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I am designing a buck converter for 600 - 10V converting (60 times lower than input).

My first question - is this even possible? I've seen a 650 - 40V converter on mouser, but that is the best that I found. Is the limitation for output is PWM duty cycle? Because it would have to be almost 0%, to get low output?

Also - would I need two stage converter then? One for 600-40 and another for 40-10?

Should I use npn or pnp? I have designed this: (PWM 10kHz, DC - 2%, gate voltage 620V, NPN with low (2 ohm) Rds) enter image description here

but it has very low (<10%) efficiency. Is it because switching ratio? I increased duty cycle, to get 50V and efficiency reached 26%. The bigger output, the better efficiency, but I still need 10V output. I've read that I need to replace diode with another mosfet, but it just made things worse. Changing LC values and DC or PWM frequency barely changes outcome.

I am green at this - I only designed and built a small boost converter so far. I do not have serious plans to make this (so dont worry about high voltage), I'm just trying to understand how this works.

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  • \$\begingroup\$ A two stage converter would probably be easier and more efficient, at least. Buck converters tend to be more efficient when the output is closer to the input, though obviously not to the same degree as linear regulators. They also tend to be more efficient at higher output currents, up to a point. Replacing the diode with another MOSFET is indeed a good way to increase efficiency, but you have to drive the FET opposite the other one or it doesn't do anything useful. \$\endgroup\$
    – Hearth
    Jun 12, 2021 at 17:57
  • \$\begingroup\$ What do you mean by "should I use NPN or PNP"? You shouldn't use either--an N-channel MOSFET is generally the best choice for the job. \$\endgroup\$
    – Hearth
    Jun 12, 2021 at 17:59
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    \$\begingroup\$ Buck is not suitable for such a high Vin/Vot ratio. More efficient is to use one of transformer converters (flyback). With buck is very hard to regulate so low duty, you will deal with duty about 1.6% and even 1% of its change will make huge Vout change. Also consider switching losses, they are not neglibigle comparing to output power you will get. \$\endgroup\$
    – user208862
    Jun 12, 2021 at 18:11
  • \$\begingroup\$ so what is the reason of low efficiency? Where does it all go? Though mosfet? How could I increase its efficiency? this dc-dc converter seems to be quite small, so there is no transformer in it. But it somehow manages to reach up to 97,7% efficiency. How could I do that? Are there some secrets to buck converters? \$\endgroup\$
    – Ri Di
    Jun 12, 2021 at 18:57
  • \$\begingroup\$ @RiDi "this" converter pretty much has a transformer inside. \$\endgroup\$
    – fraxinus
    Jun 12, 2021 at 20:10

2 Answers 2

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The main efficiency hog in a switched converter of any type is the switching time of the switch.

If the switching time gets comparable to "on" time, you get low efficiency - the power transferred is proportional to the "on" time and the switching loses are proportional to the switching time.

Here, you have 10kHz and 1/60 duty cycle, or about 1.6us "on" time. A quick browse of the datasheet gets a total of ~0.6 us switching time in the normal use case of this transistor. Not good.

Be also aware that the diode has switching time, too. You probably did simulate the circuit with an ideal diode, but real diodes will make you even less happy.

You can increase the overall efficiency by lowering the frequency and using an impractically big inductor - and it will emit an audible sound as well (bad in itself no matter if you are there to listen to it).

On the other hand, if you use an 60:1 transformer, you will get near 50% duty cycle and a possibility to use way smaller transistor with better switching properties (capacitances, etc...) at lower current to switch in the first place. You could then use a Shottky diode at the output as well, losing only 5-7% of the efficiency in it. Synchronous rectification is also an option, getting even these 5-7% back.

As an added bonus, you will get the transistor at the lower side with its emitter/source at ground and the whole gate controlling circuit at some reasonable voltage, leaving at these 600V only the drain/collector of the transistor and the transformer primary. Since you are not separating the input from the output, you may power your control circuit from the output as well.

p.s. bipolar transistors are so 1990s for switching power supplies, but since you are only practicing, you can use them as well. A reasonable SMPS circuit in 2021 will probably feature a GaN FET transistor.

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  • \$\begingroup\$ microsecond is μs, not μS. And I doubt most SMPS use GaN at this point in time; it's far too expensive. Mostly still plain old silicon, I suspect. \$\endgroup\$
    – Hearth
    Jun 12, 2021 at 21:58
  • \$\begingroup\$ @Hearth Microseconds fixed, thank you. Regarding GaN: sure, one can always use time-honored silicon designs. On the other hand, new elements are to be used in new designs, aren't they? Almost all SMPSs (ranging from phone chargers to welding equipment) I see in the last 2 years are GaN based. The GaN transistor price is probably completely offset by the smaller radiators, inductors, transformers and capacitors, as well as the lower shipping/handling costs. These things got suspiciously power-dense and not very hot when used. And I am sure I am not limited to only looking at high-end gear. \$\endgroup\$
    – fraxinus
    Jun 12, 2021 at 22:25
  • \$\begingroup\$ I may be biased because I work with SiC, but I wouldn't want to use GaN transistors in an application like this--they're not very robust if anything goes wrong. But either way, if it's for the consumer market, it's still silicon almost always. When you're working on a design that's going to be produced in the millions and sold for $5 or less, you can't afford to spend more than a few cents on your switching transistor. \$\endgroup\$
    – Hearth
    Jun 12, 2021 at 22:27
  • \$\begingroup\$ The sub-$5 range it is probably exactly as you say - you work with established, generic mass products well below the technology limits that benefit from both the competition and the economy of scale. GaN technology is yet to see the most of both types of price improvements. \$\endgroup\$
    – fraxinus
    Jun 12, 2021 at 22:37
  • \$\begingroup\$ I suspect the higher-end stuff is going to go SiC as the price of SiC parts falls, too--it doesn't require as much care and protection against avalanche as GaN; you can generally design with SiC transistors just like you would with Si ones. But like I said, I'm probably biased. I'm sure GaN will stick around for its microwave-frequency performance, though. \$\endgroup\$
    – Hearth
    Jun 12, 2021 at 22:42
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Theoretically it should be possible to achieve high efficiency with the right parts. I reproduced your design in LTspice using 'real' components. The calculated efficiency was 81.2%. Here's the circuit:-

enter image description here

I chose an STP11NM80 simply because it was the highest voltage NMOSFET in LTspice's database. However this FET has much lower RDSON and total Gate charge than the IRFPG50. The importance of this becomes apparent when you look at the Drain current waveform:-

enter image description here

When the FET is switched on it passes a current equal to the (average) load current, which in this case is 5 A. With the STP11NM80's RDSON of less than 0.4 Ω the conduction losses aren't bad. The IRFPG50's RDSON can be as high as 2.0 Ω so its conduction loss will be much higher. However with a duty cycle of just under 2% the average conduction loss is still very low (the diode conducts the output current for the other 98% of the time and the inductor 100% of the time, so their conduction losses are far more important).

More concerning is the huge current spike you see at switch-on, mostly caused by the flyback diode's reverse recovery as the voltage across it goes from ~+1.5 V to -600 V. The 38 A peak is just within the STP11NM80's peak current rating, but well over the 25 A rating of the IRFPG50.

The individual component power losses in my simulation were:-

  • FET: 1.8 W
  • diode: 7.2 W
  • inductor: 2.7 W
  • capacitor: < 0.01 W

So in this case we see that the diode is the major loss contributor, followed by the inductor and then the FET. With different component choices the rankings could change and total loss could be higher or lower.

So much for the simulation. In the real world you have more problems. You need to control the FET somehow, which requires another power supply just for the controller - which has to boost the Gate voltage above 600 V to drive the NMOSFET. If for any reason voltage regulation fails the output could rise to 600 V with no current limit.

In a practical circuit the FET's switching speed will be limited by the driver, and perhaps also to reduce EMI. This will increase FET power loss during switching when it has both high voltage across it and high current going though it.

These problems can be solved solved by using a transformer, which improves efficiency by running at a higher duty cycle, can have an auxiliary low voltage winding to power the controller, and provides galvanic isolation. You might prefer an inductor to save space, but a 500 mH inductor rated for 2 A is not small either.

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  • \$\begingroup\$ hi, could you tell me how you added STP11NM80 to LTspice's library? I would try to do designing with it instead. \$\endgroup\$
    – Ri Di
    Jun 13, 2021 at 13:36

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