6V - 2V Using resistors? Small Parts

Thread Starter

sonnyangell

Joined Mar 1, 2017
30
Hello!

I have a simple circuit that I need to modify.

Toy Spy Glasses using 4 AAA batteries (6v) currently powers 2 Blue LEDs

My modification removes the Blue LEDs. And replaces them with (2) Red Laser Diodes and (2) 3mm Red LEDs

I need to dampen the voltage from 6V > 2V for the LEDs or else they burn out respectively. I had experimented with using 2 resistors as a voltage divider, but that doesnt seem to work well enough and upon research is a bad idea. (10k ohm + 4.7k ohm)

I can simply rewire it so it takes half the battery power for the lasers and half for the LEDs

Please send me some suggestions..
 

MrChips

Joined Oct 2, 2009
21,634
You don’t use a voltage divider. LEDs are driven by current. You need a current limiting resistor in series with the LED. Try 100-470 ohm. Experiment.
 

MrSoftware

Joined Oct 29, 2013
1,860
Do you need to limit current, or only voltage? If you only need to limit voltage, a simple linear voltage regulator will do the job. There are a bunch on digikey and mouser, they surely have 1.8V as that's a common voltage, if there's no fixed 2v then you can use a variable regulator. Double check your laser module though, if you need to limit the current then you're going to need something else or additional.
 

Thread Starter

sonnyangell

Joined Mar 1, 2017
30
You don’t use a voltage divider. LEDs are driven by current. You need a current limiting resistor in series with the LED. Try 100-470 ohm. Experiment.
Youre right, thats why it didnt work well.

Im curious, what are proper uses for voltage dividers if you shouldn't use them for loads?

Also, theoretically, if you use resistors to dampen voltage to a load, does that mean the batteries will last longer? or die quicker?

Duh... I was thinking way too much - thank you!
 

ElectricSpidey

Joined Dec 2, 2017
1,191
Voltage dividers are used to provide a voltage reference to high impedance input devices.

When you place a series resistor in a circuit, it in fact acts as a divider but with the load as part of the divider.
 

dl324

Joined Mar 30, 2015
11,231
My modification removes the Blue LEDs. And replaces them with (2) Red Laser Diodes and (2) 3mm Red LEDs
Connect them in two series strings. Both laser diodes and both LEDs and use an appropriate current limiting resistor for each string.
Im curious, what are proper uses for voltage dividers if you shouldn't use them for loads?
Voltage dividers make poor power supplies. The rule of thumb is that you want the current in the divider to be 10 times the load. That wastes a lot of power.

They teach us to build a power supply that way so we can see the disadvantages and then you move to something more appropriate. If you don't have any formal training, you missed out on that learning exercise.
Also, theoretically, if you use resistors to dampen voltage to a load, does that mean the batteries will last longer? or die quicker?
Resistors or diodes, they all dissipate power. The batteries will not last longer.

Toy Spy Glasses using 4 AAA batteries (6v) currently powers 2 Blue LEDs
And make sure you use those laser diodes in a safe manner. Laser diodes are not toys.
 
Last edited:

MrChips

Joined Oct 2, 2009
21,634
[QUOTE="sonnyangell, post: 1533457, member: 421750"
Im curious, what are proper uses for voltage dividers if you shouldn't use them for loads?
[/QUOTE]
Voltage dividers work as voltage dividers. so long as you do not take too much current from the divider.
The rule of thumb is the load should not take more that 1/10 the current going down the resistor chain.
Hence to drop 6V to 2V you need a voltage divider with resistors with 1/2 ratio.
20Ω + 10Ω = 1/3 x 6V = 2V
Current = 6V / 30Ω = 200mA
Don't take more than 20mA at the load.

Power dissipated by the voltage divider = 6V x 6V / 30Ω = 1.2W
 

ci139

Joined Jul 11, 2016
1,671
  • there's no point of having 2 narrow laser beams reflecting from a distant target
    in clear atmospheric path the rays cant be seen (from off the optical path) ... not by the "emitter" nor by anyone else
    . . .
  • the beams however can be dispersed --and/or-- de-focused ((they still remain invisible from side-ways)) but they will now give a better illumination of the "far" target . . .
    . . . depending the particular scheme of de-focusing -- they may become more safe for the user of the glasses --and/or-- for the random "subjects" nearby
    however such requires quite expensive prototype lens/diffuser
 

Tonyr1084

Joined Sep 24, 2015
4,889
"Toy Spy Glasses" and "Lasers" do not belong in the same sentence. Toy spy glasses suggest a child using them. Which leads to irresponsibility. You KNOW the child is going to want to look down the barrel of a laser diode. Heck! many adults might do that as well.

To answer the question about how to control the lighting of an LED you take the starting voltage and subtract the forward voltage of the diode. Example: 6 volt battery and a typical red LED with a Vf (forward voltage) of 2 volts. 6V - 2Vf = 4 volts. Next you need to know the maximum current the LED can handle. 30 mA is a whole lot of current for a typical red LED. Very bright. 20 mA is also pretty darn bright. 15 mA should be plenty bright. 10 mA is still bright enough to be easily seen. 5 mA and the LED can still be seen, though not bright. So if you settle on 15 mA for brightness you divide the 4V by 0.015 (15 mA) and come up with a target resistance of 267Ω. You're not going to find such a resistor. Common value resistors are available in 300Ω, so 4V ÷ 300Ω = 13.3 mA. Close enough.

As for how to power laser diodes, I'm not sure what to say regarding them. Except be damned careful. And don't put lasers in the hands of a child.
 
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