Simple Resistance nightmare. - Page 1
Forum section: electronics / beginners

author: iamnothim
author_rank: Regular Contributor
timestamp: March 09, 2013, 11:27:38 pm
content: Resistance ProblemHi,I am totally confused.SImple goal: Lower the voltage off the AC transformer secondaries from 7.8 V AC to 6.3 V AC. To support a vacuum tube that needs 6.3 V AC and .400 A. Thats it. Simple loop. one component. Vacuum tube.DMM Measurements: If the secondaries are not loaded I measure 7.87 V AC. The transformer specification is 6.3V center tap .600A , when the primaries are 115V AC 3.78 VA. My wall is 120V so I think that's why I am getting 7.78V rather than the 6.3V I need. That said, I can't make the ratios pen out.My resistance calculations for a voltage drop from 7.87V to Vacuum Tube required 6.3V and .600 produced a 3.92 ohm resistor rated for .622W or better.All calculations were based upon the Tube specification of 6.3V and .400A calculating a resistance of 15.75 ohm.I plugged the following into LTspice. Source voltage 7.87 V DC. R1 3.92 ohm, load resistance (tube) 15.75 ohm. It spit out perfect values. 6.3 V .400AThe closest resister I have is 4.3 ohm 5%, 1W. At .400 and 7.87V between 6.236V and 6.04V SERIESProto-board nightmare. Nothing made sense.I had not taken actual measurements up to this point except for 7.87V AC from unloaded secondaries.I measured the heater circuit on the vacuum tube (pins 4 and 5) 2.29 ohm. As above the 6.3V .400A spec calculates a 15.75 ohm resistance.I measured resistance of the secondaries 1.00 ohm. If I take the 7.87V measured and the .600 rating I would get 13 ohm resistance.The only thing that checked was the 4.3 ohm resister.When I plugged it in….I cooked the resister.I measured the following:2.30 ohm tube0.99 ohm secondaryI plugged it in and measured the following:7.083V AC.4465AThose values calc a 15.86 ohm resistance. That matches the Tube resistance used in the calculations. It would also yield 3.16W I thought I only needed a .6W rating.Next I added a trim resister 10 micro H .5w. I dialed it down tp 1.7 ohm but it had no effect on voltage. It did, however have an effect on current. I could dial down the current from .4465A to .400 spot on. I brought it down and measured the resistance on the trim pot at 3.01 ohm. Total resistance Trans 1 ohm, Tube 2.31 ohm, Trim Pot 3.01 TOTAL: 6.17 ohmThe voltage didn't budge. In fact it went up.***** From 7.083 above to 7.08 with the Trim POT*** How can this be a parallel circuit ??6.17 ohm with .400 A equals 2.46 V *****7.08V with .400A equals 17.7 ohm ****I know it must be obvious…..Perhaps because it's AC Voltage.Perhaps transformers work differently than I thought.

author: Unknown
author_rank: Unknown
timestamp: March 09, 2013, 11:43:29 pm
content: It's not entirely clear to me what you're doing and measuring at each step (schematics or diagrams indicating connections and voltages would help), but let me give you a few hints. Keep in mind I'm not a tube person.The transformer voltage is specified under full load, so the transformer should deliver 6.3V at 0.6 A. The extra 5% due to the higher line voltage is insignificant. Try loading the transformer with a 400 mA load (eg. change resistors until it delivers 400 mA, or use a constant current sink) and measure the voltage, it should now be much closer to 6.3 V.The heater resistance will vary with temperature, and temperature will vary with current. This screws up any estimations of current from cold heater resistance. Putting 6.3 V across it for a while and measuring the current will give you an indication of the resistance when warmed up. Not that you need it to solve this.I would first measure the transformer voltage under 400 mA load. If this is still to high, add resistance so R * 0.4 A is equal to the voltage you want to drop. You can estimate the dissipated power with 0.4^2*R, but keep in mind that the power draw will be higher during warm up, so add some headroom.

author: hammil
author_rank: Regular Contributor
timestamp: March 10, 2013, 12:01:52 am
content: Try just putting the heater with no series resistor. It'll probably work fine. Otherwise, just increase the power of your resistors.

author: westfw
author_rank: Super Contributor
timestamp: March 10, 2013, 02:36:31 am
content: QuoteFilament resistance varies with temperature. When the filament has 6.3V across it and .4 amp going through it, it should be approximately hot enough that it would measure 15.75 ohms.QuoteThe secondary resistance doesn't have a lot to do with the transformer's power capability, at least not directly. A 6.3V/.6A rating means you can put a load of 10.5 ohms on there, and the voltage will be 6.3V and the current will be .6A. At that point, 0.36W of power would be wasted in the transformer secondary because of its 1-ohm resistance, and it would get somewhat warm. Plain transformers are not regulated, so higher resistance loads means the voltage will also be higher. (OTOH. magnetics and such also come into play, and the primary voltage IS "regulated", so it doesn't act quite fully unregulated, either.)It's also possible that your meter doesn't correctly measure RMS AC voltages.Unless your tube is extremely valuable, you can get a better idea of what is happening by actually connecting it (directly) to the transformer, and measuring the "live" circuit. As others have said, you shouldn't need any resistor. 6.3V tube filaments and 6.3V transformers were DESIGNED to be connected together.

author: vk6zgo
author_rank: Super Contributor
timestamp: March 10, 2013, 02:56:52 am
content: Ditch the resistor!6.3 volt filaments are quite forgiving.In this case,the difference in primary voltage amounts to less than 5%,which will be reflected in the secondary.Tubes with these types of filaments were used in car radios,where the applied voltage varied much more than that.

author: amspire
author_rank: Super Contributor
timestamp: March 10, 2013, 03:44:14 am
content: First transformers are usually rated for output voltage at full load, so a 6.3V winding will probably have a higher voltage with no load. Typically, the smaller the transformer VA rating, the poorer the transformer regulation.The transformer filament has a low cold resistance as you have seen, so it usually draws a large current until it warms up. The resistor is preventing the filament from warming up, and so you have a cool filament and a cooked resistor, instead of a hot filament and a cool resistor.Moral of the story - the resistor is a bad idea.If you really want to reduce the voltage to prolong the tube life, see if the transformer has a gap between the windings and the core. If it does, you can add a few extra turns of insulated wire to make an extra low voltage winding. Do about 3 turns, turn the power on and see what AC voltage you get out of this new winding. If you measure 0.45 volts and you want to drop the filament to 6.0V, then you want a reduction of 7.083V - 6.0V = 1.083 V. So the turns you need will be 3T x 1.083/0.45 = 7.2T (ie 7 turns).If you put this new winding in series with the 6.3V winding, it will either increase the voltage or decrease it depending on which way around the winding is connected.Richard

author: iamnothim
author_rank: Regular Contributor
timestamp: March 10, 2013, 04:18:22 am
content: That explains a lot. But I still have a lot to digest.I did not know that the secondary voltage would change as the transformer reaches full load.It also blows my mind that "higher resistance loads means the voltage will be higher" That is baffling. When I added the trim resistor (set to 3.01 ohm) the voltage changed a tiny bit but the resistance did too. From .440A to .400…. In this case that is the desired level for the 6BQNA.What I am reading…. "ditch" the resister. As the filament reaches operating temp the voltage should be 6.3V ... Even though the primaries are getting 120 vs 115.Nothing else is in the loop to load / "draw current" Just 6.3 volts to pins 4 & 5 (heater element) of the tube. Static. It's purpose is to heat the cathode. Since I want to test the tubes for emissions, etc. I want to be as accurate as possible. I am taking my measurements with a Fluke 287.I doubt 7V will bother the heater much, so I have no problem trying it with different tube types….. Except my 1964 E188CC'sI have several varieties of 9-pin tubes I want to use with this tester and I wanted to be as accurate as possible. ie. Dial in the stable voltage of 6.3V. ( PCC88 has a 7V heater).The ratings for these tubes are .300A, .370A, .365A and .400AI thought I would use a resister to get to that point. As such I calculated for each "tube type" a 2.7 ohm, 4.3 ohm and 5.1 ohm respectively . I was going to use a rotary switch to select the resistor based upon the tube I was testing.Here's the schematic. And thanks for all the input.I am going to read it more closely.

author: iamnothim
author_rank: Regular Contributor
timestamp: March 10, 2013, 05:01:23 am
content: I hooked everything back up.Just the tube and the trans.The voltage started a tad higher, but over 3 minutes it averaged 7VThe current stayed solid at .436A during a 1 minute period.EDITThe problem appears to be where I was putting my test points. I believe that was mentioned.Photos below were with the probes in the Wrong Place.I removed the trimmer that was dialed to 2.68 ohm and replaced it with a 2.7 ohm 5% 2W metal film.I am not getting5.933V and .395A That's pretty dang good for me.I hope I am finished with this part (the easiest) .

author: IanB
author_rank: Super Contributor
timestamp: March 10, 2013, 06:26:46 am
content: Quote from: iamnothim on March 10, 2013, 04:18:22 amYou were told that earlier, as it happens:Quote from: amyk on March 02, 2013, 05:25:40 am

author: iamnothim
author_rank: Regular Contributor
timestamp: March 10, 2013, 07:06:44 am
content: Yes. I am a beginner, had some issues with "load". I also had issues where I in the circuit I took measurements. Overall, I knew I had to add "load" / resistance, in series, to lower the voltage. I wanted to get to 6.3v. But I did not have the correct starting point. So I used the unloaded 7.87V to calculate the load I needed to get to 6.3vIt turns out that with the tube in the circuit the voltage ...... I want to say across the load but I can't remember.... was /is 7v. Again, I created confusion for myself by measuring at the wrong points in the circuit.When I use 7v as a starting point and measure across the load, the measured results match the calculated results. Interestingly 7v used in a ratio with 120v primary is close to 6.3 over 115. At 7.87 it wasn't close.

author: iamnothim
author_rank: Regular Contributor
timestamp: March 10, 2013, 07:14:29 am
content: Question:I have this unknown voltage. The secondary under load voltage.I put the load on the circuit and measure voltage and current. Ok... But then it doesn't match the calculations.Are I not supposed to calculate the values prior to building the circuit? From the transformer?

author: vk6zgo
author_rank: Super Contributor
timestamp: March 10, 2013, 09:24:14 am
content: Forget the open circuit voltage of the transformer.The only reading that should matter to you,is that across the pins of the tube,after a period in which the filament(heater) has warmed up to operating temperature.From the stuff you have posted,you are trying to test these tubes under specific conditions.Apart from the fact that it is unlikely to matter,anyway,there are a number of ways you can do this.(1) Series resistor---as you have found out,this is fraught with difficulties.(2) Use a variable transformer (Variac) feeding the primary of your transformer---a bit fiddly,& may not be very stable.(3) Use an AC voltage regulator to feed your primary---huge,& expensive.(4) Forget about operating the heaters from AC,& use a regulated DC supply,adjusting it till the voltage across the tube heater pins are as required.----I would suggest this,as the tube heaters don't give a damn if they are fed AC or DC,

author: iamnothim
author_rank: Regular Contributor
timestamp: March 10, 2013, 10:59:05 pm
content: vk6zgo,Thank's a bunch for the suggestions.So now "get it" Secondaries are no different than a battery. Put a load across it and the voltage drops. I'm very slow. I've read it dozens of times, why it didn't stick whenI read it and experienced it...….The voltage on the secondaries settled in at 7.1V AC making the ratio of primary to secondary much closer.I put different tubes with different current ratings in the circuit in series with resisters calculated to get the heaters close to 6.3 V. I measured voltage across the potential, load and resistance for each valve type. The expected current for a given tube type stayed constant and the voltages and resistance calculations added up, (within a few 100th's) In short Ohm's law still works. However I do see some of the difficulties with series resisters you spoke to.A 4.3 ohm resister rated at 1W got quite hot while having a .620 W load. Is that too close?Option #4 Although it sounds great, the only viable choice is a wall wort and I don't want to kludge one inside the enclosure. There isn't anything with a fixed mount. Mouser has LED power supplies but the output voltage is variable and I'm back to adding resistance. The good news is my PCC88 valves are 7V .300A and they worked perfectly.The voltage on the AC Trans is indeed sketchy. Simply as grist, I considered getting a 115V to 120V trans and flipping it. I am stuck with AC for the anode, but since the voltage changes by design, I AM THINKING…. I'll just follow the schematic build, test and learn. 6 iterations should do it.Still I wonder:For example. I'm testing an ECC88. Spec says 6.3V & .365A The resistance is calculated at 17.26 ohm. Why, when I hook up a DMM does it only read 2.73 ohm? When I energize the circuit and take measurements, the measured R=V/I yields 15.27 ohms a lot closer to that from the spec. Why doesn't the tube act like a component resister and yield a value out of the circuit that is the same as one calculated from measured voltage and current?Not looking for you to educate me, perhaps just point me in a direction I can research.Thanks for the suggestions.

author: David_AVD
author_rank: Super Contributor
timestamp: March 10, 2013, 11:11:58 pm
content: Quote from: iamnothim on March 10, 2013, 10:59:05 pmCheck the responses again. It's been pointed out several times that the valve heater is much lower resistance when cold. Just like a light bulb.

author: iamnothim
author_rank: Regular Contributor
timestamp: March 10, 2013, 11:20:50 pm
content: I've run them for 15 minutes, just valve to transformer, no resistor. It stays constant at 7.1 VThat said, I will hook it up now and run it again for a longer time with several different valves.And, I do appreciate your comments. I value them.

author: AlfBaz
author_rank: Super Contributor
timestamp: March 10, 2013, 11:32:25 pm
content: Quote from: iamnothim on March 10, 2013, 10:59:05 pmMost conductors have a positive temperature coefficient (PTC). This means as the temperature rises the resistance increases. Light bulb filaments are a great example. Other materials exhibit negative temperature coefficient (NTC) where the temperature increases the resistance decreases. A lot of insulation materials have this property.

author: iamnothim
author_rank: Regular Contributor
timestamp: March 10, 2013, 11:36:02 pm
content: Didn't read your response correctly as well......I'm thinking the voltage across the heater circuit would drop as it gets hotter.Now I'm reading it......The resistances of the heater, like a light bulb..... @westfw, indeed provided the explanation.I will re-read.

author: IanB
author_rank: Super Contributor
timestamp: March 10, 2013, 11:37:39 pm
content: Quote from: iamnothim on March 10, 2013, 11:20:50 pmBut it won't make any difference whether you run it for 15 minutes or 15 hours. It will still be 7.1 V.The heater is just that: it gets hot. When it is not powered it is cold, and when it is powered it is hot. It even glows orange to tell you it is hot, just like a light bulb.When the heater is cold, with no power, it has a low resistance. When the heater is powered and glowing orange it has a much higher resistance.In order to measure the resistance of the heater you have to power it up and measure the voltage across it, then measure the current going through it (two DMMs will help), then divide the voltage by the current. That will give you the operating resistance of the heater filament.

author: iamnothim
author_rank: Regular Contributor
timestamp: March 10, 2013, 11:38:29 pm
content: Thanks AlfBaz.That's where I was curious.I'll see if I can read and understand the math in NTC.

author: iamnothim
author_rank: Regular Contributor
timestamp: March 10, 2013, 11:49:14 pm
content: Cool Ian,I was seeing the phenomenon but I didn't know why. (and I didn't read carefully)Now with that and NTC I should learn something.I deserve the Hot..Cold ....Hot....Cold thing

author: AlfBaz
author_rank: Super Contributor
timestamp: March 10, 2013, 11:59:17 pm
content: hmm.. aren't you seeing PTC?

author: iamnothim
author_rank: Regular Contributor
timestamp: March 11, 2013, 12:02:29 am
content: ..... Not getting PTC..... I'm missing way too much.I just reread Richards post on transformer windings. (missed that too)This is what my trans looks like. Is it possible to modify they windings?

author: iamnothim
author_rank: Regular Contributor
timestamp: March 11, 2013, 12:12:35 am
content: Hammond makes transformer with a 115V primary and Secondary taps @ 90, 100, 110, 115, 120 & 130VAC.Couldn't I invert it using the 120V tap as the primary and the 115V as the secondary?

author: AlfBaz
author_rank: Super Contributor
timestamp: March 11, 2013, 12:18:06 am
content: As vk6zgo said the filament doesn't care wether its being supplied AC or DC. That being the case get your self a bridge rectifier (or make one up out of 4 diodes)A bridge rectifier package will have 4 leads, 2 for AC in (from your tranny) and to for DC out (to the heater)As the DC out value for a full wave bridge rectifier is 0.9xVAC you get (for approx 7VAC) 6.3VDC out

author: iamnothim
author_rank: Regular Contributor
timestamp: March 11, 2013, 12:48:49 am
content: AlfBaz.Ive read about rectifiers and wanted to see them in action.That sounds like both something that will work, not cost a lot, and fun to try.I'm trying my 6.3V 6A trans out and paying attention to what Richard said about cool filaments and hot resistors.I set my meter for min/max before plugging in the trans.Indeed it jumped to 1.25A before settling down to 0.44A It did falloff quickly, nonetheless almost 3x

author: kxenos
author_rank: Frequent Contributor
timestamp: March 11, 2013, 03:01:12 am
content: Put in series a 2,2?, 1W resistor and it will be fine

author: iamnothim
author_rank: Regular Contributor
timestamp: March 11, 2013, 03:52:39 am
content: I realize the heater isn't too picky about the voltage and the current. The reason I'm trying to be accurate is I want to test a valve and compare it to the factory specifications. (see attached example)In my small amount of reading, the heater heats the cathode resulting in electrons being emitted. My reasoning is that variations in the power dissipated by the heater might have a result in the cathode giving up electrons.That's why. Might be crazy, but I'm trying to have an instrument in the end. That's the fantasy.All but the PCC88 are 6.3V but the current varies between .300A and .600 so I'd like to "dial it in" based on the valve I am testing.Here's the bridge rectifier I'm looking at. All the others had much higher voltages.Purely a guess...

author: David_AVD
author_rank: Super Contributor
timestamp: March 11, 2013, 03:54:48 am
content: Maybe something with a higher current rating considering the inrush (surge) current when the valve is cold. 4 Amp bridges are still cheap.

author: IanB
author_rank: Super Contributor
timestamp: March 11, 2013, 03:57:59 am
content: Quote from: iamnothim on March 11, 2013, 12:02:29 amHere is a plot that I just made for a 12 V bulb. I was interested in what numbers I would discover.Looking at this, the first data point is 0.1052 V, 0.107 A giving effectively 0.983 ohms. The last data point is 14.0 V, 1.076 A giving effectively 13.01 ohms. So the effective resistance increased by a factor of about 13x from cold filament to hot filament.Another interesting number is the incremental or localized resistance between 13 and 14 V. The numbers are 13.0 V and 1.035 A, 14.0 V and 1.076 A. So the local resistance at this point is about (14 - 13) / (1.076 - 1.035) = 24.4 ohms. We could say that the "stiffness", or regulation of the bulb at this operating point is 41 mA/V.

author: iamnothim
author_rank: Regular Contributor
timestamp: March 11, 2013, 04:17:18 am
content: Ian, Outstanding!I love graphs. This one illustrates what Richard was writing about and what I had a glimpse of with my max/min.It's very interesting.Through the first volt you really see the surge in current.Since I caught the 1.5A burst quickly, I'd love to see it over time.I'm playing the idiot card here. I've been at this 8 weeks tops... probably only tested continuity and voltage but not across anything...This would be a worthwhile diversion for me.... Did you produce the graph from a simulation or actual data?I'd like to try that with my valves.......

author: iamnothim
author_rank: Regular Contributor
timestamp: March 11, 2013, 04:19:23 am
content: Quote from: David_AVD on March 11, 2013, 03:54:48 amDavid,So the one above won't work. It's 50A. I need 2 max.I probably not looking at the right component.

author: IanB
author_rank: Super Contributor
timestamp: March 11, 2013, 05:35:33 am
content: Quote from: iamnothim on March 11, 2013, 04:17:18 amThat was actual data. I used a variable voltage power supply and recorded the voltage and the current for each point.QuoteMy graph shows steady values with time not involved.It would be difficult for me to plot a graph showing the current surge at power on as the cold resistance is 0.98 ohms. If I connected a 12 V supply to the bulb the initial current would be over 12 A and I don't have a power supply readily to hand that is capable of that. I would have to make such a supply using a high power battery. However, given that it would be a fun experiment I might give it a try sometime.

author: David_AVD
author_rank: Super Contributor
timestamp: March 11, 2013, 05:53:11 am
content: Quote from: iamnothim on March 11, 2013, 04:19:23 amThat 50A figure is a peak rating (for one AC cycle) and not continuous. Your valve surge will be brief, but still much longer than the time allowed for in the peak rating of that 1A bridge linked to.

author: iamnothim
author_rank: Regular Contributor
timestamp: March 11, 2013, 06:27:00 am
content: I think I hear calculus. We're not mutiplin' and divid'n anymore. Curves, acceleration, n stuff.My only hope is electronic toolbox pro on my iPad.

author: westfw
author_rank: Super Contributor
timestamp: March 11, 2013, 05:41:31 pm
content: QuoteSince when? AC will typically be measured in RMS, which is 1/sqrt(2) timeshalfthe peak-to-peak voltage. For the no-load case, rectified and filtered voltage is 1.4 * VRMS, or nearly 9V for a 6.3V transformer. (minus diode drops, though...)Now, for something like heating a filament, the relevant piece of data should be the POWER. And the way RMS voltage is defined, the (averaged over time) power you get from a 6.3V RMS AC voltage (~9V peak), is supposed to be the same as you'd get from a 6.3V DC voltage (constant 6.3V)

author: IanB
author_rank: Super Contributor
timestamp: March 11, 2013, 06:14:16 pm
content: I guess with a full wave bridge there will be two diode drops of 0.5 to 1.0 V each, so that would drop 7 V AC down to about 6 V rectified DC. But of course there is no precision about this voltage drop, it is just depends on the choice of diodes.

author: iamnothim
author_rank: Regular Contributor
timestamp: March 11, 2013, 06:57:54 pm
content: Hello.I thought the no-load tests weren't important.In case they are needed for a rectifier bridge, I just took no load measurements of my Hammond mfg. transformer, Model 166G6. 115V AC 3.78 VA 6.3V C.T. 0.600 A. I have another Hammond that is 6AMy Fluke 287 measures "True RMS"With the meter set to "Peak" and probed across unloaded secondaries it recorded a Peak Max of 10.67V after that it was 7.87VWhen set to "min/max" mode it recorded a max of 7.809EDIT. With a 6BN8 valve in the circuit. 6.3V and 0.600APeak E across the load was 9.11V Peak I was 1.49A wow.E settled in to 6.71VI settled in to .649Very good numbers wo added resistance.

author: iamnothim
author_rank: Regular Contributor
timestamp: March 12, 2013, 05:48:56 am
content: Update: (should anyone care)Thanks guys for pointing me in the right direction. I think I'm close. I actually read your posts this time. Went back several times as a reference.David, your comment on surge current helped. I get it. I used the fluke in record mode and saw the 1.4A surge. I lasted 2 seconds.As you said…. One cycle surge…… math….. 2 sec. = more than 1 cycle.Now I just have to find a rectifier that will do that. ( stocked, that's priced reasonably.)I also calculated the smoothing capacitor and a regulator.This site helped me get the concept and the math:http://www.electronics-tutorials.ws/diode/diode_6.htmlI made the calculations from measurements using the Peak function of the Fluke 287.According to Fluke the voltage measurements are "True RMS" From what I have learned Peak is more useful than min/max.That's my biggest concern. Am I using the right measurements for Vmax and Vrmns.Even if it doesn't work I want to tinker with it. (When I find the right component.)

author: AlfBaz
author_rank: Super Contributor
timestamp: March 12, 2013, 12:38:39 pm
content: Quote from: westfw on March 11, 2013, 05:41:31 pmAt low voltages the diode's non-linearity is sufficient to maker sinusoidal RMS shortcut formula's inaccurate. I'll admit the formula I gave is a kludge at best but at these values is closer to the real RMS value than 1/sqrt2. At higher voltages the latter is almost spot on.QuoteI don't think I mentioned anything about filtering

author: iamnothim
author_rank: Regular Contributor
timestamp: March 12, 2013, 09:56:56 pm
content: I ordered a single-phase Rectron rectifier that supports 3AI also got the smoothing Cap and Linear Regulator so I have something to look at with my scope.