the switch is a FET connected to a Buck Converter):


In a typical buck, the following happens (to the best of my layman's knowledge):

  • The Buck Converter applies power to the FET's Gate Pin and the allows current to flow from the cells through the LED's and an Inductor.
    • The Inductor absorbs excess voltage and current that the LED's don't use.
    • As voltage in the Inductor builds,the current leaving the Inductor increases.
    • The Buck Converter monitors the voltage differential on each side of Sense Resistors that are between Ground and the FET's Source Pin.
    • Once enough current is flowing through the sense resistors to gets the voltage dropout to the Buck Converter's threshold (.25V for the QX9920), the Buck Converter cuts power to the FET's Gate. This cuts off the Ground connection to the LED's.
  • With the circuit to the cells open, the energy stored in the Inductor is released and flows through the LED's and the Freewheeling Diode.
    • The diode allows this current to flow through it because the voltage on the Ground side is now positive.
    • A timer in the Buck Converter switches the FET Gate feed back on. The amount of time can be adjusted in the QX9920 by using different capacitor values between Pin 5 and Ground


So my question is:

Why can't we use a FET instead of a diode (Let's call it a DFET below for clarity)?

Here is how I propose connecting the DFET:

  • The DFET's Gate and Source Pins are both connected to Drain Pin of the normal Buck FET (This Pin is also connect to the LED Negative Lead).
  • The DFET's Drain Pin is connected to Batt + and the Inductor.


Here is what I think happens:

  • When the Buck is applying power to the normal FET's Gate Pin, The DFET's internal "switch" is closed as it's Gate Pin is getting no positive charge. So current is flowing through the normal buck FET.
  • When the Buck cuts power to the normal buck FET, the remaining positive voltage flowing from the LED's is no longer going to Ground. So a positive charge now makes contact with the DFET's Gate Pin. This opens the Gate.


Where I get confused how is the DFET letting current flow because it would seem that seem that the Drain Pin would no longer have a more positive charge than it's Source Pin. If that is the case, why did my HX-1175 work with a DFET? I'm missing something about how the Inductor is acting once the cells are taken out of the circuit.

EDIT: Fixed some typos and clarified some wording.

Check out this wikipedia entry.

An inductor stores energy in the magnetic field it produces. When the cell is disconnected, it collapses, and that energy is released into the circuit.
(A capacitor stores energy in the form of electrostatic charge)

I couldn’t follow your thoughts completely, how that tidbit helps. :8)

Also the diode would get hot because when passing current the power dissipated (heat) would be the current times the voltage drop across the junction, usually 0.6V for silicon. FET’s don’t have junctions so less heat.
Hope THAT helps.

dthoang,

Thanks for the link. I skimmed it. Seems relevant, but I will need some time to read it slow and understand it. It's deeper than my electronics knowledge. So it will take me some time to digest.

dchomak,

I was aware of those facts, but sure appreciate you making sure I was. What I'm wondering is what charge is present on both connections to the Inductor while the magnetic field is collapsing. I picture the Inductor basically having momentum in it from the flowing electrons. So it would seem the side that was connected to Batt + and the Diode being less positively charged than the side that is connected to the LED + Lead. I guess even though the Batt + side is less positive, it must be still higher than what the side of the Diode that connects to the LED Neg Lead is. It would have to be for my "DFET" to have worked because a N-Channel FET has to have a more positive charge on the Drain Pin than the Source Pin for it to allow current to flow through it.

The short answer is that it can and is, but to get the full advantage of it, it must be switches on/off exactly at the correct time. This is called synchronous rectification. You can get chips that is designed to control a FET for synchronous rectification.

When the field collapses the current will flow in the opposite direction of the current that created it in the first place.

I just realized that a “freewheeling” diode is another name for a “flyback” diode.

Question, what is an “ImA4wheeling” diode? :bigsmile:

Probably a big one. 4 devices in one package.

Yes, an inductor acts like momentum. That is why I like the term freewheeling diode. A buck converter is like an old Saab car or an old American car with overdrive that saves gas because when you don’t need power the engine drops to an idle.

dchomak wrote:

Question, what is an “ImA4wheeling” diode?

I don't know, but I imagine folks would say that is one cool diode.

HKJ wrote:

The short answer is that it can and is, but to get the full advantage of it, it must be switches on/off exactly at the correct time. This is called synchronous rectification. You can get chips that is designed to control a FET for synchronous rectification.

Thank for that info. So it sounds like in the case I mentioned in the OP that I may have lucked out and it worked fine. Maybe not ideally, but good enough.

I wonder if, for this very simple application, it works fine without switches to time on/off. The voltage change to the "back side" of the "DFET" when Ground is disconnected would allow current to flow, which is what is needed at that moment (Maybe not fast/early enough?). I need to get handy with an oscilloscope. I will convert a HX-1175 today and try to figure out what is going on here.

The body diode in a FET is a rather poor/inefficient diode. That is why FETs are not used as simple diodes. If you can control the gate properly, a FET makes a very good synchronous rectifier.

Just a suggestion:
http://electronicdesign.com/power/mosfet-design-basics-you-need-know-part-2