Hi all....I am a radio collector with limmited electronics ability I have a Simplex five tube radio ..no model #(tubes= 25Z5 77 43 39) (I am able to fix caps and such) ...but I am not sure how to replace a resistance line cord (under the chassie) I know resistors are not the solution..they get to hot...any suggestions...ANY HELP WOULD BE GREATLY APPRECIATED....(web sites??)

:Hi all....I am a radio collector with limmited electronics ability I have a Simplex five tube radio ..no model #(tubes= 25Z5 77 43 39) (I am able to fix caps and such) ...but I am not sure how to replace a resistance line cord (under the chassie) I know resistors are not the solution..they get to hot...any suggestions...ANY HELP WOULD BE GREATLY APPRECIATED....(web sites??)

Glen,

You may want to check out

http://www.antiquewireless.org/otb/resto801.htm

Radiodoc

My preference is to use a series capacitor to drop the voltage. It doesn't generate heat.

Agree with Doug, using a cap doesn't generate heat but you will need protection for a dial lamp if one is used in your radio. It will flash bright and burn out from a turn on surge. Zener diodes will work for protection.

A wire wound resistor is similar to a resistance line cord but it will get hot and you may not have a place to mount the resistor.

Norm

:My preference is to use a series capacitor to drop the voltage. It doesn't generate heat.

:Agree with Doug, using a cap doesn't generate heat but you will need protection for a dial lamp if one is used in your radio. It will flash bright and burn out from a turn on surge. Zener diodes will work for protection.
:
:A wire wound resistor is similar to a resistance line cord but it will get hot and you may not have a place to mount the resistor.
:
:Norm
:
:
:
:
::My preference is to use a series capacitor to drop the voltage. It doesn't generate heat.

::Agree with Doug, using a cap doesn't generate heat but you will need protection for a dial lamp if one is used in your radio. It will flash bright and burn out from a turn on surge. Zener diodes will work for protection.
::
::A wire wound resistor is similar to a resistance line cord but it will get hot and you may not have a place to mount the resistor.
::If your equipe is powered by 220 CA you cant use a diode (you reduce la Ief a 50%)You need in addition un little resistor serie but it genera little heat
::Norm
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:::My preference is to use a series capacitor to drop the voltage. It doesn't generate heat.

Never done the capacitor trick except in experimentation. Unfortunately I do not own any line cord resistor radios. I'd like to add some to my collection in the future. I'd also like to have one or two with working line cord resistors just to see what they're like. At any rate, I think that it would be a good idea for all to fuse their filaments with a fuse that can just handle turn-on surges (filaments cold). Modern capacitors are less likely to short, but if one does, it'll blow all of your tubes. With a capacitor and a fuse, the set-up should be ideal.

Thomas

The link in Doc's posting tells how to compute the value of the required dropping cap. The answer, for me, has always been 7-9uF. You could probably just go with about 8uF, and then check the voltage across the tube heaters. (The smaller the cap's value, the larger the voltage drop.)

You should use a non-polarized film cap. I have used Solen caps which can fit under the chassis nicely. I have used 400-VDC Solens with no problem, but Thomas's suggestion of adding a fuse is good. Solen also sells 630-VDC caps.

The dropping cap goes in series in the heater ckt only, not the 120-VAC that feeds the rectifier. There will be an initial surge through the heater ckt because the filaments have low resistance when cold. I don't quite understand Norm's concern about any pilot lamp: I wouldn't expect the surge through the heater ckt to be any greater with a dropping cap than with a resistor. Well, maybe for a quarter of a cycle if the power switch is turned on at the peak of the 60-Hz AC waveform. Maybe Norm can elaborate on his concern?

Many sets with resistance line cords are somewhat cheapos to begin with, and maybe most don't have pilot lamps. I can't recall encountering one for sure.

Resistance line cords, even if in good condition, get noticably warm. They would not be permitted under present-day safety codes, and I think they should all be replaced on general principles. To my knowledge, new replacement resistance line cords are not available.

Everything posted so far in this thread applies to U.S. AC/DC radios designed for 120VAC. I don't know if there were non-U.S. 220-V, 50-Hz sets designed with resistance line cords. If so, the calcs for the required dropping cap values are different.

Here are sources for Solen caps:

http://www.thetubestore.com/capsbymanuf.html

http://www.partsexpress.com/webpage.cfm?&DID=7&WebPage_ID=71&CFID=2265874&CFTOKEN=33712535

Hi Doug

In an AC/DC radio a lamp will have a resistor or section of tube filament in parallel. Although the lamp often flashes at turn on this prevents burn out.

A lamp filament heats much faster than a tube and will burn out from the initial surge. Original resistor line cords will have a tap for a lamp if one is in the radio. You don't have this using a cap for dropping voltage. Some people add Zener diodes across the lamp to limit voltage at turn on.

Here is something else to think about. If a cap is used and a person turns the radio on on off fast it may blow something. Best to add some resistance across the cap for discharge. When a radio is turned off than back on it can double the voltage.

If the radio is turned off during a positive half of the cycle the cap will be charged positive. Turned back on as another positive half cycle is coming along voltage will double. Hard to explain but try it on the bench with a lamp. It will blow the lamp.

Norm

:Here are sources for Solen caps:
:
:http://www.thetubestore.com/capsbymanuf.html
:
:http://www.partsexpress.com/webpage.cfm?&DID=7&WebPage_ID=71&CFID=2265874&CFTOKEN=33712535
:

Thanks, Norm. I understand now.

I thought I had you there for a minute. But, yes, I can understand if you turn the power ON and then OFF quickly, the pilot lamp could be blown.

Maybe a CL-90 inrush limiter might help?

After I typed that last sentence, I realized that once it's hot, the CL-90 wouldn't help on a quick turn-off, turn-on.

Norm, you're the man!

You might be able to use a transformer. If you connect the two 25.2V tubes in parallel, you can run them off a 25.2V (or 24V) step down transformer. As for the other two tubes, you may able to run them with a voltage regulator chip that knocks the 25V down to their voltage (most likely 6V or 12V). If 12V you could connect them in series and direct connect them to the 25V transfomer as well. A second transformer that supplied their voltage would also work, too.

Another idea is to add a ballast tube.

:Hi all....I am a radio collector with limmited electronics ability I have a Simplex five tube radio ..no model #(tubes= 25Z5 77 43 39) (I am able to fix caps and such) ...but I am not sure how to replace a resistance line cord (under the chassie) I know resistors are not the solution..they get to hot...any suggestions...ANY HELP WOULD BE GREATLY APPRECIATED....(web sites??)

Try to get the combined filament voltage up to 60-65 volts, possibly by subbing a 12 volt tube for a 6 volt tube, note: check a tube manual, it must have the same filament current, which I beleive is .3 amps for your set, then insert a 1 amp 400 volt diode into the series string. The diode will reduce the effective heating energy available to the tubes by 50%, allowing the string to run on 120 v. If you cannot sub the tubes, use the diode anyway, it will allow you to reduce the heat that you must waste in a droping resistor by 50%. The diode will only dissipate about a 10th of a watt, as derived from the formula: .6v X .3a X 50% duty cycle.
As long as you only run the set on AC, it does not make any difference which direction the diode faces.

One other note, resistance line cords are usually packed full of asbestos

Paul: I know that it seems counter-intuitive, but a diode will reduce the RMS current of an AC waveform by just 30%, not by 50%. See: http://www.antiquewireless.org/otb/resto801.htm

To prove this, you can use integral calculus to compute the RMS of a half-wave output. Integrate sin^2 from 0-pi, divide by 2pi, and then take the square root.

I've made the same mistake.

Even my Fluke "true RMS" meter will not give the correct answer since it's firmware assumes that the measured voltage waveform is symmetric about the zero axis, which is not the case for a half-wave.