Dear All,

Its been a while since I last posted, hope everyone is keeping well.

This question may be based on the naieve notion that IR diodes act pretty much the same as LE diodes.


I purchased 10 IR diodes from farnell (next day delivery - wow!) http://www.vishay.com/docs/83780/tsks5400.pdf

According to the documentation the maximum forward (continuous) current (If) is 100mA and the basic forward voltage (Vf) is between 1.3 and 1.7 volts.

So I set up a basic circuit with a 5 volt source, an IR diode in series with a 39 ohm resistor, because:-

(5 Volts - 1.3Vf)/ 100mA = 37 (39 E12)


Ok, with a 39ohm resistor I would expect to see current of around 94mA which is close enough.

But the measurement I get is closer to 75mA.

The only way I may be able to account for this is if the Vf of the diode isn't 1.3, actually it wouldn't be 1.7 either - more like 2 Volts.

I have checked the resistor and it is bang on 39ohms so no problem with tolerance there.


Have I mis-read the datasheet, or am I making some bad assumptions?

Best Wishes

Mark

 

Why can't you just measure it using your voltmeter instead of guessing?

Oh, I see. Your DVM is wired up to measure current, do you?

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L.Chung

 

This is why buying a couple cheap meters is a good idea. I just pulled up a datasheet from google for a random 100ma IR LED and it said the median voltage drop was 1.5 volts but its max was 2.1, so this doesn't sound all that unusual. I don't know how much bulk resistance there is in diodes of that type but at 100ma of current bulk resistance should cause some amount of that voltage drop.

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Actually I did (I dug out my other DVM), I measured the voltage at the IR Diode\resistor junction and ground and it measured 3.66 Volts.

But thats just proving that it lost around 1.3 getting past the diode, so I am having a hard time accounting for the current drop. unless as Sceadwian suggest these things have an apreciable resistance - but wouldnt that be in the datasheet....

Mark

 

If you have a scope you can check for excess ripple on your 5V regulator.DMM's dont take average of non-sinusoids accurately.You can also check if your 5 volt rail is actually maintaining 5volts. If you are pushing the regulator to it's max output power it may drop it's output voltage,or it may just be a cheap PSU and have poor output regulation.You really need a scope to check these though.This assumes you've already eliminated your instrument / measurement as a the source error.

Oh and if you are using a standard carbon film resistor your R reading will be different when it's cold or hot. A +/- %5 tol resistor will vary %5 over it's temperature range (2 ohms for your 39 ohm R).

Good luck

 

Thanks for the response, the power from my regulated supply if 4.96V

I have run the scope on it, and I can't see anything unusual.

 

Do you know the tolerance of your resistor I'm guessing it's at least a 0.5W resistor.The resistance will increase with temperature rise of the resistor. So if it's a %5 resistor its hot resistance maybe approaching 41 ohms.

If you have as you say 5V in and 3.66V across the resistor but are only measuring 75mA and nothing else is in the circuit. I would guess it's a combination of your meters accuracy (OR LACK OF) as well as the voltage drop your test instrument is contributing (This should be at most a couple hundred mV),combined with the increased resistance as the resistor heats.

Try a different DMM?

You really shouldn’t be driving an LED with it’s max current (if you are) those specs are for 25 deg Celsius.

IR LED’s are usually pulsed.

 

Go figure the power dissipation of the 39Ω resistor.

It's about 370mW!

Use a 1/2W resistor to limit the current.

A standard 250mW resistor won't stay cool under these conditions.

Boncuk

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Proper Planning Prevents Piss Poor Performance