About a month ago my LPL 4500 blew a bulb, so I replaced it and have been printing since with no problem. Today I made one print and the bulb blew, I replaced it and the replacement immediately blew. I’m going to get a friend over who is good with electronics but was wondering if anyone has had similar happen to their 4500 and might point me to what the culprit might be. It is hooked up to a Analyser pro.
LPL 4500 blowing bulbs
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When the power supply on mine failed, it wasn’t getting full voltage to the bulb. But, things could go either way. Checking things out with a voltmeter will help tell the tale.
A string of bad luck with bulbs is a possibility too.
A string of bad luck with bulbs is a possibility too.
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I’ll have to find my manual but I think mine might be the model that takes 200w bulbs, I can only get 250w bulbs so that may be the issue, who knows.
I just realised that the 2nd lamp I blew today was a 100w lamp that I bought as a spare for my little LPL enlarger! My last spare, another 250w works without blowing! Can anyone point me to a source for the 200w lamps? B&H only have the 250w lamps and Vanbar in Australia only have 250w. KHB in Canada have the LPL 200w ones but their price with shipping is off the charts.
If OP's power supply is similar in rating to mine, it's rated to output up to 250 Watts. And, the XL version of the enlarger uses a 250 W bulb. So, in theory the bulb should be compatible. I bought some 250W bulbs for mine a while back, but haven't tried using them yet, as I still have a couple 200's.
If yours takes the 200w 82v EYA lamp, Adorama seems to have them;
https://www.adorama.com/lmeya.html
If yours takes the 200w 82v EYA lamp, Adorama seems to have them;
https://www.adorama.com/lmeya.html
The lamp socket may be bad. Heat over the years weakens the connectors causing the bulb to draw more current at the socket than it should.
If someone could explain to me how bad contacts can make a load draw more current than with good contacts, I'd be very willing to be blown away.
I put my money on the somewhat more logical mechanism of bad contacts essentially making the bulb flicker, which implies rapid heating & cooling of the filament causing severe stress and reduced bulb lifetime. Ever noticed how bulbs tend to blow as you turn them on, but not while they've been on already for a few seconds or more? There you go.
Maybe the OP just experienced a lamp going out and erroneously reported this as "blown".
Anyway, bad contacts should be taken into account at any failure. As inapt mains voltage in case the power supply is not stabilized.
On my Omegas the triac setting the 80v is connected to the mains (120v).
A bad lamp socket deranges the voltage control circuit , allowing the socket to get the whole 120v sine wave and this blows the lamp when it arcs across the bad connection.
A bad lamp socket deranges the voltage control circuit , allowing the socket to get the whole 120v sine wave and this blows the lamp when it arcs across the bad connection.
A 250 watt bulb will pull 3.125 amps of current at 80volts. At 78 volts it will draw 3.2 amps. At 60 volts it will draw 4.17 amps.
A bad/weak socket will not provide the proper power to the bulb which will try to draw its rated power anytime power is applied to it.
A 20% tolerance increase in current is 3.7 amps which is quite likely at or exceeding the filament limit of the bulb.
Power = Volts * Amps. Amps = power / volts. The higher the current gets the faster the bulb burns out and the hotter it gets until it burns out.
A higher voltage from a malfunctioning power supply will exceed the bulbs components limits and cause them to break down resulting in the bulb burning out.
Edit: If you take standard 18 gauge lamp cord wire and 3 strands are cut off its still usable as it will carry the current it is rated for, cut off 4 or more strands it now only carries the current load of a 20 gauge wire.
Basic electronics 101 for electricians and service techs.
A bad/weak socket will not provide the proper power to the bulb which will try to draw its rated power anytime power is applied to it.
A 20% tolerance increase in current is 3.7 amps which is quite likely at or exceeding the filament limit of the bulb.
Power = Volts * Amps. Amps = power / volts. The higher the current gets the faster the bulb burns out and the hotter it gets until it burns out.
A higher voltage from a malfunctioning power supply will exceed the bulbs components limits and cause them to break down resulting in the bulb burning out.
Edit: If you take standard 18 gauge lamp cord wire and 3 strands are cut off its still usable as it will carry the current it is rated for, cut off 4 or more strands it now only carries the current load of a 20 gauge wire.
Basic electronics 101 for electricians and service techs.
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Not at all.
The voltage is needed to make the current run through the filament. Thus with lower voltage the current will go down.
Enlarger lamps are already designed to run beyond the limits applied for standard lamps. This already drastically reduces their liftemine. There is hardly any more tolerance against even higher current.
I was referring to Incandescent bulbs not CFL or LED bulbs.
Here you got 2 graphs:
The upper one shows the case of cold (classic) resistors. The second one the case of incandescent resistors, where the current is steeply rising at the start.
Both graphs show what I stated above.
(Current at the vertical, voltage at the horizontal)
http://dschirdewahn.de/Gluebirne_Ohm.pdf
The upper one shows the case of cold (classic) resistors. The second one the case of incandescent resistors, where the current is steeply rising at the start.
Both graphs show what I stated above.
(Current at the vertical, voltage at the horizontal)
http://dschirdewahn.de/Gluebirne_Ohm.pdf
Unfortunately that’s not the 24v one that I need.
shutterfinger, your electronics 101 is a little different from mine. In the time I spent composing this post a number of other people have chimed in. The below is in response to post #11.
A light bulb does not draw constant power. By your argument, that 250W bulb would draw 250 amps with 1V across it, which is crazy. A more realistic value is around half an amp for a traditional incandescent (non-halogen, like what Thomas Edison invented) bulb with a tungsten filament. The below discussion relates to standard incandescent bulbs. Halogen bulbs may have different behavior (or may not, I simply don't know).
Tungsten has a temperature coefficient of around +4500ppm/degreeK. The operating temperature is around 3000K, room temperature is around 300K, so the filament resistance increases by roughly a factor of
1 + ((3000-300)*4.5e-3) = 13.15.
That 250W bulb has a filament resistance of 25.6 ohms at operating equilibrium when powered with 80V (either 80VDC or 80Vrms), so working backwards gives a room-temperature filament resistance around 2 ohms. Put 1V across it, there's half an amp, like I said above. These numbers are obviously rough approximations, but they're not unrealistic.
The actual behavior of lightbulbs is surprisingly complex and nonlinear. (I saw a full derivation of this in a lecture when I was in EE grad school back in the early 1980s but those notes are long gone. I wish I still had them, it was a thing of beauty that I've never seen reproduced anywhere else.) In simplified form, it goes like this: The filament is "cold" (at room temperature, 300K) when power is first applied, so it has low resistance and draws a lot of current, much more than it does when it reaches operating temperature. For a very brief time following turn-on the power is actually really high - that 250W bulb with 80V applied will initially draw ~40 amps, an instantaneous power of ~3200W. This huge initial surge causes the filament temperature to rise very rapidly. Filament resistance and power both decrease as the temperature rises. Eventually (and surprisingly quickly, maybe 1/20 of a second) the filament reaches its equilibrium operating temperature with the bulb consuming its rated current of 3.125 amps and dissipating its rated power of 250W.
Decreasing the voltage decreases the operating temperature. This has two effects - it prolongs bulb life, and it lowers the color temperature, making the light more yellowish. While it's true that the filament resistance goes down with reduced voltage, which will increase the current slightly, the effect isn't enough to keep the power dissipation constant. Power and temperature both decrease, giving the color and lifetime changes I mentioned.
You can see this in printing if you have an enlarger with a standard bulb and no voltage regulator, and if you power it from a variable power transformer (variac). Make a black-and-white print on VC paper using the rated supply voltage (120V, or whatever is your local normal). Now turn the voltage down 10% and make a matching print - you'll need to use a higher-contrast filter to do so. In color printing the color balance will change as you change lamp voltage; color enlargers often use regulated power supplies to keep the voltage constant and avoid this problem.
I don't see how a bad socket can cause a bulb to blow. It could cause other problems, like shorting out the supply, though, or increasing in resistance so the bulb can't draw full power.
None of this addresses the OP's issue, but the "explanation" in post 11 is so wrong that I couldn't let it go unchallenged.
A light bulb does not draw constant power. By your argument, that 250W bulb would draw 250 amps with 1V across it, which is crazy. A more realistic value is around half an amp for a traditional incandescent (non-halogen, like what Thomas Edison invented) bulb with a tungsten filament. The below discussion relates to standard incandescent bulbs. Halogen bulbs may have different behavior (or may not, I simply don't know).
Tungsten has a temperature coefficient of around +4500ppm/degreeK. The operating temperature is around 3000K, room temperature is around 300K, so the filament resistance increases by roughly a factor of
1 + ((3000-300)*4.5e-3) = 13.15.
That 250W bulb has a filament resistance of 25.6 ohms at operating equilibrium when powered with 80V (either 80VDC or 80Vrms), so working backwards gives a room-temperature filament resistance around 2 ohms. Put 1V across it, there's half an amp, like I said above. These numbers are obviously rough approximations, but they're not unrealistic.
The actual behavior of lightbulbs is surprisingly complex and nonlinear. (I saw a full derivation of this in a lecture when I was in EE grad school back in the early 1980s but those notes are long gone. I wish I still had them, it was a thing of beauty that I've never seen reproduced anywhere else.) In simplified form, it goes like this: The filament is "cold" (at room temperature, 300K) when power is first applied, so it has low resistance and draws a lot of current, much more than it does when it reaches operating temperature. For a very brief time following turn-on the power is actually really high - that 250W bulb with 80V applied will initially draw ~40 amps, an instantaneous power of ~3200W. This huge initial surge causes the filament temperature to rise very rapidly. Filament resistance and power both decrease as the temperature rises. Eventually (and surprisingly quickly, maybe 1/20 of a second) the filament reaches its equilibrium operating temperature with the bulb consuming its rated current of 3.125 amps and dissipating its rated power of 250W.
Decreasing the voltage decreases the operating temperature. This has two effects - it prolongs bulb life, and it lowers the color temperature, making the light more yellowish. While it's true that the filament resistance goes down with reduced voltage, which will increase the current slightly, the effect isn't enough to keep the power dissipation constant. Power and temperature both decrease, giving the color and lifetime changes I mentioned.
You can see this in printing if you have an enlarger with a standard bulb and no voltage regulator, and if you power it from a variable power transformer (variac). Make a black-and-white print on VC paper using the rated supply voltage (120V, or whatever is your local normal). Now turn the voltage down 10% and make a matching print - you'll need to use a higher-contrast filter to do so. In color printing the color balance will change as you change lamp voltage; color enlargers often use regulated power supplies to keep the voltage constant and avoid this problem.
I don't see how a bad socket can cause a bulb to blow. It could cause other problems, like shorting out the supply, though, or increasing in resistance so the bulb can't draw full power.
None of this addresses the OP's issue, but the "explanation" in post 11 is so wrong that I couldn't let it go unchallenged.
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MattKing
Moderator
That is an EJL lamp, is it not?
MattKing
Moderator
Wide tolerance in electronics is 20%, tight tolerance is 1% or less. The power formula applies in this range. The materials used in the lamps plays an important part in their operating characteristics.
I do not have a variable AC or DC power supply to create a video with to prove my point. A socket that does not make proper contact with the bulb will shorten its life.
Open your enlarger and cut the wires going to the bulb so that only half of their diameter is now supplying power to the bulb and see what happens. A socket whose metal contacts have fatigued from heat and age will respond similarly.
I speculate that most enlargers are designed with a 5% voltage/current variance.
I do not have a variable AC or DC power supply to create a video with to prove my point. A socket that does not make proper contact with the bulb will shorten its life.
Open your enlarger and cut the wires going to the bulb so that only half of their diameter is now supplying power to the bulb and see what happens. A socket whose metal contacts have fatigued from heat and age will respond similarly.
I speculate that most enlargers are designed with a 5% voltage/current variance.
You must mean Sir Joseph Wilson Swan...
He also patented the Carbon Printing Process we use today.
Look, it's nice that you're trying to help and all, but when it comes to electronics, you're evidently clueless. I'm not being mean, it's just what it is. There's nothing bad about not knowing much about a topic, but could you please just stop writing such blatant nonsense? I don't really care myself, but future visitors might stumble across your posts and be led astray (possibly dangerously as illustrated in the quote above).
The one thing we can agree on is that a faulty socket can create problems in enlargers, so it's worthwhile checking it.
I was a top notch electronics technician for 30 years.
I have intimated more than one engineer with solid troubleshooting/problem solving skills.
What is your experience?
I have intimated more than one engineer with solid troubleshooting/problem solving skills.
What is your experience?
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