dim my bright ass power bulb? (6.3V AC)

[impetus maximus]

howdy,

my 6.3V AC power bulb is too bright on my FireFly. how should i go about
taming it?

thanks

 

[Hollowbody]

Just a guess. Ceramic resistor.

 

[impetus maximus]

thanks, i was thinking a resistor. don't know what kind of value to try

 

[TradAmpGuy]

Let me check which bulb is in there. Then I'll give you a number

 

[impetus maximus]

thanks brother

 

[chuckharmonjr]

The damned blue bulb on my delay pedal was killing me it was so damned bright...I went after it with a black sharpie...much tamer now

 

[impetus maximus]

there's a resistor that determines the brightness of the LED in stomp pedals.
one of the mods i installed on my EQ pedal came with a resistor to brighten
the LED the kit came with. i used it and like it because it's easy to see.

 

[TradAmpGuy]

OK.... It's a .15A 6.3V lamp. That works out to .95W and 42 Ohms. But, lamps have less resistance and draw more current when they are cold so you cant just throw a 42 Ohm resistor in there and expect 1/2 the brightness... So.... with a handy little java app I have, I played around with a few numbers and came away with 15 Ohms in series with the lamp should drop it to about a 1/2W... maybe... I dunno. OH, and the resistor is dissipating about 1/4W of heat so a 1/2W resistor would be the minimum value I'd use.

 

[impetus maximus]

awesome man. would something like this work?

 

[TradAmpGuy]

awesome man. would something like this work?

Should be cool, or you can add a 1W to your cap order if you havent placed it already

 

[impetus maximus]

i ordered the caps around noon from Antique Electronics.
having trouble logging in to see if they shipped them yet.

 

[impetus maximus]

good news i can finally log into Antique Electronics.
bad news they already shipped my caps.

went to the shack and asked for a 1W 25Ω resistor. brought it home and it's
got 4 bands; red, violet, black, and red. it's a 27Ω. note to self double check values at the store.

Brian, remember we ran the heater wires from the transformer to the light.
then ran wires from the light to the sockets. i pulled two wires on one
side of the light and paired them together. soldered the resistor to them, put
heat shrink over that connection. then wired the resistor to the bulb.
the bulb is perfect now

thanks bro
/max

 

[TradAmpGuy]

Had to read that twice, but yep, thats perfect :cool2:

NP man

 

[Kunnz]

Wear shades !

 

[Kunnz]

Ah ok keep the shades on.

Right ok lets dim this bright ass bulb another way, A BETTER WAY!
Does capacitive reactance ring a bell?

Remember resistance uses up I squared R and we avoid that in a lot of circuits.

Capacitors and inductors are both passive and active. They DON’T dissipate power and return all energy to the source in an AC network.
They won’t load your pilot light/heater circuit either so take that resistor out throw it away and use a capacitor to drop the voltage.

To dim it a bit I would try about 4 volts to start with so using Tradampguy’s calculations that the light works out to be 0.15A, 6.3V lamp. That works out to .95W and 42 Ohms, then that makes it a breeze to calculate although they may be wrong, But who cares as the maths never lie!

The capacitance required is calculated the following way using Kirchoff and linear algebra.
Lets try 4v, so

6.3v squared = 4v squared + cap voltage squared (Vc)

39.69v = 16v + Vc

= 39.69v – 16v = 23.69 Vc squared

then take the sqr root of 23.69 so Vc = 4.86v

now if it’s a .95W that Tradampguy has said then

.95W / 4v = 0.237A

current I = 2 x pi x f x C x Vc

so to isolate C from the equation we do a wee bit of algebra so;

C = 0.237 / 2 x pi x 60 x 4.86 = 129.35uF

A (circa) 129.35 uF capacitor will drop the voltage down to 4v but if you go buy a trimmer capacitor with a slightly higher capacitance then you will be able to solder it in series with the light and adjust to your hearts content. lol.

In small circuits where we have no space that’s what we do.


Wait !

There is one more thing to consider, it’s not that important here as we are not running an industrial estate here and power factor correction wont be an issue, but anyhow lets proceed.

The current will lead the applied voltage by 90’ in a totally capacitive circuit but the resistance of the bulb here shall bring it back somewhat.

anyhow.

4v / 6.3v = 0.634

take the reciprocal of the cosine of 0.634

= 50’.58” degrees or 0.882 radians if you prefer.


Ár mór.

 

[TradAmpGuy]

Very nice. Have to give that a whirl.

 

[Kunnz]

Very nice. Have to give that a whirl.

Yea it's the way it's done on micro power supplies as well as power electronics that neglect thyristor s for good reason.

enjoy.

 

[JohnH]

129.35 uF capacitor will drop the voltage down to 4v but if you go buy a trimmer capacitor

If you believe that a trimmer capacitor exists with that kind of magnitude, then I doubt your advice

 

[MajorNut1967]

I have a question here, "is this the standard 6.3 V bayonet type of lamp?" Because if it's any standardize bayonet or screw in type pilot light, all you have to do to reduce the brightness is go get the 7 V, 8.63 or a 12 version of it, which are quite readily available.

 

[TradAmpGuy]

Yeah, its a bayonet Fender style jewel lamp. CE Distribution lists several voltages and amperage draws. The one in his amp is a standard Fender 6.3V/.15A bulb. 7.5V or 14.4V might do the trick.