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 Post subject: Re: need help
PostPosted: April 5th, 2022, 4:17 pm 
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Joined: January 15th, 2015, 7:19 am
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Location: Baltimore MD
ok good to know


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 Post subject: Re: need help
PostPosted: April 5th, 2022, 6:42 pm 
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Roscoe Primrose wrote:
B+ will be a lot higher w/o tubes installed, possibly enough to be over the cap ratings....

That is absolutely correct. Heed this warning! :o

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 Post subject: Re: need help
PostPosted: April 6th, 2022, 2:01 pm 
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Location: Potomac, MD
Ridiculous amount of extra power to waste in that 25k resistor. Why not just break one of the legs of one of the 100k cap balance resistors and add the LED in series with it. No extra power wasted. Just make sure you have the correct polarity on the LED. It won't be real bright because there would only be about 2.5 mA through the LED. But it should be bright enough and it would be unlikely to fail.


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 Post subject: Re: need help
PostPosted: April 6th, 2022, 3:51 pm 
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Location: Baltimore MD
I traced out the circuit that is in place
Attachment:
B+ supply for LED.png [11.1 KiB]
Not downloaded yet


I am going to go in and check voltages


Attachments:
B+ supply for LED.png [11.1 KiB]
Not downloaded yet
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 Post subject: Re: need help
PostPosted: April 6th, 2022, 4:05 pm 
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Location: Baltimore MD
I get 460V off the main filter caps
0 volts across the LED


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 Post subject: Re: need help
PostPosted: April 7th, 2022, 11:50 am 
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I know you guys are probably bored with this simple problem, and I apologize for my ignorance but this is where I am at now.
The LED is bad. The two 200k 2watt resistors in parallel off the B+ are open.
So I need to increase both resistance and heat dissipation.
Attachment:
LED B+ indicator.jpg
LED B+ indicator.jpg [ 30.07 KiB | Viewed 12075 times ]


so besides a new LED how much should I go with the voltage drop resistors. the two 200k 2 watt in parallel is 100k 4watt.
I can double the wattage, there is plenty of room


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 Post subject: Re: need help
PostPosted: April 7th, 2022, 12:12 pm 
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Location: Parkville, Maryland
Pelliott321 wrote:
I know you guys are probably bored with this simple problem, and I apologize for my ignorance but this is where I am at now.
The LED is bad. The two 200k 2watt resistors in parallel off the B+ are open.
So I need to increase both resistance and heat dissipation.
Attachment:
LED B+ indicator.jpg


so besides a new LED how much should I go with the voltage drop resistors. the two 200k 2 watt in parallel is 100k 4watt.
I can double the wattage, there is plenty of room

I am wondering why the 100-kOhm resistor across the LED? For me it seems that a 400-kOhm resistor in series with the LED dropping down from 480-volts limits current to 12-ma.

Again -- not all LEDs are created equal and if 12-ma is too much then use more resistance. When I had this issue I just put a pot in series -- I adjusted for the desired light output than substituted a fixed resistor

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 Post subject: Re: need help
PostPosted: April 7th, 2022, 2:43 pm 
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I was curious about that 100k in parallel also. Research calls it a current limiter also.
I am ordering a bunch of 430k 5watt that I can try four in parallel and keep the original design just beefier
and I can try just one 430k in series without the parallel 100k.

Just trying to get this ready to sell. The Carys do sound good but I just do not need them. They are selling anywhere from $2400/pair to $3500/pair.
Hope to sell them locally for less. I really do not want to have to build rates and ship.


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 Post subject: Re: need help
PostPosted: April 7th, 2022, 3:03 pm 
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Location: Parkville, Maryland
Pelliott321 wrote:
I was curious about that 100k in parallel also. Research calls it a current limiter also.
I am ordering a bunch of 430k 5watt that I can try four in parallel and keep the original design just beefier
and I can try just one 430k in series without the parallel 100k.

Just trying to get this ready to sell. The Carys do sound good but I just do not need them. They are selling anywhere from $2400/pair to $3500/pair.
Hope to sell them locally for less. I really do not want to have to build rates and ship.

I have no clue where you got that information, but a LED (or any diode for that matter) is a dead short compared to the parallel 100K-ohm resistor. ???

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 Post subject: Re: need help
PostPosted: April 7th, 2022, 3:59 pm 
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Location: Baltimore MD
Yes I do not understand why the parallel resistor is there. I must have misread the research.


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