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Can I run a diode with just a resistor?

IsaacT

IsaacT

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I have been trying to build a laser for my gf, but the host is fairly compact and no common drivers come in the current I want. So would it be possible for me to just use a couple resistors?

Thanks,
Isaac
 





Yeah you can if you want, but its not ideal.

What diode, what battery, and what current do you need? Maybe I can think of something better to use.
 
Diode would be THIS preferably.
Battery would be either 1x18650 or 2x18340's depending on which I need.
Current I need is 300mA-400mA. Anywhere in that range would be fine.

I am not experienced at all with modifying circuit boards or testing currents....in fact I don't even own a test load. So something like the Blitzbuck 290-970mA round boards would be unwise for me I think.

Thanks,
Isaac
 
You could use a resistor and one battery, buck with 1-2 batteries, linear with 2 batteries, very low drop out linear with 1 battery.

If you have an led or something you could use it as a test load for a blitzbuck. You could also use a mohgasm 428ma from cajunlasers, a little high but would probably work. You could also use an AMC7135 (350ma) if you isolate the module. If le-quack is selling them, you could use an E-drive. You could ask lazeerer for an x-drive at that current.

If you want to use a resistor 5 ohm should work well.
 
What current would 5 ohm give? I am not sure how to calculate it without using an LM317...
 
But what is V? 3.7V from the battery? Or does that fluctuate since I don't have a voltage regulator?
 
wannaburnstuff said:
But what is V? 3.7V from the battery? Or does that fluctuate since I don't have a voltage regulator?
Click to expand...

An 3.7V nominal 18650 is 4.2V fully charged and usually are run down to 3.6Vish before recharging. This will mean the current to the diode will go down during the battery cycle as it gets low.

From what I understand the reason it is not ideal to use just a resistor is that when a battery is going from no load to load there can be small surges of voltage which may not be huge but could throw the math off enough to cause start up current spikes that could damage or degrade the diode over time.
 
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wannaburnstuff said:
But what is V? 3.7V from the battery? Or does that fluctuate since I don't have a voltage regulator?
Click to expand...

V is your input voltage. This is going to change over time... since R is constant, while the battery voltage drops so will the current.
 
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wannaburnstuff said:
But what is V? 3.7V from the battery? Or does that fluctuate since I don't have a voltage regulator?
Click to expand...

If you use a single #18650 cell . . .

the peak voltage will be ~ 4.1 VDC -2.7 VDC (for red laser diode) = 1.4 VDC.

1.4 VDC/ 5.1 ohm resistor = 275 ma

Measure voltage across 1/2 watt 5.1 ohm resistor to insure it is stable.

Current will be stable, if you provide a good heat sink to limit temperature rise of the diode.

LarryDFW
 
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