A little trivial EE help please

I am a little embarrassed that I’m having trouble figuring out this problem with a high level of confidence. I have this LED that I’d like to power with 2 AA batteries. Here is a link to the data sheet for the LEDs I have: http://www.vishay.com/docs/81011/tsal6400.pdf

I think I need a 68 ohm resistor but it sounds too high.

R = (VS - VL) / I

I guess my question is: am I reading the data sheet correctly?

VF has two test conditions at 100 mA and 1 A. Can someone explain the difference is between these? (yeah yeah 900 mA :wink: It is saying that it can handle a lot more if you are using very short pulse durations? That would make sense. However I am going to have it on continuously for several minutes at a time.

Can I go anywhere between 18 and 68 ohms given my usage profile? I would like this thing to be as bright as possible but obviously not burn out. How does one properly read and interpret these data sheets?

Thanks!

I have some experience with high-power LED’s.

The forward drop will vary with current, like any diode.

For high power output you are limited by heat. Most LED’s suitable for continuous high-power output have packages suited to transferring heat - like any high-power semiconductor. Just looking at the data sheet - the LED you have is not in this type of package. So, you can get high power out of it, but only for brief periods - hence the spec for 100 us pulse duration.

You need IR spectrum?

Most high-power continuous stuff I’ve worked with are visible light diodes that are actually UV diodes with a phospher coating.

http://www.philipslumileds.com/

Check out these folks.

The 100mA is the maximum steady state current dissipation. The 1A is a maximum pulse or inrush current. You’ll note that the 1A is only rated for 100uS pulse.

If you’re going to keep it on for minutes at a time, you need to keep under the 100mA maximum current. That, plus your supply voltage, will establish a minimum Rs.

At that current, you’ll have a Vf of ~1.3V. When powered by two AA’s in series, your Vcc will be ~3V, which leaves a drop across Rs of 1.7V. To keep 100mA max across 1.7V, you’ll need a minimum R of 17 ohms.

From Fig. 4, it looks like you need 1V to get it to tun on at all, and you’ll have 1mA If. That leaves 2V across your Rs, which spots its max value at ~ 2kOhm. The more current the brighter, so you want to be on the low end of that range.

You’ve also got to choose standard values. 18ohm would be pushing it - no room for high voltage batteries, warm temperatures, resistor tolerance, etc. 68 ohms would still give a nominal 25mA at 25C. That should also give you about 100 hours of use on a set of batteries.

If you want to play around with it, put an 18ohm resistor in series with a 2k pot, and dial the pot to get the best brightness that doesn’t fry the LED. I’d only do this if you’ve got some spares. Otherwise, go with maybe 33ohm or 47ohm fixed.

This advice is of course worth what you paid…

I should have backed up. What problem are you trying to solve?

Yeah, me too…

Thanks for the help guys. I ordered a few extra LEDs so I might experiment a bit. I’m also considering running it off of my usb port as well but I like the idea of wireless. Thanks again.

Warning - I make my living writing software, but did some EE a long time ago. Internet advice is worth what you pay for it…

The data sheet says the maximum rated forward current is 100mA, for which the voltage drop is guaranteed less than 1.6V.

It is rated for 1.5A for a 100us surge.

You will get the most light out of it at 100mA, but will obviously drain your batteries faster than if you used a lower current.

Assuming a slightly conservative Vf of 1.3V, and 3V from you battery source, you would need 1.7V/100mA of series resistor in your bias circuit, i.e. 17 Ohms. The resistor needs to dissipate 17*(0.1)^2 watts, or 170mW, so a 1/4 watt resistor should be sufficient. I think there are standard 18 Ohm resistors you can find.

You probably already know this, but just in case, 940nm is in the infrared, so you won’t be able to see the light from it…

I’m playing around with using my wii controller to make a multi-touch virtual whiteboard. ;):wink:

So you are attemping a serial link using a pulse-coded modulation like many standard IR links?

You could start here: http://www.maxim-ic.com/quick_view2.cfm/qv_pk/1947/t/al

In general, these applications do not drive the LED continuously. That’s why you have data sheets specd for pulse operation.

Sorry to bail, I was putting the munchkins in bed. :wink:

Actually in this case it isn’t pulse coded (well this part at least). The wii works in a unique way. The sensor bar on the TV is actually what is emitting IR while the wii remote is the IR sensor. That’s how it tells where on the screen you are pointing the wii remote for certain games. I am pointing the wii remote at a projected screen be it on the wall or table and using my little pen as the IR source. I plan to add a momentary switch in order to be able to turn it on or off for writing. I’m kind of fixing the remote and moving the tv around if you think about it. :wink:

Anyway the wii remote is bluetooth enabled so I can talk to it with C# code and get the position of my IR pen in relation to the wii remote. I’m not really using the accelerometer or gyrometer part but the IR sensor is quite accurate.

I saw it on a TED talk a while back and have been playing around with it ever sense. :wink:

It is not clear to me how the bar and the remote establish relative position - and this is key to your application.

Speculating - perhaps the remote has a narrow acceptance angle so that it sees only a portion of the emitter bar. The LEDs on the emitter bar are illuminated in sequence, each sending a narrow angle of light. So, you get a signal when your remote is pointed at a particular portion of the emitter bar. See? Seems like what you want to do is emulate the emitter bar. Looks like sequenced pulses that are co-ordinated with the application that determines where the remote is pointed relative to the target.

The sensor bar actually has 10 IR LEDs and the image sensor in the wii remote detects the location of the two bright dots within it’s field of view. That’s why I want my pen to be pretty bright. In the whiteboard example you flash 4 known corners of the screen in order to calibrate the fixed IR sensor’s field of view. From that point everything is fixed except the movement of my IR source which the location can be determined via triangulation.

http://upload.wikimedia.org/wikipedia/commons/thumb/2/20/Nintendo_Wii_Sensor_Bar.jpg/250px-Nintendo_Wii_Sensor_Bar.jpg

… The resistor needs to dissipate 17*(0.1)^2 watts, or 170mW, so a 1/4 watt resistor should be sufficient. …

Standard protocol is to double the power requirement for resistors. A 1/4 watt resistor is too low. I would go with a 1/2 watt resistor.

Hmmm. you are counting on the sensor picking up the reflected ( scattered ) light that your IR pen provides? It won’t be much compared to an LED shining right at it.

Seems to me you could accomplish what you want with a simple web cam and a laser pointer - the laser should look very bright to the web cam and a simple image processing job to average successive images together to get the persistence effect of ‘drawing’.

yeah I have been thinking about that. I may angle the LED out of the pen pointing back towards the sensor. I got all of my stuff today at radio shack. I’ll do some tests this weekend and report back.

An IR laser would appear as a brighter point source since the reflection would be specular.

Sweet! It is totally working great with a 1/2 Watt 68ohm resistor. It puts just about 25mA across the circuit.

I have a fully functional wii whiteboard. I can point the wii remote at my laptop screen or a projected version of my screen and with my IR pen I can write, click around and totally control my machine just like it is the mouse. How fun! :wink:

Thanks for everyone’s help!

Awesome! You’ll need to put this somewhere we can see how you did it…

The form factor of my pen is pretty ugly right at the moment. Once I get it into a decent pen-shaped case I’ll post some pics for sure. I’m using AA batteries right now but if I switch to AAA I think I can fit it all inside one of those giant magic marker cases really nice. I got the smallest momentary switch I could find and I think it will fit. :wink: