[Art Hoch]

Does anyone know the value of the resistance power cord on this radio?  I assume that's what "cordohm" means on the schematic in Riders. 

I don't think anything is mentioned other than that label on the chassis top view diagram...unless it's referring to a candohm that's attached to the chassis.  Perhaps it's a regular type power cord with a candohm----- a 50 and 135 ohm candohm on the schematic----a very "bare bones" type of schematic.  No parts list is given for this 1936 model.

Thanks for your help.

[morzh]

Art

Just calculate it.

The way to do it is this:

1. take the current of any filament that is used. All of them should be the same, so if one is say 300mA, all of them are.
2. Add the voltages of all filaments together. Like 4x6.3V+25V=50.2V.
3. Subtract this number from your Mains voltage (say 120V):
120V - 50.2V=69.8V. (use 70V)
4. Divide this number by the current:
70V/0.3A = 233 Ohm.

THIS IS YOUR RESISTANCE.

5. Using UI calculate the dissipated power:70*0.3=21W
Double that to find your power rating.

So, you need 50W 233 Ohm resistor (use the closest value you can get).

6. If the radio is not intended to work with DC, calculate a capacitor instead to not dissipate any power.

[Art Hoch]

a 50 W or a 5 W?

[morzh]

50, Art, 50. Those cords dissipated 20-30W.

This is why today most of us use capacitors and not resistors to limit the voltage. Capacitors do not dissipate power.

I used one when restoring Kirk's Century radio. It came to about 6.8uF cap in series with filaments string.

[Radioroslyn]

You could use a diode to chop your ac line voltage in half and then feed it to the heater string. It's small and no heat.
Terry

[morzh]

The filaments will run much hotter.

The heat is produced by Vrms, which is 120V in case of the MAINs full power.
The 1/2 wave rectifier produces Vrms=1/2*Vpeak=0.707*120V=85V.

We need 50V.
You still have 35V to drop, or your ilaments will run at almost 3 times (1.7*1.7=2.89) hotter.

Dropping 35V requires 4 times less power to dissipate than 70 (120-50), 10.5W, but still respectabe heat.

[Mondial]

Morzh, why do you want 50 V across the series heater string? The total heater voltage is 68 V ( 25 + 25 + 6 + 6 +6 )

http://www.nostalgiaair.org/PagesByModel...016853.pdf

[Art Hoch]

Does anyone know the correct knobs for this radio I refinished?   I believe it to be the top right knob but can't say that for sure.

[Art Hoch]

I need a little confirmation or clarification on the schematic for a 1936 Silvertone model 1953.  Does the positive lead on the 8mfd capacitor connect to the 25Z5 tube?  What does "SF" by the coil symbol mean?  And where does the negative lead connect?   I think I know but want to be sure before I power up. Thanks.

[morzh]

Mondial,

Yes I forgot 43 tube is also a 25V one. Makes it a bit easier. So you need to drop 17V. At 300mA it is 57 Ohm. About 5W dissipation. 10W resistor will probably do.

[Art Hoch]

When I do the math I get: adding tube heaters=68 volts; 120 Mains minus 68= 52 volts to drop. 52 divided by .3amp = 173 (ohms). Am I missing something?

[Mondial]

Morzh's calculation includes a series diode which drops the RMS voltage to 85 V. Then you only need a 57 ohm resistor to drop the remainiing 17 V.

 If you use the diode much less power is dissipated in the resistor, so you can use a lower wattage one.

[Art Hoch]

Think I won't use a diode. 

[Art Hoch]

This radio has 2 dial lamps.  Do I need to take into consideration both of them when calculating the voltage drop or just 1 since they are so small?

[morzh]

You do...but then it affects the voltage within small margin.

What you can do is use a Variac to make the input V suh that you do not need to drop anything and then measure the exact effect your dial lamps have on the total load, then factor it in when calculating the dropout.