DC to DC ?

Discussion in 'The Projects Forum' started by rougie, Sep 17, 2012.

  1. rougie

    Thread Starter Active Member

    Dec 11, 2006
    410
    2
    Hello,

    I have a bread board which requires two voltages. The first is 5VDC and the second one is 3.3VDC.

    Please view the attachment for a simple schematic.

    The problem is that when I adjust my power supply to exactly 5VDC in order to satisfy the 5Volts part of my board, the DC/DC modules shown in attachment deliver 4.27VDC instead of 3.3VDC.

    Now I imagine this is because the DC/DC modules have little load on them, but this is really annoying because my components connected to the DC/DC modules really require 3.3VDC or at worst 3.5 MAX!!!

    My DC/DC modules are "NKE0503SC". I also realize that the more components I connect to the DC/DC modules the more I get cloeser to the expected DC/DC module voltage of 3.3VDC


    My question is, what can one do to temporarily achieve a steady 3.3VDC while building up the board?

    I was thinking of putting a 50 ohm resistor between the 3.3VDC rails to temporarily attain the desired approximate 3.3VDC.

    all feedback is appreciated!
    r
     
  2. KCHARROIS

    Member

    Jun 29, 2012
    292
    1
    Hello,

    Why not use 2 resistors and form a voltage divider to give yourself 3.3V.

    R1 being 3.4K
    R2 being 6.6k
     
  3. rougie

    Thread Starter Active Member

    Dec 11, 2006
    410
    2
    Hello KCHARROIS,

    questions:

    Q#1:
    If,
    It = 4.27/10K = 0.427ma
    VR1 = 0.427ma x 3400ohms = 1.4518V

    Voltage divider will be 4.2 - 1.4518 = 2.74Vdc ???

    Is there a reason why you set the voltage to 2.7 instead of 3.3

    Q#2)
    Will your voltage divider solution be able to hold 3.3VDC at 150ma????

    thanks for your help!
    r
     
  4. t06afre

    AAC Fanatic!

    May 11, 2009
    5,939
    1,222
    I think you need a bleeder resistor connected to the ground. To ensure a minimum load that will give 3.3 volt out from the converters. By the way. Why are you not using a low dropout regulator instead of those DC to DC converters. Do you need isolation?
     
  5. Dodgydave

    Distinguished Member

    Jun 22, 2012
    4,986
    745
  6. rougie

    Thread Starter Active Member

    Dec 11, 2006
    410
    2
    Okay,

    The 4.27 volts was wrong. I disconnected all my DC/DC modules and I was still reading the 4.27VDC on the secondaries of the DC/DC modules???? I realized, I had a wire connected from the 5V system to a diode which connected to the 3.3VDC rails .... LOL!!!

    So still though, I read 4.08VDC instead of 4.27VDC which really isn't that trivial... but still I would prefer 3.3 exactly.

    So figured out some temporary circuit!

    Thanks for your help!
    r
     
  7. t06afre

    AAC Fanatic!

    May 11, 2009
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    Have you read the data-sheet here http://www.murata-ps.com/data/power/ncl/kdc_nke.pdf
    It states clearly
    You need to have a bleeder resistor that ensure about 20% of max current draw. Max current for your device is 303mA. Se the TOLERANCE ENVELOPES figure. If you do as I say everything should work. But it is your call;)
     
  8. rougie

    Thread Starter Active Member

    Dec 11, 2006
    410
    2
    Hello t06afre,

    I have made an adjustment circuit that I think does pretty
    much what you have recommended.... Based
    on the load I adjust the circuit so it corresponds
    to a stead 3.3 VDC.

    This is all temporary because eventually the circuit on
    it's own will draw at least approximately150ma when I will
    be done with it.

    When you say a bleeder resistor, do you mean a resistor
    from +3.3VDC to ground ensuring a load of
    20% of max DC/DC current module?

    Thanks for your help?
    r
     
  9. strantor

    AAC Fanatic!

    Oct 3, 2010
    4,302
    1,988
    yes thats what he means
     
  10. rougie

    Thread Starter Active Member

    Dec 11, 2006
    410
    2
    Thanks guys.

    r
     
  11. rougie

    Thread Starter Active Member

    Dec 11, 2006
    410
    2
  12. Bernard

    AAC Fanatic!

    Aug 7, 2008
    4,172
    397
    Could use a Si diode, 1N4002, between 4.08 V & #.3 module for a +_ .7 V drop & a 110 Ω bleeder.
     
  13. WTP Pepper

    New Member

    Aug 1, 2012
    21
    6
    Because as soon as you draw current, the voltage will fall.
     
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