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So last year I did a Christmas lights project and learned that cutting off a couple christmas bulbs off a $2 string of 100 bulbs is not so straightforward, it's practically rocket science. Maybe not "rocket science" per se but it takes some serious knowledge of things like ohms and voltage and resistors and diodes etc - which if you don't have a formal engineering education may be challenging to grasp without learning the fundamentals first.

You can't cut even a few bulbs off because if you do it will increase the load on all the remaining bulbs which will significantly shorten their lifespan, and potentially increase fire risk due to overloading voltage on each bulb. This is because each bulb effectively functions as a resistor and the whole string is designed with the bulbs as resistors, specially made so the string works. But it isn't so easy to shorten or lengthen.

I learned about using resistors and successfully used some, however resistors get extremely hot and for that reason I decided to take down the shortened strings.

It was suggested that I could use diodes to step down the voltage rather than use resistors to "soak up" (my term) the excess heat and voltage from the missing bulbs.

However I still didn't really understand what diodes are or why they don't get hot or how I can actually use these, in for example a string of christmas lights to remove say 15 bulbs from a string of 100.

I previous asked this question last year: Is a diode an effective way to replace a resistor in an application where heat should be avoided? for which I accepted an answer however I still don't fully understand diodes and if and how to use a diode in such a project.

For example, the answer given was:

Using the diode however switches off the voltage and power by 50% If this reduced power is acceptable consider reversing the diode in one string to balance the load current.

However, I don't really understand this. So here are my questions:


======= Questions ========

  • How does the diode reduce the voltage without getting hot while the resistor gets extremely hot?
  • In what way can I use a diode in such a project? It still isn't clear to me how to use a diode in this kind of project.
  • What are some other practical usages for small projects in which a diode can be used (for a beginner)?

Christmas is over again but for purely educational reasons I would really like to have a comprehensive grasp of the best solutions for such a project. I would like to try more projects in the future.


Update:

I understand it more now (although not entirely). With the answer below and the following two youtube video I just watched now I understand a little better.

Basically, a resistor is more like a sponge whereas a diode is more like a wall. While a resistor absorbs current and releases it as heat, a diode on the other hand will simply reject all the current going in one direction (up to the capacity of the diode).

If you use a diode too small, it is like driving a car into a wall made of 2x4. The car will go through because it has more power than the wall (actual diodes will explode when the current is too high for it, as can be seen in various youtube videos). Likewise, a diode that is big enough is like a large brick wall. A car hitting it will be stopped and the wall will be pretty much unaffected.

This isn't a perfect analogy but it's helpful to me (I just made up this analogy).

Here are the two videos on diodes I watched which helped:

https://www.youtube.com/watch?v=6lep5e3KMdY

https://www.youtube.com/watch?v=TFgWDcBp-uY

These don't answer all my questions but they do mostly answer why a resistor heats up but a diode does not. A diode doesn't absorb the energy it simply blocks it. A resistor absorbs the energy which is then released as heat.

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  • \$\begingroup\$ what kind of a light string are you talking about? \$\endgroup\$ – jsotola Dec 31 '17 at 23:45
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    \$\begingroup\$ You should not encourage that user by accepting gibberish answers that goes above the head of the people asking. You should not feel pressured to accept an answer that doesn't solve your problem. \$\endgroup\$ – pipe Dec 31 '17 at 23:50
  • \$\begingroup\$ @jsotola it was a 100 bulb string of standard christmas lights from walmart. \$\endgroup\$ – hbsrnbnt Dec 31 '17 at 23:58
  • \$\begingroup\$ @pipe you're probably right, while I kind of understood their answer it didn't fully answer that original question either. \$\endgroup\$ – hbsrnbnt Dec 31 '17 at 23:59
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    \$\begingroup\$ @karlschmid Why do you think it's a LED string? So far the user only indicates that it is a string of incandescent bulbs. \$\endgroup\$ – pipe Jan 1 '18 at 0:17
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The suggested diode method cuts off another direction current in the alternating current circuit. That reduces the effective power 50% because half of the time the current is is blocked as effectively as the wire was disconnected. As commented, that's not 50% reduction of the effective voltage, it's less because the power is proportional to the square of the voltage in resistors.

You probably do not see anything more than some brightness loss because the frequency probably is 60Hz where you live and every bulb will get current half of every 1/60th second. The filaments in the bulbs do not get cold that fast.

To keep the brightness intact you should remove half of the bulbs of one series if you add a diode. It's difficult to pretend how the lifetime of the bulbs suffers due the changed shape of the current pulses.

The diode must stand the peak AC voltage and the current.

The diode does not get hot because it works like a switch. That's true, if the diode is selected for the load current and can be cooled freely by the air.

Serial resistor doesn't block the current totally, both halves of the AC get through, but part of the total power isn't dissipated in bulbs, it's dissipated in the resistor.

The resistor generates heat. If the resistor is selected properly using Ohm's law it dissipates electric energy to heat exactly as much as the bulbs which are replaced by that resistor. If the resistor has about the same dimensions as one bulb (=same ability to get cooled by the air), but dissipates worth of ten bulbs, it surely gets hot because its temperature must rise 900% higher over the ambient temperature than one bulb.

This text is valid only for incandescent bulbs which have a filament. Led lights work totally differently.

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  • \$\begingroup\$ I wouldn't say right out that "the diode does not get hot", it could still dissipate to a watt or two on average. That would get uncomfortably hot for a small diode. \$\endgroup\$ – pipe Jan 1 '18 at 0:15
  • \$\begingroup\$ can you fix this which appears to be a typo: but it dissipates worth of the removed bulbs thanks, I didn't understand this part. \$\endgroup\$ – hbsrnbnt Jan 1 '18 at 0:21
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    \$\begingroup\$ A diode does not reduce the RMS voltage by 50%, it reduces it by 29%. \$\endgroup\$ – Spehro Pefhany Jan 1 '18 at 0:35
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    \$\begingroup\$ @SpehroPefhany thank you, I understand why some people thought 50%, it is because it blocks the current in one direction. However, what you are saying is that even though half the current is blocked, this only reduces the voltage by less than a third (29%). Right? \$\endgroup\$ – hbsrnbnt Jan 1 '18 at 0:57
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    \$\begingroup\$ @pipe Yes, the power is reduced by 50% so the voltage must be reduced by \$1-\frac{1}{\sqrt{2}}\$. You can get the same answer by calculating RMS analytically. \$\endgroup\$ – Spehro Pefhany Jan 1 '18 at 1:28

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