# 72 led circuit

Discussion in 'The Projects Forum' started by JonoMc, Jan 12, 2018.

1. ### JonoMc Thread Starter New Member

Jan 12, 2018
5
0
Hi all. I'm a complete newbie to all of this but willing to give it a shot. I have a project where Ive built a 1.2m lego display truck and trailer that holds the centre speaker for the home theatre in the entertainment unit. I have placed inside the sleeper a wireless door bell and it works great.
What I'm trying to do now is light it up by either a push of a button or from the doorbell ringing. the door bell has a green led that lights while it is activated so I'm thinking this as a trigger through a relay to power the lights. Also wanting to have a simple on/off switch. the led im using are 0603 for their size to fit inside a lego stud. Wire gauge is 27 for the mains and 32 to each led to fit between brick plates. I have 52 orange, 16 red and 7 white ones to light up but not to their full brightness. How would I go about this and what is the smallest power required to run them. Do I put them in a parallel or series circuit and do I wire each colour separately or can I put them all on the same power wire. Someone told me that I could run 32 in parallel with a 2.2ohm resistor with 2xAAA batteries?. I have room under the truck for up to 4 AAA or one 9v. Not sure how to start all of this.
Every bit of help is much needed.
Cheers Jono

2. ### smooth_jamie Member

Jan 4, 2017
99
18
Hi Jono,

You place all your LED's in parallel. Each LED should have a resistor connected to it to limit the current otherwise they will burn out. You can calculate the required resistor using Ohm's law R =V / I. For example, if you are using a generic LED which is 5V 20mA then with a 5V power supply you will require a resistor of at least 250 Ohms (closest resistor size you can buy is 270 Ohms so always use the higher available value).

Decide what battery you want to use for your truck (either x4 AAA in series i.e. 6V or, a PP3, 9V) and then you can calculate the correct value of your limiting resistors. Activating them with a push button will be simple, I will leave that up to you to figure out. If you want to activate the circuit when it "hears" a certain tone or frequency of sound then that becomes a more interesting project

3. ### LesJones Well-Known Member

Jan 8, 2017
1,795
476
This is not quite right. the forward voltage of the LED needs to be considered.

R = (Vs - Vled)/I

I = required LED current.
R = reqired resistor value.
Vs = supply voltage
Vled = Forward voltage of LED (For example for a red LED this would be about 1.8 volts. For a white LED it would be about 3.3 volts. Consult the data sheet for the LED being used.)
So for a red LED with a 5 volt supply and a required LED current o 20 mA (0.02 amps) R = (5 - 1.8)/0.02
= 3.2/0.02 = 160 ohms The nearest preferred value is 180 ohms. It is better to run LEDs at less than their maximum current ratings.

Les.

4. ### ElectricSpidey Member

Dec 2, 2017
479
101
Using a resistor at every LED is a waste of power, I would try to series as many of them together to match the available voltage.

Example:

At a forward voltage of 2.5 volts you could series 3 red/orange LEDs on the 9 volt battery, with a very low value resister in series.

But understand, your batteries aren’t going to last very long lighting LEDs, better to get a safe low voltage power supply, say about 12 volts, then you could string more LEDs together, with a single resistor.

Also I don’t know what kind of LED Jamie is referring to as any “standard” 5 volt LED would already have the resistor installed. LED voltages are more like 2.2-2.4, 3.2-3.4…etc.

You can buy LEDs with internal resistors, or prewired.

COBs are a different story.

5. ### dl324 AAC Fanatic!

Mar 30, 2015
7,802
1,861
Welcome to AAC!

Please use paragraphs to organize your thoughts. It looked like you might have tried to use two paragraphs with no white space between them, but you expressed more than a couple thoughts. Review what you posted after posting to see if any changes should be made to improve clarity or correct grammar; I do it all of the time.

Some members, myself included, don't like being subjected to a rambling "wall of text".

When discussing circuits, you should include schematics and/or block diagrams.

As has already been mentioned, LEDs shouldn't be operated in parallel unless each parallel LED (or string) has it's own current limiting resistor. This will avoid issues with current hogging as no two LEDs will be identical.

A higher voltage source has the advantage of requiring fewer current limiting resistors which will result in less power dissipated solely as heat. But your choice of a 9V "transistor radio" battery or 4 AAA gives you limited current, e.g. runtime.

You need to give us part numbers for the LEDs, as each color will have a different forward voltage drop and some LEDs will tolerate more/less current than those we might have in mind.

Since you have already made an attempt at wiring them which didn't give satisfactory results, you need to post a schematic of how you wired them. Otherwise, we can't comment on any issues you're seeing.

-live wire- likes this.
6. ### smooth_jamie Member

Jan 4, 2017
99
18
Speaking from experience alot of the LEDs I use for my displays are 5V 20mA. Of course there are alot of colours/types you could use but if you buy a random pack off ebay it's a safe bet that it's of the 5V variety

7. ### smooth_jamie Member

Jan 4, 2017
99
18
Depends on alot of things but without a resistor on each LED you are leaving the LED to limit current on it's own. It's better to have one on each LED as some may draw more current than others.

8. ### Bernard AAC Fanatic!

Aug 7, 2008
4,952
551
Sum of LED wattage about 2 W, a heavy load for 4 alk. AAs. Might use a 12 V @ 2A wall wart.

9. ### ElectricSpidey Member

Dec 2, 2017
479
101
Show me...

ebeowulf17 and Bernard like this.
10. ### ElectricSpidey Member

Dec 2, 2017
479
101
I never said anything at all to the tune of leaving off the resistors, simply pointed out that series is more efficient.

ebeowulf17 likes this.
11. ### Bernard AAC Fanatic!

Aug 7, 2008
4,952
551
Using 5 ma , drain about 1W, or on 6V about 200 mA. I would use 4 AA NiMH rechargeable @ around 2200 mAh. , about 10 h. ; with a boost convertor to make longer strings ?

12. ### LesJones Well-Known Member

Jan 8, 2017
1,795
476
Some of the cheap boost converters on ebay can easily be modified to behave as constant current sources.
This is how I modified one.

Les.

Last edited: Jan 13, 2018
Bernard likes this.
13. ### Bernard AAC Fanatic!

Aug 7, 2008
4,952
551
Constant V might be better with multiple strings. This worked well for a holiday costume with 4 strings of 7 flashing LEDs, boost, & 3 AAA alk. batteries.

14. ### dl324 AAC Fanatic!

Mar 30, 2015
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1,861
The OP appears to have vacated this thread. I see little point in continuing to comment until he comes back with more information. Even the number of LEDs is questionable. Is it 72 or
75?

15. ### JonoMc Thread Starter New Member

Jan 12, 2018
5
0
Sorry for the late reply and the bad grammer on my first post. I have changed slightly the layout of the truck and attempted an attached scematic for it.
The led's to be used are:
14 Red SMD 0603, forward voltage 2.1, forward current 20 mA, intensity 150-180 mcd
14 White SMD 0603, forward voltage 3.0, forward current 20 mA, intensity 330-360 mcd
60 Orange SMD 0603, forward voltage 2.0, forward current 20 mA, intensity 180-210 mcd.
They all have a view angle of 120-125 deg.

The sleeper which has the relay and input signal is separated to the rest by a plug because it houses the doorbell, so I wanted to be able to remove it easily.

What would be the minimum source voltage required made from AAA's or 9v batteries?
Will this work?
Where and what type of resistors will I need and where would I put them?
Is there a way to wire a relay in off a 1v signal wire to activate the whole circuit using the power from the alarm input of DC 6v 200mA that is stepped down from 240v 50hz.
Thanks everyone for their help so far. I'm learning as I go.

16. ### ElectricSpidey Member

Dec 2, 2017
479
101
Well. Since you have already decided to place all of the LEDs in parallel, I guess you will need a resistor in series with each one.

Very energy wasteful.

17. ### JonoMc Thread Starter New Member

Jan 12, 2018
5
0
Sorry here is the pic as an attachment.
Havnt dedicated yet.
How do I put them in series?
From + (source) to - (led) + to -(led)+ etc then finish back to - (source) making a complete loop.
If you think series is better, then I'll figure out how to redraw it like that.
Does that mean a series circuit is more energy efficient and uses less resistors.

18. ### WBahn Moderator

Mar 31, 2012
23,396
7,102
Now it seems we are up to 88 LEDs. If they are all in parallel and all are pulling 20 mA, then your power source needs to supply a total of 1.76 A of current. If it's a 6 V supply then you need to deliver a total of 10.6 W of power and if it's a 9 V source that ups it to nearly 16 W. Your LEDs would consume about 4 W of that and the rest would need to be dissipated as heat in resistors. 6 W to 12 W may not sound like a lot, but it is when you are trying to get rid of the heat it produces.

Typical AAA alkaline batteries have about 1000 mAh of capacity so at 1.76 A you would drain them in roughly half an hour. A typical 9 V alkaline has a capacity of about 500 mAh and so you would drain it in about 15 minutes.

But that's assuming you could get either to deliver nearly two amperes in the first place. A typical current draw for a AAA battery is something like 50 mA and for a 9 V is about 25 mA. You can probably get twice that without too much trouble, but the capacity will start dropping pretty quickly at some point.

No matter what you do, it is going to be difficult to power that many LEDs off of those kind of batteries, either at all or for very long.

So what can you do?

First, see if you REALLY need 20 mA of current in those 0603 LEDs. You can almost certainly find LEDs that will be bright enough to serve your purpose at somewhere around 2 mA. That's a game changer right there. Now instead of nearly 2 A you are at something under 200 mA. That's at least getting into the ballpark of what those battery sources can be pushed to. You have also increased your battery life to a few hours.

The next thing is run them in series strings as much as possible. At 6 V there's not much you can do for a 3 V LED. You want at least a volt or two of overhead for decent current limiting. You could possibly run a white and a red in a string, but that is only giving you a volt of overhead when the battery is fresh and that's not leaving you much to work with, especially as the battery drains. But you can run two of the red and orange in each string. That will nearly cut your current needs in half.

On a 9 V battery you can run three red and orange and two white in each string. That would be 32 strings and, at 2 mA per LED, that would put you at a total draw of about 64 mA, which is a quite doable draw for that 9 V battery, which will probably last about 5 hr or so.

19. ### ElectricSpidey Member

Dec 2, 2017
479
101
Yes series is better because the energy wasted in the resistors can be used to light LEDs.

Only drawback to series is using such a low voltage source, because you can't string very many of them before you run out of voltage.

But @ 6 volts you can probably get at least a few to light in series using a single much smaller resistor in each loop.

Series would be + to the + lead of the LED (usually the longer one, but I have never used SMD so I cant really say) then the negative lead to the next positive lead, etc...then that last lead to the negative battery contact and anywhere in the loop you can place the resistor.

20. ### dl324 AAC Fanatic!

Mar 30, 2015
7,802
1,861
You can determine minimum acceptable current for each LED color by starting with a resistor value that will give you 20mA and increase resistance until they become too dim for your application.

How are you planning to solder wires to 0603 LEDs? They're tiny and wires won't have any strain relief.

Bernard likes this.