Lowering CE 20A circuit voltages

[wa2tak]

Read some various ways on this great forum of getting the proper 6.3 and HV voltages with 120 VAC line .

Curiously, did not see this one: voltage dropping resistor in the 120 line input.
Closest was using a external variac.

Sooooooooooo let's put one ON the rig.

Here's the math, as I see it:

My filaments are at 7 vac as expected.
Tubes want 6.3... 7 - 6.3 = 0.7 vac drop....10% drop.

10% x 120 = 12 vac line drop.

My 20A in CW was at 1.5 ac amps.

Need resistor value of 12 vac / 1. 5 amps = 8 ohms.
Need resistor power rating of 1.5 amps x 12 vac = 18 watts.

Thus; use wirewound variable resistor...cost about $10 to $20.

Can us chassis as heat sink...or use extruded heat sink mounted to chassis.

Some vintage radios used resistance wire IN the AC line.
They were called "curtain burners" as they caused fire.

Any comments?

Steve
WA2TAK

[K8DI]

The issue is, in standby, where the rig will be 90% of the time, including between the dits and dahs, the current will be less, the resistor drop will be less, and the filaments will still run high…90% of the time.  Better to do resistance in both filament lines, if you don’t mind the raised B+ levels. 

Ed

[wa2tak]

VERY astute observation, Ed.

Using just tube filament currents ( per tube data sheets )...total = 4.7 amps.

Power = 4.7 x 6.3 = 30 watts

30 watts @ 120 vac = 0.25 ac amps

10 ohm voltage drop = 0.25 ac amps x 10 ohms = 2.5 vac voltage drop.

120 vac - 2.5 vac = 117.5 volts AC to rig xmfr. = STILL TOO HIGH with rig in "idle".

Your idea of using dropping resistors seems DOABLE for only $2.63 each leg...size is 1 x 0.4 inches:

7 - 6.3 = 0.7 voltage drop
resistor value = 0.7v  / 4.7amps = 0.15 ohms
resistor wattage = 3.5 watts

Please see this one:
https://www.mouser.com/ProductDetail/TE-Connectivity-Holsworthy/SMW7R15JT?qs=zW32dvEIR3tzutB0RL4gyw%3D%3D

I also used surface mount caps to fix silver mica disease in IF cans.

Next...getting that 400+ B+ to proper value.

[DMOD]

VERY astute observation, Ed.

Using just tube filament currents ( per tube data sheets )...total = 4.7 amps.

Power = 4.7 x 6.3 = 30 watts

30 watts @ 120 vac = 0.25 ac amps

10 ohm voltage drop = 0.25 ac amps x 10 ohms = 2.5 vac voltage drop.

120 vac - 2.5 vac = 117.5 volts AC to rig xmfr. = STILL TOO HIGH with rig in "idle".

Your idea of using dropping resistors seems DOABLE for only $2.63 each leg...size is 1 x 0.4 inches:

7 - 6.3 = 0.7 voltage drop
resistor value = 0.7v  / 4.7amps = 0.15 ohms
resistor wattage = 3.5 watts

Please see this one:
https://www.mouser.com/ProductDetail/TE-Connectivity-Holsworthy/SMW7R15JT?qs=zW32dvEIR3tzutB0RL4gyw%3D%3D

I also used surface mount caps to fix silver mica disease in IF cans.

Next...getting that 400+ B+ to proper value.

My  suggestion:

1) Measure Key Up voltage at F and F,

2) Add up all filament currents from Franks Tubes, we'll call that Ift,

3) (Key Up Filament Voltage - 6.3 Volts)/ Ift = Rdrop,

4. Place Rdrop/2 = RLeg in each filament supply leg.

For example: Let's say Key Up voltage = 6.8Vac.
My total filament current = 4.5A.
My preferred filament voltage = 6.3V.

Rdrop needed = 0.5Vac/4.5A = 0.1 ohms

RLeg = 0.05 ohms for each leg from F,F, at 2W each.

Phil
 

[W7TFO]

I'd use a filament transformer in 'buck' mode to achieve the desired result.  This is a linear type of E reduction, not so dependent on current flow under varied load.

Inexpensive, too with very little loss due to heat.

73DG

[N8ETQ]

Dennis has the correct approach.  You gotta "buck"..
Also consider that in 1959 AC Line was closer to 110.
On my "Burn in Bench" I have a big 12.6VAC Filament
xfmr wired to Buck line to a 5 position power strip.

I have not seen  120VAC around here for years.
Just checked and currently at 122.8VAC.  Smokin'
hot...  Goes up in the summer too.

GL es 73
Dan

[W7TFO]

Motors are happy with slightly higher Voltages, as they pull less I, according to the PF.

Everything else (with exception of resistive space heaters) not so much.

So box up an appropriate piece of iron, affix cords or receps, and proceed.  If you want to include a VM, ping me---some here.

73DG

[Jim, W5JO]

Just looked at a mil spec sheet for a 5814 (12AU7) which says the minimum filament voltage is 5.7 and the max is 6.9

Eimac says + 5% and - 5% for their tubes.

[wa2tak]

Good Job Jim!

I wish to CONFIRM your findings.

Per Mil Spec: Mil-E-1D which lists various military tubes and their SPECIFIC TUBE SPECS:

tube      filament voltage range
6X4        5.7 - 6.9
6CL6         same
6AU6         same
6CL6         same

6L6WCB - ruggedized tube  5.7 -6.9

Thus +/- 10 % variation of filament voltage WAS acceptable for the MILITARY.

Now...my CE 20A:

120 VAC

B+ = 430 vdc 'idle'
B+ = 345 vdc on transmit

Insert 400 ohms into B+ >> 305 vdc on transmit

Schematic B+ voltage in SB1 mode = 300 vdc

[wa2tak]

PS....here is the actual 6L6WGB spec sheet: attached

[K9MB]

I'd use a filament transformer in 'buck' mode to achieve the desired result.  This is a linear type of E reduction, not so dependent on current flow under varied load.

Inexpensive, too with very little loss due to heat.

73DG

As always, you go to the best solution. Adding a filament transformer in series with the primary of the power transformer will reduce the voltage across the primary and put you back into the 50s and early 60s grid voltage levels that was what the CE was designed to use.
Everything will cool down if you are not putting in 10-15% more voltage than the transformer was designed to use. It can be a 2 amp winding so it can be a little one and will take little space. Jameco sells them and other places.
Elegant solution-

https://www.jameco.com/z/29225-R-Jameco-ReliaPro-Power-Transformer-12-6VCT-2A-115VAC-Wire-Leads_29226.html

[W7TFO]

Thanks for the kudos, Mike!

73DG

[K9MB]

Thanks for the kudos, Mike!

73DG

Always a pleasure…
My only reservation is that you may have caused Rube Goldberg to roll over in his grave, because there are much more complicated solutions out there that will never be built up.… 😉 MB