Why does the "dim bulb test" limit current instead of adding load to the supply? - Page 1
Forum section: electronics / beginners

author: Beamin
author_rank: Super Contributor
timestamp: August 18, 2018, 12:59:19 pm
content: This always seemed backwards to me. I always though that if you took a 100 watt load (light bulb) and put it in series with say another 100 watt load electronic device that you would pull 200 watts out of your power supply. But in reality the most it pulls is 100 then when the load is added the current stays at 100 but dims the bulb? To figure out which device; bulb or load gets what % of the hundred watts you would have to measure the ohms of the bulb while its hot and your electronic device? Then E=IR it? Do bulb behave as temp controlled variable resistors? Or is there something special about the bulb where its resistance changes in proportion to filament temp? I always thought 100 watts + 100 watts = 200 watts draw. With no real light bulbs to test I can't do this.Also instead of a 100W light bulb that's illegal can I substitute it with a CFL with a really high CRI that says it's as bright as a 100 watt on the box?Will I get arrested if I plug in a 100 watt filament bulb by the signals it sends to the power company?

author: Rerouter
author_rank: Super Contributor
timestamp: August 18, 2018, 01:17:07 pm
content: a 100W bulb when it is cold draws more than 100W, as the coppers resistance is lower while its cold,The bulb is there because it limits the maximum, and if you see the bulb go bright, you know right away something has gone wrong.It also lets you test things with a decent startup current, again they have a lower resistance while cold. VS just a fixed load resistor which will droop much more.

author: sokoloff
author_rank: Super Contributor
timestamp: August 18, 2018, 01:31:00 pm
content: A dim bulb is a series resistor that has low resistance at low filament temp (low current) and higher resistance as current increases (and filament heats up).To figure the power, it’s a simple series resistor problem. With a short circuit in-series with the bulb, the system draws 100 watts (just like when the short circuit is the mains wiring in a normal lamp install). That’s the most power the circuit will draw.

author: Audioguru
author_rank: Super Contributor
timestamp: August 18, 2018, 01:40:55 pm
content: Quote from: Rerouter on August 18, 2018, 01:17:07 pmThe filament is not copper that would melt, it is tungsten. The resistance of tungsten is also lower when cold.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: August 18, 2018, 01:41:12 pm
content: There are two different problems here. A one hundred watt bulb consumes one hundred watt when line voltage is applied and when it is fully heated. If you put two of them in series neither will get line voltage, they each get half. If they were simple resistors, cutting their voltage in half would drop power by four. Each bulb would consume 25 watts, the total would be fifty watts.The answer is complicated by the temperature varying resistance of the bulb, but if the bulbs are balanced the result is the same.Putting a bulb in series with a unit under test allows almost full current for startup. But if a short or other high current fault exists the bulb will heat up rapidly and limit current, hopefully to a non damaging level. A low wattage bulb limits the current to a lower level and might impede operation of a properly functioning device so the wattage of the bulb used must be selected thoughtfully.

author: Zero999
author_rank: Super Contributor
timestamp: August 18, 2018, 01:47:31 pm
content: Quote from: Audioguru on August 18, 2018, 01:40:55 pmYes and the resistance of all metals has a positive temperature coefficient.

author: IanB
author_rank: Super Contributor
timestamp: August 18, 2018, 01:50:07 pm
content: It is a curious thing that a filament bulb can act as an approximate constant current regulator. It was once used as such in lead acid battery chargers, limiting charging current when the battery was empty and dropping out when the battery was in the CV stage.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: August 18, 2018, 03:37:49 pm
content: Quote from: Beamin on August 18, 2018, 12:59:19 pmI hate to say this to someone with over 800 posts in an Electrical Engineering forum ... but are you trolling? Really?Because your post, if taken seriously, reveals some really fundamental misunderstandings about the relationship between current, voltage, resistance, and power, and also between such basic circuit concepts as series and parallel connections.While others have responded in some detail above, all those answers are "assuming" that you have the necessary fundamental understanding of the concepts... but your post, if taken seriously, reveals that you may not.The short answer to the question in the Topic Subject is: Because the "dim bulb" is connected in series with the DUT, not in parallel with it.May I respectfully suggest that you review the basics of Ohm's Law? And work out some parallel and series resistor networks, voltage dividers, power dissipation in resistors, and such like problems, just for practice. Hopefully some "lights" will brighten up as a result.(By the way the "Dollar Stores" around here have plenty of incandescent filament bulbs in stock... you might try checking your local...

author: ArthurDent
author_rank: Super Contributor
timestamp: August 18, 2018, 03:51:16 pm
content: "I always though that if you took a 100 watt load (light bulb) and put it in series with say another 100 watt load electronic device that you would pull 200 watts out of your power supply."That is not how it works. If you have a power source and put a 100 watt 120 volt incandescent bulb across the output, that is a 100 watt load only IF the output voltage is 120 volts. Putting a 100 watt 120 volt bulb across a 12 volt source is not a 100 watt load. The 100 watt load is only at the rated voltage.If you have a 120 volt source with a 100 watt 120 volt incandescent bulb wired across the output so there is a 100 watt load, then you wire a second 100 watt 120 volt incandescent bulb across the output so you have two 100 watt 120 volt incandescent bulb wired in PARALLEL across the output, you will have a 200 watt load.Using a CFL will only confuse things more because they may not draw any current until the voltage across them reaches perhaps half the line voltage then they will draw some percentage of the replacement wattage rating.If you have a small metered bench supply you can safely prove this with 2 equal value resistors. Wire one resistor across the properly adjusted supply and record the current reading. Next wire the second resistor in series with the first and the current reading will be 1/2 of the first reading. Finally wire the 2 resistors in parallel across the output and the current reading will be 2 times the first reading as Ohm's law predicts.I've answered this question as serious and if it was you may want to do a lot more reading on the basics of electricity so you fully understand the very basics.

author: ebastler
author_rank: Super Contributor
timestamp: August 18, 2018, 04:00:56 pm
content: alsetalokin4017 made the relevant points in his post above. While the prior comments on temperature dependence of a filament's ressitance etc. are all correct, they do not seem to address Beamin's real problem.Beamin, if these were serious questions and not an attempt at trolling: Please read up on Ohm's law (relationship between voltage, current, resistance), and on the relationship between power, voltage and current.

author: IanB
author_rank: Super Contributor
timestamp: August 18, 2018, 04:04:57 pm
content: Also, nobody seems to have directly mentioned that two 100 W, 120 V bulbs in series becomes equivalent to a 200 W, 240 V bulb. So then you can just obviously ask the question, what happens if you attach a 200 W, 240 V bulb to a 120 V supply?(But, really, the OP's question is so far out of left field that I have no clue where it came from. I think Beamin just asked it as a joke to wind everyone up. I seem to recall she has done the same kind of thing before.)

author: amyk
author_rank: Super Contributor
timestamp: August 18, 2018, 04:50:42 pm
content: I agree with the others here than this feels more like a bad trolling attempt.But in an attempt to enlighten(!), consider this: if you think a 100W bulb will always consume 100W of power, regardless of the voltage across it, then what happens at 0V? Do they light when disconnected from a socket and have their terminals shorted to each other?Quote from: IanB on August 18, 2018, 01:50:07 pmhttps://en.wikipedia.org/wiki/Iron%E2%80%93hydrogen_resistor

author: james_s
author_rank: Super Contributor
timestamp: August 18, 2018, 06:09:37 pm
content: Quote from: IanB on August 18, 2018, 01:50:07 pmThey are used as ballasts in self-ballasted mercury vapor lamps too. Not nearly as common anymore as they used to be but they are still available. They contain a standard mercury arc tube in series with a tungsten filament in the same envelope. When power is first applied the arc tube voltage is low and the filament lights brightly but as they warm up the filament gradually dims as the arc tube gets brighter. At one point they were an economical replacement for incandescent lamps that did not require the installation of expensive ballasts. Efficiency was not much better than incandescent but they did last 10 times as long.

author: Beamin
author_rank: Super Contributor
timestamp: August 19, 2018, 04:30:38 am
content: Quote from: IanB on August 18, 2018, 04:04:57 pmI always just kind of built circuits from examples and never really got into the basics that much. It's a problem of the Arduino generation where to blink an LED requires writing lines of code. But sometimes I get these seizures and it will wipe out some bits of my memory. Like I know more about RF and making antennas then some of the more simple concepts. I tend to focus on really specific things like radio stuff. Most of my circuit building is just guestimating the parts from a schematic and physically trying different values. The soldering and putting in a nice box is the fun part so I haven't spent much time recently relearning ohms law again for the who knows how many times. Unfortunately with my condition I never know what I am going to forget. I was also thinking the light bulb was drastically changing in resistance from orange to white that just happened to be a really good way to current limit. One of those "It just happens a light bulb works by a fluke (27) of nature". I try not to troll this forum even though 90% will find it funny 10% will click report. Where would find a data sheet for something like tungsten's temp coefficient? I tried light bulb manufactures but their customers really only care about on and off.So what is physically different in a 12 and 120Vlight bulb? They are all tungsten do they use longer lengths or thicker filaments?Quote from: Hero999 on August 18, 2018, 01:47:31 pmAre there some material with negative temp coefficient that results in runaway? Isn't that's what transistors and PN junctions do?I know I have asked this before but what are good examples of beginner versions of the spice program? There was one where you would see red and green lines that show current flowing through them. Are there any that you can download rather then web base? I have lost the URL but since you can't save your circuits it gets annoying. I learn best by seeing visual like in those youtube videos by that Russian guy using whats sounds like a computer voice. Potential is shown as raised lines and current is dots that move around.https://youtu.be/G3H5lKoWPpYSpice is too technical where it gives you the box with numbers in it, very useful if you are doing complicated stuff but I need something in between that and the web based tool. One where you could simulate poking a scope probe around or DMM.So "series string TV sets" use the filaments that all total 120 volts each one dropping down. So the first tube sees the full 120 volts then the next 90 and so on or is it an average; say 10 12V tubes all in series. But if it has a transformer then each heater is run in parallel.While we are on heating up wires: If I want to build a 120V isolation transformer using a torroid I would just take 10 turns of 12 ga wire on one side and 10 turns 12ga on the other? When this is plugged in is there enough magnetic field induced that the wire on the line side wouldn't turn into a heater when the secondary/load side is disconnected? Does the creation of the magnetic field make it efficient enough to carry away most of the current energy into magmatism to stop it from burning up in a hurry?Thanks for the patience in all that answered seriously rererelearning things is frustrating as hell but I often have luck rereading my own threads here.

author: james_s
author_rank: Super Contributor
timestamp: August 19, 2018, 06:32:55 am
content: A thicker filament requires higher current, a longer filament requires higher voltage. If you take 10 12V bulbs that draw 1A and wire them in series, you are stringing those 10 filaments together and you end up with a 120V bulb that draws 1A, ie a 120W bulb. The filament diameter is the same as any bulb that draws 1A, the length will depend on the rated voltage.If you wanted to make a bulb of a given wattage, you would decide on the voltage you want it to operate from, then use Ohms law to calculate the current required for the desired wattage at the selected voltage, then you would select a filament diameter corresponding to that current, then select the length for the voltage you want.The reason many specialized lamps such as those for slide projectors and other applications requiring precise optical control use odd voltages is that they need to use a short filament for optical reasons and this dictates using a low voltage. There are other factors too, a long thin filament is mechanically fragile and suffers a greater amount of thermal losses, especially convective losses for gas filled lamps. This is why 240V incandescent lamps are substantially less efficient than 120V lamps. You can eliminate the gas fill and use a vacuum but then your tungsten evaporation rate is much higher resulting in shorter life. Everything is a compromise.

author: ebastler
author_rank: Super Contributor
timestamp: August 19, 2018, 08:46:42 am
content: Quote from: Beamin on August 19, 2018, 04:30:38 amGive TINA-TI a try -- free download athttp://www.ti.com/tool/TINA-TI.It is a pretty friendly Spice front-end. A simple schematics editor, and while it does not color-code the wires to show current flow as you describe, it lets you connect virtual voltmeter probes and a virtual oscilloscope to the circuit.

author: Seekonk
author_rank: Super Contributor
timestamp: August 19, 2018, 11:13:51 am
content: I used a lamp as a constant voltage source! We had a low voltage detection board. Production needed a cheap tester for it and the board had to trip at the same value of mv AC as DC. I used a small lamp in a bridge circuit and either switched AC or DC into the bridge. Line voltages of +- 20% did not phase it, spot on within 1mv. Though it doesn't respond quickly to line changes. All the tech needed to do was to cal the test box on DC with a cheap meter. Switching to AC would be the same mv RMS as the set DC value.I first learned about the dim bulb test in about '64. EICO had an ad for an amplifier kit. It pictured a lamp next to the kit and said the only piece of test equipment needed was a lamp.

author: Beamin
author_rank: Super Contributor
timestamp: August 19, 2018, 11:33:43 am
content: Quote from: Seekonk on August 19, 2018, 11:13:51 amHow did that work? Put the bulb in series with it and as the voltage changed the resistance of the bulb would change? Less voltage cooler filament and it would go down in ohms? Or am I mixing up current with voltage? My ignorance of E=IR reminds me of a louis rossman video where he says he never learned BlkBrROYGBV out of spite for a teacher he had. That's a brilliant idea to paint rainbow colors on a beige or blue back ground

author: SG-1
author_rank: Regular Contributor
timestamp: August 19, 2018, 11:39:08 am
content: Ohm's law states that the electrical current (I)in a closed circuitis proportional to the voltage (V) and inversely proportional to the resistance (R). When the voltage is increased, the current will increaseprovided the resistance of the circuit does not change.Most versions of ohms law, especially the simplified math version leave out two critical statements that allows one to know if the formula application is valid. Those two statements are in bold text above. If the measured value does not match the calculated value those statements can provide clues as to why the desired results were not achieved.

author: Beamin
author_rank: Super Contributor
timestamp: August 19, 2018, 07:15:25 pm
content: Quote from: SG-1 on August 19, 2018, 11:39:08 amAfter reading that it all comes back to me. Its frustrating it's like I know I know something but can't quite reach it. This hobby is use it or lose it or at least with me the math is. Not to long ago a friend's son was over my house and he saw math problems out on my table. He asked if I was in school I said" no I had to figure something out( building an antennaat a different wave length)I was board" and he said "you mean you do math when you don't have to?" Yes math makes life easier.

author: Vtile
author_rank: Super Contributor
timestamp: August 19, 2018, 08:45:33 pm
content: Quote from: Beamin on August 18, 2018, 12:59:19 pm(V:100Vdc)---------(GND) Voltage over the two resistors are 100V.(V:100Vdc)---------(GND) Voltage over the two resistors are 100V.(V:100Vdc)---------(GND) Voltage over the two resistors are 100V.(V:100Vdc)------(GND) Voltage over the one resistor is 100V.The trick is that the total power will be the inverse sum of the impotences in series connection (in voltage dominant network). However in current dominant network the total impotence would be the inverse sum of powers.

author: Vtile
author_rank: Super Contributor
timestamp: August 20, 2018, 09:10:08 pm
content: One of our member kindly pointed out that I made a typo with the electrical 'dances', lets go through some of them...ResistanceConductanceImpedanceReactanceAdmittanceSusceptanceImpotenseUh...Impotense = 1/P🤣

author: Beamin
author_rank: Super Contributor
timestamp: August 21, 2018, 06:17:27 pm
content: Quote from: Vtile on August 19, 2018, 08:45:33 pmI think I see what you getting at but I'm losingsomething in translation.Also my original question was more then just the simple ohms law question but I was thinking that a tungsten filiment had some special property where the change in temp vs resistance was coincidentally proportional to the way it regulated voltage. Like how the diameter of the moon just happens to make a perfect eclipse, or the earth is just the right size to resovle mm wavelengths like they are using in the event horizon telescopes; a flukeof nature. I also like how meters nicely match frequency vs wave length in 10's and 3's due to the 300,000 km/s of light. 300Mhz = 1 meter 100mhz = 3m 30Mhz= 10 meters how easy is that?

author: sokoloff
author_rank: Super Contributor
timestamp: August 21, 2018, 06:48:28 pm
content: Taking only your subject line question, the answer is "for the same reason that a dropping resistor on an LED limits current rather than adding load to the supply".

author: ebastler
author_rank: Super Contributor
timestamp: August 21, 2018, 07:49:40 pm
content: Quote from: sokoloff on August 21, 2018, 06:48:28 pmOr: "Because it is connected in series with the device unter test, not in parallel."

author: mmagin
author_rank: Frequent Contributor
timestamp: August 21, 2018, 10:06:24 pm
content: Quote from: Beamin on August 18, 2018, 12:59:19 pmYou can still buy certain industrial / rough service incadescent 100W bulbs likehttps://www.amazon.com/Bulbrite-100A-RS-2PK-Incandescent-Service/dp/B017ULQIZWAlso, there are somewhat lower wattage halogen replacements (halogen tube inside an outer A-series envelope) you can use.