| Dave Gonshor | 3/97 |
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(Note: Several years back, the FLASH! published an excellent article by Larry Weide on how to deal with the nasty problem of a bad resistance line cord. This article is a follow-up to Larry's article.) Here's how to replace resistance line cords on AC/DC series filament sets, four or five tubes, with a voltage dropping capacitor. The hardest part is finding a non-polarized capacitor, 7 to 10 MFD, at least 100 VAC, preferably more. Polarzied electrolytic capacitors won't work because you are working with AC filament current. There will be no heat generated in addition to the filaments because there is essentially no ohmic loss in the capacitor. Therefore, this method has a great advantage over a big fat power resistor dissipating a lot of heat. Here's how it works. Calculate the effective resistance (R) of the tube filaments by summing the filament voltages and dividing by the filament current. Calculate the reactance of the capacitor by XC = 1 ÷ (2p60C) where PI = 3.14. The series impedance magnitude of the filaments and the dropping capacitors is; Z = ÖR2 + XC2 So, the filament current is; I = V ÷ Z where V is the rms AC line voltage. For example, C = 7.5µfd (from my local surplus store, rated at 230 VAC, non-polarized) V = 117VAC Filament current of 0.3 amp (all series tubes will have the same filament current rating) Five tubes consisting of two 25 volt filaments, and three 6.3 volt filaments, R = 68.9 Volts ÷ .03 Amps R = 230W XC = 1 ÷ (2p60 x 7.5E-6) XC = 354W Z = Ö68.92 + 3542 Z = 422W I = V ÷ Z I = .28A (slightly low, so the filament volgates will be slightly low) The voltage across the tube filament is; V = I x Wfil V = .28 x 230 = 64.4 VAC The voltage across the capacitor is; V = .28 x 354 = 99 VAC Note the total loop voltage (64.4 + 99 = 163.4 VAC) does not add up to the 117 VAC line voltage beacsue the voltages across the tube filaments and the capacitor are not in phase (the voltage across the capacitor lags the voltage across the filaments by 90 degrees). The total line voltage is the vector sum of the voltage across the tube filaments and across the capacitor. Don't make the mistake of thinking that the voltage across the tube filaments is (25+25+6.3+6.3+6.3) = 68.9 VAC, so you need a voltage across the capacitor of (117 - 68.9 = 48.1) VAC. The voltages are not phase so this doesn't work! If all this math has you worried just buy a non-polarized capacitor in the 7 to 10 MFD range and try it. It will probably work just fine. In reality, you probably won't have much of a selection of this type capacitor to choose from anyway. |
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