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Help 15vdc too much

Discussion in 'Troubleshooting and Repair' started by Snipervision, Jan 5, 2012.

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  1. Snipervision

    Snipervision

    6
    0
    Jan 5, 2012
    I have a 15vdc 10a power supply that is putting out 15.15vdc & of course the item I am powering says it will not work with more than 15vdc. What can I do?
    Here is a link to my power supply: http://www3.towerhobbies.com/cgi-bin/wti0001p?&I=LXVCL0&P=7

    MRC #RB982 power supply. It is powering a ICE radio Control battery charger.

    Thx in advance for any help.
     
  2. duke37

    duke37

    5,364
    772
    Jan 9, 2011
    Welcome to the forum.

    I doubt if 0.15V is significant but a silicon diode in series will drop about 1V. You will need a diode that can take the current that the charger demands and may need a heat sink.
     
  3. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    It can probably be adjusted but I'd need to see its guts in detail to say if & how.
    Not that I believe that the 0.15V overvoltage will matter for your item.
     
  4. Snipervision

    Snipervision

    6
    0
    Jan 5, 2012
  5. Snipervision

    Snipervision

    6
    0
    Jan 5, 2012
    Here are some pics. If you need different angles or close ups just let me know. Thx again!!

    [​IMG]

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    [​IMG]

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    [​IMG]

    [​IMG]

    [​IMG]

    [​IMG]
     
    Last edited: Jan 7, 2012
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,505
    2,849
    Jan 21, 2010
    Stick a diode in series with your incoming power. It's a very cheap fix.

    Ensure that the current rating is sufficient, the voltage rating will not be important.
     
  7. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    Wow, 10 pic's.. Still it was hard to get a glimpse of the vital parts..
    First of all, you'll need to service that PSU anyway - since C7 is on its way out.
    Next, IC3 is determining the output voltage, together with a couple of resistors (possibly surface mounted, on the solder side).
    If you're not up to smd modding the series diode trick Duke & Steve mentioned will also work.
     
  8. Snipervision

    Snipervision

    6
    0
    Jan 5, 2012
    Would I solder the diode on the 2 outside points & not the high center one where power cord plugs in for inconing 120v?

    [​IMG]

    Could you be a little more specific on the exact diode when I stop at Radio Shack?
     
  9. davelectronic

    davelectronic

    1,087
    12
    Dec 13, 2010
    You have what looks like a bad capacitor in there, next to the red ZD markings, in the final output i think. :)
     
  10. duke37

    duke37

    5,364
    772
    Jan 9, 2011
    You will need a 10A diode in series with a wire from the power supply to the charger.

    The diode will dissipate 10W at full current so will need a heat sink. You may be better off getting a 10A or 25A bridge rectifier which can be bolted directly on a heat sink. You can use one diode to get a drop of about 1V or two diodes to get a drop of 2V (20W).

    Voltage rating is not important since the diode will be forward biased.
     
  11. davelectronic

    davelectronic

    1,087
    12
    Dec 13, 2010
    Then wait for the bad capacitor to go pop, as if you dont replace it then it will vent, possibly doing damage. :D
     
  12. Snipervision

    Snipervision

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    0
    Jan 5, 2012
    Can you elaborate a little more on the 10a bridge rectifier & where it bolts? If I go into Radio Shack & ask for a 10a rectifier is that all they need to know?
     
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,505
    2,849
    Jan 21, 2010
    Ask for a 10A (or greater) bolt-on bridge rectifier. If they press for a voltage rating, say 24 volts or higher.

    (One reason for getting this rather than a simple diode is that you can be almost totally sure a bridge rectifier will be isolated from where the bolt goes.)

    Any reputable electronics retailer should find this a sufficient description. It may or may not be sufficient for Radio Shack though.

    Connect the +ve output of your power supply to one of the legs marked "~", and then your new +ve output will be the "+" leg of the rectifier. You may need to use spade connectors if it is not possible to solder directly on to the terminals.
     
  14. duke37

    duke37

    5,364
    772
    Jan 9, 2011
    Radio Shack have a bridge rectifier, 25A 50V #276-1185 $3.49, which should do. Remember that this will dissipate quite a bit of heat so will need boltng to a heat sink if used at full current.

    While you are at the shop, get a capacitor to replace the dodgy one as others have said.

    Have you thought about putting a light load on the power supply to see if the voltage drops within your specification?

    Have you had a look to see if there is a trimmer (potentiometer) to set the voltage of the PSU?

    Have you checked your voltmeter to see that it is accurate?
     
    Last edited: Jan 8, 2012
  15. Snipervision

    Snipervision

    6
    0
    Jan 5, 2012
    I purchased a 50v, 25a, 1.7v drop, 100ua bridge rectifier as per duke37 from Radio Shack. The opposite corners are marked + & - & the other opposite corners are marked ac & ac. If I am correct you take the positive output of my power supply & connect it to one of the corners marked ac & the rectifier + is my new positive output?? How did you know the capacitor was bad...the one by the "ZD" symbol. It looks just fine. When I check the output with my meter it read 15.35vdc. Thx again everyone.
     
  16. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,505
    2,849
    Jan 21, 2010
    The capacitor has a domed top. they should be flat. This indicates that bad things are going on inside it.

    You won't be able to measure any change on a multimeter (or frequently even on a capacitance meter). An ESR meter will show a degradation though.

    Connect the +ve output to one (either, or indeed both) lead marked AC and the + lead is your new +ve output.

    If this is insufficient, connect the +ve output to the - lead on the bridge and use + as your output. This will give twice the voltage drop.
     
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