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

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

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  1. Jan 11, 2013 #1
    RobertEagle

    RobertEagle

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    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. Jan 11, 2013 #2
    BobK

    BobK

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

    Bob
     
  3. Jan 11, 2013 #3
    RobertEagle

    RobertEagle

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    Aug 24, 2012
    This is the circuit:

    [​IMG]


    RobertEagle
     

    Attached Files:

    Last edited by a moderator: Jan 12, 2013
  4. Jan 11, 2013 #4
    BobK

    BobK

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    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. Jan 11, 2013 #5
    RobertEagle

    RobertEagle

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    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. Jan 11, 2013 #6
    BobK

    BobK

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    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. Jan 12, 2013 #7
    (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    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. Jan 12, 2013 #8
    RobertEagle

    RobertEagle

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    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. Jan 12, 2013 #9
    davenn

    davenn Moderator

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    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. Jan 12, 2013 #10
    RobertEagle

    RobertEagle

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    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. Jan 12, 2013 #11
    davenn

    davenn Moderator

    13,717
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    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. Jan 12, 2013 #12
    RobertEagle

    RobertEagle

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    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
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