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Best way to reduce Filament voltage?

 

11/12/2003 12:47 AM

Major Pain

Best way to reduce Filament voltage? Well after using a hammond power trannie on a new build Ive got 7.0vac on my filaments and was wondering the best way to reduce the voltage?     Im guessing that 7.0 will damage the tubes and they would be better on the 6.3 vac.     Would using 2-100 ohm 5 watt resitors ,one resistor on each filamnet leg do the trick? or is there   another way to do this?     Thanks     M.P.

 

11/12/2003 1:31 AM

Don Symes

First - filament voltages are spec'd +/-20% (iirc), whch goes ofer 7.5V. Even 6.3*1.1 = 6.9V.     Yes, you will slightly shorten your tube's life, but not by an amount you're likely to notice.     Next - the resistors method is cheap and reliable provided you add up all the heater currents - they're parallel, right? - and do a bit of algebra to get the resistor values and wattages, then use the nearsest standard value and the next bigger wattage for reliability.     You want two resistors to drop .7V.   Each one needs to drop .35V     0.35V/HeaterCurrent = ResistorValue     HeaterCurrent^2 * ResistorValue = Wattage     HTH!

 

11/12/2003 2:23 AM
Matt H	also, measure the filament winding once you've got all the tubes in before you stick resistors in there... it's pretty normal to get a higher than expected voltage with no load.

 

11/14/2003 3:43 PM
Major Pain	Don,   Would this be adding a resistor in between the filament string to drop the voltage?     Im using 3x12AX7 and 2x6L6 in this amp. and getting 7.5 under load which is kinda hi     Im also algebra challenged too.....slept thru most of hi school.....     M.P.

 

11/16/2003 1:32 PM

Don Symes

Well, you have 3*.3A for the 12ax7s ==> .9A and 2*.9A for the 6L6s ==> 1.8A so total heater current is 2.7A.     Now, you want to drop 7.5 - 6.3 = 1.2V total, or .6V per resistor (one in each side of the heater pair), so     .6V/1.8A = 0.3 ohms and .6V*1.8A = 1.08W.     Grab 3 1-ohm 1/2-watt resistors per side and put 'em in parallel, then put them in series with the heater wires.     That ought'a do it.

 

11/17/2003 2:59 AM
Major Pain	Re: ? Hey Don,     Grab 3 1-ohm 1/2-watt resistors per side and put 'em in parallel, then put them in series with the heater wires.     That ought'a do it.       I guess Im still a little fuzzy on this, do you mean put the three 1-ohm resistors in parallel then   then insert this into on side of the filament chain, Then do exactly the same for the other side too?       M.P.

 

11/18/2003 1:56 AM

Don Symes

1 Stack 3 resistors together.     2 Wrap the 3 leads on one end together and solder.     3 Wrap up the 3 leads on the other end and solder.     Repeat 1-3 to make a second set.     a Unsolder a heater wire from the first tube they hit.     b Slip a bit of heatshrink tubing over the wire. Trim the resistor leads to a convenient length and solder to the tube pin you just took the wire off of.     c Solder the wire to the other resistor leads. Shrink the tubing to insulate that junction.     Repeat a-c for the other heater wire to that same tube.

 

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