LED to replace 6.3v pilot lamp

they make drop-in replacements for amp bulbs using LEDs

like these

i've switched most of my bulbs out for the LEDs. they'll probably outlive me

 

I'll try and someone can fill in due to the AC.

6.3vac thinking it will increase with rectification (led)
So let's say 7V? You'll need to get actual data.

So for this particular LED (see pic) high bright from Radio Farters Shack
FW current 25mA FW supply 3.3V

So R= 7V-3.3V / .025
3.7/.025
=148

Maybe a 200ohm 1/2 watt would be a good start???

I could be wrong though d/t the AC factor.

 

they make colored LED bulbs too for some fun mix and match options. i think i got a 5 pack of them for about a dollar a piece

 

jhundt,

Since the 6.3 V transformer is A.C. and the 6.3 V value is RMS (the average heating value of the sine wave), you have to consider the peak voltage of the transformer when doing the calculation Andy provided, otherwise you may burn up you LED prematurely. Andy used 7 V instead of 6.3 V which is a good idea to preserve your LED when you run a chassis with some or all tubes pulled because the transformer output is higher than nominal when lightly loaded. But instead of 7 volts, you need to use 7(rms) * 1.414 = 9.9 V (peak). Then you calculate the resistor value as was demonstrated above:

(9.9V - 3.3V)/25 ma = 264 ohms (270 ohms is closest standard value)

Resistor wattage = I^2 * R = 0.025*0.025*270 = 0.17 watts (1/2 watt will be good)

Now you must consider that the LED will only work on the positive half of the AC cycle unless you put a diode bridge ahead of the current limit resistor so you get both halves rectified into positive polarity. This will double the light output.

Remember to use the voltage and current ratings of the LED you want to put in your amp when doing the calculations. There are a lot of different LEDs on the market and the specs vary, but the operating voltage and the current are somewhat critical to performance and longevity.

If you need a schematic I'll try to scratch one for you.

Bob

 

I wasn't sure about the 1.4 factor,

thanks Bob!!

 

Great post Bob.

 

jhundt,

If the bulbs are 8 VDC and in series, they are fed by 16 VDC - true?

Two 3.3 V LEDs in series need 6.6 V to light. Since they are in series, the current consumed would be the current for a single LED...say 25 ma.

The voltage drop required would be 16 V - 6.6 V = 9.4 V

The resistor required to provide 6.6 V to the LEDs at 25 ma is R = V/I
9.4 V / 0.025 Amps = 376 Ohms (390 Ohms is the closest common value)

Power for the resistor is 0.025^2 * 390 = 0.23 Watts ...so a 1/2 Watt resistor will work.

You should measure the DC feed to the lamps to see what the no-load voltage really is in case the supply is unregulated and sits higher than 16 Volts with a little load like 25 ma. In that case the resistor will calculate to a higher value of both resistance and power.

Bob

 

jhundt,

'Glad thing work out and you're amps are all lit up with only small damage to your person. I tend to want to scratch my forehead when I have a soldering iron in my hand...so I get lots of offers for bit parts in Frankenstein movies :>)

Bob

 

Congratulations, sometimes we get such a nice satisfied feeling from even a simple little bit of DIY!

Traditionally (i.e. 20 years ago) most LED's needed 10 -> 20 mA to light them up brightly. This has changed drastically in recent years - there are a lot of extremely efficient LED's out there now, and some of them are reasonably bright if you feed them just one single milliampere of current! That means these super high efficiency LED's need a resistor ten to twenty times bigger (!) than the "traditional" numbers Bob gave you for "classic" LEDs.

Bottom line, if it's bright enough, and the current is not big enough to damage the LED, you're using the right resistor. Sounds like you're using the right resistor!

There is one other thing I'd like to add. LED's don't tolerate reverse voltage well. They can be destroyed by too much of it - and in many cases, "too much" is anything more than about 5 volts.

This becomes an issue when you run them off AC, like the 6.3V AC heater power supply you were discussing, which can put up to 9 volts (peak) reverse voltage across the LED, 60 times each second.

There is a simple fix - you use an ordinary silicon rectifier diode to prevent reverse voltage from appearing across the LED. The simplest way is to wire the diode backwards, across (in parallel with) the LED. So you connect the LED's positive lead to the diode's negative lead, and vice-versa.

If you do this, the diode will clamp the reverse voltage across the LED to a maximum of about 0.7 volts, which is completely harmless. It won't affect the brightness or normal operation of the LED in any way. It also won't affect the tube heaters in any way - those see the normal 6.3V AC, just as they always have.

You can use just about any ordinary silicon rectifier diode for this, such as any of the 1N400x diodes (1N4001, 1N4002, et cetera). They are cheap and easy to find - even Rat Shack should have them.

99% of the time LED's are used with DC supplies, which never reverse polarity. That's why you rarely see this sort of protection diode used with LEDs. But they really should be considered mandatory when powering an LED from AC voltage.

-Gnobuddy

 

It shouldn't hurt anything, usually they fail open-circuit, so they just quietly go out (stop emitting light), without hurting anything else.

Even if an LED failed short-circuit (which I've never seen happen), it wouldn't hurt anything - the series resistor you added will prevent the short from affecting anything else.

Murphy's Law being what it is, I'm gonna predict your LED will now proceed to outlive me, reverse voltage and all!

-Gnobuddy

 

Gnobuddy,

I'm sure I'm 20 years behind the times, but if you look at the Mouser catalog online for standard, through-hole LED's (white diffused,e.g.) over 95% of 'em are listed as 20 mA forward current (three pages worth). So my advice wasn't too incorrect for the context here. However, I'll put a fossil warning on any future design calculations so no one feels misled... :>)

Bob

 

Bob, no worries, I'm about 70 years behind the times myself - I'm currently designing and building a little tube amp, a technology that was already becoming obsolete around 1955!

I'm also far behind the times with solid-state components: I've never soldered an SMD component, and I'm not particularly keen to try, which immediately puts me a couple of decades out of touch.

As for LED's, I knew that there had been a big push towards developing LED's with higher and higher power handling ability for years, but until quite recently I hadn't realised there had also been a similar push for improved efficiency. It's pretty shocking to see one little 5 millimetre LED rated to put out 500 millicandela, and right next to it see a second one rated for 15,000 millicandela - thirty times more light, usually for about the same rated current.

And that's where the rated current becomes a little misleading: many of those super-efficient LEDs will in fact happily run at 20 mA, but they become eye-searingly bright at those currents. Quite literally intolerable to look at directly, and in some cases, actually bright enough to cause eye damage if you look straight at them.

I found some of those very efficient modern LEDs are bright enough to work as power indicators at one single milliamp of current. Very nice for DIY effects pedals, because the old 20 mA LED's would kill a 9V flat battery in no time all by themselves!

The low current consumption hardly matters when running off a hefty filament power supply, but excessive brightness is still very much an issue. There have even been commercial products (such as the old Alesis M1 Active studio monitor) that caused a lot of customer complaints because the blue power-on LED indicator was painfully bright!

-Gnobuddy

 

See if you can find the entire led pilot light assembly out of a champ 600 or Excelsior. I hate em.........