# Designing a high efficiency Buck converter ~2kw

Discussion in 'The Projects Forum' started by sroddier, Sep 5, 2015.

1. ### sroddier Thread Starter New Member

Sep 5, 2015
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Hello everyone,

I'm a professor in a french university, every year our student are working on a wind turbine (urban) competition.
Our wind turbine is working very well, everything is done by us (electronics, generator, blades etc...)
We are still strugling on the electronic converter, we need a buck converter in order to allow the wind turbine to work in a good condition.
We use an arduino with a 33000hz switching frequency, we need to vary the output voltage between 0-100%.
The input is between 0V-80V, the current is 0-25A.

Our actual buck converter is not good enough, we used a mosfet at the ground side to be able to switch it, and a shotky diode. It is only 85% efficiency, we want a much higher efficiency (the cost is not a problem).

Our new design is looking like this:
MOSFET IRFP4110
Self: Epcos, 0,57 mH, 35A, 100kHz, 1,4mΩ Rdc
Capacitor: 22µF (any idea the best techno choice?)
Driver: MIC4424YN 3A
Diode: STPS61150CW

Any idea or help would be great. Thanks

Stephan

2. ### ronv AAC Fanatic!

Nov 12, 2008
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What is the desired output voltage?
What to do when the input falls below the desired output?
Is it charging a battery? Or?

Sep 5, 2015
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4. ### sroddier Thread Starter New Member

Sep 5, 2015
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The output is depending of the wind speed (between 0-80km/h), so the maximum power available is between 1W to 1500W.

It is a competition so there is 2 kind of load, one very exotic (doesn't exit in the real world) that follow this equation: I=U²/156
The second one is a battery of 48V.

The arduino program will be looking to get the maximum power from the wind turbine, like a mppt for solar panel.

5. ### Papabravo Expert

Feb 24, 2006
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A schematic diagram of your circuit would be most helpful. A buck converter has an output voltage lower than the input voltage. It is just not practical to specify that the converter continue to work at very low input voltages. A practical range might look like 12-80VDC. Is the input current limited to the 0-25 Ampere range?

At these power levels of 2 kWatts you need to pay careful attention to at least three things.
1. rds(on) of the MOSFET switch. Even 0.25 ohms will consume 6.25 watts at 25 Amperes. To achieve high efficiency you cannot afford this.
2. Lowest possible ESR(Equivalent Series Resistance) in the capacitors. Don't take the manufacturer's word for it you need to measure this.
3. Inductor must have a core material that will not saturate at twice the potential output current (~50 Amperes), and be wound with wire large enough to limit the temperature rise at twice the output current (~50 Amperes)
BTW 85% efficiency for a first time design is pretty awesome; I'm impressed. Getting to 95% will be a challenge.
bonne chance!

Edit: I see you posted the diagram while I was composing my response.

6. ### Papabravo Expert

Feb 24, 2006
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The component you list in your schematic "IRFP4110" does not appear to exist in the wild. Perhaps you meant the :IRFB4110"
The rds(on) of this part is specified as 3.7 mΩ which is better than I expected. There is a problem however with turning the MOSFET fully on. The Vgs(th) is specified as 2.0 V(min) to 4.0V(max). this means you are just barely turning the part on. The maximum specification on Vgs is +20V, and I'm guessing you need +10-12V to really turn that part on. Connecting Vs of the MIC4423 to the Arduino's +5V just isn't going to get you where you need to be.

7. ### sroddier Thread Starter New Member

Sep 5, 2015
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1. For the rds on, we tried to find the lowest that can suit our goal: ~0.004 ohms difficult to go much lower

2. We will look into the ESR
3. In the specification of the "Epcos, 0,57 mH, 35A, 100kHz, 1,4mΩ Rdc" do you think 35A rate is the maximum current for the wire or for the magnétic field?

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8. ### sroddier Thread Starter New Member

Sep 5, 2015
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Yes you are right, we need to get an external alimentation for the MIC4423, we would probably put something like 15V. Do you know if it is better to put the mosfet on the highside?

9. ### sroddier Thread Starter New Member

Sep 5, 2015
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At very low voltage, when the wind turbine is starting, we want it to be completely off, then around 10-12V we want to start from getting power from it. The 0-25A is not current limited, 25A is the maximum current that we will have at 80km/h under the 48V load.

10. ### crutschow Expert

Mar 14, 2008
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You should consider a synchronous rectifier design which minimizes the losses of the free-wheeling diode.
At a minimum you should use a Schottky diode for that purpose. You show a standard silicon diode, which will have a high loss (the schematic shows a 1N4001 but that's only a 1A diode so I'm confused there ).

Putting the MOSFET in the return side is fine.
If you placed in in the high side then you would need a gate voltage supply 10V above the output (transistor source) voltage.

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11. ### sroddier Thread Starter New Member

Sep 5, 2015
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Thanks,
Sorry for the schematic, i didn't have a schottky in fritzing (i didn't find one, to be exact), so we would use at least a schottky.

The synchronous rectifier needs a mosfet on the high side, isn't it? Can we get 100% ratio (full on)? I've looked into it, and it seems we need a boost cap (which need at least 5-10% to get working).

12. ### Roderick Young Member

Feb 22, 2015
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I agree with crutschow, and think the largest boost to efficiency will be going to synchronous rectification, that is, replacing the diode with a MOSFET (possibly another 4110, or several).

The 500 uH main inductor suggests to me that your switching frequency is very low. That's good in terms of switching losses - less power burned because the transistor spends less times per second in the linear region. I also see that you elected to use a MIC4423. Very smart to use a prepackaged MOSFET driver. In general, the more quickly you can bring charge and take it away from the MOSFET, the better it will perform. You should experiment with the gate drive voltage. Conventional wisdom says to use 10 volts, but you may find an operating point that is slightly more efficient.

If you are getting 85% efficiency now, the inductor is probably fine, but it's worth looking into whether you can use heavier wire for less DC resistance, and whether you can make the inductor physically larger to get even less saturation of the core.

I'm not sure if it's just a single MOSFET that you are using, or whether that stands for several MOSFETs in parallel (with individual drive circuits). A 2 kW design typically doesn't have just one single switching element, it has many. Especially at low voltage, you will be switching high currents. With a single MOSFET, that would put all the current through a single trace on a PC board. I think it would be better to distribute the current.

With a low switching frequency, you will need large capacitors. Even if the load will tolerate a lot of ripple, maybe it can't absorb power well with huge current surges, and the capacitors will absorb those. It's tempting to use electrolytic capacitors to get the high capacitance you need, but if you can use all ceramic capacitors, the ESR will be much lower, and you will not be dissipating power in the ESR. That is, assuming that your PCB layout doesn't nullify the advantage of low ESR with trace resistance and inductance.

I would love to hear how your project turns out after everything is done. Perhaps you could post either details or a link to details here after the competition is over?

sroddier likes this.
13. ### crutschow Expert

Mar 14, 2008
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Yes, the drive for the synchronous rectifier would be in the high side.
Perhaps it would be better to put the series N-MOSFET in the high-side with a bootstrap driver for that.
Then the N-MOSFET synchronous rectifier would be in the common side (source to common).

14. ### sroddier Thread Starter New Member

Sep 5, 2015
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Thank you very much again everyone,
I surely will post the result and documentation once it will be working.

Do you know if there is any tricks to take care when we use several MOSFET, i know by experience that real life can be hard compared to the black board... the 2kw will be only if the input is aroud 100V, the 25A is the maximum amp we will have to deal with.

Do you know what might happen if we get near the 100% duty cycle with the bootstrap driver?

The switching frequency is 33khz from an arduino UNO

15. ### crutschow Expert

Mar 14, 2008
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The bootstrap capacitor charges quite rapidly since it only has to provide the gate charge to the MOSFET, so even if the duty-cycle is near 100%, as long as it still goes to zero for a short period, you should be okay.

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16. ### sroddier Thread Starter New Member

Sep 5, 2015
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I think we will try both achitecture, so we can compare the behavior and the efficiency of each solution.

Do you know how to calculate the value for the bootstrap capacitor?

17. ### crutschow Expert

Mar 14, 2008
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It just has to be large enough to supply the gate charge without significant droop of the capacitor voltage.
I would say if it stores a charge at least 100 times the gate charge as stated in the data sheet (giving a 1% droop), you should be fine.

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Nov 12, 2008
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Sep 5, 2015
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Thank you!