Connecting a Diode: Tips for Li-Ion 18500 Battery

[richleyance]

I've done my background research on this but i want to double check with others since the last time i did this i blew $30 because i shorted my diode. I got a new one but I'm really nervous about starting it up. These are the two parts https://www.amazon.com/gp/product/B00HFHJBDI/?tag=pfamazon01-20
http://www.ebay.com/itm/like/231594023994?lpid=82&chn=ps&ul_noapp=true
and I'm using a Li-Ion 18500 Cylindrical 3.7V 1400mAh Battery

This is what it looks like right now
http://imgur.com/wHcVu0Q
http://imgur.com/SlGESEl
http://imgur.com/wOfVpPP

And yes i know its a lousy soldering job, but that's the best i can do with two hands and no solder stand.

 

[davenn]

Hi there
welcome to PF

and I'm using a Li-Ion 18500 Cylindrical 3.7V 1400mAh Battery

so why are you using a 3.7V battery when it specifically states a 5V input for the driver ?

these are the connection points for the power to and from the driver board ...

so why have you put your wires in totally different places ?
particularly referring to the + and - 5V input

the connection locations are very clearly labelled. DONT try and do your own thing
that's what leads to equip failureDave

 

[Integrand]

It states <5V and looks like a boost circuit so using a 3.7V cell could make sense. It's hard to say without more information about the design than that photo shows.

About the connections, at first glance they look fine, if you check both sides of the board. What am I missing?

 

[ilik]

strictly follow the guides. solder +5V as directed on picture and nowhere else(do not exceed 5 volts). if diode burns that means current regulation is not working properly. do not use 3.7V as source. use +5V because driver circuit is engineered the way that it expects +5V. one more thing, unfortunately your driver does not come with a datasheet. never buy products without datasheet. because the statement +5V -5V might mean 2 things . one: (+5v,0v) and two: (5+V, -5V). some regulators work on bipolar voltage only. try to use the microscope (or magnifier )to read out model number of 6 pinned chip that is located in the center. that thing is a voltage regulator, possibly wired up in a current regulation mode. and tell us the model number. we will try to think something then...

 

[Integrand]

I remain unconvinced that <5V indicates 5V. Curious to know more, I just tried an ebay search and found numerous variations of this driver. It is indeed a boost circuit decided to run off cells like those the OP has chosen. It seems the output voltage of the boost circuit is 5.5V. I remain convinced that the OP has not erred by soldering the battery wires to the same nodes on the other side of the board - check the vias.

 

[davenn]

I remain convinced that the OP has not erred by soldering the battery wires to the same nodes on the other side of the board - check the vias

there is NO obvious/visible via where the +V input is supposed to go. From the indicated pad, it goes directly to the inductor and the cap

 

[Integrand]

there is NO obvious/visible via where the +V input is supposed to go. From the indicated pad, it goes directly to the inductor and the cap

I agree with you. I was basing my comment on the apparent vias to the inductor on the second of the pictures in the OP's ebay link to the driver.

We still need more information. Even if these connections are correct, we still can't say whether it is a correct circuit without knowing the current-limiting capability of the driver, for example, as ilik said.

 

[davenn]

We still need more information. Even if these connections are correct, we still can't say whether it is a correct circuit without knowing the current-limiting capability of the driver,

all the info is given for both the LD and the driver, can't see any reason why they wouldn't be compatible
it's a constant current source of 1.8A and states its suitability for the given diode he is interested in Dave

 

[CraigHB]

Whatever the case, test your circuit initially with a current limited supply. Using a Li-Ion battery for testing circuits is not a good idea since they can deliver currents high enough to fry things, or even the battery itself. Typically you would use a benchtop DC supply with a current limit set appropriately. If all you have to supply power is your Li-Ion battery, an inline fast blow fuse can do the job. A PTC fuse can do the job as well. It's a small effort to avoid blowing expensive parts.

 

[ilik]

all the info is given for both the LD and the driver, can't see any reason why they wouldn't be compatible
it's a constant current source of 1.8A and states its suitability for the given diode he is interested in Dave

well, correct me if i am wrong but; constant current source would definitely fry a laser diode. because as the diode heats up it increases it's resistance.
so if heated diode would rise it's load impedance to say 1 ohms,then 1.8A*1 Ohm = 1.8volts. so, 1.8volts (drop)* 1.8A = 3.24 watts. = FRIED DIODE.
idea in driver circuit is not to give a constant current, but to give constant voltage. so that when diode rises it's resistance - current goes down. and this way power dissipation of a diode never exceeds nominal 2 Watts.

 

[davenn]

well, correct me if i am wrong but; constant current source would definitely fry a laser diode. because as the diode heats up it increases it's resistance.
so if heated diode would rise it's load impedance to say 1 ohms,then 1.8A*1 Ohm = 1.8volts. so, 1.8volts (drop)* 1.8A = 3.24 watts. = FRIED DIODE.
idea in driver circuit is not to give a constant current, but to give constant voltage. so that when diode rises it's resistance - current goes down. and this way power dissipation of a diode never exceeds nominal 2 Watts.

yes, you are wrong

the driver is specifically designed for the LD

 

[CraigHB]

LEDs are current driven devices and they are rated in terms of current. Forward voltage does change with temperature, but not enough to create a substantial difference in power consumption. As long as current is within tolerance the LED will operate nominally. The effect of temperature on forward voltage will not cause an increase in power consumption resulting in failure. That's assuming operating temperatures are within tolerance.