7808 or 7806 regulator for Rasmussen tube gainclone ?

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sandro600

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Joined 2003
#1
On Joe's pages

http://members.ozemail.com.au/~lisaras/ps.htm

The voltage regulator is 7808 in the schematic but when I test the output voltage of the circuit the voltage is 8V not 6.3V
as recommended for the 6922

Other sites like Peter Moore and decdun use a 7806 with different capacitance pre and post regulator

http://www.petemoore.pwp.blueyonder.co.uk/GC Pics/Filament Supply.jpg

Any thoughts/ recommendations ?

im measuring the output of the regulator with no tube attached if that makes any difference . Thanks
 
#3
thanks for the speedy reply. I should expand and say nothing , not the circuit or the tube is yet connected to the power supply.

I was checking the output of all the power supplies ive just finished building for the buffer and the only one that comes above spec was the filament supply.

Still ok to connect to the tube ?
 
#6
Re: Re: 7808 or 7806 regulator for Rasmussen tube gainclone ?

moamps said:


Hi,

Use 7806, standard application (or Peter's).

Regards
Click to expand...


...or use a 7805 with a pair of 1n4001 diodes in series between 0V pin of the 7805 and the 0V rail . Diodes drop a total of 1.2V makes the output closer to the ideal 6.3V

316a
 
#7
Jax said:
Use the 7808. The 5.6 ohm resistor drops the voltage to 6.3V when the tube is installed. The resistor also serves as a soft start for the heater which is nice.
Click to expand...

Hi,

filament resistance is about 4ohms. With 7808 and 5R6 starting current is above 0.8A. With 7806 is 1A (current limit). No big deal.
Second, Joe's filament PS looks slightly underrated to me. With 10VAC on secondary, 1.4V drops on bridge diodes, only 1000uF for 0,3A current, 7808 can't work correctly (IMHO). It isn't low-drop regulator.

Regards
 
#8
Hi,

With 10VAC on secondary, 1.4V drops on bridge diodes, only 1000uF for 0,3A current, 7808 can't work correctly (IMHO). It isn't low-drop regulator.
Click to expand...

10 VAC * 1.4 = 14 VDC.

The 6N1P heater requires 0.6 A at 6.3 V +/- 10% as far as voltage is concerned.

So the 7808 needs to see a minimum voltage difference of 5V between input and output, a condition that is met, and needs to be a 1A type.

The resistor will drop down the voltage to 6V+ and will provide for a gentle start-up.

Frankly, I can't see anything wrong with it.

Cheers,😉
 
#9
fdegrove said:

10 VAC * 1.4 = 14 VDC.
The 6N1P heater requires 0.6 A at 6.3 V +/- 10% as far as voltage is concerned.
So the 7808 needs to see a minimum voltage difference of 5V between input and output, a condition that is met, and needs to be a 1A type.
Click to expand...


Hi,

In a real world rectifiers diodes aren't ideal (1.2 to 1.4V drop) and Vripple on cap isn't zero (about 2.5V for C=1000uF, I=0,3A); minimum input voltage on regulator 7808 can be in this case about 10V what isn't enough for 8V output voltage.
Regulators output will be 8V with added 50 or 60Hz drop on it.

Regards
 
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