Supply voltage for 12v6 12B4 heaters - diyAudio
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Old 27th December 2005, 07:03 AM   #1
Jay is offline Jay  Indonesia
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Question Supply voltage for 12v6 12B4 heaters

I'm building a 12B4 preamp using LM317. For the heaters I'm using power supply I used for my old preamp using 3x 12AT7 per channel. This power supply uses LM340 (7815) followed by CRC (470uF-5R1-220uF). So the output (unloaded) is 15V!

I don't remember how the 12AT7s heaters were wired, but I checked that 12AT7 and 12B4 are both 12v6. Do you think that the supply is safe? I can't imagine that the loaded voltage will be reduced to 12v6 from 15v...
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Old 27th December 2005, 07:57 AM   #2
Tweeker is offline Tweeker  United States
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Heater current for 3 12AT7 at 12.6V is .45 amp.
.45*5R1=2.3V drop. Gives 12.7V. 5R6 would be 12.5V.
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Old 27th December 2005, 08:28 AM   #3
Jay is offline Jay  Indonesia
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Hi Tweeker,

following your way of calculating it, here's mine for 12B4...

One 12B4 draws 0.3A at 12.6v.
To get 12.6v from 15v supply, a resistor is required to drop 15-12.6=2.4volt.
Thus the resistance R=v/i=2.4/0.3=8 ohm.

Is that calculation correct?

Using 8R2 will get 12.54, which is fine. How about 10R to get 12V (in case I can't find 8R2 of suitable size). Will 12v degrade sound quality?
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Old 27th December 2005, 08:46 AM   #4
Tweeker is offline Tweeker  United States
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8R2 would work if average current is 300ma. 12V is within 5% of 12.6V.

Wirewound resistors are fine here, 15 watts + recommended.
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Old 28th December 2005, 02:40 AM   #5
Jay is offline Jay  Indonesia
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Thanks Tweeker. Intutively I think 15W is too big. I planned to try 8R2/10W I used to use in tweeter crossovers. It's already too big This preamp should've been completed last night, but another new project came up
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Old 28th December 2005, 03:19 AM   #6
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Maybe you can replace the 7815 with a 7812 that will supply 12 volts. If you want 12.6 add a diode in series with the ground lead of the 7812.
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Old 28th December 2005, 03:34 AM   #7
Tweeker is offline Tweeker  United States
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Power dissipated in resistor= current squared * resistance (P=I2R). Its best to double this value for part used. You can also use series and or parallel resistors to get what you need.

Whoops, I misplaced a decimal point. 2 watts would be fine.
.3I*.3I*8.2R = 0.82 watts.
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Old 28th December 2005, 03:36 AM   #8
Jay is offline Jay  Indonesia
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What diode (rating) can I use? So I don't need R anymore? If this diode is a zener, without R it should be 0.6v?
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Old 28th December 2005, 04:44 AM   #9
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Any generic silicon diode would work. A 1n4001 or 1n4007. The purpose is to use the .7 volt drop to raise the 12 volt regulator .7 volts above ground. Since the current through the diode is only about 10mA you usually get about .6 volts. The regulator case must be insulated from ground.

No resistors would be needed since the output voltage will be close to 12.6 volts. Measure before connecting your expensive tubes.
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Old 28th December 2005, 06:29 AM   #10
Jay is offline Jay  Indonesia
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Okay thanks a lot! I will try this on the other channel. I guess without CRC the sound difference will be audible. But, well, I don't know, it's only heaters
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