[Radioroslyn]

Hi Guys I found a little Kadette Jr to play with. It's a 1933 boy's radio, two tube job. First tube is a 6F7 which is the RF amp and Det stages. The second tube is a 12A7 it's the rectifier and audio output. Now for the RS part to achieve the heater voltage/current it use a resistor line cord. I need to lose 112vac @ 300ma what value condenser do I need to replace the resistor in the cord???
http://www.nostalgiaair.org/PagesByModel...009608.pdf
Here's a pic of one, not mine it hasn't gotten here yet. http://www.radiolaguy.com/Showcase/Plast...tte-Jr.htm
TNX Much!!
Terry

[morzh]

Terry

What is the summary voltage of the heaters, as it dpends whether 12A6 has the halves of filament paralleled or not. Is it 18V or 12V?

[morzh]

Meantime consideing you are dopping a lot, roughly your resistor should be just that, R=112/0.3 Ohms.

Then do C = 1/(6.28*60*R) in Farads (7 uF roughly).

[Radioroslyn]

No tap on the 12A7 heater so it's 18v @300ma.
Terry

[Eliot Ness]

The following link has the formula and near the end another link to a spreadsheet that will calculate it for you:

http://www.vintage-radio.com/repair-rest...calcs.html

[morzh]

6.7 uF.

[OldRestorer]

As a true rocket sciantest I agree with the replies above.

[morzh]

Terry

Here's how you do it.

You have 120V.
In order to drop what you need you do not care how much you drop, you care to keep the current 300mA, this way you keep your voltage 18V and drop the rest.

So

1. your current is your V over Z, where Z is total impedance.
I=V/Z=120/Z
I= 300mA= 0.3A
Z=V/I=120/0.3=400 Ohm


2. Express Z via Rc (capacitive impedance) and the R active (filament).
Z=SQRT(R*R + Rc*Rc) =400

3. Your R active
R= Vf/0.3= 18V/0.3A = 60 Ohm

4. Find cap impedance
R*R + Rc*Rc= Z*Z= 400*400= 160000
Rc*Rc=160000-3600=156400
Rc=395

5. Find Capacitance
Rc=1/(6.28*f*C)
C=1/(Rc*60*6.28 )=1/(395*60*6.28 )=6.71*e-6 F= 6.71 uF

[Eliot Ness]

That little Kadette doesn't have very much spare room inside for a dropping capacitor so you might want to consider running it off a variac or set-up a transformer for it. If you do that I'd put something other than a standard AC plug on the cord so someone doesn't plug it into an outlet by mistake.

[morzh]

I was doing Kirk's little PT something Philo, really small one, I got 6.2uF dropping cap inside.

[Eliot Ness]

It might be able to be squeezed in there somewhere, but take a look at how small this little guy is:

http://www.radiolaguy.com/Showcase/Plast...tte-Jr.htm

They're really neat radios, especially for 1933, but there isn't much free space inside one of these guys.

[morzh]

You are correct, it is tight. If there is such thing as 250V cap of such a value, it might be smaller than the one I was thinking off.
Terry will have to look into it.

[Radioroslyn]

Probably not a good idea to use two electrolytic cap back to back? It is very cramped in there.
Terry

[morzh]

No...not a good idea. The thing you are trying to do is to create a non-pol lytic.
This is possible ONLY if you pull the centepoint to a potential that is ALWAYS above (if it is the two plusses) the voltage you are passing through. But, if you have a transformer the two voltages are not referenced to each other, and if it is an ACDC set, then such potential may or may not be present in the schematic.


See if it is even remotely possible to cram the cap inside without touching something hot.
Or if you are willing to sacrifice some aesthetics and dissing by purists, if you could put a film cap in a box and attach it to the cord somehow to make the third wire for filaments; a 3-wire cord could be used.

[exray]

Here's a source for suitable caps. 7.5 uf shown, go to links to see other options. Small and relatively inexpensive.
http://www.digikey.com/product-search/en...itor+7.5uf

These are true AC rated caps....not just DC rated in some derated configuration. Dielectric differences, etc, etc