40vdc to 5vdc

Discussion in 'General Electronics Chat' started by cmartinez, Apr 20, 2015.

  1. cmartinez

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    Is there a simple way to go from 40vdc to 5vdc without the need for transformers or inductors?
    I was going to use an LM317, but then found out that its datasheet specifies a maximum of 30vdc at its input.
    Maximum power drawn will be about 20mA @ 5vdc....
    Or maybe I should just use a zener for this?
     
  2. Hypatia's Protege

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    you could 'series' a number of rectifiers such that their collective Ef 'drop' brings the input EMF within 'range' of the linear regulator...:D

    Although this may achieved with a single reverse-biased Zener, the power (and, hence, reverse current) specification may be 'disappointing'...

    Best regards
    HP
     
    Last edited: Apr 20, 2015
  3. #12

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    There is a suffix H that goes to higher voltage.
     
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  4. MikeML

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    The TI datasheet says that the LM317A has a max input-to-output differential of 40V, so you are within that.
     
  5. Hypatia's Protege

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    Yes -- Be certain, however, that you are within the 'safe area' RE: [Ei-Eo] @ I_out...

    Best regards
    HP
     
  6. crutschow

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    I would go with the LM317HV, that #12 suggested, to give you some margin, particularly if you accidentally short the output.
     
  7. Alec_t

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    Or you could use an 'amplified diode' to drop some volts to supply the standard LM317.
     
  8. jpanhalt

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    I would not discard the use of a switching buck regulator. There are lots of them available with quite simple applications. Here is just one: http://www.ti.com/lit/ds/symlink/lm2825.pdf

    It doesn't even use an external inductor. NB: That is actually quite an old device. So, I suspect similar and smaller regulators are now available.

    John
     
    Last edited: Apr 20, 2015
  9. dl324

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    Hi John,
    I was thinking this way also until I noticed the "no inductor" requirement. Power dissipation in the regulator will be low at the current the OP is talking about.
     
  10. OBW0549

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    Is 40V your maximum input voltage, or is that the nominal? If nominal, what is the maximum? What is the minimum input voltage?
     
  11. ScottWang

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    If you using zener then count the current about 30mA including the current of zener and load, the formula as below:
    Vin = 40V
    Vout = 5V, 5.1V zener
    I_total = 30mA
    R_limit = (Vin-Vout)/I_total
    = (40V-5V)/30mA
    = 35V/30mA
    = 1.167 K
    Choosing 1.2K
    R_Watts = 35V*30mA = 1.05W, choosing the current limiting resistor as 1.2K/5W.
     
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  12. MaxHeadRoom

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    One reason I usually go for Toroidal transformers when building supplies, only takes a small overwind of ~10 turns for a 5v supply.
    Max.
     
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  13. cmartinez

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    it's exactly 40V, regulated
     
  14. OBW0549

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    Ah. In that case, either a Zener or the LM317HV would do.
     
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  15. cmartinez

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    I'll check into it, thanks!
     
  16. Dodgydave

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  17. #12

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  18. crutschow

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    I see no reason to use a switcher for only 20mA of current unless it's a battery source and you need maximum efficiency.
     
  19. MaxHeadRoom

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    Quite a while back I had to design some equipment for a mobile generator application running 240vdc.
    The Op-amp circuit was probably no more than 20Ma, I ended up using a resistor, zener and electrolytic there are many of these units in place with no problems to date.
    Max.
     
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  20. cmartinez

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    Ok... maybe I've just been "drowning in a glass of water" as we like to say down here.

    Here's the complete picture:
    I want to power up a LED in an optoisolator (H11L2M) that works at a maximum of 30mA. I've used this opto before with a 5V power source and connected it to a 270 ohm resistor to limit its current to below 18mA, and it's been working just fine all this time. I don't know what the emitter's voltage drop is for this device, but I'm guessing it's not too much.

    Anyway, I'd now like to feed the same optocoupler's emitter with a 40VDC regulated supply (never mind for now what the conditions on its output are going to be, I've got that covered). So I've been thinking about simply using a resistor voltage divider.

    Now according to this site, I've plugged in the values of
    • Vin=40, R1=2200, R2=1800, RL=330.
    And the results are:
    • Vout, no load = 18V
    • Vout, with load= 4.5V (sounds good to me)
    • Current through the load = 14mA (sounds good too)
    • Power dissipation at the load resistor = 0.061W
    So it seems I'm safe, if I just use this circuit considering the previous values, and I connect the LED in series with the load resistor. I know that the current through the LED will be a little less due to its voltage drop, but I doubt it will reach a level below 10mA, which is its threshold current, according to the opto's datasheet.

    Any thoughts anyone might want to share?
     
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