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New to Electronics, Need Purchasing help/direction
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New to Electronics, Need Purchasing help/direction

by BellezaDeDiosa on Wed Oct 10, 2018 7:17 pm

Hi, I am new to using electronics, I've only ever added twinkle lights to halloween costumes, and I really want to up my game... So here I am... My son want's to be a zombie for halloween with an "eyeball falling out." I was going to construct an eyeball using a "pingpong ball" (I've found ones that are so cheap and thin they would probably dent the first time they are hit, and a standard Christmas light glows nicely thought it) . I wanted to give it a light coat of neon/black light reactive paint and I thought inserting a UV/UVA 400nm Purple LED 5mm Clear Lens inside the ball would give it a radio active kind of glow. My question... I've see you connect these LED's directly to a breadboard, but I'm wondering if you can attach a lead wire to them so that I can be powered/controlled not directly at the site of the LED, but off to the side and be able to be tucked out of site? I apologize for my lack of knowledge and terms. Any impute is greatly appreciated.

https://www.adafruit.com/product/1793

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Re: New to Electronics, Need Purchasing help/direction

by adafruit_support_mike on Thu Oct 11, 2018 5:25 am

No worries.. we all start by learning the basics, and this is a friendly place to do it.

Yes, you can mount an LED on a pair of wires, with the battery and current-limiting resistor somewhere else. The breadboard is a fast way to test circuits without having to solder things together, but wire and solder are more common in an actual build.

These tutorials cover the basics of working with LEDs, and putting LEDs in costumes:

https://learn.adafruit.com/all-about-leds/overview
https://learn.adafruit.com/lets-put-leds-in-things

For the kind of thing you've described, you'll probably find silicone-insulated wire easiest to work with:

https://www.adafruit.com/category/473

It's very flexible, so the eye can flop around instead of looking like it's on the end of a stick. It's also a good idea to cover the solder joints and exposed leads with heat-shrink tubing:

https://www.adafruit.com/product/1649

That does two things: first, it covers any exposed metal and prevents short circuits. Second, it provides strain relief for the solder joint. Silicone-insulated wire may be flexible, but solder joints aren't. The point where the wire meets the solder is a hinge, and it will break if it flexes too many times. Heat-shrink absorbs the strain and keeps the hinge point from moving too much, making the joint last much longer.

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Re: New to Electronics, Need Purchasing help/direction

by BellezaDeDiosa on Thu Oct 11, 2018 3:42 pm

Thank you so much for your help! I have the UV 5mm led's in my cart, along with the heat shrink, and red and black wire that was listed.

Just a few more questions. I have this battery case in my cart as well. https://www.adafruit.com/product/770 . I'm hoping 2AA batteries will be enough to power one LED for a few hours of trick or treating.

Now.... for the resistor. I believe I have KVL figured out. I have 2AA batteries at 1.5V each for a total of 3V, then my LED that uses 2.2V, so I need a resistor that uses at lease.8V. Correct? I get totally lost on Ohm's Law, so I'm stuck and don't know what resistor I need :-( . If you could put a link in to what resistor I need that would be wonderful, but if you could also show me how to figure out that part so hopefully next time I'll be able to do it on my own.

Thank you so much!
Last edited by BellezaDeDiosa on Fri Oct 12, 2018 1:57 am, edited 1 time in total.

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Re: New to Electronics, Need Purchasing help/direction

by adafruit_support_mike on Fri Oct 12, 2018 1:41 am

You'll need a slightly bigger battery pack.

The UV LEDs won't light until the voltage across them is about 3.4V.. you can think of that as a "you must be this tall to ride" requirement. Any battery pack you use has to be able to provide at least that much voltage.

AA cells run about 1.5V, so two of them in series will give you about 3V.. still to short for the UV-LED ride. You need at least one more cell to cross the minimum voltage. This 3-AA pack would work:

https://www.adafruit.com/product/3287

and so would this 3-AAA pack:

https://www.adafruit.com/product/727

The output voltage from those will be about 4.5V, which is taller than the 3.4V UV-LED requirement. The excess voltage will be 4.5V-3.4V=1.1V, and that's what you use to choose a resistor.

To understand Ohm's Law, it helps to learn a couple of underlying ideas first.

The basic unit of stuff in electronics is 'charge'.. usually electrons, since those are the easiest charged particles to move. The unit of charge is the Coulomb, and is about 6.25e18 electrons. That's a big number, but electrons are really small.. there are about 1C of free electrons in a cube of copper the size of a grain of salt (0.3mm per edge).

The property we call 'current' is the rate at which electrons (technically 'charge carriers') move from one place to another. If you imagine the electrons moving through a window, current is the total number of electrons that move from one side of the window to the other in a given amount of time. As a mechanical analogy, it behaves like the momentum of a stream of water.. you can have a wide stream moving slowly, or a small stream moving fast. The unit of electrical current is the Ampere, and 1A means you have 1C of electrons moving through a fixed surface per second.

(#27 wire has about the same cross-section as an 0.3mm cube. If 1A of current is flowing through the wire, it takes about 30 seconds for 1C of electrons to move 1cm).

It takes energy to move electrons from one place to another, and that energy is the property we call 'voltage'. The unit of energy is the Volt, and 1V means you've applied 1 Joule of energy (about as much as an apple falling off a table) to 1 Coulomb of charge.

The property we call 'resistance' makes it harder to move electrons from one place to another. It's equivalent to the mechanical property called 'friction' because it converts energy to heat. The unit of resistance is the Ohm, and Ohm's Law says it takes 1V to push 1A through 1 Ohm.


So.. running the numbers for the resistor to use with a UV LED and a 4.5V battery pack:

The LED needs 3.4V to make the electrons passing through it release energy as photons of UV light, so the voltage across the resistor will be 4.5V-3.4V=1.1V.

Most LEDs are designed to operate at 25mA, but that's considered full brightness. You can use less current if you want less light, and that's a matter of experimenting.

25mA is 0.025A, and we can rewrite the V=IR of Ohm's Law as R=V/I.. we need R much resistance to make it cost V much energy to send I much current from one place to another.

1.1V / 0.025A = 44 Ohms. It takes 1.1V to push 25mA through a 44 Ohm resistor.

Once you have a basic resistor value, you can scale it up and down to change the current.. if you want half as much current, you need twice as much resistance.. 88 Ohms. If you want 1/10th of 25mA you need 10x44 Ohms = 440 Ohms.

Playing with the values like that is a normal part of circuit design. The end result is that you probably want a resistor value somewhere between 44 Ohms and 440 Ohms.


That leads us to another issue called 'preferred values'. There are an infinite number of possible resistor values, but manufacturers have to choose a set of values to actually make. Experience shows that it's most useful to choose values that have a constant ratio from one to the next.. it makes scaling designs up or down easier.

The set of ratios we've chosen are called the E-Series of resistor values. In the E6 series, there are six values, each about 50% larger than the previous one: 1.0, 1.5, 2.2, 3.3, 4.7, and 6.8, with 10 being about 50% larger than 6.8, and then we can go to 15, 22, 33, etc. We call each group a 'decade' of resistor values.

The E12 series has twelve values, each about 20% larger than the previous one: 1.0, 1.2, 1.5, 2.2, 2.7, 3.3, 3.9, 4.7, 5.6, 6.8, 8.2. You'll notice that every other E12 value is also an E6 value, so we only get six new values. That process continues for the E24, E48, E96, and E192 series, but E12 values are the ones we tend to use most. Circuit designers learn to think in E12 values the way computer programmers learn to think in binary and hexidecimal.

So taking the range from 44 Ohms to 440 Ohms, the resistors you're most likely to want to play with will be: 47, 56, 68, 82, 100, 120, 150, 220, 270, 330, 390, and 470. Lower values will be brighter, and higher values will be dimmer.


To estimate how long the battery pack will last, we need to deal with the units called milliamp-hours. They mean exactly what it says on the tin: drawing 25mA for 1 hour is a total of 25mAh. The unit actually describes the amount of energy stored in the battery (and used by a load), but the relationships aren't immediately obvious. I'll work through the math in a bit.

AA cells usually hold 2000mAh of energy, and AAA cells usually hold about 1200mAh. To find out how long the battery will last, you divide the milliamp-hour value by the amount of current you want to use.. 2000mAh/25mA=80h, and 1200mAh/25mA=48h.

So at the maximum current you want to send through the UV LED, the 3-AAA pack could run for a couple of days. If you use a higher resistor value and less current, the battery pack would last even longer.

Now to explain what 'milliamp hours' actually means..

If we go back to the basic definitions of the units, 1 Ampere is 1 Coulomb per second. There are 3600 seconds per hour, so "1 Amp-hour" is another way of saying "3600 Coulombs". 1mA is 0.001A, so 1mAh is another way to say 3.6C.

Coulombs just count charge, not energy, but every battery also produces voltage.. and Volts are Joules of energy per Coulomb of charge. When we say "1mAh @ 1.5V" it means we have enough energy to add 1.5J to 3.6C of charge.. 5.4 Joules.

Scaling that up to AA and AAA cells, an AA holds 10.8kJ and an AAA holds just under 6.5kJ. Putting three AAs in series gives us 32.4kJ and a 3-AAA pack gives us 19.5kJ.

A UV LED operating at 3.4V pulls 3.4J of energy out of each Coulomb of charge that flows through it. 1mAh is 3.6C, so each milliamp through the LED uses about 12.2J of energy per hour. The resistor with 1.1V across it uses about 4J per hour, for a total of about 16.2J per milliamp per hour.

Running the LED-and-resistor at 25mA would cost 405J per hour, and at that rate 32.4kJ of stored energy in a 3-AA pack would last about 80 hours. The 19.5kJ stored in a 3-AAA pack would last about 48 hours.. the same values we calculated before.

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Re: New to Electronics, Need Purchasing help/direction

by BellezaDeDiosa on Sun Oct 21, 2018 2:06 am

Thank you so much for all your help. It just needs to be soldered together! When my son sees this he will be so excited, his vision of being a "zombie worrier with an eyeball hanging out" is one step closer to completion.
IMG_0668_polarr.jpeg
Zombie eyeball
IMG_0668_polarr.jpeg (483.48 KiB) Viewed 249 times

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Re: New to Electronics, Need Purchasing help/direction

by adafruit_support_mike on Mon Oct 22, 2018 11:55 pm

Looking good! Thanks for posting the photo.

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Re: New to Electronics, Need Purchasing help/direction

by BellezaDeDiosa on Sat Dec 22, 2018 8:16 pm

Thank you so much again for all you help, I could not have done this without you. My son absolutely loved his costume, and I had such a blast making it come to life for him. My 2.5 year old daughter now wants to be a Zombie Worrier next year.
Attachments
IMG_0887.jpeg
Final Costume at night
IMG_0887.jpeg (604.22 KiB) Viewed 206 times
IMG_0867 2.jpeg
Final costume in the light
IMG_0867 2.jpeg (844.05 KiB) Viewed 206 times
IMG_0840.jpeg
Eyeball and Shield
IMG_0840.jpeg (819.81 KiB) Viewed 206 times

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Re: New to Electronics, Need Purchasing help/direction

by adafruit_support_mike on Sun Dec 23, 2018 2:03 am

That is awsome! Thanks for posting the photos!

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