Not getting expected Voltage Drop with diodes in series

Discussion in 'The Projects Forum' started by jellytot, May 13, 2015.

  1. jellytot

    Thread Starter Member

    May 20, 2014
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    I have a wall adapter that converts AC power to 12V output. It powers a load that draws 300ma. I want to drop the voltage closer to 9V. I read that diodes have a voltage drop of between around 0.6V-0.8V per diode, so I hooked up 4 diodes in series and expected a new voltage of between 9.6V-10.2V, but my multimeter reads it's around 11 Volts.
    Did I misunderstand how this works?
    I do have a 9V regulator so I can use that instead but was wondering why this didn't work.
    Previously, I remember I had the same issue when I tried to drop 3V to 1.5V using 2-3 diodes, but voltage was still higher than expected. 1.5V regulators are harder to get so I was hoping to use this diode method, if I can get it to work.
     
  2. Reloadron

    Active Member

    Jan 15, 2015
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    Was this under a load?

    Ron
     
  3. jellytot

    Thread Starter Member

    May 20, 2014
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    Yes, in my 12V example, it powers a 300ma load.
     
  4. Papabravo

    Expert

    Feb 24, 2006
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    Yes, well the output voltage may not be regulated and the diodes may be dropping less than you expect at the 300 mA load current. If you have a regulated +12V output then your scheme will work a little better. Measue the unloaded output and tell us what diodes you are using. For all we know they might be Schottky diodes!
     
  5. crutschow

    Expert

    Mar 14, 2008
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    The diodes need to be loaded when you measure their voltage drop. It's not clear if you did the measurement under load.
    What is the diode part number?
     
  6. Reloadron

    Active Member

    Jan 15, 2015
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    When the wall wart is under a load, as you mention it is, just measure the voltage drop across each diode or all the diodes in series. Just remember this needs done while the wall wart is driving a load. Also, per crutschow, what are the diodes?

    Ron
     
  7. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    Don't forget to place high value resistors (~1-10Mohm) in parallel with each diode. Otherwise some will work harder than others and may burn out.
     
  8. crutschow

    Expert

    Mar 14, 2008
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    That may be needed if the diodes are reversed biased, but here they are only forward biased.
     
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  9. jellytot

    Thread Starter Member

    May 20, 2014
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    The diodes were under load when measured. The multimeter reads the load at 250ma, at 10.90V. Diodes are standard rectifier diodes 1N4001.

    ---(+power)----[+1N4oo1-]---[+1N4oo1-]---[+1N4oo1-]---/[Multimeter]\-------(-power)---

    With the diodes removed, multimeter reads 11.9V
    ---(+power)------------------------------------------------------/[Multimeter]\-------(-power)---

    Which means a voltage drop of ~0.33V per diode. Nowhere near 0.6-0.8V I read it's supposed to be.
     
  10. joeyd999

    AAC Fanatic!

    Jun 6, 2011
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    I don't see a load in your "schematic".
     
  11. bmuigai

    New Member

    Sep 13, 2012
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    have you checked the datasheet for your diodes. Germanium should drop less voltage, although the 1N400x's i have used, ...not sure... but I think they are silicon n should drop according to your expectation. Also pardon me for asking stupidly, but if your arrangement is as indicated above, where is the load in this case?
     
  12. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Why using series diodes to drop voltage is a bad idea: Note the output voltage vs current to load.

    212.gif
     
  13. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    The resistors are there to equalize the voltage drops across each diode, since they will not all be the same. It is still recommended for forward-biased diodes, for the same reason it is recommended when using multiple diodes in series for rectification to increase their overall voltage rating.
     
  14. wayneh

    Expert

    Sep 9, 2010
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    On the other hand, if you limit the x-axis to a practical 10X range of current, for instance in the 10-100mA range, your plot shows why diodes work fairly well. Poor compared to a regulator, but plenty good for things with steady loads.
     
  15. crutschow

    Expert

    Mar 14, 2008
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    1-10 megohm resistors will have little effect on the forward voltage drop of each diode except perhaps at very low currents.
    Those resistors are normally used when a reverse voltage is applied and you want to equalize the reverse voltage across each diode.
    I don't see how any of that applies here. :confused:
     
  16. crutschow

    Expert

    Mar 14, 2008
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    According to my LTspice simulation you should be getting over 0.8V drop across each 1N4001 diode at 250mA, giving an output of about 8.6V from a 12V supply.

    So to explain your results, here are some choices:
    A. Your circuit is wired incorrectly.
    B. You don't have a 250mA load.
    C. The 12V supply is greater than 12V.
    D. The diodes are faulty.
     
  17. MrChips

    Moderator

    Oct 2, 2009
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    You are performing your test incorrectly.
    Replace the voltmeter with a 47Ω 5W resistor.
    Measure the voltage across this resistor.
     
  18. MrChips

    Moderator

    Oct 2, 2009
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    It would be better to use 9V 5W zener and a 10Ω 3W series resistor.
     
  19. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    On the third hand, and the reason I did the sim, is to show what happens if you load the diodes with only 10 or 20megΩ (the input impedance of the voltmeter)!
     
  20. neonharp

    New Member

    May 10, 2015
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