I'm now working with a string of 140 Christmas lights that sequence. I'm now going to use a partial string as to completely replace a lighting scheme in an old jukebox. I have cut it down to only 60 lights. Now each one is a 3.5 volt bulb, and measuring with my little Radio Shack multi-meter on ohms, I get 2.9 across the leads of the individual bulb.<p>I tried to drop the input "voltage"(??) into the sequencer unit with a household dimmer. It worked only to either power the bulbs from dim to near burn-out brightness, or off. But NO sequencing. So I guess I have to keep the input as designed from the wall outlet so the circuitry will work (sequencing).
I've figured the working properties of the original string of 140 lights that lamps one and five, two and six, three and seven, etc. are the sequencing pattern. So with that pattern, I've cut it down to 60 lamps for my 59 lamp application, which means I have five "strings" of 12 lights on each string.I'm not completely sure what the output is for each string from the little sequencine unit, so I'm going to measure the outputs from it next, to see what actual voltage is coming from there. (I just assumed it would be 110)<p>Surely I can just add some kind of resistance so that each string of 12 can burn without burning out? What do you think?<p>I've been reading as many books as I can get my hands on, but you know the old saying "hard to teach an old dog new tricks"...any equations please explain a little more in depth, to ensure my understanding.<p>Again, I thank you for your help.
Again, reducing voltage for Chrismas lights..
Re: Again, reducing voltage for Chrismas lights..
Messr Rob:
The first thing that comes to mind is the old Wurlitzer’s with the bubble lites.
‘Course it also could be a Seeburg or Rockola.When U specified 3.5 v on your lamps, that could take 35 lamps in series to make one complete string across 120v or account for four separate strings to be utilized across 4 output legs from your sequencer. Are these lamps socketed or just leaded lamps hardwired ?
Certainly I would expect your sequencer to require a constant 120vac supply to its electronics power supply and timing circuits….thus the explanation for the squirrelly operation on reduced supply voltage. If the lamp to lamp spacing is correct for you to string out around the J/Bx wouldn’t the simplest method be just to cache the end of the excess length internal to the J/Bx housing and leave the string alone.
If for some reason that wouldn’t work for you it should be possible to cut down on the lamps in each sequencer string and sub in a fixed wire wound or possibly metal film res to simulate the missing lamps…..so the sequencer still thinks its business as usual. One thing though, I’m a bit fuzzy in your mentioning 5 strings, as I was figuring above, there only being 4 strings of 35 lamps fed from the sequencer at four outputs. However, your included reference to lamps 1 and five being on at the SAME time would account for the three other strings to be on lamp 2 (et al) another string on lamp 3 (et al) and the last string on lamp 4 (et al) thus confirming 4 strings fed from the sequencer The uninitiated might experience some difficulty in computing the value of resistors to utilize in place of the missing lamps on each of the strings .Any attempt to measure lamp resistance in computation would have to take into the consideration, the wide variance in lamp filament resistance at a hot or cold condition. Also, specifically in ref to, “little multimeter”, that tends to infer no capability to measure AC current, as we would need to know the lamps current consumption. A viable alternative would be to reinstate one string back into the initial lamp count and then cut and insert a 1w 1 ohm resistor in series in that line and then slow down the seq speed enough so that a peak reading of the AC voltage reading across that resistor may be taken. Then the lamp current can be computed and utilized in finding an appropriate ballast resistor for a string of fewer lamps. RSVP. <p>73's de Edd
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<p>[ January 29, 2003: Message edited by: Edd Whatley ]</p>
The first thing that comes to mind is the old Wurlitzer’s with the bubble lites.
‘Course it also could be a Seeburg or Rockola.When U specified 3.5 v on your lamps, that could take 35 lamps in series to make one complete string across 120v or account for four separate strings to be utilized across 4 output legs from your sequencer. Are these lamps socketed or just leaded lamps hardwired ?
Certainly I would expect your sequencer to require a constant 120vac supply to its electronics power supply and timing circuits….thus the explanation for the squirrelly operation on reduced supply voltage. If the lamp to lamp spacing is correct for you to string out around the J/Bx wouldn’t the simplest method be just to cache the end of the excess length internal to the J/Bx housing and leave the string alone.
If for some reason that wouldn’t work for you it should be possible to cut down on the lamps in each sequencer string and sub in a fixed wire wound or possibly metal film res to simulate the missing lamps…..so the sequencer still thinks its business as usual. One thing though, I’m a bit fuzzy in your mentioning 5 strings, as I was figuring above, there only being 4 strings of 35 lamps fed from the sequencer at four outputs. However, your included reference to lamps 1 and five being on at the SAME time would account for the three other strings to be on lamp 2 (et al) another string on lamp 3 (et al) and the last string on lamp 4 (et al) thus confirming 4 strings fed from the sequencer The uninitiated might experience some difficulty in computing the value of resistors to utilize in place of the missing lamps on each of the strings .Any attempt to measure lamp resistance in computation would have to take into the consideration, the wide variance in lamp filament resistance at a hot or cold condition. Also, specifically in ref to, “little multimeter”, that tends to infer no capability to measure AC current, as we would need to know the lamps current consumption. A viable alternative would be to reinstate one string back into the initial lamp count and then cut and insert a 1w 1 ohm resistor in series in that line and then slow down the seq speed enough so that a peak reading of the AC voltage reading across that resistor may be taken. Then the lamp current can be computed and utilized in finding an appropriate ballast resistor for a string of fewer lamps. RSVP. <p>73's de Edd
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<p>[ January 29, 2003: Message edited by: Edd Whatley ]</p>Re: Again, reducing voltage for Chrismas lights..
Hmmm...thanks Edd..some good insight there, although it will take my feeble mind a bit to decipher it...<p>I believe you are right. After again examining the very small circuit box, there are two of what I call power wires, then four strings wires, then one wire that ran by itself all the way until connecting up with the last three lights in the original 140 set.<p>On the idea to "hide away" the additional lights...I'd like to keep this as clean as possible, since my projects seem to get messy more than not. This is just an old seventies jukebox--long, rectangular. But it does have an interesting light array, where the original bulbs (14v) are mounted in little "cups"--the leads sticking out over the sides and into "boards". These boards are in groups of four, five, and such, whereas all the lights are "in a row" that looks like three long-legged arches on the top of the unit, and three upside-down arches on the bottom of the unit. I've also considered replacing the few burnt out original bulbs, and repowering the whole thing with another supply. But the way the boards are circuit-wired (traces), the sequencing scheme is a random one where groups come on and off. I want to convert it to the "marquee" effect. I think I could hard wire the originals to duplicate the desired effect, but then I have to come up with a way to power the 59 14v bulbs, with this Chrismas light sequencer unit. (I'm sure you know the kind I'm talking about--the cheap sets you buy at Walmart and the like, with the little plastic box with one small knob to adjust the chase effect)<p>More thoughts?<p>Again thanks.
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Chris Smith
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Re: Again, reducing voltage for Chrismas lights..
Light dimmers require a MINIMUM Wattage which is why you probably cant get the adjusted voltage? Also some brands don’t have the ZERO cross over which allows the device to go below the 50 % voltage rating. Try placing a 25 watt or greater bulb in line with the string [parallel] to see if the minimal load is met and go from there? <p>A minimal load of around 10 to 15 watts is usually the lowest needed to hold the function of the IC from failing.
Re: Again, reducing voltage for Chrismas lights..
Modrob:
Your description of your J Box didn’t register as to brand. I was more familiar with the “Seeburger” line back from ’55-68 but worked on many others also. You might see if your unit is pictured on :
http://www.juke-index.co.nz/seeburg.html
From Ur description, no chance it was a Sunstar was it...one classy unit.
On the Wally World chaser lights the AC line power goes into 2 conns of the unit and then the lamp series strings (4) go to the 4 drive outputs on the sequencer and the last long line you mentioned is the common line, tied to the end of all the strings. Before modifying the string you need to get the data mentioned above (Vac across the 1 ohm test/shunt resistor)to compute lamps series current . If your inter lamp spacing out is ok then seems like all you need to do is shorten each string so there would be 15 X 4 = 60 lamps and then supply data mentioned so the value of 4 series ballast resistors could be computed.
If the lamps have to be inserted in the old lamps socketing , a hot glue gun and some freeze spray could make a pretty fast/simple job of it. Rather than even considering the old system…and those 14V incandescent lamps don’t come cheap nowadays. You’re going to have ~80 spare lamps on hand anyhow for the chasing lamp lines. .<p>
73's de Edd
[email protected] .......(Interstellar~~~~~Warp~~~~~Speed)
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Your description of your J Box didn’t register as to brand. I was more familiar with the “Seeburger” line back from ’55-68 but worked on many others also. You might see if your unit is pictured on :
http://www.juke-index.co.nz/seeburg.html
From Ur description, no chance it was a Sunstar was it...one classy unit.
On the Wally World chaser lights the AC line power goes into 2 conns of the unit and then the lamp series strings (4) go to the 4 drive outputs on the sequencer and the last long line you mentioned is the common line, tied to the end of all the strings. Before modifying the string you need to get the data mentioned above (Vac across the 1 ohm test/shunt resistor)to compute lamps series current . If your inter lamp spacing out is ok then seems like all you need to do is shorten each string so there would be 15 X 4 = 60 lamps and then supply data mentioned so the value of 4 series ballast resistors could be computed.
If the lamps have to be inserted in the old lamps socketing , a hot glue gun and some freeze spray could make a pretty fast/simple job of it. Rather than even considering the old system…and those 14V incandescent lamps don’t come cheap nowadays. You’re going to have ~80 spare lamps on hand anyhow for the chasing lamp lines. .<p>
73's de Edd
[email protected] .......(Interstellar~~~~~Warp~~~~~Speed)
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Re: Again, reducing voltage for Chrismas lights..
AHHH! Before you set something on fire, let's try to fix this right.<p>About the best you can do with an Ohm meter and measuring light bulbs is confirm if the bulb is open or not. The filament of an incandescent lamp changes resistance when the wire becomes incandescent. The filament WILL NOT HAVE THE SAME RESISTANCE WHEN LIT AS WHEN OFF!!! Thus, you CAN'T use the 2.9 Ohm value you found earlier.<p>The internal driver circuit on these lights can be varied. I have seen 4017 CMOS sequencers as well as custom silicon inside Christmas light controllers. The 4017 circuits are usually quite simple and have only a pot to change the speed of the motion. The using a voltage divider it lowers the AC voltage and this is then rectified and fed to the ICs. There are then SCRs/transistors on the output of the IC chip that actually drive the lights. If you lower the voltage feeding the AC plug, you will also lower the voltage running the IC and get what you have already seen, a circuit that does not function correctly.<p>You need to put a power/current limiting resistor in series with the light string, like you suggested. But you did your "reverse engineering" wrong. You said you have five circuits of lights going in sequence. But you also said you see lights 1 and 5 and 2 and 6 etc. are together. You got that part right but figured wrong. Since you repeat at light 5, you have circuits 1, 2, 3, 4, then you repeat circuit 1 at light 5. Get it? If not, think it through.<p>Roughly you have cut your string down to 42% of the original length. This means you must cut your current down by an equal amount. On the original string you had FOUR sets of lights with a total number of 140 lights. Each string or circuit, therefore, has 35 lights on it (divide 140 lights between 4 circuits; 140/4 = 35). You stated that these are 3.5 volt bulbs. Here is WHY you need 3.5 volt bulbs. Each bulb (electrically) is very close, thus each will have the same characteristics when they drop voltage.<p>So the voltage per bulb is 120 Volts divided by the number desired on the string (35) to get:<p>120 Volts/35 bulbs = 3.43 volts per bulb. Thus round off to 3.5 Volt bulbs.<p>
Now you have cut the string down from 140 bulbs to 60 bulbs. If you did this correctly, you will have shortened each string by the same number of bulbs. Thus you will now have 15 bulbs per string. You can do one of two things; replace all the bulbs with a higher voltage bulb (not likely) or add a resistor to limit current. So how do you calculate the resistor? Use Ohms law. I'll work you through it. This is a series circuit. You will drop voltage across each bulb equally (per string). You removed 20 bulbs at 3.5 volts each which means you have to drop 70 volts across a resistor (20 bulbs * 3.5 volts/bulb = 70 volts). Using Ohms law you have<p>E=I*R <p>Rework until you get
R = E/I = 70 volts/I <p>Note that you have one equation and TWO unknowns. This is not good. So what you need to do is figure out what the original current was. You can do this one of two ways; hook up a string of 35 lights and measure the current or hook up 3.5 volts and one bulb and measure the current. (HINT HINT HINT: Two D batteries will provide 3.0 volts and is much safer to work with). If you go the battery method (much safer) you will come close. While the light will not be "quite" as bright, 3 volts VS 3.5 volts is not that big a deal and will get you close enough current value. <p>
So divide the 70 volts by the current to get the OHMS value of your resistor. BUT WAIT; there's more. Multiply the voltage across the resistor by the current to get the minimum WATTAGE rating of the resistor.<p>Go with a larger wattage value of resistor and slightly higher Ohm value.<p>If you are confused, measure the current of either one bulb on 3.5 volts or a full string of 35 bulbs and report your findings. If I do not respond, send me a message directly and I will help out.
Now you have cut the string down from 140 bulbs to 60 bulbs. If you did this correctly, you will have shortened each string by the same number of bulbs. Thus you will now have 15 bulbs per string. You can do one of two things; replace all the bulbs with a higher voltage bulb (not likely) or add a resistor to limit current. So how do you calculate the resistor? Use Ohms law. I'll work you through it. This is a series circuit. You will drop voltage across each bulb equally (per string). You removed 20 bulbs at 3.5 volts each which means you have to drop 70 volts across a resistor (20 bulbs * 3.5 volts/bulb = 70 volts). Using Ohms law you have<p>E=I*R <p>Rework until you get
R = E/I = 70 volts/I <p>Note that you have one equation and TWO unknowns. This is not good. So what you need to do is figure out what the original current was. You can do this one of two ways; hook up a string of 35 lights and measure the current or hook up 3.5 volts and one bulb and measure the current. (HINT HINT HINT: Two D batteries will provide 3.0 volts and is much safer to work with). If you go the battery method (much safer) you will come close. While the light will not be "quite" as bright, 3 volts VS 3.5 volts is not that big a deal and will get you close enough current value. <p>
So divide the 70 volts by the current to get the OHMS value of your resistor. BUT WAIT; there's more. Multiply the voltage across the resistor by the current to get the minimum WATTAGE rating of the resistor.<p>Go with a larger wattage value of resistor and slightly higher Ohm value.<p>If you are confused, measure the current of either one bulb on 3.5 volts or a full string of 35 bulbs and report your findings. If I do not respond, send me a message directly and I will help out.
No trees were harmed in the creation of this message. But billions of electrons, photons, and electromagnetic waves were terribly inconvenienced!
Re: Again, reducing voltage for Chrismas lights..
modjob:
And if all the composite info supplied by us in the above posts still has you "befuzzled", send me 2 lamps (breakage ! ..?) in a bubble wrap mailer and I'll compute it for you and send to you the data required for the ballast resistors ,or them...if I have in stock.Looks like the unit is depicted at:
http://www.jukebox-world.de/Forum/Archi ... tainer.htm
Thats also a colorful unit.
Hit me att my atti addee for more info.<p>
73's de Edd
[email protected] ...(Interstellar~~~~Warp~~~~~Speed)
[email protected]...(Firewalled-Spam*Cookies*Crumbs)
And if all the composite info supplied by us in the above posts still has you "befuzzled", send me 2 lamps (breakage ! ..?) in a bubble wrap mailer and I'll compute it for you and send to you the data required for the ballast resistors ,or them...if I have in stock.Looks like the unit is depicted at:
http://www.jukebox-world.de/Forum/Archi ... tainer.htm
Thats also a colorful unit.
Hit me att my atti addee for more info.<p>
73's de Edd
[email protected] ...(Interstellar~~~~Warp~~~~~Speed)
[email protected]...(Firewalled-Spam*Cookies*Crumbs)
Re: Again, reducing voltage for Chrismas lights..
This is just getting better and better, and drives me to learn more...again I appreciate more than you'll know how you guys come forward to offer your help to us that are less knowledgeable.<p>After perusing over your comments again, I find a couple of things. Now that I know there are only four strings of lights, I believe I cut them a little short (12 instead of 15 per string) and will have to remedy that right away.<p>Now, I offer this that I have done...
I hooked up one 3.5 volt bulb to the output wires of a 3 volt 300ma wall wart (DC..does this make a difference?). Just checking both contacts of wall wart by itself, I read 4.2 volts when the wallwart (adjustable one) is set on 3volts. Then I connect each leg of the bulb to each leg of the wallwart. I then place one probe on one side of the bulb lead, and the other probe on the other side of the bulb lead. I read 3.6 volts. I get this reading only by setting the meter on the DC output section. I get nothing anywhere else. And further still, when measuring "amperage", what exactly does a small meter have to be "set" on? My little Radio Shack digital meter does not have an "amps" setting. So, do these figures help? Or am I still lacking something else? In reading various posts on the Ohms law, I take it that the "*" (asterisk) is to mean "multiply" in all cases?<p>Tim<p>[ February 02, 2003: Message edited by: ModRob ]</p>
I hooked up one 3.5 volt bulb to the output wires of a 3 volt 300ma wall wart (DC..does this make a difference?). Just checking both contacts of wall wart by itself, I read 4.2 volts when the wallwart (adjustable one) is set on 3volts. Then I connect each leg of the bulb to each leg of the wallwart. I then place one probe on one side of the bulb lead, and the other probe on the other side of the bulb lead. I read 3.6 volts. I get this reading only by setting the meter on the DC output section. I get nothing anywhere else. And further still, when measuring "amperage", what exactly does a small meter have to be "set" on? My little Radio Shack digital meter does not have an "amps" setting. So, do these figures help? Or am I still lacking something else? In reading various posts on the Ohms law, I take it that the "*" (asterisk) is to mean "multiply" in all cases?<p>Tim<p>[ February 02, 2003: Message edited by: ModRob ]</p>
Re: Again, reducing voltage for Chrismas lights..
So you know, the reason you get 4.2V from just the two contacts of the wall-wart is that in a power supply circuit, the voltage is slightly higher when there is no load (resistance across the contacts). The meter has an extremely high load (resistance) across the contacts, so it can give you an accurate reading without the effects of placing a low resistance on the power supply.<p>Your "little" meter probably has a mA setting, which is a milli-amps reading. Chances are it is internally fused, and will only read up to 400mA. I'm not sure if that will be exceeded by testing the current (amps) through the lights.<p>You have this adjustable wall-wart, which brings me towards a suggestion.
Instead of being extremely in-efficient and dropping the voltage with a series resistor (disadvantages: resistance will change over time, wastes energy, generates heat), I suggest you use a few transistors to switch the voltage.<p>I'm not sure if you are famillier with the use of transistors, but in your jukebox transistors would make your life a lot easier. On transistors there are three pins: Collector, Base, and Emitter. For your application we'll be talking about an NPN transistor, a part number reference could be 2N3904.<p>The sequencer circuit you have could "drive" the transistors. We'll need a total of 4 of them. I'm assuming the output of the sequencer circuit in the juke has 5 connections, 4 for the differenct circuits of lights, and one common ground. The 4 outputs for the circuits would be connected through a 2.2k ohm resistor to the base of the transistor for that circuit. There would be 4 of these connections, one for each channel of the sequencer.<p>The emitters of the transistors would all be connected through a 10k resistor to ground, and the collectors would be connected to the 3 volt supply. Outputs to the lights would be from the collectors of the transistors.<p>Somebody correct me if I'm wrong with that hook-up, sometimes I get the basic transistor connections mixed up.<p>
You would probably want to choose the series resistor solution, because it is much easier. I'm just letting you know there is an option to use transistors, or even MOSFETS.<p>To measure the current draw of the lights, you would connect one terminal of your meter to the positive wall-wart connections, and the other terminal to the connection of the lamp. The ground terminals of the wall-wart and lamp would be connected together. Of course, your meter is on the mA setting. Chances are you'll probably have to take the black probe out of the COM jack and put it in the mA jack on your meter.<p>Doing that will give you the current draw of one lamp. Multiply this value by 15 bulbs per string:<p>R = 70 volts/I (current)
R = 70 volts/ (measured mA * 15)<p>Remeber that 1 mA is .001 A, and the R=V/I equation is in A, so if you have 100mA it will be plugged into R=V/I as .1<p>
Hopefully this helps, once again, somebody correct me if my thinking on the BJTs is wrong.<p>~Kyle
Instead of being extremely in-efficient and dropping the voltage with a series resistor (disadvantages: resistance will change over time, wastes energy, generates heat), I suggest you use a few transistors to switch the voltage.<p>I'm not sure if you are famillier with the use of transistors, but in your jukebox transistors would make your life a lot easier. On transistors there are three pins: Collector, Base, and Emitter. For your application we'll be talking about an NPN transistor, a part number reference could be 2N3904.<p>The sequencer circuit you have could "drive" the transistors. We'll need a total of 4 of them. I'm assuming the output of the sequencer circuit in the juke has 5 connections, 4 for the differenct circuits of lights, and one common ground. The 4 outputs for the circuits would be connected through a 2.2k ohm resistor to the base of the transistor for that circuit. There would be 4 of these connections, one for each channel of the sequencer.<p>The emitters of the transistors would all be connected through a 10k resistor to ground, and the collectors would be connected to the 3 volt supply. Outputs to the lights would be from the collectors of the transistors.<p>Somebody correct me if I'm wrong with that hook-up, sometimes I get the basic transistor connections mixed up.<p>
You would probably want to choose the series resistor solution, because it is much easier. I'm just letting you know there is an option to use transistors, or even MOSFETS.<p>To measure the current draw of the lights, you would connect one terminal of your meter to the positive wall-wart connections, and the other terminal to the connection of the lamp. The ground terminals of the wall-wart and lamp would be connected together. Of course, your meter is on the mA setting. Chances are you'll probably have to take the black probe out of the COM jack and put it in the mA jack on your meter.<p>Doing that will give you the current draw of one lamp. Multiply this value by 15 bulbs per string:<p>R = 70 volts/I (current)
R = 70 volts/ (measured mA * 15)<p>Remeber that 1 mA is .001 A, and the R=V/I equation is in A, so if you have 100mA it will be plugged into R=V/I as .1<p>
Hopefully this helps, once again, somebody correct me if my thinking on the BJTs is wrong.<p>~Kyle
Re: Again, reducing voltage for Chrismas lights..
Whew! I'm stepping out here without a safety net fellas...here goes:<p>Now, after using a different cheapie meter, (it had a DCA catagory, and a "ma" probe jack) I measured again. See if this makes sense to you. Using the 3volt dc wallwart, I measured the lamp with the meter selector knob on DCA 200m setting. This allowed the bulb to now light, and I came up with a figure on the meter of 136.4.
Using your equations, I divide 70 by 136, and end up with .51, multiplied by 15 which gives me 7.65...does this mean rounding up to a 10ohm resistor, or am I still way off? And then if I multiply the voltage by the .51 current figure to find the wattage needed, I get 35.7? That sounds like an odd figure to me....???
Using your equations, I divide 70 by 136, and end up with .51, multiplied by 15 which gives me 7.65...does this mean rounding up to a 10ohm resistor, or am I still way off? And then if I multiply the voltage by the .51 current figure to find the wattage needed, I get 35.7? That sounds like an odd figure to me....???
Re: Again, reducing voltage for Chrismas lights..
I think your confusing yourself.<p>Try drawing a set of lights as you go.<p>Now lets look at this a different way and we'll say hello to that nice man Mr Kirchoff (whose motto was ohm sweet ohm, don't worry it gets worse).<p>You have already discovered that one bulb @ 3v approx, draws 137mA (round up it gets you a safety margin)<p>#1 1*3v Bulb uses 137mA<p>Mr K says that in a series circuit (which you have in one string) the SAME current will flow through all resistances (in this case your bulbs) and there will be a voltage drop across each of your resistances/bulbs. All the voltage drops when added back up will equal the voltage going into the circuit - a case of what goes in must come out! <p>Now to work out your dropper resistor from 120v.<p>As you have already seen 15 bulbs times 3.5v=53v
Your mains power supply is 120v therefore you need to drop <p>#2 120v-53v =67v <p>Now as we have just scaled the voltage up by 15 we also have to do the same with the current<p>#3 15*0.137A or 137mA = 2A<p>Dropper resistor = Vdrop/I<p>#4 67/2 =33.5 nearest standard value 33 or 35 ohms<p>Power dissipated = Pdis=v**/R or I** *R or V*I<p>#5<p>67**(v)/33(R) = 136W
2**(I) *33(R) = 132W
67(v)*2(I) = 134W<p>The differences are due to rounding in the above figures. **=squared.<p>So you need a resistor of 33 ohms at 136W good grief!<p>You will need to parallel some resistors to get that wattage.<p>Assuming you can get hold of 20W resistors divide 136W/20W = 6.8 resistors round up 7 resistors and you have some spare capacity.<p>Multiply 33 ohms by your 7 resistors and you'll get after rounding 231 ohms.<p>Probably better off having 10 resistors at 330 ohms rated at 15 or 20W.<p>A Transformer would be ebtter from a heat point of view as well as isolation.<p>Colin<p>MR Kirschoff one of those damn people who insisted on laying down a law or two<p>PS If you're feeling rich www.rockby.com.au have packs of 10 560R 20W resistors for A$2.80 plus P&P<p>[ February 02, 2003: Message edited by: bodgy ]</p>
<p>Now to work out your dropper resistor from 120v.<p>As you have already seen 15 bulbs times 3.5v=53vYour mains power supply is 120v therefore you need to drop <p>#2 120v-53v =67v <p>Now as we have just scaled the voltage up by 15 we also have to do the same with the current<p>#3 15*0.137A or 137mA = 2A<p>Dropper resistor = Vdrop/I<p>#4 67/2 =33.5 nearest standard value 33 or 35 ohms<p>Power dissipated = Pdis=v**/R or I** *R or V*I<p>#5<p>67**(v)/33(R) = 136W
2**(I) *33(R) = 132W
67(v)*2(I) = 134W<p>The differences are due to rounding in the above figures. **=squared.<p>So you need a resistor of 33 ohms at 136W good grief!<p>You will need to parallel some resistors to get that wattage.<p>Assuming you can get hold of 20W resistors divide 136W/20W = 6.8 resistors round up 7 resistors and you have some spare capacity.<p>Multiply 33 ohms by your 7 resistors and you'll get after rounding 231 ohms.<p>Probably better off having 10 resistors at 330 ohms rated at 15 or 20W.<p>A Transformer would be ebtter from a heat point of view as well as isolation.<p>Colin<p>MR Kirschoff one of those damn people who insisted on laying down a law or two<p>PS If you're feeling rich www.rockby.com.au have packs of 10 560R 20W resistors for A$2.80 plus P&P<p>[ February 02, 2003: Message edited by: bodgy ]</p>
On a clear disk you can seek forever.
Re: Again, reducing voltage for Chrismas lights..
Good grief is right! So glad you guided me through the figuring...there is a whole lot more to it than just buying a couple of cheap resistors and two minutes work...wow...<p>Transformer? Hmmm...Yes, I've been reading about them too, and have the basic understanding of it...lets look into that method...<p>Again, any ideas and instructions appreciated...
Re: Again, reducing voltage for Chrismas lights..
I do think a really good way would be using transistors, like I said earlier.<p>If you want to use a bunch of resistors in parallel and series, then AllElectronics would be a good source. They are really inexpensive for those types of things.<p>www.allelectronics.com<p>4 TIP140 NPN transistors could be used on the outputs of the sequencer circuit to switch the voltage if you want to go that way.<p>Otherwise, a transformer could be used to reduce the voltage from 120V to what was it....53V?
Re: Again, reducing voltage for Chrismas lights..
modjob:<p>You probably would have been better off using a current stable power source of 2 series “D’ cells as a solid 3Vdc power supply to feed your lamp under test. FIO, the typical fresh alkaline D cell is capable of short time bursts of in excess of 8 amps into a load. This in contrast to the specified Wall Wart with its outputs variance between the unloaded and loaded states.<p>Hopefully you didn’t have your own computed/or/ referenced,elsewhere, ballast resistor(s) available, as you would have been blinded by the flash by virtue of them being of a too low of a resistive value.
Consult the labeled/molded specs on the green controller unit to verify the units power consumption. I would expect its initial specs to be in the order of the equivalent consumption of the controller electronics plus one string being on continually (as power is just being walked, in order, thru four different strings and then repeating the seq.)
Ergo: 1 series strings of 35 lamps would be across ~120 V and IF your computation of 137 ma per lamp is right, that would be a 1 string consumption of 120 V X .137 A = 16.44 W plus whatever the controller itself uses<p>Looking to the figures that were supplied to you ,all figures are in accordance, that is, until just after the correct 67 V to drop figure… then the 134W !!!…now either Reverse Polish Logic or Fuzzy Math must have crept in, as 67V /.137 A= 489 ohms or use a 500 ohm for lamp life consideration…if they’re bright enough for you.
Since that resistor value is also going to have 137 ma running thru it, like ALL of the the series lamps. Its power rating would be its dropping voltage of 67 X its current passing thru it of .137 A = 9.17W or use a 10W to keep cool. The latter reasoning is the fact that the res is not in continual steady on use, ,only 25% of the time with the time sharing with the other strings. Effectively each string alternatively is only seeing an equivalent 2.5 Watts of dissipation.<p>So overall….only 15 lamps in each of the four strings and a ballast res at the end of each string and the loose ends of the four resistors joined and connected to the common wire that was at the end of each string when it was 35 per string.<p>As for the solid state, series variable PRF PRT technique , don’t think it would justify a paltry 2.5 W savings, particularly on line powered equipment.<p>As U said “Good Grief…Charlie Brown” this simple query/post will soon be approaching the hits of the infamous “ eccentric wheel “<p>73's de Edd
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Consult the labeled/molded specs on the green controller unit to verify the units power consumption. I would expect its initial specs to be in the order of the equivalent consumption of the controller electronics plus one string being on continually (as power is just being walked, in order, thru four different strings and then repeating the seq.)
Ergo: 1 series strings of 35 lamps would be across ~120 V and IF your computation of 137 ma per lamp is right, that would be a 1 string consumption of 120 V X .137 A = 16.44 W plus whatever the controller itself uses<p>Looking to the figures that were supplied to you ,all figures are in accordance, that is, until just after the correct 67 V to drop figure… then the 134W !!!…now either Reverse Polish Logic or Fuzzy Math must have crept in, as 67V /.137 A= 489 ohms or use a 500 ohm for lamp life consideration…if they’re bright enough for you.
Since that resistor value is also going to have 137 ma running thru it, like ALL of the the series lamps. Its power rating would be its dropping voltage of 67 X its current passing thru it of .137 A = 9.17W or use a 10W to keep cool. The latter reasoning is the fact that the res is not in continual steady on use, ,only 25% of the time with the time sharing with the other strings. Effectively each string alternatively is only seeing an equivalent 2.5 Watts of dissipation.<p>So overall….only 15 lamps in each of the four strings and a ballast res at the end of each string and the loose ends of the four resistors joined and connected to the common wire that was at the end of each string when it was 35 per string.<p>As for the solid state, series variable PRF PRT technique , don’t think it would justify a paltry 2.5 W savings, particularly on line powered equipment.<p>As U said “Good Grief…Charlie Brown” this simple query/post will soon be approaching the hits of the infamous “ eccentric wheel “<p>73's de Edd
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Re: Again, reducing voltage for Chrismas lights..
Hmmm, something seems wrong here. I see what you measured 136.4 mA, call it 137 mA or better yet 140 mA. Then bodgy did some calculating that lost me. I do not see why bodgy scaled up the current; this will not change in a series circuit. The "resistance" of each bulb at 3.5 volts and having a 140 mA draw will be:<p>R = E/I
R = 3.5V/0.14A = 25 Ohms<p>Thus you must sub in a 25 Ohm resistor for EACH light you removed. Since you removed 20 lights in each string, you need 20 resistors. This becomes a 500 Ohm resistor.<p>Power can be expressed in a number of ways. But the main formula is P=I*E (remember PIE). But you can sub in values of Ohms law. For instance, E=I*R so you can get:<p>P = I * (I*R)
P = I^2 R
P = (0.14)^2 * 500 = 9.8 = 10 watts.<p>Think of it this way. A one hundred watt light bulb puts out quite a lot of heat; so much so you can't even hold it. You can hold on to all those little lights. But let's check our calculations.<p>We know the series current is 137 mA rounded to 140 mA.<p>We know that since the original string contained 35 bulbs and we shortend this to 15 we lost 20 bulbs. Thus we lost 58 percent of the string. Mulitplying out we find that 58% of 120 volts is 69.6 or 70 volts. So we have a 70 volt drop and a 0.14 Amp draw. This is:<p>R = E/I = 70v/0.14A = 500 Ohms.<p>We can still use PIE or P = I*E as we have all of this solved for us.<p>P = I*E
P = 0.14A * 70 Volts
P = 9.8 Watts.<p>This IS a good sized resistor, but not impossible. You could put ten one watt 50 Ohm resistors in series. This would be a marginal design but would work.<p>There are some other possibilities. Take a C7 or similar Christmas light and put it in series with the mini lights. This will limit the current. If one is not enough, use two. <p>The transistor idea is excellent, but I did not want to confuse you with this. But it gave me another idea that you might use. Simply put a two amp diode in series with the lights. Do NOT PUT THIS on the input to the AC line but to the output to the string. In fact you could proably get away with using ONE diode in the common line. If you do this, use two two amp diodes in parallel. (This is not a very "proper" thing to do but would be okay for just Christmas lights.) Be sure to get a four hundred peak voltage diode. This will cut the voltage in half as you would get only one half of the AC cycle. Depending on what type of device is switching the lights on and off in the controller, you may need to reverse the polarity of the diode to get the circuit to work. Something like Radio Shack # 276-1114 should work nicely.<p>[ February 03, 2003: Message edited by: Jolly Roger ]</p>
R = 3.5V/0.14A = 25 Ohms<p>Thus you must sub in a 25 Ohm resistor for EACH light you removed. Since you removed 20 lights in each string, you need 20 resistors. This becomes a 500 Ohm resistor.<p>Power can be expressed in a number of ways. But the main formula is P=I*E (remember PIE). But you can sub in values of Ohms law. For instance, E=I*R so you can get:<p>P = I * (I*R)
P = I^2 R
P = (0.14)^2 * 500 = 9.8 = 10 watts.<p>Think of it this way. A one hundred watt light bulb puts out quite a lot of heat; so much so you can't even hold it. You can hold on to all those little lights. But let's check our calculations.<p>We know the series current is 137 mA rounded to 140 mA.<p>We know that since the original string contained 35 bulbs and we shortend this to 15 we lost 20 bulbs. Thus we lost 58 percent of the string. Mulitplying out we find that 58% of 120 volts is 69.6 or 70 volts. So we have a 70 volt drop and a 0.14 Amp draw. This is:<p>R = E/I = 70v/0.14A = 500 Ohms.<p>We can still use PIE or P = I*E as we have all of this solved for us.<p>P = I*E
P = 0.14A * 70 Volts
P = 9.8 Watts.<p>This IS a good sized resistor, but not impossible. You could put ten one watt 50 Ohm resistors in series. This would be a marginal design but would work.<p>There are some other possibilities. Take a C7 or similar Christmas light and put it in series with the mini lights. This will limit the current. If one is not enough, use two. <p>The transistor idea is excellent, but I did not want to confuse you with this. But it gave me another idea that you might use. Simply put a two amp diode in series with the lights. Do NOT PUT THIS on the input to the AC line but to the output to the string. In fact you could proably get away with using ONE diode in the common line. If you do this, use two two amp diodes in parallel. (This is not a very "proper" thing to do but would be okay for just Christmas lights.) Be sure to get a four hundred peak voltage diode. This will cut the voltage in half as you would get only one half of the AC cycle. Depending on what type of device is switching the lights on and off in the controller, you may need to reverse the polarity of the diode to get the circuit to work. Something like Radio Shack # 276-1114 should work nicely.<p>[ February 03, 2003: Message edited by: Jolly Roger ]</p>
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