Hi all,

I've just finished making a reproduction of a Sith wayfinder from the latest Star Wars movie, The Rise of Skywalker. Looks great as it is, but it would be even better if I could light it up, just like in the movie! The problem is, I know almost nothing about electronics.

What I'd like to do is have two LEDs - one bright white (or maybe bright green) to provide the main illumination and one red (flashing, to represent the navigational beacon) - connected to a small battery pack to be housed within the wayfinder. I was thinking of using it occasionally like a night-light or display object, and was considering using rechargeable batteries (like Eneloops).

Can anyone help me out with a simple circuit?

 

Hello there . This is as simple as it gets.
They sell LEDs with the integrated chip that blinks the LEDs and this circuit will also work for a regular LED.

 

If you are in North America then Eneloops from Japan are expensive. Energizer or Duracell Ni-MH batteries are just as good, are sold in many stores but cost less.

 

The LED colors and the batch change Vd or the amount of voltage the LED drops. The eye also has a sensitivity. So, you will have to play with the resistors to get the intensities "right".

R<=(10mA/(3-Vd) where in this case 10mA (1e-3)is the nominal operating current. Vd is generally between 2.1 and 3.5V altough you can buy 12V leds that have the built-in resistor and some do not. LED's can be placed in series, but the ntensity can vary somewhat within the lot. Don't parallel LED's.

 

The problem is, I know almost nothing about electronics.

Hello. I just re-read your inquiry.
Nothing anyone of us has said means anything to you!
Do you already have your L.E.D.'s(Light.Emitting
Diode's.)?
If not ,we can suggest
What & wear you can acquire the components.One more question...Do you know how to solder?
Correction: any one of us has said."EXCEPT" Mr. Audioguruagain.

 

Thanks everyone.

@Delta prime: No, I don't have any components yet, as I wasn't quite sure what to get. I'd like the white LED to be reasonably bright at all angles, as it will be the main source of internal illumination for the wayfinder, whereas the red flashing LED can be more directional (in fact, I might even put a housing on it so that it only shines through one of the translucent panels). Two questions: (1) in your circuit diagram, does the bottom symbol represent the white LED, and (2) how fast do the LEDs with integrated chips flash (or are they available in different "speeds")?

 

Would it be as simple as connecting a non-flashing white LED, a flashing red LED and a suitable resistor in series?

 

Yes but, don't place the white LED in series with the red one or they will both flash.

2 LEDs 2 resistors, series and parallel.

.............LED............resistor............
.............LED............resistor............

 

For an LED to to be reasonably bright at all angles then you must look for a wide "viewing angle" on its datasheet. Most cheap LEDs have no datasheet and say, "super bright" but have a dim LED in a case that focusses the light beam into a bright narrow angle.

I have some LEDs from cheap solar garden lights that produce "Colors Changing" that flash the different colors and others slowly fade into the different colors.

 

Thanks @ElectricSpidey and @Audioguru again, much appreciated.

@ElectricSpidey, you say "2 LEDS 2 resistors, series and parallel". I thought series and parallel were mutually-exclusive? Having said that, I think I understand your circuit diagram though.

Might check my local Jaycar store and see what they have in terms of bright white LEDs with a wide viewing angle and flashing red LEDs with an integrated chip, and then work backwards to see what resistors I need. How much run time could I expect out of two AA batteries? Would I be better off with three?

 

A white LED might need 3.2V or a little more but two AA cells produce only 3V when new then their voltage quickly drops lower than a white LED needs so it will simply dim for a few minutes until it produces no light any more.

Three AA cells produce 4.5V then the diming will take much longer.

 

Back again...

I've attached a circuit diagram of what I think is required to have one flashing LED and one always-on LED. Not sure if I've got it right though.

Okay, so let's say:

- D1 is a white LED with a forward voltage of 3.3V and an operating current of 20mA;
- D2 is a red flashing LED with a forward voltage of 5V and an operating current of 20mA; and
- the power source is a 9v battery (someone offline suggested that this might be the way to go).

Given the above, what value resistors would I need at R1 and R2? Also, if I went for a rechargeable 9v battery with an actual voltage somewhere between 7.8 and 8.4 volts, would that have a significant impact on the value of the resistors required at R1 and R2?

Thanks in advance

 

Nobody makes a 3.3V LED. A white LED has a range of voltages, maybe 3V to 3.6V and you get whatever they have.
I assume that the flashing red LED has a circuit in it that needs 5V at 20mA.

Use Ohm's Law to calculate the resistor value. Calculate with the highest battery voltage and the lowest LED voltage so that the LED does not get a current so high it gets fried.

9.4V new alkaline battery minus 3V white LED= 6.4V across the resistor. Then the resistor is 6.4V/20mA= 320 ohms. Use a standard value of 330 ohms. If the battery voltage drops to 6V and the LED is actually 3.6V then the current is (6V - 3.6V)/330 ohms= 7.3mA, dim.

You do the calculation the resistor for the 5V flashing LED, and/or the rechargeable battery.

 

Thanks @Audioguru again, much appreciated. The forward voltages were sourced from the sellers' websites, but I'm not knowledgeable enough to know how accurate they are. You are correct that the flashing red LED has a circuit in it that needs 5V at 20mA.

For the flashing 5V LED, I assume the calculation would go like this:

9.4V (battery) - 5V (LED) = 4.4V across the resistor
4.4V/20mA = 220 ohms (not sure what the closest standard value is - I tried Googling it, but I found the results confusing)(*Edit* OK, so it looks like 220 ohms is a standard value, as far as I can tell).

Have I got the calcs right, or do things change given that the two LEDs are running in parallel?

 

By the way, if the stated forward voltage of 3.3V for the white LED is not accurate, is there a way I can work out what the actual value is?

 

Yea, measure it.

 

The Chinese "Seller" on ebay, Amazon, AliExpress or Banggood knows nothing about the cheap junk he finds then sells to you.
The manufacturer's datasheet has all the details.
The seller does not know who is the manufacturer, and says Unbranded.

I quickly measure the forward voltage of the LEDs I use.

 

Thanks guys, but to be honest, as a non-electronics person with what I thought was a basic project for lighting a home-made movie prop, the replies haven't helped me progress much beyond the queries in my opening post. Really, I was just after a simple circuit layout for one always-on LED and one flashing LED, and some help with calculating the resistor values required for running the two LEDs in parallel (which I assume is the required set-up, although that's not abundantly clear from the replies).

Unfortunately this forum seems to assume a certain level of knowledge that I just don't have. I might try a prop-building or cosplay forum, and see if I have better luck there.

 

Maybe start with a datasheet. http://www1.futureelectronics.com/doc/EVERLIGHT /334-15__T1C1-4WYA.pdf I've never seen one like this before, BUT Vf and If are the two numbers yo need to know. Vf has a range and it also determines the relative intensity. That info is NOT in the datasheet. It's well-known. Color changes Vf.

You can't put LEDS in parallel. You have to put a LED and a Resistor in series. Two LED's, two resistors.

You have 3 separate circuits connected to some battery, say with a 9V battery.

Find a flashing LED.

Aside from the flashing LED, say Vf=2.1 and If = 10 mA; battery is 9V

R>= (9-2.1)/10e-3; done. That's 690. 690 doesn't exist. Pick close, but higher.

Then there are the power formulas. P=I^2*R; 10mA is close enough. P>0.069W
1/4 W will work and so would an 1/8W work.

 

here https://www.jameco.com/z/LFH3360-LED-Blinking-Red-Diffused-5mm-T-1-3-4-697nm-2-8mcd_34690.html is a LED that blinks that operates between 3 and 12V,

If you need a different color, different rate, different intensity, longer battery life that's another problem.