Back lit Chess Board, Led Question

[Frost]

Hey guys i plan on making a back lit chess board with some led's under a piece of acrylic with transparent colored plastic on it so that it will color the light without having to have 20+ different colored leds in their own compartment and i have some questions about resistors. The main question is, is there anything i need to know that wont have the leds blow up.

Here is a drawing (im not five years old even though my paint skills say otherwise) Do you see any problems with it? Where and what resistors should i need? i do not know what power source im going to use right off the bat, probably a 2 double a battery box but that might be subject to change.


Thanks for the help

 

[Arouse1973]

A couple of things first.
1. What LEDs are you going to use?
2. What power supply are you using?
3. How much power do you want the LEDs to use I.e how bright?
4. How many LEDs are you going to need, I noticed 3 LEDs in series in your drawing, is this the maximum?

Adam

 

[Scotophor]

Also, if you're using 2 x AA cells for your battery (3.0 volts), this will barely be enough to light one white LED (or a few in parallel, if they are well-matched). Most white LEDs really want 3.5 volts or so to light at their full brightness. Putting multiple white LEDs in series on a 3 volt supply will simply not work. You need to have a supply with higher voltage than the LED wants, so you can drop some of it across a current-limiting resistor. LEDs are current-dependent and not self-limiting, so a series resistance is usually mandatory.

 

[Scotophor]

Another thing I forgot to mention is that when you wire multiple LEDs in series, their voltages add. So, using your drawing as an example, a series string of 3 white LEDs will need a minimum of 10.5 volts (3.5 x 3). This would probably mean a minimum supply of 12 volts, leaving 1.5 volts to be dropped by the current-limiting resistor. For running from batteries, this is not ideal; when the battery voltage declines, that extra 1.5 volts to drop across the resistor will go away rather quickly, meaning that the LEDs will dim significantly long before the battery should be considered fully depleted. When running LEDs from batteries, try to drop at least 30% of your nominal supply voltage across the current-limiting resistor.

Using your lighted chess board as an example, and assuming you want it to be battery-operated for portability, I would suggest running on 4 "C" or "D" cells for a nominal 6 volt supply, and putting all of your LEDs in parallel, each in series with its own current-limiting resistor. While this is a relatively simple build for a newcomer to electronics, it is not optimal from an efficiency standpoint.

For better efficiency (more run time from a given size of battery supply), I would design a circuit to use a boost converter to produce a high DC voltage, and wire the LEDs in the minimum number of series strings. For example, it might be enough to have 20 LEDs to light the board, and wire them in a single string (requiring about 70 volts). A boost converter can be 80% efficient or better, and if it is set up as a regulated current source to provide the typical ~30 mA, no series resistor will be needed.

 

[Frost]

Another thing I forgot to mention is that when you wire multiple LEDs in series, their voltages add. So, using your drawing as an example, a series string of 3 white LEDs will need a minimum of 10.5 volts (3.5 x 3). This would probably mean a minimum supply of 12 volts, leaving 1.5 volts to be dropped by the current-limiting resistor. For running from batteries, this is not ideal; when the battery voltage declines, that extra 1.5 volts to drop across the resistor will go away rather quickly, meaning that the LEDs will dim significantly long before the battery should be considered fully depleted. When running LEDs from batteries, try to drop at least 30% of your nominal supply voltage across the current-limiting resistor.

Using your lighted chess board as an example, and assuming you want it to be battery-operated for portability, I would suggest running on 4 "C" or "D" cells for a nominal 6 volt supply, and putting all of your LEDs in parallel, each in series with its own current-limiting resistor. While this is a relatively simple build for a newcomer to electronics, it is not optimal from an efficiency standpoint.

For better efficiency (more run time from a given size of battery supply), I would design a circuit to use a boost converter to produce a high DC voltage, and wire the LEDs in the minimum number of series strings. For example, it might be enough to have 20 LEDs to light the board, and wire them in a single string (requiring about 70 volts). A boost converter can be 80% efficient or better, and if it is set up as a regulated current source to provide the typical ~30 mA, no series resistor will be needed.

Ok thanks, lke i said i am new so all ideas are welcome. I am not completely stuck on the battery power source so going to a wired connection might be better. Now that i think about it 9 led's might not be bright enough so i think i will be going with 20. Could you link me a power source that you would recommend? I think these are the leds im going to use https://www.sparkfun.com/products/531

 

[Gryd3]

Ok thanks, lke i said i am new so all ideas are welcome. I am not completely stuck on the battery power source so going to a wired connection might be better. Now that i think about it 9 led's might not be bright enough so i think i will be going with 20. Could you link me a power source that you would recommend? I think these are the leds im going to use https://www.sparkfun.com/products/531

How bright do you want?
Have you looked at EL panel?
https://www.sparkfun.com/products/10799

 

[Frost]

How bright do you want?
Have you looked at EL panel?
https://www.sparkfun.com/products/10799

It needs to be pretty bright. EL panels would be too expensive to use since that one is only 10cmx10cm and its 15$ in order to get it to the size of a standard chess board it would run $100+

 

[Gryd3]

It needs to be pretty bright. EL panels would be too expensive to use since that one is only 10cmx10cm and its 15$ in order to get it to the size of a standard chess board it would run $100+

Understood. There are other sources, but the result is still in the $100 ball-park.

Now as far as LEDs and powering them is concerned, lets keep this simple and stick with a constant voltage source.
Lets decide if battery or plug-in is required.
How long do you want the batteries to last?

You can use batteries, (rechargeables or not) or an AC-DC adaptor that you can plug in the wall.
Don't worry too much about voltages and currents right now, because the layout of how the LEDs will be wired will define this. You can wire the LEDs to match the power supply you use... or pick a power supply if you have a specific LED wiring plan in mind.
Each LED is 20mA and has a forward voltage drop of up to 3.4V. You plan to use 20, so the required 'power' sould be 0.02A * 3.4V * 20(LEDs) = 1.4Watts

The 'most' important power supply issues are:
-The voltage must be higher 3.6Volts
-The amperage must be higher than 0.02A (More important for wall adaptors)
-The amperage multiplied by the voltage must be higher than 1.4
-The capacity of the battery pack must be acceptable...*

*A single alkaline AA can hold up to 2600mAh... but only provides 1.5V... So in order for the first rule to be valid, we must use 3 of them in series for a 4.5V battery pack. Because they are in series, the 2600mAh does not add. The voltage is also too low to use any more than 1 LED in series, so each of the 20 LEDs will need their own resistor. This battery pack is estimated to last at most 6 hours IF the batteries chosen are high end alkaline, and there is no additional wasted power that is not accounted for. (Calculated at 1.8Watts to include waste in resistors) Using other AAs could result in times closer to the 2 hours mark... (You could increase this time by adding batteries, but how big of a battery pack is too big?)

 

[Arouse1973]

I think you need an LED with a wide half angle say 85 degrees. They are not normally that bright because otherwise you might get spotting. This is where you can see the high brightness part of the LED. I think you want something that's going to defuse nicely over the coloured plastic your going to use.
Bob K might have a better solution, I am guessing he will be along shortly.
Adam

 

[BobK]

I think you need to use diffused LEDs, and probably a lot more than 9 in order to get reasonably uniform illumination.

bob

 

[Gryd3]

I think you need to use diffused LEDs, and probably a lot more than 9 in order to get reasonably uniform illumination.

bob

Op changed his mind to 20 LEDs.
You and @Arouse1973 both mention the spotting, I have seen users in the past take some sandpaper to rough up the LED lense to make their own 'diffused' LED.
It's that or using an additional layer of material to help diffuse the light.

 

[Frost]

Understood. There are other sources, but the result is still in the $100 ball-park.

Now as far as LEDs and powering them is concerned, lets keep this simple and stick with a constant voltage source.
Lets decide if battery or plug-in is required.
How long do you want the batteries to last?

You can use batteries, (rechargeables or not) or an AC-DC adaptor that you can plug in the wall.
Don't worry too much about voltages and currents right now, because the layout of how the LEDs will be wired will define this. You can wire the LEDs to match the power supply you use... or pick a power supply if you have a specific LED wiring plan in mind.
Each LED is 20mA and has a forward voltage drop of up to 3.4V. You plan to use 20, so the required 'power' sould be 0.02A * 3.4V * 20(LEDs) = 1.4Watts

The 'most' important power supply issues are:
-The voltage must be higher 3.6Volts
-The amperage must be higher than 0.02A (More important for wall adaptors)
-The amperage multiplied by the voltage must be higher than 1.4
-The capacity of the battery pack must be acceptable...*

*A single alkaline AA can hold up to 2600mAh... but only provides 1.5V... So in order for the first rule to be valid, we must use 3 of them in series for a 4.5V battery pack. Because they are in series, the 2600mAh does not add. The voltage is also too low to use any more than 1 LED in series, so each of the 20 LEDs will need their own resistor. This battery pack is estimated to last at most 6 hours IF the batteries chosen are high end alkaline, and there is no additional wasted power that is not accounted for. (Calculated at 1.8Watts to include waste in resistors) Using other AAs could result in times closer to the 2 hours mark... (You could increase this time by adding batteries, but how big of a battery pack is too big?)

Ok it seems like a wall adapter will be a better idea now. Would this wall adapter do it or would it be over kill? https://www.sparkfun.com/products/298 this is the female that im going to use https://www.sparkfun.com/products/119 if that will fit

 

[Gryd3]

Ok it seems like a wall adapter will be a better idea now. Would this wall adapter do it or would it be over kill? https://www.sparkfun.com/products/298 this is the female that im going to use https://www.sparkfun.com/products/119 if that will fit

Looks like it would be fine to me.

Lets go over some details to determine the proper wiring arrangement and resistor values.
You have a 9V source, and each LED will drop at most 3.6V, so you can only chain two LEDs together at a time.
That would mean that for each pair of LEDs you string together you need 1 resistor with a value of:
9V - (3.6V*2) / 0.020A = 90Ω, so we will round up and use 100Ω resistors.

This will require you to build 10 strings. Each of two LEDs and one resistor in series.
All 10 of these will then be connected in parallel to make your LED array. (You can always add more strings in parallel if your light is not bright enough... Your limit for this supply would be 30 strings connected in parallel which would be 600mA (Best to undercut it a little... don't push 650mA.

I have attached a diagram for you from an online wizard, but I wanted to explain the numbers instead of blindly slapping a solution on here.

 

[Frost]

Looks like it would be fine to me.

Lets go over some details to determine the proper wiring arrangement and resistor values.
You have a 9V source, and each LED will drop at most 3.6V, so you can only chain two LEDs together at a time.
That would mean that for each pair of LEDs you string together you need 1 resistor with a value of:
9V - (3.6V*2) / 0.020A = 90Ω, so we will round up and use 100Ω resistors.

This will require you to build 10 strings. Each of two LEDs and one resistor in series.
All 10 of these will then be connected in parallel to make your LED array. (You can always add more strings in parallel if your light is not bright enough... Your limit for this supply would be 30 strings connected in parallel which would be 600mA (Best to undercut it a little... don't push 650mA.

I have attached a diagram for you from an online wizard, but I wanted to explain the numbers instead of blindly slapping a solution on here.

Thank you for explaining as well as giving the solution, helps a bunch, however, the solution you posted here looks like the light will not be distributed evenly. What if instead of one array of 10 strings we wire two arrays of 5 strings? would that be a problem?

 

[Gryd3]

Thank you for explaining as well as giving the solution, helps a bunch, however, the solution you posted here looks like the light will not be distributed evenly. What if instead of one array of 10 strings we wire two arrays of 5 strings? would that be a problem?

Positioning the LEDs can be done in any way you want.
The important part is that each string must be 2 LEDs and one 100Ω resistor.
The strings themselves can be connected in 2 or 5 groups and it will not make a difference.
The diagram shown is merely for the electrical connections, so as long as the + side for each string is connected together, and the - side for each string is connected together they can be anywhere.

 

[Frost]

Positioning the LEDs can be done in any way you want.
The important part is that each string must be 2 LEDs and one 100Ω resistor.
The strings themselves can be connected in 2 or 5 groups and it will not make a difference.
The diagram shown is merely for the electrical connections, so as long as the + side for each string is connected together, and the - side for each string is connected together they can be anywhere.

Ok sweet. It all sounds good. This switch should be able to handle it though right? https://www.sparkfun.com/products/9276

I appreciate al the help you have given me.

 

[Gryd3]

Ok sweet. It all sounds good. This switch should be able to handle it though right? https://www.sparkfun.com/products/9276

I appreciate al the help you have given me.

That switch is overkill
It will handle it.
You need a switch that can handle at least 9V and 650mA because that is what your power supply is rated for. You could get away with smaller because your LEDs will only draw 200mA, but it helps to oversize things a little.

 

[Gryd3]

I think this: https://www.sparkfun.com/products/11138
or this: https://www.sparkfun.com/products/10727
would be a better fit for you though.
It would look cleaner than the big bulky heavy duty switch

 

[Frost]

Yea but then i cant fit this on it https://www.sparkfun.com/products/9278 XD one last question, would i wire the power straight to the switch or what?

 

[Gryd3]

Yea but then i cant fit this on it https://www.sparkfun.com/products/9278 XD one last question, would i wire the power straight to the switch or what?

Once your array of LEDs is wired, you will have a single + and - wire for the LEDs.
You will also have a single + and - wire for the DC adaptor. All you need is to put the switch in between the LED array and DC adaptor.
The switch will interrupt the flow of electricity... so if you take the + from the DC adaptor and connect it to one end of the switch, and the + from the LED array and connect it to the other end of the switch. It is not going to be able to control when power does and does not flow. (The - side of the DC adaptor, and LEDs can directly connect to each other... because the + side has the switch on it)