Norm:

Thanks for your help in explaining the resitance cord
problem I have with the RCA radio

If I can clarify, you are suggesting using a diode to
cut the voltage in half and a choke and a capacitor,
in a filter design, to further reduce the voltage.

If the have it right: AC --|>|---|choke|--- to filiment
|
Capacitor
|
chassis ground

Your formula, which I think is for capacitive reactance, would say a 60 hertz signal ( F ) with, let
say 30 ufd electrolytic capacitor, would give me a
capacitive reactance of 90 ohms.

Is this what you had in mind for an alternative to
a resistance cord or high wattage resistors?

Thanks for all you help

Randall

You can use any combination in series with the filaments to give the proper voltage drop. A capacitor needs to be non polarized. An 8 mfd cap gives the proper voltage drop without heating. You should be able to find motor starting caps with reasonable ratings.

You don't need these items as a filter, just to drop filament voltage. In your radio I would look for something other than a resistor as there is a lot of voltage to drop.

Norm

:
: Norm:

: Thanks for your help in explaining the resitance cord
: problem I have with the RCA radio

: If I can clarify, you are suggesting using a diode to
: cut the voltage in half and a choke and a capacitor,
: in a filter design, to further reduce the voltage.

: If the have it right: AC --|>|---|choke|--- to filiment
: |
: Capacitor
: |
: chassis ground

: Your formula, which I think is for capacitive reactance, would say a 60 hertz signal ( F ) with, let
: say 30 ufd electrolytic capacitor, would give me a
: capacitive reactance of 90 ohms.

: Is this what you had in mind for an alternative to
: a resistance cord or high wattage resistors?

: Thanks for all you help

Hi Randall, Norm has explained it in his usual thorough way, however there's one point you haven't got quite right. If you place a diode in series with the heaters it doesn't halve the voltage, it halves the power. On the half cycle it conducts there is the normal power disipated into the load, the other half cycles are blocked and so the average power is halved. The power is equal to the voltage squared, half the power is equivalent to the square root of 2, 0.707. So the effective supply voltage when a diode is introduced is 0.707 X 120 = 85 volts. there is almost no power loss in the diode so this cuts down on the heat produced. You will then need extra resistance or reactance to drop the remaining voltage back to the total heater voltage. If you use a capacitor, its voltage is 90 degrees out of phase with the current and the voltage drop across it won't be exactly equal to the supply minus the heater voltage, it will be the vector difference. Don Black.
:
: Norm:

: Thanks for your help in explaining the resitance cord
: problem I have with the RCA radio

: If I can clarify, you are suggesting using a diode to
: cut the voltage in half and a choke and a capacitor,
: in a filter design, to further reduce the voltage.

: If the have it right: AC --|>|---|choke|--- to filiment
: |
: Capacitor
: |
: chassis ground

: Your formula, which I think is for capacitive reactance, would say a 60 hertz signal ( F ) with, let
: say 30 ufd electrolytic capacitor, would give me a
: capacitive reactance of 90 ohms.

: Is this what you had in mind for an alternative to
: a resistance cord or high wattage resistors?

: Thanks for all you help