Voltage drop resistor for 5 20LF6 on 120VAC

[357]

I need to figure out what resistor to drop the line voltage to match the 20V required for the 20V heaters.

there are 5 tubes fed with 120.

I need to get the volts down to 102V or so.

 

[linearone]

I need to figure out what resistor to drop the line voltage to match the 20V required for the 20V heaters.

there are 5 tubes fed with 120.

I need to get the volts down to 102V or so.

Well... if you have 5 of these tubes and the heaters are 20v each at .6 amps, fed in series it would be 100v 3 amp load... I think... You need to drop 20v.
using ohms law... 6.7 ohms sounds about right. but... it will need to be 60 watts yikes...

is my math right?

(RNI) NH250-13.7 13.7 ohm 0.1% 250w NH-250 "NH" style Aluminum Body Type A

65.00​

Add​

surplussales of nebraska has a 13.7 ohm you could use two in paralel the wattage rating is correct.
look around you might find something

 

[ken white]

I need to figure out what resistor to drop the line voltage to match the 20V required for the 20V heaters.

there are 5 tubes fed with 120.

I need to get the volts down to 102V or so.

You are going to need to either measure the filament current or look up the filament current specification and go from there. The resistor size would be equal to:

R = 20V/Current in Filaments assuming your series connection and the resistors power rating is equal to I^2*R

If it were me, I would use a capacitor of the same resistor impedance value since it won't dissipate the energy and turn it into heat.

C = (R*377)^(-1)

 

[ken white]

Well... if you have 5 of these tubes and the heaters are 20v each at .6 amps, fed in series it would be 100v 3 amp load... I think... You need to drop 20v.
using ohms law... 6.7 ohms sounds about right. but... it will need to be 60 watts yikes...

is my math right?

Your math is correct if the filament current is 3 amps...

If the filaments need 0.6 amps as you stated just before the 3 amp statement, then R = 33.33 ohms at 12 W.

 

[linearone]

why not just feed them in parallel and use a cheap easy to find transformer instead. it will be easier cheaper and smaller. all you need is a 20v 3 amp transformer

 

[linearone]

You are going to need to either measure the filament current or look up the filament current specification and go from there. The resistor size would be equal to:

R = 20V/Current in Filaments assuming your series connection and the resistors power rating is equal to I^2*R

If it were me, I would use a capacitor of the same resistor impedance value since it won't dissipate the energy and turn it into heat.

C = (R*377)^(-1)

thats stated in farads right? I got
0.000395898492 which looks like 395.898492uf to me.

 

[ken white]

thats stated in farads right? I got
0.000395898492 which looks like 395.898492uf to me.

The result is in farads. For the 33.33 ohm impedance it would be approximately 80 uF.

 

[linearone]

The result is in farads. For the 33.33 ohm impedance it would be approximately 80 uF.

how did I screw that up?

 

[linearone]

Your math is correct if the filament current is 3 amps...

If the filaments need 0.6 amps as you stated just before the 3 amp statement, then R = 33.33 ohms at 12 W.

sorry .6 amps each heater.

 

[ken white]

how did I screw that up?

You didn't, you calculated the value at 6.7 ohms which would be for each filament needing 3 amps, not 0.6 amps.

 

[linearone]

You didn't, you calculated the value at 6.7 ohms which would be for each filament needing 3 amps, not 0.6 amps.

In my head I saw a string of light bulbs all needing 20v each each using .6amps of current each, I saw 100v total in series and 3 amps total current.

 

[paws264]

Well... if you have 5 of these tubes and the heaters are 20v each at .6 amps, fed in series it would be 100v 3 amp load... I think... You need to drop 20v.
using ohms law... 6.7 ohms sounds about right. but... it will need to be 60 watts yikes...

is my math right?

(RNI) NH250-13.7 13.7 ohm 0.1% 250w NH-250 "NH" style Aluminum Body Type A

65.00​

Add​

surplussales of nebraska has a 13.7 ohm you could use two in paralel the wattage rating is correct.
look around you might find something

Nope, I believe that your understanding of Ohm's law is a little off.

The current (amps) in a purely series circuit is the same when measured anywhere within the series circuit. Voltage drops across loads in a purely series circuit combine to equal the total applied voltage.

Current is divided in a purely parallel circuit with the total current equal to the sum of the value of the currents flowing in the parallel branches. Voltage applied to the legs of a purely parallel circuit are of equal potential however, the current will be of different values (added up to get total current).

I am not sure what amp this applies to but, the Kendrich Golden Eagle uses 5 of the 20lf6 tubes with no dropping resistor; the engineers counted on there being a voltage sag when a load is placed on the line. Voltage measured at the mains and voltage measured at the plug or, inside the amplifier at key-down will be different.

.

 

[paws264]

In my head I saw a string of light bulbs all needing 20v each each using .6amps of current each, I saw 100v total in series and 3 amps total current.

Only if the tubes were strung in parallel, not series would this be true.

.

ETA: And fed with 20 volts each tube from a 20 volt power supply (Do Not parallel 20lf6 tubes across a 120 volt line), but you knew that already, right?

.

 

[linearone]

Only if the tubes were strung in parallel, not series would this be true.

.

yup.
I think the real problem was trying to use ohms law at 7am.

 

[HiDef]

You might want to measure actual heater current on the bench with a warmed up tube before ordering a few ohms.

Otherwise you might consider two stacks of diodes. IIRC 1N4007 or similar. Make 2 strings of 35 or so in series. Hook the ends of the 2 strings up but in opposite polarity. Feed one juncture with the AC line. You will have adjustable voltage drop depending on where you put the load.

The capacitive reactance idea is neat too.