I want to replace this 25ohm resistor with a capacitor because I have a 5 watt resistor in there but it still gets to hot for my liking. I can’t figure out what size it needs to be.

Start over, you asked about the 25 ohm not the whole cord.

It's really not practical to replace the 25 ohm with a cap because the cap would have to be a very large value.
You can remove the pilot lamp from the filament string and drive it with a 5uF, 250VAC cap directly from the AC power.

I'll add a schematic.

Are you using an 8uF capacitor to replace the resistor power cord?

Jay

Sorry I screwed up also it is this 86 ohm resistor. I put the wrong radio schematic up. I am working on two different radios at the same time.

http://www.nostalgiaair.org/pagesbymode ... 014918.pdf

If you are using a 35L6, then the cap should be 7.5uF, 250VAC rated to replace the 86 ohm resistor.
If you are using a 50L6, the resistor should be removed.

Jay

Thanks Jay sorry for my screw up

7.5 uFd 60 Hz with a current of 0.15 amp flowing through it will have a voltage drop of

(1/377 * 7.5 E-6) 0.15 = 354 volts.

The tubes will have a drop of 35, 35, 12.6 12.6 12.6 = 107.8

Adding by pythag means it will take 370 volts AC just to light the filaments.

Tell me my calcs are wrong!

Tube string resistance is 107.8V / 0.15A = 718.7 ohms
Total impedance needed is 120V / 0.15A = 800 ohms
Since the we are adding capacitive reactance and resistance, we need to compute the other leg of the triangle.
Square root of ((800 x 800) - (718.7 x 718.7)) = 351.4 ohms capacitive reactance
Capacitor value is 1 / (377 x 351.4) = 7.55uF

I think you will find the above is the correct way to calculate the impedance of a resistor and capacitor in series but I am getting old and make mistakes.

Jay

Yes!

108 volts for tubes @.15 amps = 720 ohms

7.5 uFd @ 60 Hz = 353 ohms

SQR 720 ^2 + 353 ^ 2 = 802 ohms


120 volts / 802 ohms = 0.15 Amps

What voltage across capacitor ?

This is the result of running the circuit analysis software called Micro Cap.
Circuit:

Voltages and current:

The cap and resistor voltages are not in phase so they do not add directly. The resistor voltage is about 152V peak or 107.5VRMS.
The cap voltage is about 75V peak or 53VRMS.
The current has to be the same in both the cap and resistor since they are in series. The current is about 149mARMS.

Jay

One of the big advantages of using the cap for the voltage drop is that the filament currents during tube warm up are lower than with a resistor.
The tube filaments will be about 100 ohms when they are cold. With the 86 ohm resistor, the initial current through the filaments will be about 650 mA.
With the 7.5uF cap instead of the resistor, the initial current will be about 325 mA or half as much as with the resistor. Warm up time is longer but the tubes and pilot lamp will have much longer life.

Jay

That is a nice program. Does the capacitor impact the cathode to heater voltages or hum
especially around the 12SQ7 ?

The program is free for the evaluation version and that's an overkill for most of what we would want to analyze. It comes with models of parts like the 1N4001 diode but not the 1N400x family. You have to modify the breakdown voltage to make the 1N4001 model work like a 1N4007. It has some tube models like the 6L6, 6V6, and 12AX7. Also a good program for drawing schematics.
You get the program here: http://www.spectrum-soft.com/demoform.shtm

The tubes should not know if the voltage drop is from a resistor or capacitor. I have never noticed any hum problems or unusual voltages when using the capacitor voltage drop. I've lost count on how many ballast tubes I have replaced with caps but it's probably more than 20.

Jay

My Warco radios have 6SQ7 in middle of string, never had hum issue with the curtain burner cord or cap. Yes this arrangement is original. I added the #1822 dial lamp to balance string current, with 50 ohm shunt resistor & lamp in series, 6K6 heater was approx 4.8v. Again was original design. A later version(based on tube code dates), uses a #43 as the output.

Because I'd done some consulting work that fluorescent lamps in the past, I had needed to find a workable SPICE model for filaments that included the time constants and cold resistance. I was able to model the filaments for those lamps and applied that to this problem and ran a comparison of the 86 ohm and 7.5uF capacitor dropper with this circuit in LTspice:

The .param statements define the filament characteristics and were chosen to give an inrush current of about 10 times the rated current (1.5A) at 107.8 volts. This is usually a good figure for Tungsten filaments. The time constants were chosen so the current would drop to about 150mA in 10 - 15 seconds. Below is the result of the comparison:

The yellow trace is with the resistor dropper and the green is with the capacitor. You can see proof of what Jay said about the cap lowering the inrush current. The schematic shows the current for the first second and the last 5 seconds of the simulation.
John

Thanks for the program url, i might try something to further look into the heater cathode
voltage issue. The only other suggestion is a discharge resistor across capacitor. Just in case
the capacitor retained a charge if the plug prongs were touched by someone unplugging the
radio, then picking it up. 470 K would do.

Great graph!!!

I've seen comments that if cap remained charged, current at turn on would be double. Even so, looks as if it'd be approx equal to resistor and would dissipate in the first few cycles. Or probably 100x faster than resistor.

There is a lot of things involved when electrolytic capacitors are charged through a low
120 volt AC source impedance.

If you want to see some, try powering up a cheap SMPS or LED lightbulb by crossing wires
instead of a switch. That spark can ruin light switches when used to turn on strings of them,
well within the steady state amperage rating.

The cap discharge resistor is a good idea. 7 or 8uF charged to over 100V would not feel good. If the radio was in operation, the rectifier current would discharge the cap to a fairly low voltage. However, plugging and unplugging before the rectifier got hot could definitely leave a high voltage on the cap.

I think 100K would be a better value for the discharge. That would get the voltage to a safe level in about 1 second. 470K would take about 5 seconds and it would be easy to touch the plug in that much time. At 100K, a 1/2 watt resistor would be OK.

I have looked at the surge current with the cap but it is so short that it shouldn't be a concern.

This isn't near as bad as the situation of charging an electrolytic without a surge limiting circuit. That can ruin switch contacts in a few cycles. The tube filaments have enough resistance even when cold to limit the peak current to the 1 amp area.

Jay