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13-10-2014 #1
Did you mount diode on a bit of board as I suggested? It's running a little hotter than I calculated. Junction temperature will be around 120 degC, still less than the maximum allowed of 150 degC, but 20 degC more than I'd like.
Last edited by irving2008; 13-10-2014 at 05:50 AM.
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13-10-2014 #2
This was only a quick test with a choc block, I'll need to cut up a piece of laminate and give it another go.
T.
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13-10-2014 #3
Hi, you should not need a diode your power supply will be polarized +ve & -ve your problem is you are trying to drop 44 volts which creates a lot of heat, if your voltage was a bit lower you could have used a LM317 or similar the best you could input to these would be 32vdc its a weird voltage drop to are trying to use my suggestion is to try to get hold of a NUD4011 http://www.onsemi.com/pub_link/Collateral/NUD4011-D.PDF
this should handle your voltage drop
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Hi and welcome to the forum. Have you read this complete thread because I think you have missed the point and purpose about the diode also all the math has been done. I am not trying to put you down in any way but this guy has stated he wants a simple way and he is short of space ..Clive
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14-10-2014 #5
Thank you for the welcome. I understand he has a problem with space but if you are using a 5Watt resistor you need a lot of space the chip i suggested can be mounted on a 15mm x 8mm pcb depends how you proceed, that would include a smd resistor also i use its brother the nud4001 so i know how small it is, the chip is a current limiting chip that is why it can be used with such high voltages up to 200Volts input as for the price i pay around 30p for the nud4001 the price for the nud4011 is a little higher..its only a suggestion he does not have to use it but many brains make light work lol i'm not sure it may even fit on the pads of a piece of vero board,
One other point " a LED is a Diode" a zener can be used to "clamp" the voltage" you just need to current limit it, I also have another chip in mind and the size of that is only 3mm looks like a smd diode 2 pins thank you clive
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15-10-2014 #6
Cropwell has a good point and running the LED at a lower current reduces the heat generated - assuming sufficient brightness.
You can't change the laws of physics - any linear analog solution needs to dissipate 0.88w @ 20mA LED current and using more esoteric devices doesn't change that. The NUD chip is a great solution for driving strings of LEDs and/or where input voltage varies. It's simply overkill for this requirement and adds unnecessary complexity. Incidentally the NUD chip would need 2sq in of copper (the same 40 x 30mm board) to stay within ratings and from the data sheet would run a junction temperature of 113degC and a lead temperature of 46degC due to marginally better thermal characteristics than the diode, but for 10degC difference its not worth the hassle.
@binary. Your assertion that a string of LEDs would generate no heat is wrong. LEDs are around 50% efficient and do generate heat in the junction. They don't emit IR like incandescent bulbs so don't feel hot through radiated heat. But a standard 5mm 2.2v red LED run at 20mA would consume 0.02 x 2.2 = .044W of which 50% is light and 50% is heat. A string of 20 would indeed drop 44v but would still generate 4.4W of heat, the other 4.4W being light energy. But bulky overkill !!!Last edited by irving2008; 15-10-2014 at 05:51 AM.
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15-10-2014 #7
Re. putting more diodes in series, maybe this is the next trend for control cabinets
http://www.custompcreview.com/review...eview/12513/3/Last edited by EddyCurrent; 15-10-2014 at 01:24 PM.
Spelling mistakes are not intentional, I only seem to see them some time after I've posted
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15-10-2014 #8
Btw this is the LED I'm using:
http://cpc.farnell.com/apem/q8f3cxxr...56?Ntt=SC09356
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15-10-2014 #9
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15-10-2014 #10
Maybe this thread would be useful ?
http://www.instructables.com/answers...er-dc-voltage/Last edited by EddyCurrent; 15-10-2014 at 01:51 PM.
Spelling mistakes are not intentional, I only seem to see them some time after I've posted
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