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 Post subject: Re: need help
PostPosted: April 11th, 2022, 2:16 pm 
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GaryB wrote:
tomp wrote:
Not true. When in the forward condition which is what happens in the normal operation of the LED, there is no high impedance related to turn on speed . . .

I probably wasn't explaining myself well. There is clearly some sort of destructive effect going on since P Elliott reported the LEDs were blowing up on turn on without the parallel 100K resistor. The question is what is the cause of this.
I was merely pointing out that the impedance of any diode is a function of the current passing through it. If you look at I-V curve of the LED that you attached a few posts earlier, the impedance of the diode is just the inverse of the I-V slope and that varies with current. When the LED diode is turned on and light is emitted (current around 20mA or higher), it's a relatively low impedance. When the voltage across the diode is lower, say <1.5v, then the current through the diode will be <<1mA and the impedance of the diode will much higher.
But as I think about it further, the more important factor is the change in current through the LED with time. When the LED turns on, there is a sudden increase in current through the diode. That in turn could give a voltage spike due to the inductance of the dropping resistors and the associated wiring. It's the standard effect where V = L dI/dt.
Experimentally, a parallel resistor solves the problem. And I suspect that Roscoe's parallel capacitor would also be highly effective. One can do both if one believes in belt and suspenders.


The formula you quoted is for what voltage is generated when the current through an inductor is interrupted such as with an automobile ignition coil. In this case, when voltage is applied to the resistors, any increase in current through the resistors would be slowed down by leakage inductance in the resistors, providing even more of a "soft start" for the LED. Remember, when voltage is applied to an inductor, the current lags. If the current was suddenly interrupted through the resistors, the collapse of the field would try to keep the current flowing in the same direction which is the forward conduction direction. As long as the current was within the limits of the diode, no problem would occur. You also talk about the LED turn on time when in fact, the LED is always "on" in the forward direction. It is not a switched device. I still cannot envision a mechanism that would cause the LED to fail. A reverse voltage on the LED would cause a problem, but from the schematic I can't see how that would happen.


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 Post subject: Re: need help
PostPosted: April 11th, 2022, 4:20 pm 
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tomp wrote:
. . . I still cannot envision a mechanism that would cause the LED to fail. A reverse voltage on the LED would cause a problem, but from the schematic I can't see how that would happen . . .


Galileo, when arguing with the Pope about the movement of the earth around the sun, famously said "and yet it moves". Paraphrasing that far more famous scientist, I say "and yet it fails". This isn't worth a long heated argument but it's interesting to think about potential mechanisms.

tomp wrote:
You also talk about the LED turn on time when in fact, the LED is always "on" in the forward direction.

I think you're missing the point. With any diode, one always has a "knee voltage" where the device suddenly start conducting and allows much more current to flow. This is how diodes work; they're non-linear devices that have a turn on voltage or knee voltage. This knee voltage is typically associated with the bandgap of the semiconductor material used in the fabrication of the diode. In a conventional silicon diode, that knee voltage is around 0.6v. Put 0.1v across a silicon diode and you'll get no current to flow. In an LED , the knee will be between 1.5v and 2.0v or higher, depending upon the color of the LED. What does it mean for a diode to be on when it allows negligible current through it?

In any case, the way an LED breaks down in the forward direction is to have too much current flowing through it, which equals too much power, which would be caused by too much voltage across it since the current is exponentially dependent upon the applied voltage. Since the LEDs are in fact breaking down, then the explanation must be related to too much applied voltage.

Hope this is interesting to people. I can stop at any time if this is boring or too argumentative.


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 Post subject: Re: need help
PostPosted: April 11th, 2022, 4:26 pm 
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These amps have SS rectification. When the power is turned on, for 10+ seconds, the B+ is much higher than at normal operating voltage since the tubes are cold and there's no load... (part of why (some) electrolytics have a voltage rating and a surge voltage rating....).

Roscoe

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 Post subject: Re: need help
PostPosted: April 11th, 2022, 4:44 pm 
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That is why there are two power switches, one for the filaments then you wait for at least 30 secs and flip the B+ switch.
The B+ red LED is still considerably brighter than the green filament PS LED I think I need to try a smaller value resistor as Gary suggested


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 Post subject: Re: need help
PostPosted: April 11th, 2022, 5:10 pm 
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GaryB wrote:
tomp wrote:
. . . I still cannot envision a mechanism that would cause the LED to fail. A reverse voltage on the LED would cause a problem, but from the schematic I can't see how that would happen . . .


Galileo, when arguing with the Pope about the movement of the earth around the sun, famously said "and yet it moves". Paraphrasing that far more famous scientist, I say "and yet it fails". This isn't worth a long heated argument but it's interesting to think about potential mechanisms.

tomp wrote:
You also talk about the LED turn on time when in fact, the LED is always "on" in the forward direction.

I think you're missing the point. With any diode, one always has a "knee voltage" where the device suddenly start conducting and allows much more current to flow. This is how diodes work; they're non-linear devices that have a turn on voltage or knee voltage. This knee voltage is typically associated with the bandgap of the semiconductor material used in the fabrication of the diode. In a conventional silicon diode, that knee voltage is around 0.6v. Put 0.1v across a silicon diode and you'll get no current to flow. In an LED , the knee will be between 1.5v and 2.0v or higher, depending upon the color of the LED. What does it mean for a diode to be on when it allows negligible current through it?

In any case, the way an LED breaks down in the forward direction is to have too much current flowing through it, which equals too much power, which would be caused by too much voltage across it since the current is exponentially dependent upon the applied voltage. Since the LEDs are in fact breaking down, then the explanation must be related to too much applied voltage.

Hope this is interesting to people. I can stop at any time if this is boring or too argumentative.


The term "on" means there is no switching mechanism. As soon as sufficient forward voltage is applied, current will flow. The point is that in the forward direction voltage is not a failure mechanism, but as you said forward current if sufficiently high can destroy the junction. As a result, time does not enter into a forward failure mechanism like it can for a switched device like an SCR.


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 Post subject: Re: need help
PostPosted: April 11th, 2022, 5:18 pm 
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Will this sell?
Just a quicky beauty shot, needs more work but a decent start
You can see how much brighter the Red B+ PS LED is than the Green filament PS LED is

Attachment:
PE_LR_worked1-8757_web.jpg
PE_LR_worked1-8757_web.jpg [ 353.36 KiB | Viewed 2051 times ]


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 Post subject: Re: need help
PostPosted: April 11th, 2022, 9:54 pm 
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Beautiful photo, Paul. Incidentally, with high-value resistors in series with the LED, the only way I can see it failing is if, under some strange start-up or power down condition there is reverse voltage on the LED it could fail. To protect put an anti-parallel diode across the LED.


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 Post subject: Re: need help
PostPosted: April 11th, 2022, 10:46 pm 
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dberning wrote:
Beautiful photo, Paul. Incidentally, with high-value resistors in series with the LED, the only way I can see it failing is if, under some strange start-up or power down condition there is reverse voltage on the LED it could fail. To protect put an anti-parallel diode across the LED.

That's cool! Any reverse voltage that could trash the LED is shorted out. NICE!

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 Post subject: Re: need help
PostPosted: April 11th, 2022, 10:52 pm 
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So I looked up anti-parallel diode… two diodes in parallel with one diode reversed.
I have to ask….. add the pair to the LED or add one diode in reverse polarity


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 Post subject: Re: need help
PostPosted: April 11th, 2022, 11:24 pm 
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Pelliott321 wrote:
So I looked up anti-parallel diode… two diodes in parallel with one diode reversed.
I have to ask….. add the pair to the LED or add one diode in reverse polarity

One diode -- reverse polarity.

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