LED Post out of LED Board - DIY Tube Amp Head

[Zarrir]

I have just finished a SS rectified 18Watt and noticed that i still have three power transformers in stock which have the 0/5v/6v secondary leads on them.In case i decide to build some more SS rectified amps, is it feasible to use those leads to light up the inside of the amp´s head with LEDs? Perhaps using a encapsulated bridge rectifier? Any schematics out there?

[HotBluePlates]

... three power transformers in stock which have the 0/5v/6v secondary leads on them. ... is it feasible to use those leads to light up ... LEDs?

Yes.

... Perhaps using a encapsulated bridge rectifier? ...

For the LEDs? A rectifier is not needed.

LED = Light Emitting Diode, so the LED itself is already a rectifier. If you're careful about your routing of the 5vac wiring, the LED plus a dropping resistor is all you need.

[Zarrir]

Great, thanks!! I thought LEDs would only take direct current

[Paul1453]

The LED is a diode that will light when it is forward biased.

At 50 or 60 cycles per second your eyes won't be able to tell it is flashing On and Off.   

I recently learned something else about LEDs.

If you just tack an LED on to the 6.3VAC your heater filaments are using.
They don't seem to like the current surge from the heaters starting up.
It seems to burn out the LED rather quickly if you don't add a resistor with the LED.

If your AC source is not powering the heaters, that might not happen even without the resistor.   

[HotBluePlates]

...
If you just tack an LED on to the 6.3VAC your heater filaments are using.
They don't seem to like the current surge from the heaters starting up.
It seems to burn out the LED rather quickly if you don't add a resistor with the LED.
...

Think about it differently: You have a diode passing current to ground; neither you, your diode (LED) or heater winding get any benefit by letting the diode free-run and pass any/all current it possibly can every time it is forward-biased. The diode (LED) might even attempt to pass enough current to destroy itself.

So you have to know something about the LED you plan to use (from either the vendor website or a data sheet). Find out the rated forward voltage for the LED, and its operating current.



Then calculate the dropping resistor needed, based on the voltage source which will power the LED. You can use the convenient calculator Hoffman has already provided.

Or you can do the math manually. Hoffman sells a super-bright white LED which runs at 3.6v forward and 20-30mA. Assume you'll use that dormant 5vac winding, and that you'll shoot for middle-of-the-range operating current of 25mA.

• Voltage to drop: 5v - 3.6v = 1.4v
• Dropping resistance: 1.4v / 0.025A = 56Ω
• Resistor wattage rating: 1.4v * 0.025A * 2 = 0.07 watts  -> Any convenient size will do (1/8w, 1/4w, 1/2w, etc)

You may not have 56Ω handy, and while "excessive" you do have a couple of 100Ω 1/2w resistors on hand for artificial center-tapping a 6.3vac heater string. So you use a pair of those in parallel, making a composite resistor of 50Ω @ 1w. The calculator says the resulting current is 28mA. Check.

Or you want to suck as little current as possible, and opt of 20mA operation. 1.4v / 20mA = 70Ω, which is a non-standard value. But 68Ω is a standard value and results in just a little extra current.

[Paul1453]

Thank you HBP.   

More of my calculation training weakness showing.
I appreciate your explanations of how to figure these things out.

As most of you know, I'm the king of reusing salvaged parts.
Or better known as a cheapskate.   

So I had a number of different colored LEDs, but no idea of their actual specs.
On my 1st Plexi build, I followed the plans and used a 200 ohm resistor with a Red LED.
That was tacked on to the 6.3VAC heater tap along with a non-elevated artificial 100 ohm CT.
That LED is still doing fine after a couple of months of usage.

I made a couple of more builds after that with both AC and DC source voltage for the heaters and the LED.
I found I could identify the + and - leads of the LED with a simple 9V battery.
So I just figured that if the LED didn't burn up with 9V I might not need a dropping resistor after all.
On the DC source supply I put the LED in with the right polarity, on AC source it doesn't matter.
The LEDs on both AC and DC sources and with no resistors worked fine at first.
But within a couple of days the LEDs burnt out on both of them.
I replaced the LED on the DC circuit without a resistor again,
and it again burnt out within a day or two.

That is how I speculated that without a dropping resistor the startup current surge was breaking down the LED quickly.   
That dropping resistor, along with setting the current draw of the LED, also helps to protect the LED from the initial surge, or not really?

[vibrolax]

That is how I speculated that without a dropping resistor the startup current surge was breaking down the LED quickly.   
That dropping resistor, along with setting the current draw of the LED, also helps to protect the LED from the initial surge, or not really?

In the conducting state, an LED presents a very low resistance, and will gulp down a lethal dose of current.  Thus it requires a series current limiting resistor to avoid death.  It's not the surge that kills it.

Here's my 2004 build that ran an LED across 2 x 100 Ohm virtual center tap resistors:http://www.frontiernet.net/~jff/pipsqueak_pentode/pentode.pdf

While you can operate an LED off a 5 VAC or 6.3 VAC filament without another diode rectifier, any LED I've encountered has a low peak reverse voltage rating.  If you try using an LED with an unrectified 12.6 VAC filament and current limiting resistor, it will die by exceeding the LED's PRV rating.

With 12.6 VAC filaments, I added a diode reversed with respect to the LED to prevent PRV failure
https://www.ibiblio.org/kuphaldt/electricCircuits/Semi/03298.png

[Paul1453]

Thank you vibrolax!

That makes sense, I think I got it now with your guys help.   

[PRR]

> die by exceeding the LED's PRV rating.

Have you done this?

The reverse voltage breakdown is not a fatal limit by itself. Trouble will come with uncontrolled (breakdown) conduction and large current. But I have run LEDs half-wave on 120VAC with a 56K series resistor, ~~2mA current each way. Assuming 5V PIV, the half-cycle dissipation is 5V*2mA, 10mW, which does not seem to be fatal.

Reverse breakdown has a reputation for degrading junctions over time. But this rig ran in a classroom 24/7 for at least 5 years. (Actually I don't remember ever taking it out...)

[vibrolax]

No, I haven't destroyed an LED by running it beyond the PRV rating.
I have an aversion to blowing things up by deliberately exceeding the ratings, so I included the diode.

I sometimes wish I thought more like Leo F and company, where the absolute maximum ratings were just the starting point...
I know things often sound best just before they explode, and I admire the people who explore that territory.

I certainly believe your results. 

[Zarrir]

Thanks for the richness of technical stuff. Can i simply assume that LEDs actually do not like AC current very much? Will add a small bridge rectifier and apropriate resistors to the circuit and see how it goes. While blue leds look cool inside heads, rock n roll asks for orange and red leds.

[Paul1453]

No, but you can recognize that they don't like excessive current like I just did.

My experience shows they burn up just as fast from excessive current on DC as they do on AC.   

[HotBluePlates]

... Can i simply assume that LEDs actually do not like AC current very much? ...

Any diode has a limit of how much current it can carry before the junction burns up.

Paul's error was not seeing an LED as "just like any diode" which will pass forward current in any amount until it burns itself up. In the graph I posted, the colored lines stay at/near zero current until some forward voltage is reached, then then bend upward (increase current conduction). The current never levels off, and may be enough to destroy the LED.

*A resistor* is purpose-made to slow/reduce current. But current through a resistor leaves some voltage drop across the resistor. Hence the use of a voltage source (a.c. or d.c.) larger than the required forward-voltage of the LED, and the calculation for finding a resistor that limits current to a safe value.

This is also a practical application of Ohm's Law; if you don't already know it, you'll need to learn it to progress in electronics.

And as for a.c. voltage supplying LEDs... I made an amp which uses LEDs to light the control panel, and that LED string is powered from the 6.3vac heater wiring. It's been going fine for 4 years now (apologies for the poor cell-phone photo).

[pompeiisneaks]

One other way to avoid reverse diode breakdown is to just put another diode in parallel with the LED to handle the opposite direction.  Problem solved.  :)  Something like the attached image.  (also current limiting resistors are pretty useful.  Depends on the diode and setting, but a 330 ohm usually is enough for soemthing like a 6v winding. )

[pompeiisneaks]

Of course I got lazy in the drawing, I didn't send the line back... anyway I think that makes sense, its like any half wave rectified circuit.  I found a picture that shows it better here:
http://fc3spro.com/TECH/HOWTO/ERRORCODES/ZENKI/led1.jpg

~Phil

[Paul1453]

As a cheapskate, and also a big fan of simplicity of design and minimum parts usage,
I would seek to use the available VAC resources as simply and cheaply as possible on your LED design.

Obviously, eliminating a $.01 resistor was not a good decision on my part.   

But it did go along with my simplicity of design and minimum parts usage principles, at the time.
In the end, I have learned a lesson worth much more than the salvaged LEDs I burned up.   

[Zarrir]

Nice one Phil, thanks!!

[super&plexi]

When you were finding out polarity with those 9 volt batteries I'd wager to say they were either really weak,  the batteries that is, or you did it  really really quick, because a fresh 9 volt can  fry them real fast, like a couple seconds, without a resistor. In fact on almost all of the effects building sites anywhere from 1K to4.7K are standard go to values for safety resistors or I guess more properly current dropping LED resistors, ( for 9.6 volt effects FX depending on how bright you want the LED 2 light) and if you got good eyes or magnifier and you look at the LED there's a few things that'll tell you polarity.  if its  squared off on one end that's usually cathode ground side, and if you look inside the larger section is usually ground and the smaller little triangle of metal inside there is usually anode+, and that's only if both legs have been cut to the same size because from the factory they almost always come with one leg longer than the other denoting Anode. from  cathode. And hey we're not Cheapskates we're just frugal! Not Penny Pinchers just Pennywise eh? Oh yeah where you getting LEDs for one penny

[PRR]

> put another diode in parallel with the LED

This can be another LED. Get more light.

[Paul1453]

>Oh yeah where you getting LEDs for one penny

I didn't get LEDs for a penny yet, but those damn Chinese did send me 100 other little diodes for $.99 delivered?
How they can do that I have no idea.

I did notice those internal details on the LEDs after testing.
9V battery was weak but definitely not dead.
I figured it had to be giving it more than the 6.3V I was going to use.
I thought the LEDs might of had a max current limit line.
Like HBP posted but with an exit line that leveled out as voltage continued to increase.

I thought if the 9V was going to kill them it would light, get real bright, and pop like a fuse rather quickly.
That didn't happen, but I'm glad I learned the real deal here.