How the hell did I burn it? - Page 1
Forum section: electronics / beginners

author: miceuz
author_rank: Frequent Contributor
timestamp: September 14, 2017, 07:35:12 am
content: I was hooking up these LDD-300 current sources to an off the shelf PWM dimmer and got them burned. I kinda know what am I doing, and it's not the first time I am doing this by any stretch, but I burned them all and I don't get why.I have recreated my system in the diagram below - outputs of RGB dimmer are bog standard N-channel mosfets, everything should just work, and have worked the first time I tried to hook up the current regulator. Then I have added longer wires and all of them went cactus as soon as I touched the dimmer potentiometer. I have started with potentiometers all to the right, so all the dimmer outputs were clamped to ground. As soon I have switched PWM via output fets the current source went cactus.I was being lazy and tried to do all of this in a mess of flying wires. I have marked wire lengths. The only reason I imagine is VCC jumping over 32V because of the wire inductance, but it's just 300mA!

author: rstofer
author_rank: Super Contributor
timestamp: September 14, 2017, 01:42:33 pm
content: I have never used one of these devices so I have no idea what I'm talking about!According to the datasheet, the LEDs should be connected between Vout+ and Vout-.So, I guess I wonder about the anode connection to the LEDs.http://www.mouser.com/ds/2/260/LDD-L-spec-47593.pdf

author: Twoflower
author_rank: Frequent Contributor
timestamp: September 14, 2017, 04:56:46 pm
content: If the FET is off, what voltage do you expect on the DIMM input? In your drawing it is just floating in the free air. Unless there's a pull up resistor somewhere.Could it be that the GND of the PWM-Source is not the same ground of the remaining part? Also it's probably better to use a second GND-wire for the FET and driving circuit to minimize ground-loops.A filter-Cap at the input DIMM of the current source won't hurt I would say.How do you drive the FET?

author: miceuz
author_rank: Frequent Contributor
timestamp: September 14, 2017, 06:05:24 pm
content: The FET is driven by the internal circuitry of the chinese dimmer - there is LM358 driving it in some comparator configuration I guess.Vout and Vin buzz out to be connected - about 0.6Ohm.When the FET is off, Dimm is left floating, yes, but datasheet says "if unused leave disconnected", so I expect it to be pulled up inside, although I could not measure it - I have only dead ones and those have output short to ground.Ground references are the same and connected to the same point.

author: Twoflower
author_rank: Frequent Contributor
timestamp: September 14, 2017, 06:46:19 pm
content: You're right, no filter C and open is OK. I should have read the Spec first. It looks unusual, so I noted it.The only thing I could imagine is that the LED connection was wrong. Pin 4 and 5 arenotidentical. The datasheet says do not connect them. So connecting the LED on pin 4 instead pin 5 could probably caused the problem. But without knowing the internal circuit it's hard to say. It could be a shunt in between, some filter, ESD and/or EMF protection circuit.Edit: Typo corrected

author: glarsson
author_rank: Frequent Contributor
timestamp: September 14, 2017, 07:01:31 pm
content: Quote from: miceuz on September 14, 2017, 06:05:24 pmThere must be a reason why these two are separate connections. Not unrealistic to expect a current shunt here - a shunt used to measure the current through the LED and output stage. With the LED connected to 24V the device is used with the current shunt bypassed (i.e. without current limiting) and the LED or output stage will let the magic smoke out. Why not follow the manufacturers instructions on how to wire up the LDD device?

author: rstofer
author_rank: Super Contributor
timestamp: September 14, 2017, 07:04:52 pm
content: Quote from: Twoflower on September 14, 2017, 06:46:19 pmYes, the datasheet is very specific out not connecting Vin- and Vout- probably because there is a current measuring resistor between them. The control, seeing no current flow, goes to full output which probably over-currents the module and the magic smoke pours out. Or some variation on that theme. This makes sense because the default output is full ON.It would be worth the experiment to wire the setup as required by the datasheet and see what happens.

author: miceuz
author_rank: Frequent Contributor
timestamp: September 14, 2017, 09:30:19 pm
content: Thanks all, it's probably me trying to be too smart - maybe I should have used Vout. Maybe I has using it when I was hooking up the first test, I don't remember now. BTW datasheet does not tell anything about connecting Vin+ and Vout+, they are talking just about Vin- and Vout-Another thing - if I have three leds, can I connect all three Vout together?

author: glarsson
author_rank: Frequent Contributor
timestamp: September 14, 2017, 09:43:30 pm
content: Unless the manufacturer tells you it is safe you shouldn't (unless you intend to burn up some more modules).

author: rstofer
author_rank: Super Contributor
timestamp: September 14, 2017, 10:31:56 pm
content: Reading the datasheet, I noticed that this module is a constant current source and the output is 300 mA. That's HUGE! There would need to be 15 ordinary 20 mA LEDs in PARALLEL to soak up that much current.Yes, you can put several LEDs in series but you still need multiple parallel strings to use up the current. But you said you have used these before so I suspect you have this under control.Just thinking about how they achieve the current source and output voltage, I wouldn't think you could do anything with the Vout +- except run them to the LEDs. Completely isolated from Vin +-

author: glarsson
author_rank: Frequent Contributor
timestamp: September 14, 2017, 10:47:09 pm
content: The current is designed for 1W LEDs (the star-shaped metal PCB with a LED mounted on it) in series. The number of LEDs in series depends on the voltage used to power the LDD device.

author: chris_leyson
author_rank: Super Contributor
timestamp: September 14, 2017, 11:14:38 pm
content: Looks like a constant current buck converter running at 1MHz, internally the device is probably something similar to the LM3404 and there is most probably a current sense resistor connected between Vout- and Vin- and the LEDs should be connected between Vout+ and Vout- pins. If you connected the LEDs anode terminal to Vin+ then you've bypassed the buck converter and that is one reason why it got fried. I also noticed that you are running 1m of cable from Vout to the LED string, connect a bypass cap between Vout+ and Vout-, start with say 1uF 50V X7R cap, that will reduce the 1MHz ripple current in the LED string but the LED average current will stay the same. Look up the LM3404 data sheet and that will give some idea as to how buck LED drivers work. And by the way, 300mA is nothing these days if you are driving something like Lumiled Luxeon or similar LEDs, the LM3404 will manage 1A without problems providing you've got adequate thermal design.