Diode as voltage regulator problem

Discussion in 'The Projects Forum' started by edvard_m, Aug 13, 2010.

  1. edvard_m

    edvard_m Thread Starter New Member

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    Hi everyone. I'm trying to build a cheap voltage regulator to have a regulated voltage of 1.4V from a 3V source. I was trying to do something similar to what was presented here: http://www.reuk.co.uk/Zener-Diode-Voltage-Regulator.htm but I couldn't find a zener diode on Digikey with a Zener voltage of 1.4V (or .7V to put in series).

    Now I'm just looking for any through hole axial diode to get a 1.4 drop across. The problem is on Digikey, the parts say Forward Voltage .7V @ 1A or .7V @ 5A etc. I know I want .7V Forward Voltage, but how do I know what current will be going through the diodes?

    Thanks.
  2. marshallf3

    marshallf3 Well-Known Member

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    Just ignore the current rating, virtually all regular silicon (but not schottky) rectifiers will drop between 0.6 - 0.7V across the junction if the current is kept within it's its operating range.
  3. Ghar

    Ghar Active Member

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    Check out the diode datasheet to see a plot of voltage vs current.
    You'll see there's quite a bit of variation even in the same part number (with temperature, current, whatever)

    You set the current using a resistor, usually called the 'current limiting resistor'. There are a lot of calculators and explanations online on how to do this.
    Last edited: Aug 13, 2010
  4. edvard_m

    edvard_m Thread Starter New Member

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    Ok. I know my load draws about 1mA. How do i know if the current will be within its operating range if I don't know what the current running through the diodes is? If it is too low, then won't i get a smaller voltage drop across each diode based on the i-v characteristics of a diode?
  5. Ghar

    Ghar Active Member

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    The current in the diode will be:

    (Vsource - Vdiode)/R - ILoad

    Vdiode won't be changing that much for small changes in current but you can follow the graph to see where you are.
    The change is something like 0.1 V for every decade of current (factor of 10), it depends on the exact diode.
  6. marshallf3

    marshallf3 Well-Known Member

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    Unless you're shooting for absolute precision I consider it to be a non-issue in most simple circuitry.
  7. edvard_m

    edvard_m Thread Starter New Member

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    Ok. If I chose a 0.7V Forward Voltage (FV) @ 1A, if I ran it at .1A, I'd already be down to about 0.4V FV.

    So I found another diode that operated at 1.7V FV @ 1A. I looked at the i-v curve and it looks like if the diode current is 10mA, I can get the FV to be 0.7V.

    http://www.diodes.com/datasheets/ds25002.pdf The UF1005

    I1 = Idiode + Iload
    (3V-1.4V)/R1 = 10mA + 1mA = 11mA

    This means I need R1 to be 14.5Ω. Is this calculation correct? I want to run as little current as possible through the diodes to save the 2AA batteries, but it looks like this is the least current I can supply from what I have to choose from on Digikey.
  8. Ghar

    Ghar Active Member

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    You're off by a factor of 10, it's 145 ohms.
    You should plug the value back in to check especially when it's such a short equation.

    How close do you need to be though? You could cut power consumption in half and lose less than 0.1V.
    The diodes themselves will be off from the datasheet number as well, so you'll probably need to adjust it if you want exactly 1.4V
    It'll also change with your load unless it's a constant 1 mA.
    Marshallf does have a good point.

    You could also look into putting 3 Schottkys in series, you might be able to find ones that work with 1.4V at a lower current.
    Last edited: Aug 14, 2010
  9. Audioguru

    Audioguru New Member

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    A diode is not a voltage regulator.
    The voltage across a diode changes when its current and temperature changes.
    You should use a low current voltage reference IC.
  10. sage.radachowsky

    sage.radachowsky Member

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  11. edvard_m

    edvard_m Thread Starter New Member

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    Thanks for the help everyone. And thanks sage for the link. The voltage regulator IC's are cheaper than I expected.
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