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LM3404 LED Driver chip

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cjurczak

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I am working with the National Semiconductor LM3404 that is a buck regulator I am using to drive a single high power LED. I threw together a small PCB to test and make sure I have everything working. And to my surprise it works the first time I turn the power on.

Datasheet: **broken link removed**

Attached to this post are some quick pics to give you an idea of what I'm working with...I know its not much, and the board is kind of messy, but it gets the job done...

Ok now for the problem...After the board runs for a few minutes at its designed output of 750mA, the current through the LED drops suddenly to ~18mA...After letting it run at for a few moments longer it settled at about 16mA and I noticed that they recirculating diode is hot to the touch...I unplugged the power everything cooled down promptly...And upon applying power everything repeated...

This time I allowed the circuit to run continuously at ~16mA the diode never got any hotter than before...So in an effort to cool it off while still running I blew onto the board...the longer and harder I blew the higher the current would go...Up to about 25mA...No where near the designed 750mA...

The diode in question is the Schottky B130LAW: **broken link removed**

It is rated at 30V and 1A in the SOD-123 package...From the LM3404 datasheet this seems to be a suitable diode for use in this situation...Unfortunatly I have poor selection of diodes on hand, and ZERO schottky diodes laying around...

I'd appreciate any comments or previous experience on this topic...This board is being use just as a proof of concept, since I have never used this chip before...Eventually I'd like to incorporate it into a larger project, but not unless I can iron out this bugs...

Thanks for all your time, and help...
 
well for starters you do not supply enough information: supply voltage, operating frequency, and LED voltage.

If, for instance, the LED is at 3.8V running on a 5V rail gives you a duty cycle of (3.8V+.3V)/5V=82% that leaves 18% on the diode...which is probably ok...

If, on the other hand, you are running from 25V (you said you are using 30V diodes) you have (3.8+.3)/25=16% or 84% on the diode. 0.25W for 50% duty cycle at 750mA per the graph. So conduction dissipation alone becomes 0.25*2*0.84=0.42W which makes the die temperature 0.42W*222C/W=93C...not much left there for switching losses.

Which brings up another question: did you go cheap on the inductor?

Dan
 
Sorry for the delayed response, I'm been on the road and I didn't have anymore specifics...but...

Supply voltage for the system is a 12VDC switching psu...Operating frequnecy is about 200kHz, set be a 150k resistor...The forward voltage of the LED is 4VDC, the forward current of the LED is 700mA, the dynamic resistance is 1.13 ohms...The inductor is a 22uH 1A inductor

The inductors datasheet is here... https://www.electro-tech-online.com/custompdfs/2008/12/pm54_series.pdf

Eventually this driver will be powering two identical LEDs in series...Which is why I have the input voltage at 12VDC...but while testing this chip out right now I'm just powering a single LED...

Would a higher rated diode with a larger package be of any use to help dissipate heat more effectivly, and hope to keep it operating normally...Or should I try to match my source more carefully to my load?

I seems dropping all this energy on the diode would be inefficient...I was hoping this design would be an energy efficient solution...

Thanks again for all your help...look forward to hearing back...
 
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