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Old 8th February 2013, 05:50 PM   #11
cotdt is offline cotdt  United States
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With fixed bias scheme of the TSE, I'm not sure the filament supply matters. It sounds very similar to AC filaments.
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Old 8th February 2013, 05:54 PM   #12
dmcgown is offline dmcgown  United States
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The circuit you have sketched is a current regulator using an adjustable regulator (317)
The resistor between the output and adjust pin on the regulator is in the incorrect location. It needs to be connected to the output pin, with the adjust pin connected on the other end. The resistor is sized to the so the current it will result in a 1.25V drop across the resistor. Obviously the current you need is what the 300B likes to see for a 5V filament, which is 1.2A, which results in a resistor value of 1.25/1.2 = 1.04 ohms

One other thing, you will need to plan for dropout voltage between 2.5 to 3 volts, in addition to the 1.25V drop across the resistor. This means your minimum DC voltage at the input pin of the regulator is 9VDC. Also plan on heatsinking the regulator, since it will be dissipating close to 4W. (3V x 1.2A)

David
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Old 8th February 2013, 05:58 PM   #13
dmcgown is offline dmcgown  United States
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I forgot to make clear that the load is connected at the same side of the resistor as the adjust pin on the regulator.
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Old 8th February 2013, 06:25 PM   #14
12E1 is offline 12E1  United Kingdom
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Originally Posted by dmcgown View Post
The resistor between the output and adjust pin on the regulator is in the incorrect location. It needs to be connected to the output pin, with the adjust pin connected on the other end.
Ouch! I missed that. And I used to be a hardware designer. I must be losing it...
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Old 8th February 2013, 06:27 PM   #15
Green77 is offline Green77  Sweden
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Ok, now i don't follow the resistor is between out and adj? I think that i draw like that in the schematic.
Thanks //Daniel
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Old 8th February 2013, 06:35 PM   #16
Green77 is offline Green77  Sweden
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Ok, thanks! I see the error now

//Daniel
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Old 8th February 2013, 07:05 PM   #17
tomchr is offline tomchr  United States
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Originally Posted by dmcgown View Post
One other thing, you will need to plan for dropout voltage between 2.5 to 3 volts, in addition to the 1.25V drop across the resistor. This means your minimum DC voltage at the input pin of the regulator is 9VDC.
Yep. You can't generate 5 V from a 6 V AC transformer winding using a linear regulator.

At low mains, you'll have roughly 0.9*1.3*6 = 7 V on the input to the voltage regulator. That does not account for the ripple voltage. At 50 Hz mains, 1.25 A, 4700 uF, you'll end up with roughly ((1/(2*50)) * 1.25)/4700E-6 = 2.7 V ripple. That doesn't leave enough voltage across the regulator to regulate properly.

~Tom
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Old 8th February 2013, 08:08 PM   #18
euro21 is offline euro21  Hungary
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Originally Posted by 12E1 View Post
If the constant current circuit does its job, then 4700uF is fine.
If you use DHT, this is not entirely true.
As Rod Coleman described (#26 pre amp topic), each injected mV hum more or less appears on the anode.

As you can see in my sketch, 200mVpp hum (BEFORE CCS!) produce 90mVpp output "noise".
In case of DHT not good enough filtration.
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File Type: jpg Hum inject to DHT heater.jpg (169.9 KB, 85 views)
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Old 8th February 2013, 09:39 PM   #19
12E1 is offline 12E1  United Kingdom
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Originally Posted by euro21 View Post
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Originally Posted by 12E1 View Post
If the constant current circuit does its job, then 4700uF is fine.
If you use DHT, this is not entirely true.
As Rod Coleman described (#26 pre amp topic), each injected mV hum more or less appears on the anode.

As you can see in my sketch, 200mVpp hum (BEFORE CCS!) produce 90mVpp output "noise".
In case of DHT not good enough filtration.
That suggests that the either the filament changes its resistance, or that the CCS does not do its job well enough, or a combination of the two. I don't doubt what you say, however. It would certainly be true that any hum component present on the filament would then appear as unwanted hum on the amplifier output.
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Old 8th February 2013, 10:23 PM   #20
dmcgown is offline dmcgown  United States
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What he needs to do is reconfigure the rectifier into a bridge so he is working with a 12V winding. That should provide enough voltage. It should work with if using fixed bias as pointed out above, although with a 2.5A rated transformer, he will be beyond the rated current. I would recommend adding another transformer with a full wave bridge, so there seperate heater circuits in case he wants to explore cathode bias, and heatsink the 317s (or 1034s) very well.

David
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