Hi all I have model V Simplex Radio (schematic on this site) I have been told may ways to replace the resistance line. I have used 2 back to back capacitors...(don't like that)took them out and resistors get to hot... So instead..(if you look at schematic) I have replaced line cord with a 4006-7 diode just before 30w resistor (on right bottom side of schematic)added jumper from pins 1-6 ...(REPLACED ALL CAPS W/ NEW ONE AND ARE SURE THEY ARE ALL RIGHT ...POLARITY AND SO...FORTH) BUT.... My tubes 25Z5 and others run to hot... HELP PLEASE.....

OOPS Typo error (I meant 30hom resistor on bottom right side of schematic)

:Hi all I have model V Simplex Radio (schematic on this site) I have been told may ways to replace the resistance line. I have used 2 back to back capacitors...(don't like that)took them out and resistors get to hot... So instead..(if you look at schematic) I have replaced line cord with a 4006-7 diode just before 30w resistor (on right bottom side of schematic)added jumper from pins 1-6 ...(REPLACED ALL CAPS W/ NEW ONE AND ARE SURE THEY ARE ALL RIGHT ...POLARITY AND SO...FORTH) BUT.... My tubes 25Z5 and others run to hot... HELP PLEASE.....

Hi Mr. Berg,
You have two 25 volt tubes and three 6.3 volt tubes in series. All require 0.3 ampere of current flow.
Add up the voltages and you get 68.9 volts. For the diode to do you good on cutting the voltage, you need a string that requires less than 60 volts. The diode circuit won't work for you unless you do some complicated rewiring.
The diode circuit you described, however, should cause the tubes to run cold. You need to check the wiring for the cause of overheating.
Maybe some of the other guys have experience with using capacitors in the filament circuts, but I handle this with a resistor. In your case that would be 145 ohm 13 watt (15 or 30 watt will be the standard size).
Be sure, too, there is adequate ventalation if this is a bakelite or plasticon case.
Unless the radio is to be used for day to day service or the cord is in poor condition, keeping the old cord is usually better.

Best Regards,

Bill

A diode will reduce the RMS voltage by only 30%. So, a diode will cut 120 VAC to about 85V. I think that is why your tubes are running too hot.

Don't try verifying this with most digital multimeters, even ones that are advertised as true-RMS. A voltage waveform that is not symmetrical about the zero axis will fool most DMMs.

More here: http://www.enginova.com/true_rms_volts.htm
Doug

:Hi Mr. Berg,
: You have two 25 volt tubes and three 6.3 volt tubes in series. All require 0.3 ampere of current flow.
: Add up the voltages and you get 68.9 volts. For the diode to do you good on cutting the voltage, you need a string that requires less than 60 volts. The diode circuit won't work for you unless you do some complicated rewiring.
: The diode circuit you described, however, should cause the tubes to run cold. You need to check the wiring for the cause of overheating.
: Maybe some of the other guys have experience with using capacitors in the filament circuts, but I handle this with a resistor. In your case that would be 145 ohm 13 watt (15 or 30 watt will be the standard size).
: Be sure, too, there is adequate ventalation if this is a bakelite or plasticon case.
: Unless the radio is to be used for day to day service or the cord is in poor condition, keeping the old cord is usually better.
:
:Best Regards,
:
:Bill

Hi Doug,
To get the 85% won't you have to have filter capacitors in the circuit?

Best Regards,

Bill

Bill, I'm not quite tracking you here.

A diode will drop the RMS voltage by 30%, which for 120VAC, would drop the AC filament voltage to 120 x 0.7 = 85V (not 85%). No capacitors need be involved (this is an AC filament string).

Here is the schematic of the radio: http://www.nostalgiaair.org/PagesByModel/446/M0018446.pdf

Here is an explanation of the 30% factor: http://www.enginova.com/true_rms_volts.htm The diode reduces the power of the waveform by a factor of half, but reduces the RMS voltage by a factor of 1/(sqrt 2).
Doug

:Hi Doug,
: To get the 85% won't you have to have filter capacitors in the circuit?
:
:Best Regards,
:
:Bill
:

Hi Doug,
After reading your well written article, I remembered this from College. The power is halved, not the voltage when the waveform is run through a diode.
Therefor Mr. Berg is overheating his tubes because he has a 30 ohm resistor in series when he really needs a 54 ohm resistor.
Thank you for explaining this.

Best Regards,

Bill Grimm