Bottlehead Forum
Bottlehead Kits => Legacy Kit Products => Quickie => Topic started by: bundee1 on February 18, 2010, 11:30:03 AM
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I finished the quickie and it sounds pretty good. Im getting signal leakage and hiss on the 2nd input but its still nice. Are the tubes supposed to light up? Mine dont even though I get sound.
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The filaments on these battery tubes are very very tiny. You may or may not see them in a dark room after your eyes have adapted to the darkness. A normal tube small tube uses 10-15 times as much power to heat the filament.
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At night, in a total dark room, I can just make out a bit of glow.
My only disapointment with the "Quickie". I like tubes that light up. Like most primates I amused by shiny things.
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Place an LED under the tube?
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I'm actually planning on doing just that - though tiny, low current orange LED's seem to be something that the local Rat Shack doesn't carry.
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The LED idea works well - just gently force a small LED in the center hole of the tube socket, add to the circuit between the B+ and plate chokes for each channel, and voila! I used yellowish-orange micro LEDs, as that was all I could get my hands on locally. They came in a $2 pack of assorted LED's from Radio Shack. A little dot of permanent marker or red stain would make them look even more authentic, as they are a bit bright and a bit too yellow (versus the reddish-orange of a "real" tube).
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So how would you wire the LEDs if you're not using a choke?
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Put them between the center pin of the power switch, where the plate resistors normally join and terminate, and whatever you're using as a plate load for each anode (be it the same resistors, choke, or PJCSS). Do this for each channel. They should go between the plate load and the B+ as opposed to between the tube socket and plate load, to avoid adding noise to the circuit.
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Ok, I went to radio shack to check out some leds but I have no idea which voltage or mA to get. They had a pretty wide range of choices.
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Let's see, IIRC, and it has been a while, there is a forward voltage on LEDs. That is usually small like 0.7V. And LEDs want a specific current, like you mentioned.
What you are going to do is put a resistor in series with the LEDs. You can use one resistor and then the two LEDs in parallel. This combination will be in parallel with either the 36V to the plates or the 1.5V to one or the other heaters. To find the resistance the formula is R=V/I where the resistance is the voltage you want the resistor to drop (V) divided by the current (I) that the LED is rated at.
If you use the heater battery, that is 0.8V/LED current. If it is the plate voltage that is 35.3V/LED current. In these two examples I used 0.7V drop across the LED. That might not be right.
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LEDs run a higher voltage than silicon diodes. The voltage depends on the color; red LEDs run about 1.6-1.8 volts drop depending on the specific chemistry of the semiconductor. Orange and yellow are around 2.4v, and blue can be as high as 4 volts. They are rated for maximum current but will operate at any lower current, with less light output naturally.
To illuminate the tube, I would get a small LED and shove it up the center post of the socket. Wire the cathode to the plate (pin 6) and the anode to the center post, and use the center post for the plate load (4K resistor or choke or PJCCS, which originally went to pin 6). Continue to use pin 2 as the output connection.
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Thanks guys. I think I understand now.
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Do you think adding the LEDs affect battery life appreciably?
Thanks
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"Do you think adding the LEDs affect battery life ..."
Yes.
"... appreciably?"
No.
The LEDs will drop a couple volts, so the 36-v set of batteries will be nominally dead when the voltage falls to 26v instead of the normal 24v. This will cost you about 12% of the nominal battery life, i.e. you'll get 175 hours instead of 200 hours.