Given: Cree XM-L LED.
Want: Up to 2A drive, PWM controled by PC via USB.
This can be two parts. ie actual LED drive and PC to LED drive interface. These may or may not be integrated.
A "very easy" approach is to
1. use an off the shelf USB to "output" device. "Output" may be analog level, PWM, 8 bit port etc to control ...
2. An off the shelf LED driver that uses analog or PWM input.
For example, the circuit below using a RT8482 requires an analog input level or PWM with a simple RC filter (to convert the PWM to analog). The analog could be provided by a USB to analog output I/O device (COTS) or by a USB to parallel port device (not a printer port per se) (COTS) with a simple R2R digital to analog converter (about 16 resistors plus maybe a cheap op-amp).
Many examples of R-2R ladders here - links live
Or a microcontroller with USB capability could have a relatively simple program written to provide PWM or analog output. A USB enabled Arduino or a Raspberry Pi would do this. (USB has to be slave not host mode).
LED drive:
(1) "Off the shelf" complete units that do the LED drive part of this job well are available at good prices from eg ebay, or Mouser and similar. Using such is a good default solution unless you have some reason to do otherwise.
(2) DIY LED driver.
Digikey LED drivers are found here. Alas the parametric search is poor in this case (which is unusual).
Searching using LED driver 2A gives better results.
There will be a nummber.
Example only: For $US1.52/1 in stock Digikey you get
1
Ricktek RT8482, buck or boost, LED driver.
Drives external MOSFET so LED current capability essentially unlimited.
Looks like a good start. 350 kHz for smallish inductors.
- High Voltage Capability : VIN Up to 36V, VOUT Up to
48V
Buck, Boost or Buck Boost Operation
C u r r e n t M o d e P W M w i t h 3 5 0 k H z S w i t c h i n g
Frequency
Easy Dimming : Analog, PWM Digital or PWM
Converting to Analog with One External Capacitor
Programmable Soft Start to Avoid Inrush Current
Programmable Over Voltage Protection
VIN Under Voltage Lockout and Thermal Shutdown
16-Lead WQFN and SOP Packages
RoHS Compliant and Halogen Free
A MOSFET suitable for use as M1 would be eg ONSEMI NTD4960 $US0.40/1 in stock Digikey, 30V, 9A, 9 milliohm on resistance nominal, logic gate - data sheet curves show good at 4V gate and say 4A.
ADDED:
Should I be looking at specific types of inductors for this sort of application
Inductors are very special for best results. If this is a one-off then off the shelf inductors from eg Digikey or similar are wise. We can give advice in this when final real spec is known.
I'm assuming all of the caps in this type of application would be ceramic?
Ceramic capacitors will work well for all capacitors shown. At least 10V rating. More or much more voltage OK.
D1 is Schottky and should have current rating equal or greater than LED max current.
Now I just need to figure out how to generate the PWM signal.
PWM is "easy" [tm] and may not be needed. Above LED controller example can use analog or PWM control.
USB to I/O
This USB to paraell FIFO I/O module](http://www.ftdichip.com/Support/Documents/DataSheets/DLP/usb245r-ds-v10.pdf) uses FTDI's FT245R USB-parallell FIFO interface IC - datasheet here .
Vast amounts of related FT245 information here
FT245 available from Digikey ~= $US4.50/1 from here
FT245 based module from Digikey for about $40/1 here
This page discusses a DIY USB printer port which, as you have complete control over the hardware and how it acts, could "easily" meet your need. Based on a PIC18F4550 microcontroller and not much else. All software PCB patterns, circuit etc free.
Typical commercial USB to analog device
This will drive a relay from a micro-controller pin: -
The relays you have described are fine and putting a contact in live and neutral is fine too. Just make sure you select your final one to switch 10A and check to see it is rated for the voltage and has a good gap coil to contacts (or is specified as being suitably safe).
The transistor in the circuit has to be able to supply the relay coil current so it may need to be rated at 200mA depending which relay you choose. Voltage rating 20V or above. If you are using a "hungry" coil and it needs more than 100mA you might want to reduce the 10k to a 2k2 to make sure the transistor turns on properly.
If the relay doesn't have an in-built diode protection circuit then you'll need to add a diode to the circuit across the relay. It's normally not conducting with the anode at the collector of the transistor. 1N400x type will do.
Best Answer
The best solution is probably a opto-coupler. I would arrange it so that the LED is lit when the signal goes low. Optos are generally faster to turn on than off, and this way will use lots less quiescient current.
The output of the opto will be the collector and emitter of a floating transistor, usually NPN. Connect the emitter to the PIC ground and the collector to a PIC INT pin or a interrupt on change pin. Either enable the internal pullup of the pin if it has one, or provide a external pullup to Vdd. The highest pullup would be about 100 kΩ if you have a fairly long time (10s of µs) between edges, don't care about the trailing edge of each pulse, and current drain is important. I'd probably use 10 kΩ unless speed under a few µs was needed.
Added about your proposed circuit
You proposed the circuit:
No, this is not what I meant. It doesn't seem you read much of what I wrote.
There is no need for Q2. As I said, connect the emitter of the opto's transistor to ground and the collector to the PIC input pin. Basically let the transistor in the opto be Q2 in your schematic.
Also, as I said, you have to make sure the PIC pin is pulled up somehow. This can be by enabling the internal pullup, but requires a external resistor if not.
Another point you seem to have missed is that it would be better to turn on the LED in the opto when the pulse goes low. It is not clear you have done that. In any case, you can't just connect a LED to a 12 V signal. Figure the LED will have about 1.8 V drop (the real value will be in the opto datasheet), and you shouldn't need more than a few mA thru the LED. Let's say that after taking the current transfer ratio, the response time, and the output load into account, you decide the LED should be driven with 2 mA minimum. A 4.7 kΩ resistor in series with the LED would guarantee that when 12 V is applied.