This what I'm using now for small LEDs and it seems to work fine for that. Where it says "low" it's hooked to + voltage on an SMD LED that acts as the trigger.
Yeah, but major differences!
* The LEDs here run from 5V, you're talking about 9V
* The 2N5088 has a max current of 200mA a 3W LED requires well over an amp (depending on colour)
* Based on some assumptions, that circuit allows 300 mA to the LEDs and is in excess of the transistor's max current.
Where is says "repeat for 24 leds", it's actually driving 100 with no problems at all.
If you don't change the resistor, the total current will be far less (and will also probably limit the immediate risk of thermal runaway -- see below)
The 2n5088 never gets warm at all,
If that transistor doesn't get hot, it's because of one of the following:
* it's blown and is now a short or open circuit.
* it's only turned on for very brief periods
* it's not a 2N5088
* you're using a resistor > 3 ohms (perhaps 33?)
In any case, it's a terrible circuit. You're driving the LEDs in parallel (thus form what is essentially a voltage source). LEDs are not like light bulbs.
A light bulb starts with a relatively low resistance, and as current flows the wire heats up which increases its resistance. If 2 bulbs are put in parallel they will share the current because if one gets slightly more current it heats up more which reduces the current it can draw.
A LED may draw (say) 20 mA for a given voltage (let's call it 3.2 volts). But another LED form exactly the same batch may draw 21 mA at that same voltage, and another 19 mA. That doesn't sound like much of a difference, but the one drawing 21 mA will heat up slightly faster than the one drawing 20mA. The BIG difference is that the LED uses MORE current at a given voltage as it heats up. So the one drawing 21 mA starts to draw 22 mA -- that leaves less for the others which cool down a little and then require even less current. The additional current flows into the first LED which gets even warmer and draws more current, and so on...
Eventually one LED draws too much and fails. Then another LED takes up its place until it too fails (although it fails faster). Soon ALL the LEDs have failed. This is a major failure mode on LED Christmas lights that often have many LEDs in parallel.
If LEDs are placed in parallel they should each have their own current limiting resistor. If placed in series, each string requires a resistor.
but I'm guessing it would with a 3watt LED, so I'd need a bigger one.
Sure would.
I have a few questions:
Overall is there a problem in this plan?
Other than excessive power dissipation, for a single LED it could be made to work.
For the power in the diagram, I use a 7805 to regulate that 9v to 5v for the circuit.
Any issue with that? Or is 3 watts to much for the 7805? If 7805 is ok, How many LEDs can it drive?
The 7805 can supply between 100mA and 5A max depending on the type and pacage. I'll assume you have the fairly common TO-220 version which is typically rated at 1A. I'll also assume that your 9V source remains at about 9V under whatever load you are using. Oh yeah, I'll assume that your LEDs are white or blue.
You could probably drive a single 1W LED or more than 15 normal LEDs with that circuit and that regulator.
It would be able to supply almost enough current for a single 3W LED (although remember that the 7805 is not actually current limited to 1A) or up to 500 normal LEDs operating at 20mA
However your question is really nonsense. I could drive 500 LEDs at 20 mA, or 1 at 1A (it had better be a 3W LED), or 1000 at 10mA, or 1 million at 1 uA. It all depends on the current because, as I've said, LEDs are driven by a current source, not a voltage source.
I have a 7805 driving 500 LEDs now and it doesn't get hot.
This means that the average LED current (if we go by your circuit) is about 0.6mA (total of 300mA for all LEDs) and the power dissipated in the 7805 is about 1.2W. It should get warm.
With your circuit you can drive as many LEDS as you require, but as you add more LEDs they will get dimmer and dimmer...
Dissipating 1.2W, a 7805 will reach (without a heatsink) a case temperature 60 degrees C above ambient. In pretty much ALL ambient conditions (excluding Siberia in winter), this would be hot to the touch, and you wouldn't be able to keep your finger in it. It may be that your LEDs are not on for very long, or that your power supply is sagging under load, or that your LEDs are actually drawing far less current. Any of those might explain why your 7805 is not hot.
With the 3 watts I'd do the same, but is there something wrong with a couple of small diodes to drop the voltage from 5v down .7v twice to get down to 3.6 volts. Then the LED resistor could be quite small or even not needed, yes?
No, you need a current source, not a voltage source.
Whilst small LEDs are fairly tolerant (especially when lightly driven), high power LEDs need to have heatsinks and carefully controlled current if they are to be driven to maximum brightness (unless you don't care about their longevity).
The 7805 will dissipate 4W (and get pretty hot without a heatsink). The specs tell me it would reach a junction temperature of 260C above ambient -- which would cause it to self destruct if it didn't have an over temperature shutdown. Incidentally the case would reach a temperature of 240 above ambient which is easily enough to start a fire.
Presuming it has a heatsink, the diodes would drop about 1V each at 1A leaving less than the voltage required to drive the LED at full brightness, so the current would be less.
HOWEVER as the LED and diodes (and possibly the transistor) heat up, the current would increase and increase until either the regulator shut down, or the LED or Transistor started smoking. If you were really lucky, you might be able to destroy both the transistor and the LED and then the regulator.
It might make a good youtube video.
On the other hand it might just sit there working...
But either case is possible and it all depends on the tolerance of the various parts, the temperature, and a number of other random factors.
But please feel free to try.
On balance, I'd say that 2 diodes *might* be enough to prevent everything from going really pear shaped.
I also think you might be safe driving at 40 mph with bald tyres on ice. Hey, it might work, you might not slide. That intersection you slide through *may* be empty. (I don't drive that way)
Why do I need a darlington pair with the bigger LEDs, if this is working with what I have? Is it just that one might assume the base voltage of Q might not be high enough to turn on a single transistor? In this case it seems to be.
Because you'll be switching a heavier load and transistors that can handle the heavier load generally have lower gain. The lower gain translates into the need for a higher base current, and this can be provided by another transistor, typically connected as a darlington pair.
Can you say why some circuits suggest a MOSFET? Is it because they use a lower "trigger" voltage, so you can skip having a darlington pair?
No, they require a lower gate current, but at a higher voltage. With 2V, there are very few mosfets which will turn on sufficiently -- most require 5 to 15 volts.
I guess I want to know why I can or can't I basically do exactly what I've done already but just beef up the transistor? And maybe power supply.
You can, but it may not work, it may cause components to fail, or get very hot.
I see all these fancy LED drivers, but I don't see why I really need that..
It's because people with far more experience that you or I have determined how to get the best out of LEDs and the available power sources.
If excessive heat is not a problem, you can probably do it in a far easier way.
I live in a hot climate. Excessive heat increases my air-conditioning bill.
My parents travel in a mobile home. Excessive power dissipation reduces the life of their batteries (or increases fuel costs)
Some people think that efficiency is good (perhaps they pay for power).
I can see the advantage though if you live in a very cold climate and don't use a heat pump (which would generate heat more efficiently than your circuit)