1. Win a Broadcaster or one of 3 Teles! The annual Supporting Member Giveaway is on. To enter Click Here. To see all the prizes and full details Click Here. To view the thread about the giveaway Click Here.

Issues: Fender Champion 600 Reissue - Pilot LED to #47 Bulb Conversion

Discussion in 'Amp Tech Center' started by Huddy, Aug 21, 2020.

  1. Huddy

    Huddy Tele-Holic

    Age:
    37
    Posts:
    703
    Joined:
    Nov 5, 2016
    Location:
    Newport News, VA
    So I may have dug myself into a hole. I've got this Champion 600 and the original Jewel light was broken. I wanted to replace it but couldn't find the proper metric sized jewel lens/cover. I looked at the schematic and say that the original LED was driven off the filament winding with a diode doing something so I figured I just had to clip one of the legs of that diode (D5) and I'd be in good shape after soldering in the leads to the lamp holder. One of the leads to the pilot as a 220 Ω resistor (R24) in series with it.

    The bulb is not illuminating. When the bulb is removed I've got 6.3 v to the lamp holder. When I install the bulb I've got 0 vac. The tube heater pins still have 6.3v when the bulb installed. Is that 220 Ω resistor my issue and I need to jumper it? do I need to look for something else?

    Thanks so much in advance.

    Screen Shot 2020-08-21 at 9.08.04 PM.png
     

    Attached Files:

  2. andrewRneumann

    andrewRneumann Tele-Holic

    Posts:
    593
    Joined:
    Mar 22, 2020
    Location:
    Cincinnati, OH, USA
    Hello,

    You need D5 to allow reverse current around the LED. I recall that LED’s can’t put up with much reverse voltage.

    Assuming the LED you procured does NOT have a resistor integrated, you will need that resistor. Some LED assemblies have a resistor built in and don’t need the extra one in series.

    What kind of LED is it and do you have the data sheet?

    Re-reading the OP—are you replacing the LED with an incandescent bulb?
     
  3. Jon Snell

    Jon Snell Tele-Holic

    Posts:
    766
    Joined:
    Aug 31, 2015
    Location:
    Jurassic Coast. UK.
    To fit an incandescent lamp, remove the diode and replace R24 with a wire link.
    The lamp is fitted where the LED was.
     
    Peegoo and Huddy like this.
  4. Huddy

    Huddy Tele-Holic

    Age:
    37
    Posts:
    703
    Joined:
    Nov 5, 2016
    Location:
    Newport News, VA
    Yup that 200 Ω resistor that I should've pulled and jumper'd was the culprit. All is right with the world now.
     
    andrewRneumann likes this.
  5. Huddy

    Huddy Tele-Holic

    Age:
    37
    Posts:
    703
    Joined:
    Nov 5, 2016
    Location:
    Newport News, VA
    Jumper'd where the 200 Ω resistor was and worked like a charm - thanks!
     
  6. andrewRneumann

    andrewRneumann Tele-Holic

    Posts:
    593
    Joined:
    Mar 22, 2020
    Location:
    Cincinnati, OH, USA
    I should have asked that first or read the title of your post! Ahhh the humiliation... glad you got it sorted.

    EDIT: I’m still scratching my head as to why you were measuring 0vac with the lamp installed.
     
    Huddy likes this.
  7. tubegeek

    tubegeek Friend of Leo's

    Age:
    60
    Posts:
    3,882
    Joined:
    Jan 31, 2020
    Location:
    Brooklyn, NY
    Because almost 6.3V was dropped across the current limiting resistor.
     
    Huddy and andrewRneumann like this.
  8. andrewRneumann

    andrewRneumann Tele-Holic

    Posts:
    593
    Joined:
    Mar 22, 2020
    Location:
    Cincinnati, OH, USA
    Ahhh so the cold resistance of the bulb must be pretty low say less than 20R?
     
    Huddy likes this.
  9. tubegeek

    tubegeek Friend of Leo's

    Age:
    60
    Posts:
    3,882
    Joined:
    Jan 31, 2020
    Location:
    Brooklyn, NY
    Exactly. Whatever it is, it's a lot less than the 200 ohms it's voltage-dividing with. 20R would give you something like .6V, maybe you can measure that, maybe you can't - depends on the meter.

    Power: W=V^2/R. Let's say it's a 1 watt bulb. 1= (6.3x6.3)/R

    So the hot resistance might be around 36 ohm.... bupkis. Cold would be EVEN LESS!

    EDIT - wow, what a guess - I oughta play the lottery. A #47 bulb is .945W, draws .15A.

    OK so that bulb will NOT heat up with 6.3 V and 200R. It'd take 30V (plus 6.3 across the bulb) to get .15A through 300 Ohms - the bulb will be current-starved, never heat up, and still have its cold resistance as measured.
     
    Last edited: Aug 22, 2020
    Huddy likes this.
  10. andrewRneumann

    andrewRneumann Tele-Holic

    Posts:
    593
    Joined:
    Mar 22, 2020
    Location:
    Cincinnati, OH, USA
    Another way to look at it is that the two lamp leads are essentially the same node (and therefore same potential) when there is barely any resistance between them. I’ve got it now Jeeves. Thanks!
     
    tubegeek and Huddy like this.
  11. frn0003

    frn0003 NEW MEMBER!

    Age:
    64
    Posts:
    1
    Joined:
    Jan 29, 2021
    Location:
    Michigan
    Does anyone know where I can get a replacement jewel screw in fixture for my champion 600? I notice the fender deluxe reverb is to big.

    Thanks
     
  12. Huddy

    Huddy Tele-Holic

    Age:
    37
    Posts:
    703
    Joined:
    Nov 5, 2016
    Location:
    Newport News, VA
    I couldn’t find one. I just replaced with the standard 6.3v bulb and holder. Did have to enlarge the mounting hole though. Just need soldering skills and $9-10.
     
IMPORTANT: Treat everyone here with respect, no matter how difficult!
No sex, drug, political, religion or hate discussion permitted here.