# How can I reduce with MOSFET from big DC to small DC

Discussion in 'General Electronics Discussion' started by RobertEagle, Jan 11, 2013.

1. ### RobertEagle

11
0
Aug 24, 2012
Hi,

I have a input voltage of 200Volts DC and I want to reduce it with the smallest energy loses to 18Volts DC.

I've already made a circuit in which I've used a MOSFET IRFBG30PBF http://datasheet.octopart.com/IRFBG30PBF-Vishay-datasheet-57413.pdf, but I got 18Volts DC without much current.

Does somebody know a circuit I can use in my situation. I prefer the most simple.

Thank you,
RobertEagle

2. ### BobK

7,682
1,688
Jan 5, 2010
What circuit did you use? You need a buck converter.

Bob

3. ### RobertEagle

11
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Aug 24, 2012
This is the circuit:

RobertEagle

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4. ### BobK

7,682
1,688
Jan 5, 2010
That is a linear regulator. Dropping 1000V (or is it 200V?) to 18V will give you an efficiency of 18/1000 = 1.8% at best. And it will needt to dissapate about 1W for each ma of output current.

As I said before, you need a buck regulator for any efficieny at all. And that is not a simple project with 200V (or is it 1000V?) input.

Bob

5. ### RobertEagle

11
0
Aug 24, 2012
I'm sorry for what I have written there. That 1000 voltage note was actually an older calculus.
Just to be sure, I've measured again, and the peak is at 300Volts.

I've get it. So the linear regulator has to dissipate (300-18)*I W in order to give that 1.8% power, while a buck regulator has somewhere between 85%-95% efficiency.

Do you know a circuit which could solve my problem?

Thank you,
RobertEagle

6. ### BobK

7,682
1,688
Jan 5, 2010
How much power do you need at 18V?

Wikipedia is always a good starting point:

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

You could use the same MOSFET you have as the switch, but you would have to power the controller logic from a lower voltage. One way to do this is to bootstrap. I.e. you have a source of low voltage DC to get you started, then you run the circuit off the 18V output. There are ICs that will act as a buck converter and operate the way. For the bootstrap source, you charge up a capcitor through a high resistance from the HV which gives the circuit enough time to start up.

But, it is unusal to have 200-300V DC as you only power source. If you have regular AC power, it is much easier to get from that to 18V DC.

Bob

Last edited: Jan 12, 2013
7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,497
2,838
Jan 21, 2010
With 200V to 300V DC the best way to get 18V DC is to purchase a switchmode power supply that can run from 90 to 265 VAC and which produces 18V DC output.

Simply feed this with the 200-300 VDC and you'll get what you want.

You'll also get it with good efficiency and isolation from the input.

Building such a power supply from scratch is possible, but isn't likely to be as cheap as buying one.

(note that 210 VAC has peaks to 300VDC, so any power supply capable of operating from 210 VAC is sufficiently engineered to operate from 300VDC)

At the other end, 200 VDC corresponds to just over 140 VAC (in terms of peak voltage).

Just don't try to use one which has to be switched manually to select the input voltage.

8. ### RobertEagle

11
0
Aug 24, 2012
I'll try to recap everything, because, my scope gave me wrong measurements.

First of all, I have 2 secondary coils. These 2 coils have a frequency of 12.5kHz, at 2kV, and are sinusoidal. This 2 coils are linked to my circuit (or were).

To be sure of the power, I took 2 resistors (2W at 459kOhm) and put one on each channel. On each channel I got 1.915W at 2.04mA, totaling a power of 3.82W.

Because of my diodes, which are limited to 1000V, the DC voltage reaches 650VDC (I don’t understand why not 1000). But because they are connected to the circuit, then the drain voltage is 250V with no load(with load it gets 65V).
So, is useless to continue with my circuit.

I know a thing. Transformers are radiating much of the energy through the EM waves. I think they are inefficient at converting.

If you can, can you tell me, what’s the best method of converting two 2kV AC 12.5kHz signals down to low DC voltages with minimal power loses? I’m not familiar with converters.

Thank you again,
RobertEagle

9. ### davennModerator

13,837
1,951
Sep 5, 2009
you have suddenly jumped from 200VDC to 2kV high freq AC

Now you have changed the whole game and everything people said above to help you has been a waste of time.

Its REALLY REALLY inportant to give all the info up front so that people know what they are dealing with when offering advice.... else we all just go around in circles

give some more complete info on the whole project and what it is your are trying to achieve

Dave

Last edited: Jan 12, 2013
10. ### RobertEagle

11
0
Aug 24, 2012
Yes, and I sorry for what I have caused. The measurements were wrong.

I want a system, converter, whatever it be, preferable not a transformer (because of it's major power loses and heaviness) that can convert this double AC signal of a voltage of 2kV, 12.5kHz, sinusoidal, down to a low DC voltage with minimal power loses.
Another problem is that I'm not familiar with converters circuits.

Here I have a snap of these 2 signals. 250V/div: http://img191.imageshack.us/img191/5092/28549146.jpg

RobertEagle

11. ### davennModerator

13,837
1,951
Sep 5, 2009
OK
but you still havent told us the source of this 12.5kHz 2kV signal and why you are trying to drop it to 18VDC ?
What is the 18VDC for ?
Im not aware of any way to drop 2kV to a much lower AC voltage without some sort of transformer

Dave

12. ### RobertEagle

11
0
Aug 24, 2012
The source of the 2 signals is a transformer.
And, 18VDC at 0.21A is made for a couple of microcontrollers.

What it matters is the ,,how". How to get down to those voltages.
Surely isn't any way of doing this without a transformer? And if there's not, how a circuit with a tranformer should look like?

Last edited: Jan 12, 2013