Powering Multiple LEDs

[mr_H]

Hi, very new to this. I would like to work on a task with my students where they can:
- use the 08M to power multiple LEDs.
- ~8 LEDs switched on/off from each output.
- 2 outputs used so 16 LEDs in total.
- The LEDs would be in the shape of their initials.

Hoping someone can advise direct on the wiring diagram and specifications required.

Many thanks

 

[eclectic]

Mr H
First quick thought.

Have a look at Manual 3, page 6.

The ULN2803 chip, with inputs wired together,
could drive eight LED/resistors.

I've attached a partially complete diagram
e.

 

[Michael 2727]

Also look at Standard Interfacing Circuits in picaxe_mamual3.pdf
and Transistor interfacing or power MOSFETS, either should work fine.

I would use a 2N7000 MOSFET, via a 1K resistor from any output.
Run all of the LEDs in parallel each with its own resistor.
You could run them in series with a single resistor, BUT you may
get varying light outputs between individual LEDs.

 

[mr_H]

Multiple LEDs simulation working

Thanks for the advice guys. I cannot find the chip ULN2803 on my circuit simulator so I tried the MOSFET option. Running 8 LEDs seemed a bit much for the 4.5 volts and it worked better with a separate power supply at 9V. Would you agree this is the way to go? I have attached a diagram of my circuit.
Is the benefit of the ULN2803 in reducing the number of components?
Cheers,
Harry (Melbourne, Australia)

 

[westaust55]

Raising the voltage from 4.5 volts to 9 volts requires that you drop a greater voltage across the series resistors which equates to greater losses. There is no current reduction for the same illumination level. I suggest that you stay with the 4.5/5.90 Volt supply.

What you do need to do is install a separate series resistor for each LED rather than one (100 Ohm) resistor for the entire 8 LED’s. That will help to achieve equal current and therefore equal light output level from each LED.

A starting point would be say a 330 Ohm resistor in series with each LED. If they are “normal red LED’s with a maximum rating of 20mA, then the 330 Ohm resistor will limit the current to avbout 12 mA which is still very readable. Others have in fact reduced the current to as little as 5mA and still retained visible LED indication but a higher current may still be worthwhile in the classroom environment when ambient light levels and viewing angles are not necessarily ideal.

Q. Is the benefit of the ULN2803 in reducing the number of components?
A. Yes, one package would replace 8 discrete transistors – but you still need a series resistor for each LED.

 

[Michael 2727]

As has been said, use a resistor on each LED for consistant results.

The voltage does not really matter as long as the safe current value is
used for the particular LED/s, (typ) 20mA for standard LEDs.
Check the data sheet on the LEDs you bought, if available.

The Forward Voltage drop (typ) for Red, Green, Yellow and Orange LEDs is 1.5V to 1.7V
The Forward Voltage drop (typ) for White, Blue and UV LEDs is 3V to 3.6V

LED Calculator links -
Single - http://led.linear1.org/1led.wiz
Multiple - http://led.linear1.org/led.wiz

Google has many more.

 

[Rickharris]

A "normal" red LED (if there is such a thing) will draw about 10 mA and use about 2.5 volts fora reasonable brightness.

LEDs in series will need a voltage supplied that is the sum of their individual forward voltages i.e. for 8 LEDs 2.5 v X 8 = 20 volts

LEDs in parrallel will only require the supply to be equal to their individual forward voltage BUT will draw much more current - Their operating current x 8 so 10 mA x 8 = 80 mA

You have to make a choice depending on your available power and current.

 

[moxhamj]

Yikes Mr_H - re post #4, please don't do that to LEDs! You can't put leds in parallel. If you put leds in parallel, then each led, (or chain of leds) needs its own dropping resistor. But you can put them in series. Red leds - you might get 2 in series with 5V. Or 5 with 10V. Put a group of 5 in series, then work out a dropping resistor for that group. Work out the voltage drop for that colour. Multiply by the number. Subtract from the working volts (you might get 3V say). Then use V=IR with I=0.02A to work out R.

Mosfet is a good idea. A BC547 will also do the job. Good for 100mA which is more than enough. 5c each so won't break the bank and cheaper than a mosfet. Drive with 2k7. Do you want a schematic for all that?

 

[hippy]

Yikes Mr_H - re post #4, please don't do that to LEDs! You can't put leds in parallel. If you put leds in parallel, then each led, (or chain of leds) needs its own dropping resistor.

Is that absolutely true or just a recommendation ?

My view is that ( LED+R) || ( LED+R ) is the same as ( LED || LED ) + R. If the LED's need 10mA, in the first each R has to pass 10mA, whereas in the second the single R has to pass 20mA, the current through FET / transistor / supply is still 20mA total. The difference being the wattage of the R, the more current the hotter it will get.

A tri-colour, common cathode LED can either have separate R's on its anodes or a single R in its cathode-- I've used them both ways.

Am I missing some important knowledge ?

 

[Michael 2727]

Like Christmas Tree lights (in series) if you blow up one LED none will work.
In parallel or series and using a Single Resistor only there is more likely to be
LEDs that will appear brighter/dimmer than the others.

LEDs in parallel with individual resistors are less likely to produce varying
brightness between other LEDs.
The one resistor method is the cheapie/shortcut version but it still works.

I have Chinese made LED torches (where else) with 9 and 10 white LEDs
every LED is slightly different, you don't usually notice this until the
battery is going flat, they use a single resistor for all the LEDs.

In some applications this would be very annoying.(spot the dodgy pixel)

 

[moxhamj]

It would very much depend on the voltage matching between the leds. If they all came from the same batch, you could get away with it. If, however, the leds came out of my parts drawer, some will be made this year, and some will be 1970's vintage. It is very unlikely the voltages (or even the colours) will match.

Maybe I'm just a purist at heart, but I like to use dropping resistors in the proper way. Technically, with a picaxe, you don't even need dropping resistors at all for driving one or maybe even two leds out of a pin.

Ballpark, I'd put 5 leds in series and see how bright it was with something like a 270 ohm resistor for that string.

 

[boriz]

I’m sure many of you know this, but for those who don’t…

“( LED || LED ) + R” should never be done. It relies on an exactly matched forward voltage drop for each LED, and even a tiny difference will significantly bias most of the current to the LED with the smallest fV, thereby heating it more than the other which reduces it’s fV even more. Thermal runaway. Same reason why you can’t use normal Diodes in parallel.

The advantage to having LEDs in series is obvious. Each LED drops it’s fV and the remaining voltage is dropped by the resistor. The higher the voltage dropped by the resistor, the more power is wasted by it as heat. So the closer you can get to the supply voltage with your series fV’s, the less power is lost in the resistor for any given LED current.

EG1:
12v supply and 3 white power LEDs, each needing 300mA with a fV of 3.5v. (1 Watt LEDs)

Using LED1+R1||LED2+R2||LED3+R3 setup, each individual LED+R circuit will use 300mA and will have 3.5v dropped across the LED and 8.5v dropped across the R. That’s (3.5v*300mA) = 1.05 Watts dissipated by the LED and (8.5v*300mA) = 2.55 Watts dissipated by the resistor. And for all three parallel circuits that’s a total of 10.8 Watts of power required to get 3 Watts of light. An efficiency of around 27%

EG2:
Same supply and LEDs. This time with the LEDs in series.

Using LED1+LED2+LED3+R setup, each LED drops 3.5v@300mA, a total of 10.5v. The R drops what’s left over, 1.5v. So (10.5v*300mA) = 3.15 Watts dissipated in the LEDs and (1.5v*300mA) = 0.45 Watts dissipated in R. A total of 3.6 Watts power required for 3 Watts of light. An efficiency of around 83%

Series is always significantly better.

 

[jsch2603]

This also is a good LED calc. link:
http://ledcalculator.net/

 

[boriz]

This is why using a simple flyback boost regulator (like this) can often be a more efficient system than using resistors.

 

[hippy]

Voltage drop across individual paralled LED's is the factor I hadn't considered so thanks for that.

In practice I've never seen the thermal runaway problems or the worse case scenarios but it does depend on how they are used so I'll accept it as a bad method of choice. The only situation where it is more acceptable is where only one LED at a time would be lit and the issue then becomes reverse voltage through the LED. That's how I usually use this configuration and at low current which is probably why I've never noticed any problems.

 

[mr_H]

Thanks

I am grateful for all assistance. There is much that I need to learn and I appreciate your help in communicating a solution for me. I have followed the discussion keenly.
I have used my circuit design software to make a schematic diagram for the picaxe using the BC548B (this seemed to be the closest to BC547 which was suggested). This works.

So I have been able to get the MOSFET IRF520 and the transistor BC548B to do the job. It will come down to a matter of price which I will find out tomorrow.

Cheers,
Harry

 

[eclectic]

Harry.

re. your switch on input 3.

Have a look at Manual 3, page 25.

e

 

[BeanieBots]

The BC548B is only good for 100mA. Probably not a good choice for driving 8 LEDs. I'd suggest the IRF520 to a be a much better choice.
Also, as pointed out by eclectic, your input switch/resistor arrangement won't work reliably.

@Boriz, nice LED driving writeup.

 

[Michael 2727]

The IRF520 MOSFET is a TO-220 package, and slight overkill being able to pass 9-Amps or so, having said that, it's unlikely to be blown up on a small battery supply.
The 2N7000 MOSFET will handle up to 200mA and a TO-92 package, but kids being kids
may manage to destroy a few them. Although they should be a lot cheaper than the IRF520s.
(MOSFETS can be destroyed by Static Electricity be aware of this when handeling them, mainly lose.)
The BJT transistors BC548/others should be selected on their current carrying capability, plus a 20% to 50% margin for good luck.

If you used the IRF520 MOSFET you could supply 100 LEDs or many more with the
same setup. A MOSFETs GATE requires very little current but 3V to 5V to turn on.

On the other hand a BJT transistor (with a gain of say 250) and 100 LEDs you would need
at least 8mA of current into the BASE for a similar result.

Just about any NPN/N-ch Transistor/MOSFET capable of 200mA to 250mA will work
with your current setup and the few LEDs you have.
Buying in bulk (100) may save you up to 90% on the single unit price.