I just bought some lasers and resistors from Amazon and I am a bit confused. First of all the seller says the resistors are 200 ohm. I am a bit confused because I have only seen resistors with number (23) with (23k) by it.

My question:

Can a 200 ohm resistors reduce a 9 v to 5 v so I can power the laser? Is this safe to do without having anything blowing up?

Picture analysis of laser

shows details for laser

  • 1
    \$\begingroup\$ Resistors are generally a bad way to drop voltage, as the voltage dropped across the resistor depends on the current through it. Your 200 Ohm resistor will only drop 4 volts when passing 20 mA. Whether this is "safe" or not depends on why you want to get 5 volts, and what current your 5 volt load requires. \$\endgroup\$ Apr 25, 2017 at 23:05
  • \$\begingroup\$ Get a LM317 configured as a current limit and you kinda have a cool circuit, but you need to know the current intake of the laser. If they said you need a 200ohm to drop the voltage from 9 to 5 to power it correctly, then 20mA is your target. Read the datasheet the congmfoguration is simple but you need to calculate the math ti.com/lit/ds/symlink/lm317.pdf Also can you provide where you got the LEDs from? \$\endgroup\$
    – Bradman175
    Apr 25, 2017 at 23:10
  • 2
    \$\begingroup\$ Resistors are usually a bad way to drop voltage, but they are okay to use to drop voltage for diodes. \$\endgroup\$
    – user253751
    Apr 26, 2017 at 0:57
  • \$\begingroup\$ I didn't know they were bad, isn't that the point of a resistor. Keep in mind I am teaching myself this stuff so I might not know everything @immibis. I have heard that a capacitor can be used to drop voltage. \$\endgroup\$ Apr 26, 2017 at 1:07
  • \$\begingroup\$ @GovaDEster The problem with using a resistor to drop voltage is that if the load's resistance changes, so does the voltage drop across the resistor and therefore the voltage across the load. You couldn't use one to drop voltage for, say, a cellphone because the cellphone's resistance will change depending on how much power it needs (e.g. it will go down if you turn the screen on). But for simple devices like a diode it's okay because their resistance doesn't suddenly change like that. \$\endgroup\$
    – user253751
    Apr 26, 2017 at 1:16

3 Answers 3



The seller says these are self regulating which means it will draw 20mA at 5V. What confuses me is the seller says it draws 30mA at 3V where these lasers are diodes thus they should always have a fairly fixed voltage drop that only changes when current changes. By this logic the diode would be brighter at 3V than at 5V. Anyways we can deduce that:

V = IR
R = V/I
R = 5/0.02
R = 250Ω

Looking like a simple device, it should be that a 250Ω resistor is already connected in series with the laser diode. If you want to power it from 9V then:

V = IR
R = V/I
R = (9-5)/0.02
R = 4/0.02
R = 200Ω

You would need to add a 200Ω resistor in series. So you're correct.

  • \$\begingroup\$ What do you mean in series \$\endgroup\$ Apr 25, 2017 at 23:59
  • 1
    \$\begingroup\$ @GovaDEster You should learn some basic theory before you try connecting things up. \$\endgroup\$
    – user253751
    Apr 26, 2017 at 0:58
  • \$\begingroup\$ I figured out what you meant in series. Its pretty basic. The thing is if I did that the first resistor in the series would start to heat. \$\endgroup\$ Apr 26, 2017 at 1:09
  • \$\begingroup\$ @GovaDEster That's not quite correct, i would recommend reading build-electronic-circuits.com/electronics-for-beginners before you start \$\endgroup\$
    – BeB00
    Apr 26, 2017 at 13:50

It depends on the current to be drawn by the laser. Using ohms law, where you want to drop 4 volts (the difference from 9v battery to 5v laser), and given the 200 ohm resistors:

I = 4/200 = 0.02 Amps = 20 mA

so 20 mA should be the current requirement for your laser load for this to work properly.

Do you have the voltage and current requirements for your laser?

A disadvantage of using a resistor to drop the voltage is that it does so by turning the excess power to heat, draining the battery faster.


I don't know what your load current is so I can't see how much power you'll be dissipating when dropping your 4 V.

If it's less than 250 mA (1 W dropped power), I'd use a 78S05 regulator on a heatsink. This takes in 7..30 V and outputs a steady 5 V, regardless of load current changes.

If power efficiency is important to you, there are plenty of switching regulators or pre-made tiny brick DC-DC converters available. The latter are more expensive but much more convenient to work with if you're breadboarding.


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