Powering LPC-815 without a driver...suggestions?

Hi,
I have a very small Cree lamp and a nice little sink with an acrylic lens that fits into it perfectly...problem is there is really no room for a driver...maybe a resistor and Diode...I've heard of powering diodes using just a resistor before. Can anyone suggest an alternate setup without a driver? I am using a 18500 3.7v cell to power...maybe add a diode or two to drop the voltage and a resistor to limit the current? If this is doable then exactly what values would be recommended? I have room for a 3w metal film resistor and a couple of diodes but that's about it.
Thanks for the help!
Jeff

You should never power a diode without using a driver. If you want to use only one 18500 battery you have to buy a flexdrive V5. Otherwise you can make a DIY Laser Driver
but you will need at least one more battery.

Jim

It is possible to use just a resistor between the diode and the battery. The problem is that the resistance of the diode isn't lineair. You have to do calculation to reduce the current.

Sure you can power a laser diode with just a battery and a resistor !

You will have to calculate the max current flow at max cell voltage and it will be down hill from there :beer:

jimdt7 said:

You should never power a diode without using a driver. If you want to use only one 18500 battery you have to buy a flexdrive V5. Otherwise you can make a DIY Laser Driver
but you will need at least one more battery.

Jim

Click to expand...

The question isn't should I...it's can I...
If I thought a V5 would fit properly I would use one...I don't think it will and I don't want to spend another $25 on this particular project.

Now that you mention it...making my own DDL is not entirely out of the question. I have some LM1084 ICs but my concern with this is the drop out voltage on a 3.7v cell. If my understanding is correct the voltage across a red diode is about 2.4. With the dropout of 1.25 that leaves me with 2.5v to work with...granted the cell will start out around 4.2v but eventually will drop below 3.65 which is my margin for the dropout. Given this...a DDL should be possible with a single 18500, correct? When the voltage drops below 3.65v the diode should just power down. Or, am I wrong in my assumptions?

Putting a LM1084 with a 3R SMD across the Vout and Vadj might fit but it will be very tight....

Go for it do (4.2- Vfwd) /R = I then know the laser will be less powerful as the batter gets weaker and has no protection.

Jmillerdoc said:

The question isn't should I...it's can I...
If I thought a V5 would fit properly I would use one...I don't think it will and I don't want to spend another $25 on this particular project.

Now that you mention it...making my own DDL is not entirely out of the question. I have some LM1084 ICs but my concern with this is the drop out voltage on a 3.7v cell. If my understanding is correct the voltage across a red diode is about 2.4. With the dropout of 1.25 that leaves me with 2.5v to work with...granted the cell will start out around 4.2v but eventually will drop below 3.65 which is my margin for the dropout. Given this...a DDL should be possible with a single 18500, correct? When the voltage drops below 3.65v the diode should just power down. Or, am I wrong in my assumptions?

Putting a LM1084 with a 3R SMD across the Vout and Vadj might fit but it will be very tight....

Click to expand...

Unfortunately, that won't work.

The v-drop will be around 1.25v when regulating voltage, but when current regulating it's more around 2.5v. 1.25v drops on the IC and another 1.25v drops on the resistors

Edit: Nevermind

You can power it with a single resistor. We strongly advise against this because

Reverse polarity will possibly kill you diode.
Testing is required since the forward voltage differs between diodes.
The current (and therefore brightness) rapidly drops with use.

Jeff;

I've done several of them.

I calculated a 4.3 ohm resistor for a:

lithium cell @ 4.1 VDC,

red laser diode @ 2.8 VDC.

(4.1-2.8)/4.3=0.302amps=302ma

This should give ~300 ma with adequate heat sinking.

1 watt resistor should be adequate.

Red laser diodes are sensitive to static discharges.

LarryDFW

The problem arises when the cell voltage drops to its nominal 3.6V. Current drops to (3.6-2.8)/4.3=180mA. It only drops further from then on.

You may want to add a capacitor just to be sure u have a long diode life, and 4Ohm resistor will work As Cyparagon said..
I experimented on this extensively and as it turns out, u wont need another thing and with a 3W resistor u wouldnt have to worry about it getting hot.. I used a 10 Ohm 2W resistor to test a LPC diode and a 18650 3.7V at about 4.1V
I was really amazed because the lpc diode was in the stock heat sink from the sled.. and it ran for about 45minutes with no signs of getting hot.. Maybe because it was running at a low current because of the 10Ohm resistance..
Here is a photo..

EDIT: Ok i just tested it with a 3.3Ohm resistor, and a 18650 @3.9V i didnt had time to charge it.. as the calculations say im getting almost 330mA~340mA
The diode is LPC 815, It worked quite nicely and better than expected imo.. I dont see any downside of this.. If you are too dumb to know which way to put the batteries, i dont see any reason why this setup is not recommended... With Aixiz module i was able to run it for a duty cycke 20min On and 1 minute Off!