As far as the cap is concerned, I can't visualize how it connected to pin three (grid) of a 6C6 would boost any voltage by 50V, or why such would be necesary.
Please give us the complete tube line-up of the set.
What is the problem you are trying repair?
This capacitor has -nothing- to do with the filament string. The filament connections on most 4, 5, 6, and 7 pin standard base tubes commence at the highest and lowest pin numbers; 1 and 4, 1 and 5, 1 and 6, or 1 and 7. Follow the wiring to and from the highest pin number and pin 1 of each tube and that will tell you how the heaters get their voltage, and what is connected to the string to drop the voltage appropriately. Back in the 30s filament dropping in AC/DC sets was typically not done with a capacitor, but, rather, with a resistor or ballast tube. A capacitor or choke could have been used, but that would make the set only operable on AC at a specified frequency, though this would be no different than an AC set with a power transformer. Actually, I'm not sure why this wasn't done, but I suppose that being an inexpensive set, as AC/DC sets typically were, it was expected to work in many locations, sometimes remote, and at odd frequencies.
You may use a capacitor in place of what ever dropping resistor is present, or you may use a resistor or a diode. I like the capacitor idea the best because it doesn't get hot or waste energy, and the tubes typically soft-start. While there is a possibility for instantaneous double voltage at turn-on, if the capacitor retains a charge of one polarity and the voltage entering the set at that moment is of another, I don't think that this is likely to cause a problem, as it will be at about 1/60 of a second. I have used the diode method without a separate dropping (or buffer) resistor, and I don't exactly like this method because the 6 volt tubes turn on quite brightly at first. This can, of course, be reduced by adding a buffer resistor, or eliminated by placing two 6.8 volt zener diodes back-to back across each 6 volt tube.
T.
Like Thomas, I prefer a dropping capacitor for replacing a resistive line cord. But you need a nonpolarized AC cap - or two polarized caps back to back.
With so many other projects to tend to, I haven't tried my paralleling diode idea, since I don't have the right capacitors at hand for a filament string, though I don't think that it would solve the internal resistance problem; only the reverse polarity problem.
On a side, for those who are new to this, though one can use either a capacitor or a diode for AC voltage dropping, they cannot be mixed, as a capacitor passes AC only, and a diode only passes current in one direction. To mix the two would yield a dark filament string.
T.
A diode will pass half of a 60-Hz AC waveform. The resulting half-wave signal is, itself, an AC signal, but it is composed of many frequencies - a series cap would attenuate them (more so the lower frequency components). But a fourier analysis would be required to figure it out. Or, you could just use the first few terms of the fourier expansion, and solve each one separately, and add up the individual results.
Or, maybe put the complex waveform through a resistor immersed in a water bath calorimiter and figure the resultant power output. Calculate RMS voltage: V = sqrt [P x R]
I'll leave it for some kid's science fair project.
::On a side, for those who are new to this, though one can use either a capacitor or a diode for AC voltage dropping, they cannot be mixed, as a capacitor passes AC only, and a diode only passes current in one direction. To mix the two would yield a dark filament string.
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You are right Thomas.
.. but those dark filament strings do run a lot cooler.
.. and it results in a much quieter speaker output too. ... which is great for use in libraries and funeral parlors though.
Good point!
....But yeah, caps don't pass DC, and so if you put a diode in series with a cap, you'll have a dead radio. You can mix caps and resistors, and diodes and resistors, but not caps and diodes.
T.
Actually Doug it is considered DC (Pulsating DC) because the current never REVERSES direction.
A.C.= "Constantly changing amplitude and periodically changing direction"
So AC means Alternating current directions.
The diode however only allows current to flow in one direction and completely blocks the current flowing in the opposite direction.
...so it is DC ....even though you may see humps...or pulses.
Yes, it does change amplitude.. as the DC rises and returns to the ZERO center... but it
My point was that if one were to connect a diode in series with a capacitor to drop voltage for a filament string, it wouldn't work, since the diode would charge the capacitor on each up pulse, but then when voltage fell back to zero, the capacitor wouldn't have anything to discharge into, since current can only flow through a diode in one direction, and not in reverse, and so current would cease to flow once the capacitor attained full charge.
...This was not in response to anything you said prior to my saying this, but, rather, simply to those new to the subject who might attempt to try this. ...It was in addition to what you had already offered as suggestions.
T.
It would be hard to figure out the correct value to drop and particular voltage as it is not a sine wave, but it would pass current through it.
:A diode in series with an AC supply will not produce DC. It will produce a half-wave AC - which, in turn, is an AC signal comprising multiple sinusoidal waves of different frequencies.
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Call it what you will. Since the current flowing out of the diode has only one polarity, it cannot continuously pass current through the capacitor.
I am sorry I ever brought this up, because people are not handling it well. If you do not believe me, try it yourself. Take a filament string, wire in series with it a diode, and then a capacitor. Pick a big capacitor. 30 MFD. 50 MFD. 100 MFD! Once the cap has fully charged, no current will flow through the circuit, and the filament string will remain dark.
Also, as I've said before, this was not aimed at Doug, but rather anyone new to the subject who might thing that they can use a capacitor in place of a resistor when using a diode as a voltage dropping device. It will not work. The only way it could work is if the capacitor has leakage and can somehow discharge between pulses. In that case the leakage would have to be about comparable to the resistance you would use to drop the excess voltage that the diode didn't drop, which negates the necessity for a capacitor in the first place.
T.
Bob Z
:If you wire a diode in series with a capacitor and then a load (say, a filament string), the diode will charge the capacitor, and once the capacitor is full, no further current will flow.
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:Call it what you will. Since the current flowing out of the diode has only one polarity, it cannot continuously pass current through the capacitor.
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:I am sorry I ever brought this up, because people are not handling it well. If you do not believe me, try it yourself. Take a filament string, wire in series with it a diode, and then a capacitor. Pick a big capacitor. 30 MFD. 50 MFD. 100 MFD! Once the cap has fully charged, no current will flow through the circuit, and the filament string will remain dark.
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:Also, as I've said before, this was not aimed at Doug, but rather anyone new to the subject who might thing that they can use a capacitor in place of a resistor when using a diode as a voltage dropping device. It will not work. The only way it could work is if the capacitor has leakage and can somehow discharge between pulses. In that case the leakage would have to be about comparable to the resistance you would use to drop the excess voltage that the diode didn't drop, which negates the necessity for a capacitor in the first place.
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:T.
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Of course, the there needs to be a load on the series diode-cap string. If there is a load, the capacitor will not charge to a constant, DC voltage.
One way to understand this is that any periodic voltage waveform, no matter what its shape or bias, can be expressed as a sum of pure sine waves (fourier expansion).
This could be demonstrated by experiment, of course, but I don't think it is necessary.
What comes out of the diode is a varying waveform. You could actually call it AC, though traditionally AC (alternating current) alternates back and forth, not just forth and stop forth and stop. In order for a capacitor to 'pass' (what is apparent to the observer) current, current must both flow into and then out of the capacitor. Since the diode only allows energy to flow through it in one way, once the capacitor has been charged, it cannot discharge and allow current to flow out of it in the other direction. Current flow will stop.
Before you guys reply to this post, go take a diode (a good diode that isn't leaky) and a capacitor (100 MFD) and an AC/DC radio (you can take any make you wish--take one that works completely and can be plugged in the wall as-is). Plug the radio into the wall. Observe it light up and operate. Place the diode in series with the radio. The tubes will glow dimly and the radio might not operate well if at all (if you chose a radio whose filament string already adds up to 120 volts), but the tubes will still light up. Now place a capacitor in series with the radio and the diode. Be sure, if the capacitor is electrolytic, that its polarity is oriented properly (+ must face cathode of diode). Turn on this assembly. The radio will not light at all. It will remain completely dark.
If you prefer, use a 25 watt light bulb, or 10 watt. The bulb will not remain lit.
T.
T.
This is pretty much basic physics and electicity. The voltage waveform needn't be a pure sinewave - it can be a half sinewave, a tiangular wave, etc.
If a cap has a half sinewave voltage applied to it, the resulting current will also be a half sinewave. We can do testing, but basic theory makes testing unnecessary, in my opinion.
Now, if there is no load on the series diode/cap arrangement, no voltage will appear across the cap, and no current will flow - because all the voltage will appear across the open, infinite-impedance "load."
If you are right... I'll bow to your knowledge ... AND I'll buy you a cup of coffee.
As I see it:
The diode will always only EVER allow current to flow one way... right?
So it will rectify the AC sine wave to pulsating DC for the load.
1.) If you place a cap in series with that pulsating DC it will partially charge up on the first pulse and stop.
This is because the cap saw a current path through the diode and the load on the first positive 1/2 cycle of the AC.
2.) On the negative 1/2 cycle of the AC the diode blocks the path to the load and the cap sits there idle.
3.) The cap remains partially charged and idle until the next positive 1/2 cycle continues to charge it some more until fully charged..
:Furthermore, measure the voltage across the 'test' radio or light bulb. There will be 0 volts across the radio/bulb. There will be from 120 to 150 VDC across the capacitor.
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:T.
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T.