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 Post subject: Idea for resistor line cord
PostPosted: Oct Tue 12, 2004 7:19 am 
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Location: Vermilion, Ohio
I've never had a "curtain burner" set in my collection but, I know that a good replacement resistor line cord is as rare hens teeth. A thought recently crossed my mind, it was a very short trip.<BR> Most pre electronic ignition vehicles had a resistor wire in the harness from the ignition switch to the ignition coil. It was a foot or so long and dropped 12V to about 9V to prervent arcing of the points. I don't remember the resistance per foot of this wire, nor do i know the resistance of radio cord wire. Is it possible that the automotive type could be used for replacement? Or, maybe this wire in series with a smaller chassis mounted reasistor? I'm sure there were many values of resistor wire, but one could compare the specs to see if it would be a viable solution. The automotive type is available in many auto supply stores. I also recall seeing it in bulk somewhere.....?<P> Just a thought, Tom<P>------------------<BR>


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 Post subject: Idea for resistor line cord
PostPosted: Oct Sat 16, 2004 3:41 am 
Silent Key

Joined: Jan Thu 01, 1970 1:00 am
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Location: Sandpoint, IDAHO 83864
I know what you are talking about, but I doubt it has enough resistance to replace a curtain burner. They usually had resistances of 100 to maybe 300 ohms, and the automotive stuff would only have a few ohms at best. <BR>Curt<P>------------------<BR>Curt, N7AH


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 Post subject: Idea for resistor line cord
PostPosted: Oct Sat 16, 2004 8:20 am 
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Location: Cobourg Ontario Canada
Saw an article about this on a German enthusiasts web site. He suggested using a small resistor in the wall plug (wall wart) and another inside the radio. And a new standard type cord of course.<P>Reason for doing it this way, one resistor puts out too much heat, you cant put it in the radio and its not safe to put it on the wall. But 2 small resistors wouldnt heat up that much, not enough to be a problem.<P>


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 Post subject: Idea for resistor line cord
PostPosted: Oct Mon 18, 2004 12:25 am 
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I'm not keen on the above idea of putting one resistor in a wall wart and one in the radio. Seems risky. Also, a "standard" 2-conductor line cord won't work, because you'll need 3 conductors - one with the resistors for the heater circuit, a 120-V line for the power supply, and a neutral. This all assumes we're talking about U.S. 120-V wiring.<P>The idea of using 12-V auto wiring isn't kosher - its insulation is not rated for 120-V usage, and without knowing what power or current it was designed for, you'd run the risk of overheating.<P>My proposed solution would be to put a 1N4007 diode inside the radio and in series with the heater string. That'll drop the power to the heater string in half, and the diode will be heated only by the rms heater current times the diode forward drop (less than 1 volt).<P>Inccidently, modern electrical codes would not permit a resistance line cord in new equipment.<P>------------------<BR>**********<BR>Doug Criner <A HREF="http://www.enginova.com" TARGET=_blank>http://www.enginova.com</A>


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 Post subject: Idea for resistor line cord
PostPosted: Oct Mon 18, 2004 1:00 am 
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Hi Doug<P> You are right a diode will cut power in half but this doesn't mean voltage. In most cases a resistor is still needed. A diode can also cause noise due to switching characteristics which will usually be cleaned up by adding a cap. If the radio has a lamp a zener diode may need to be added for protection at turn on.<P>------------------<BR>Norm


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 Post subject: Idea for resistor line cord
PostPosted: Oct Mon 18, 2004 9:17 pm 
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Norm, here is my math, which hopefully will clarify my diode idea:<P>R = resistance of series heater string<BR>V = RMS line voltage = 120 V<P>With no diode, I = V/R, P = V^2/R<P>With diode, everything is the same, except there is only one-half an AC waveform, so:<P>With diode, I = 0.5 V/R, P = 0.5 V^2/R; effective RMS voltage with diode = 0.5 x V = 60 V.<P>Now, for comparison, let's consider a dropping resistor equal to R. Then, the RMS voltage across the heater string will be (R/2R) x V = 0.5 V = 60 V, I = 0.5 V/R, and P = (0.5 V)^2/(0.5 R) = 0.5 V^2/R (same heating power with the diode).<P>Therefore, I conclude that a diode will accomplish the same as a dropping resistor equal to the series resistance of the heater string, and the heaters will run at the same temperature either way.<P>Whether either option will produce the desired effect depends upon the resistance of the heater sting. If there are 10 tubes rated at 6.3 V, 63 V total, then I'd tend to think either the diode or a dropping resistor equal to series heater resistance should be OK. If less than 10 6.3-V tubes, then an additional dropping resistor could be in order.<P>Another approach is to put a capacitor (instead of a diode) in series with the heater string. The math then involves some trig, but the "dropping" cap can be sized to give the exact power we need on the heater string.<P>The advantage of either a diode (if half-power is close enough) or a dropping cap is that neither the diode or cap will get hot, and they can be mounted inside the chassis. Also, only a 2-conductor line cord is needed.<P>Your point about noise from diode switching is very valid, even though theoretically a diode will switch near the voltage zero crossing point - which for a resistive circuit, should also be at the zero current point. I'd experiment with a cap-resistor snubber circuit across the diode if interference is a problem.<P>With regard to protecting a lamp with a zener, could you elaborate on that? Certainly, the lamp will see an inrush at turn-on, but would that inrush be any more severe with the diode than a dropping resistor at turn on? If you're looking at half a 60-Hz cycle, then yes, but it seems to me that the heating of the temperature of the lamp's filament shouldn't shouldn't vary too much during 1/120th of a second. I don't thing you would notice a visible flicker with the diode. But a zener would certainly be good for the lamp, in any case--although it will switch at it's rated breakdown voltage, not at the zero crossing. This might also require a snubber circuit to hold down RFI.<P>The absolutely preferred way to replace a resistance line cord is to use a step-down transformer rated at sufficient VA. This is what should have been installed in these "curtain burners" originally. Manufacturers were looking for a short-cut to implement a transformerless AC/DC radio before the All American 5 or other similar tube complements were available.<P>Disclaimer: I haven't tried either the diode or the dropping cap in a curtain burner, but theoretically, I would prefer either to a haywired dropping resistor. <P>------------------<BR>**********<BR>Doug Criner <A HREF="http://www.enginova.com" TARGET=_blank>http://www.enginova.com</A>


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 Post subject: Idea for resistor line cord
PostPosted: Oct Tue 19, 2004 1:41 am 
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Regarding my "dropping" capacitor concept: a dropping cap will avoid any diode switching transients and will supply full-wave AC power to the heater string (including the dial lamp, if any). The cap can be sized to supply 0-100% of full 120-V power to the heater string.<P>I would want to use at least a 600-V cap. If it were ever to short, it would blow one or more of the tubes. (One thing about a dropping resistor, it's hard to imagine in shorting.)<P>The smaller the cap, the larger the voltage drop across it, but the voltage across the resistive heater string is NOT 120V minus the cap voltage because the cap voltage will be 90 degrees out of phase with the heater voltage. If anybody wants to know exactly how to size the cap, let me know.<P><BR>------------------<BR>**********<BR>Doug Criner <A HREF="http://www.enginova.com" TARGET=_blank>http://www.enginova.com</A>


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 Post subject: Idea for resistor line cord
PostPosted: Oct Tue 19, 2004 2:51 am 
Silent Key

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Doug- the thing that bothers me with using a capacitor, is the old bugaboo about using electrolytics with AC impressed on them. Also what about capacity tolerance of the normal electrolytic? Maybe they have tightened specs on the newer ones, but they used to have very wide tolerances, like -20% to +50% or something like that.<P>I am not saying a capacitor will not work, but I just need to be convinced that it is safe. You would not want the series resistor to have that broad of a resistance range if you were using a resistor.<BR>Curt<P>------------------<BR>Curt, N7AH


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 Post subject: Idea for resistor line cord
PostPosted: Oct Tue 19, 2004 3:17 am 
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Curt - good points.<P>I wouldn't want to use an electrolytic cap! I think it would blow apart. Now, whether a film-type cap of suitable size is available would need to be explored. Of course, a few .22 uF caps could be paralled if necessary.<P>The "gold standard" for replacing a resistance line cord, in my opinion, is to use a step-down transformer.<P>Side Note: I've heard "old timers" say that you could connect two same-size electrolytics back-to-back, and get by using them across an AC voltage. I've never believed or understood why that would be acceptable.<P><P>------------------<BR>**********<BR>Doug Criner <A HREF="http://www.enginova.com" TARGET=_blank>http://www.enginova.com</A>


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 Post subject: Idea for resistor line cord
PostPosted: Oct Tue 19, 2004 5:56 am 
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The effective power output (heating equivalent) of a diode is NOT half the line voltage. It is the RMS of the line voltage, or equivalent to 85V of full wave AC. It is also a stiff voltage source, not a current source like a resistor so the tubes don't get a "soft" start. See:<BR> <A HREF="http://www.antiquewireless.org/otb/resto801.htm" TARGET=_blank>http://www.antiquewireless.org/otb/resto801.htm</A> <P>------------------<BR>


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 Post subject: Idea for resistor line cord
PostPosted: Oct Tue 19, 2004 6:43 pm 
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Hi Ken: Thanks for the useful link to more info on reistance line cords. You're correct that it's the rms value of AC voltage that is used in power calculations.<P>However, in the U.S. at least, the standard convention is that the nominal utility voltage, e.g. 120 V, is always quoted as an rms value. It is actually a sinewave of magnitude: 120 x sqrt 2 = 120 x 1.414 = 170 V.<P>Likewise, most AC voltmeters are internally calibrated to read rms - either "true" rms like my Fluke meter or just 0.707 x the peak voltage.<P>A diode in series with an AC supply will block one half of the sinewave, much like a halfwave rectifier, and will therefore cut the supplied rms voltage value by half. The peak voltage (line to neutral) will still be 120 x sqrt 2, so if you measure the AC voltage with a non-rms voltmeter, it will still show 120 V, but the true rms value is 60 V.<P>What I'm saying could be demonstrated by a little experiment with a diode circuit feeding a resistive load, and two voltmeters - one true rms and one not. The first will read 60 and the second will read 120. If you look at the waveform on a 'scope, it will be a half sinewave ranging from zero to 120 x sqrt 2 = 120 x 1.414 = 170 V. (The peak voltage across a nominal 120-V line is actually 170 V.)<P>There are a couple of practical examples of all this. The fact that nominal 120-V AC voltage actually has a peak of 170 V enables the B+ voltage on an AC/DC radio to exceed 120 V. Also, since nominal 120-V is an rms value, we can then say that the rms current to a 60-W light bulb will be: 60/120 = 0.5 A. (If we put that same light bulb in series with a diode, it will only consume 30 W.)<P>But power itself, watts, is not an rms value - it's the "real deal" that heats our tubes' filaments. <P>------------------<BR>**********<BR>Doug Criner <A HREF="http://www.enginova.com" TARGET=_blank>http://www.enginova.com</A>


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 Post subject: Idea for resistor line cord
PostPosted: Oct Tue 19, 2004 8:13 pm 
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Hi<P> All I will add, a diode itself will not reduce tube brightness enough in a radio with a 60 volt tube heater string. A small value resistor is still required. <P> <P>------------------<BR>Norm


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 Post subject: Idea for resistor line cord
PostPosted: Oct Tue 19, 2004 9:43 pm 
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There is still some confusion here. The output of a diode is .707 of the RMS input - not half. Most voltmeters will not measure accurately due to waveform. My Westinghouse dynamometer voltmeter reads 85V. The Simpson, Fluke and Triplett VTVM all give varying results, none accurate.<BR>I confirmed my calculations by enclosing a light bulb and a photocell in a box and measuring the photocell output with a diode in series with the bulb. I then removed the diode, connected the lamp to a variac, adjusted it for the same output and measured the voltage on the bulb - 85V.<BR>We are not actually concerned with voltage, but with power. That's why we use the RMS value of the full wave AC. It corresponds to the power into the load. The 85V power equivalent value is what the tubes see and is what we must use in calculating the value of a supplemental resistor if needed.<BR>I had a long discussion with an MIT EE prof after the article came out. He thought the same thing as you, but eventually agreed.<P>------------------<BR>


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 Post subject: Idea for resistor line cord
PostPosted: Oct Wed 20, 2004 2:30 am 
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Ken, I can explain some of your findings, but not all.<P>I set up a silicon diode in series with a 5K resistor, supplied from a household recepticle. With my Simpson 160 analog meter (non-RMS), I measured: about 126V across the line and about 80V across the resistor - fairly close to your readings.<P>Then, with my Fluke 179 true RMS meter, I measured: 125.1V across the line and 67.1V across the resistor - a little more than half the line voltage (62.6V).<P>A non-RMS meter will display unreliable meaurements across a non-sinusoidal voltage. My 80V reading on the Simpson and your 85V reading are not relevant indicators of the effective RMS voltage applied to the resistor.<P>OK, if I'm right, then why wasn't my RMS voltage measurered with the Fluke exactly half the line voltage, as I've been claiming? There are two possible explanations.<P>First, the difference between 67.1V and 62.6V is within the specified accuracy for the Fluke when measuring a non-sinusoidal waveform. Secondly, there would be some reverse-bias leakage through the diode, which might also explain some of Norm's observations.<P>Your photometric experiment is interesting. Think about this: with or without the diode, the PEAK current during each cycle will be the same. Perhaps you were measuring a portion of the lamp's peak flicker. (The same situation could apply to a string of tube heaters in series with a diode.)<P>But here is a fact of physics about which there can be no doubt: The power disipated by a periodic AC voltage applied to a resistor is equal to the square of the RMS voltage divided by the resistance. This is true no matter what the shape of the waveform - whether sinusoidal, halfwave, square, triangular, you name it. The only thing that needs to be agreed upon is how to compute the RMS voltage of a non-sinusoidal waveform.<P>We are so lucky to have the time and wits to concern ourselves about all this!<P>PS - I have digital photos of my experiment, showing my voltmeter readings, etc., with my hands nowhere in the pictures!<P><BR>------------------<BR>**********<BR>Doug Criner <A HREF="http://www.enginova.com" TARGET=_blank>http://www.enginova.com</A>


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 Post subject: Idea for resistor line cord
PostPosted: Oct Wed 20, 2004 5:27 am 
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Hi Doug<P> Here is an easy way to show what happens. Series wire enough tube filaments to add up to around 62 volts. This would be a typical radio with 300 ma tubes. <P> With your variac set voltage to 62 volts or half line. Observe filament brightness. Now run the filament string off the line with a series diode. Notice brightness. <P> This is close to Ken's photocell experiment but we are concerened with radio. You will see a diode in series with the line causes filaments to be much brighter.<P> We just need to know voltage out of a diode, off the AC line, is .707 times input. 120 VAC X .707 = 85 volts. <P>------------------<BR>Norm


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 Post subject: Idea for resistor line cord
PostPosted: Oct Wed 20, 2004 7:24 am 
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I've used both caps and diodes for droppers over the years and really have no preference one over the other. They both create the dial lamp burnout scenario but that is easily fixed.<BR>In many of the little 'mantle' radios there simply isn't space for the cap(s) and the diode gets the nod even if a bit of resistance needs to be added. Most of those little sets seem extremely tolerant of less than ideal filament voltage anyway.<BR>There's a neat little calculator on Robert Casey's website for figuring the correct capacitance. I'll dig up a link if anyone isn't familiar with the site. You can get the AC caps in a variety of values at places like Digi-Key, and yes Curt, the tolerance is plenty adequate.<BR>Regarding the "shorting" aspect. I have never heard of either a cap or diode shorting in this application but Murphy is always around the corner. On a practical level I don't think you'd see an instantaneous burn out of all the tubes just by having their heater voltage jump up by say 50%. Seems it would continue running until the first one finally opened up???<BR>-Bill<P>------------------<BR>


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 Post subject: Idea for resistor line cord
PostPosted: Oct Wed 20, 2004 4:54 pm 
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Norm & Ken:<P>Yes, you are correct. I was wrong about the voltage. I'm sorry to have led this thread on a goose chase.<P>With a series diode, the RMS voltage across the heater string will be 0.707 times the line voltage (120 x 0.707 = 85V). This is not enough drop to replace a resistance line cord without a supplemental resistor.<P>I'm still wondering why my Fluke "true RMS" meter gave the wrong answer, but I should have known better.<P>------------------<BR>**********<BR>Doug Criner <A HREF="http://www.enginova.com" TARGET=_blank>http://www.enginova.com</A>


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 Post subject: Idea for resistor line cord
PostPosted: Oct Wed 20, 2004 10:04 pm 
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Meters are calibrated to read bi-directional AC (full wave). VTVMs and VOMs rectify the AC input to read it. Since the output of a diode is already rectified, we can't expect reliable readings. The DVM has an RMS to DC converter intended for bi-directional AC. The diode output is uni-directional AC, so you can't expect a reliable reading although you may get one.<BR>The lesson is that the common meters are useless for measuring this kind of waveform.<P>------------------<BR>


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 Post subject: Idea for resistor line cord
PostPosted: Oct Thu 21, 2004 10:42 pm 
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The amount of voltage obtained with the diode method can be adjusted (upward) with the apropriate size cap installed after the diode, to ground, the same as the input filter in a power supply but there is a problem arriving at the propper value for the dial lamp resister due to the difference in voltage between the initial surge when the set is first turned on and the steady state voltage after it warms up. I have burned out a lot of zener diodes trying to arrive at the correct values for a particular set. <BR>The best solution I have found for dropping the voltage is to use a series capacitor of 5 to 8 mfd depending on the amount of drop needed, NOT AN ELECTROLYTIC CAPACITOR. This also reduces the initial surge and makes the value for the pilot lamp resister easier to establish. <P>------------------<BR>


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