Design a circuit to light the LEDs?

Thread Starter

pinkyponky

Joined Nov 28, 2019
338
Hello!

As I know that if I want to lit single LED, we need a battery, resistor and LED. Let's take a 2V of LED forward voltage and 20mA of LED forward current.

How the resistor value is calculated?.

R = (5V-2V)/20mA = 150 ohm => 200 ohm (Just I have chosen to reduce current for safe side). Is the calculation is correct?.

5V -> Battery Voltage
2V -> Diode Forward Voltage
20mA -> Diode Forward Current

The main aim of this project is to write my name on PCB with LEDs, just for name display. So I'm guessing 1000 LEDs are needed for this project. I would like to know from you that, is the below circuit will work if I design the circuit as shown below.

If it is not the way to build the project for displaying the name, please suggest me other circuit diagrams?.

1682945557825.png

Thank you!
 

AnalogKid

Joined Aug 1, 2013
10,574
Yes, your calculation is correct.

Yes, your schematic is correct.

Beyond that, you will get the same result with a higher power supply voltage and fewer resistors. For example, with a 12 V supply, you can have 5 LEDs in series with a single resistor. The cumulative Vf would be 10 V, and you use the same equation.

If you are stuck with a 5 V source, you still can have 2 LEDs per resistor. Note that now there will be only 1 V across the resistor, as opposed to 3 V in your original schematic. This means that small variations in resistor value from one part to another will have a larger effect on brightness. IOW, there might be a perceivable unevenness in brightness across the entire display.

This is a trade-off. On the plus side, the more LEDs you have in series, the longer the battery will last. In your schematic, you are dissipating more power in the resistors than in the LEDs. The ratio is 2:3, LEDs : resistors. If you move to 2 LEDs per resistor, the ratio is now 4:1.

Beyond that, you will get an even better result with a higher power supply voltage and fewer resistors. For example, with a 12 V supply, you can have 5 LEDs in series with a single resistor. The cumulative Vf would be 10 V, and you use the same equation. In this case, the power ratio would be 5:1

ak
 
Last edited:

Ian0

Joined Aug 7, 2020
8,386
Look at how LED tape is constructed. 12V LED tape has three LEDs and one resistor in series.
With a 5V supply you will need a lot of current and a lot of resistors.
If you rectify 230V mains, you can probably put a hundred LEDs in series with one current-limiting resistor.
 

DickCappels

Joined Aug 21, 2008
10,067
Your calculation for resistance was done correctly.

Your design will light up, and it might even burst into flames because it might take close to 100 watts, so be careful.

20 ma x 5V x 1000 = 100W.
 

Ian0

Joined Aug 7, 2020
8,386
The figure of 20mA to light an LED is out of date. 20mA was the figure I used when I learned electronics in the 1980s. LEDs are at least ten times as efficient now as they were then. So, not only might it burst into flames, it might burn out your retinas as well.
 

MrChips

Joined Oct 2, 2009
29,217
Your calculation is correct but actual testing gives better results.
Start with 10kΩ and gradually reduce the value of the resistor until you get the brightness that suits you.
 

Gruvytune

Joined Sep 11, 2022
1
Hi. I'm mostly an amateur hobbyist. A question I've wondered which I'm sure is answered easily, is aren't we wasting lots of energy in the heat given off by resistors? LED bulbs very quite warm although the LED itself isn't very hot.
Or is an incandescent bulb so much less efficient in general and so much more heat energy wasted?
When I read bulb packages saying yearly cost is like pennies to run, I imagine it's higher just in heat given off my Cree bulbs. Or is it insignificant.
Thank you
 

BobTPH

Joined Jun 5, 2013
7,540
The LED bulbs do not use series resistors to control the current. They either use a capacitor, or a more complicated circuit. In AC circuits, a capacitor can drop voltage without producing heat.
 

WBahn

Joined Mar 31, 2012
29,163
Hello!

As I know that if I want to lit single LED, we need a battery, resistor and LED. Let's take a 2V of LED forward voltage and 20mA of LED forward current.

How the resistor value is calculated?.

R = (5V-2V)/20mA = 150 ohm => 200 ohm (Just I have chosen to reduce current for safe side). Is the calculation is correct?.

5V -> Battery Voltage
2V -> Diode Forward Voltage
20mA -> Diode Forward Current

The main aim of this project is to write my name on PCB with LEDs, just for name display. So I'm guessing 1000 LEDs are needed for this project. I would like to know from you that, is the below circuit will work if I design the circuit as shown below.

If it is not the way to build the project for displaying the name, please suggest me other circuit diagrams?.

View attachment 293243

Thank you!
You current will be about 15 mA per LED, so your 5 V supply will need to deliver 15 A (and 75 W).

That's a pretty healthy supply, but not undoable.

How are you planning to power this? If it is being powered by a main-powered supply, that's one thing. But if you are planning to run this off of a battery, that's going to be a challenge as 15 A will drain even large batteries is a short amount of time.

It's also a lot of heat that must be dissipated off your PCB. Each component isn't dissipating much individually, but if they are packed in pretty tight, you might have a hard time getting the cumulative heat carried away.

It looks like all of the LEDs will either be on or off, correct?

If so, let's see if we can do a bit better.

If you use a laptop power supply, the typical output voltage is 19.5 V. If we allow for 3.5 V across the current limiting resistor, that would give you 16 V for your LEDs. If your LEDs are truly 2 V forward drop, you can put eight in series. Your 200 Ω resistor would then produce about 17.5 mA. You would need 125 of these strings, bringing your total current needs to just under 2.2 A and total power to just under 45 W. Even small laptop supplies are happy delivering that.

But you can do much better than that by making a better choice of LED. Today you can get LEDs that are likely to be plenty bright at a fraction of the current, perhaps even in the 1 mA range. If you went with something that needed 5 mA to give you the brightness you want, you would reduce you current needs to a bit over 0.5 A and your power to approximately 10 W. If a 1 mA LED is sufficient, your are now down to about 125 mA and just about 2.5 W.

As an aside, have you considered cost? If your LEDs cost you a dime apiece, you are looking at $100 just for the LEDs. Is that worth it just to have your name up in lights? If it is, go for it.
 

Ian0

Joined Aug 7, 2020
8,386
Hi. I'm mostly an amateur hobbyist. A question I've wondered which I'm sure is answered easily, is aren't we wasting lots of energy in the heat given off by resistors? LED bulbs very quite warm although the LED itself isn't very hot.
Or is an incandescent bulb so much less efficient in general and so much more heat energy wasted?
When I read bulb packages saying yearly cost is like pennies to run, I imagine it's higher just in heat given off my Cree bulbs. Or is it insignificant.
Thank you
Filament lamps are about 5% efficient. LEDs are about 50% efficient.
If you had one 3V LED run with a series resistor off a 30V supply, the combination would be about as efficient as a filament lamp.
 
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