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 12th June 2015, 04:29 AM #1 JimBanville   diyAudio Member   Join Date: Jul 2014 Need help reducing voltage going to led I have a subwoofer plate amplifier. It is putting out 14v to its "power on" LED. I have a LED built into a new subwoofer enclosure. That LED is 4-6v forward voltage and 20 mA. I want to reduce the 14v from the amp down to for the enclosure's LED to (a) not burn up the LED and (2) to dim it further as well. It glows too brightly. I want to do this an cheaply as possible. Maybe a buck or two. I need very simple/basic instructions. I actually have a few random resistors laying around. Unknown values. Putting them in series with the LED when powered from a 9v battery, they seem to dim the light by half, but the measured voltage remained constant. I fuss that are just reducing the amperage only? Thanks for your help
 12th June 2015, 05:17 AM #2 Enzo   diyAudio Member   Join Date: Sep 2004 Location: Lansing, Michigan The measured voltage across the LED itself will remain whatever the LED's natural voltage is. The way to power any LED is to determine the CURRENT you want. 10ma is plenty for most LEDs for example. So if we start with 14v, and your LED wants 10ma, then we have to pick a series resistor that will drop 10v at 10ma. My 10v comes from the 14v minus the 4v. Ohm's Law then tells us the resistor would be R = V/I, which is in my example 10v/0.01A, or 1000 ohms. You can start with any voltage, and with the LED natural voltage, determine what the drop must be, and with the current calculate the resistance. So yes, the voltages will stay about the same, it is the current we care about. You have some unknown resistors? If you cannot read the color code, just look it up online. Since you are taking voltage readings, I assume you have a meter, so you could measure the odd resistors you have. But in the absence of resistors, do you have any spare controls? Potentiometers. If you have something like a 5k or 10k pot and some clip wires, then connect the pot wired as a variable resistor in place of your mystery resistor. Now you can turn the control until you like the brightness of the LED. The unhook the pot, an measure the resistance across it. Now you know the approximate value resistor you need.
 12th June 2015, 05:23 AM #3 jlind54   diyAudio Member   Join Date: Nov 2012 Location: Kansas Great explanation above^ Using a single resistor in series with the led will effectively reduce the current and and give you a voltage you desire across the led. do you have the full datasheet of the LED you are planning to use? Sample 14V- 4 volts = 10 volts to drop across the resistor 10v/.02amps= 500 ohms, or if you want it a bit dimmer go for say 10v/.015amps=660ohms Looks like I was too slow. Last edited by jlind54; 12th June 2015 at 05:36 AM.
JimBanville
diyAudio Member

Join Date: Jul 2014
Quote:
 Originally Posted by Enzo The measured voltage across the LED itself will remain whatever the LED's natural voltage is. The way to power any LED is to determine the CURRENT you want. 10ma is plenty for most LEDs for example. So if we start with 14v, and your LED wants 10ma, then we have to pick a series resistor that will drop 10v at 10ma. My 10v comes from the 14v minus the 4v. Ohm's Law then tells us the resistor would be R = V/I, which is in my example 10v/0.01A, or 1000 ohms. You can start with any voltage, and with the LED natural voltage, determine what the drop must be, and with the current calculate the resistance. So yes, the voltages will stay about the same, it is the current we care about. You have some unknown resistors? If you cannot read the color code, just look it up online. Since you are taking voltage readings, I assume you have a meter, so you could measure the odd resistors you have. But in the absence of resistors, do you have any spare controls? Potentiometers. If you have something like a 5k or 10k pot and some clip wires, then connect the pot wired as a variable resistor in place of your mystery resistor. Now you can turn the control until you like the brightness of the LED. The unhook the pot, an measure the resistance across it. Now you know the approximate value resistor you need.
Thanks. I may have a pot somewhere. I'll check. Here's the info on the 5v led I want to power from 14v source and dim...
Size 3mm Pre-Wired Round Top
Color Blue
Luminosity/Brightness 10,000-12,000 mcd
Forward Voltage 4v-6v (volts)
Wire Length Approximately 15cm (6 inches)
Resistor Built In Yes
Shrink Wrapped Connections Yes
Current 20mA
Wavelength 460-465nm
Viewing Angle 20-30 Degrees
Mount Style Custom
Lens Color Clear
LED Brightness Class Super/Ultra/Extreme
Operating Voltage 4v-6v (volts)

 12th June 2015, 05:55 AM #5 JimBanville   diyAudio Member   Join Date: Jul 2014 Can't find the pot. Of the resistors I have from old projects, there are 4 soldered in series for about 11k. It's much dimmer thru them from the 9v battery. Didn't power up the amp to check the 14v source. So are u saying it's okay to power the 5v led from the 14v source and to just concern myself with getting the brightness I want with whatever value resistor I wind up getting my hands on? I was under the assumption I needed to get the power supply voltage down to around 5v, to match the led. Thanks again.
 12th June 2015, 10:58 AM #6 DF96   diyAudio Member   Join Date: May 2007 The resistor will drop the voltage to whatever voltage the LED develops at that current. LEDs are almost always used with either a constant current supply or a resistor from a voltage supply. Directly connecting to a voltage supply will normally mean no light or a very bright light for a short time then no light.
 12th June 2015, 11:03 AM #7 AndrewT   diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders LEDs are essentially a constant voltage device. Vf, the voltage across the LED, hardly changes once the device has "turned on". turn on in this case means having reached a Vf that is near the "normal Vf" when light is emitted. This first level of turn on may not emit light that is visible to us. A typical red LED has a Vf around 1.8Vf. A particlular LED will vary by around 100mV, i.e. from 1.7Vf to 1.9Vf from very dull, to very bright. @ 1.6Vf there may be no light and @ 2Vf, it may blow up. The brightness responds to CURRENT. It is the CURRENT that needs to be controlled, not the voltage. A typical red LED may have a working current range of 0.5mA to 20mA. This is what needs to be adjusted/controlled to determine the level of light that the user requires. __________________ regards Andrew T. Last edited by AndrewT; 12th June 2015 at 11:13 AM.
 12th June 2015, 11:11 AM #8 AndrewT   diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders A blue LED normally has a Vf in the range 3Vf to 4Vf. Seeing a specification stating 4V to 6V makes me suspect it has a current limiting resistor built into the LED package, i.e. it is not a LED, but is actually a LED+RESISTOR. The 4V to 6V is the applied voltage over which the integrated resistor can control the current without the LED, or resistor, blowing up. You would need to add an extra resistor if you want to operate from a higher supply voltage. The sum of the two resistors is what now controls the operating current. __________________ regards Andrew T. Last edited by AndrewT; 12th June 2015 at 11:13 AM.
JimBanville
diyAudio Member

Join Date: Jul 2014
Quote:
 Originally Posted by AndrewT You would need to add an extra resistor if you want to operate from a higher supply voltage. The sum of the two resistors is what now controls the operating current.
Ok people, I truly appreciate the information

The specs say the led does have a built in resistor, is 4-6v and 20mA.

So as I understand it, I need a resistor to absorb the excess 10v or so of the 14v power supply that is powering the led, and if I want it dimmer, I should add more resistance. I can determine the extra needed dimming resistance with a pot, which could then be permanently replaced with a resistor.

Now, what value resistor should I use?
For 14v supplying a 4-6v 20mA led, would 500 ohm, 1/2 watt resistor be good?
If so, then I will be greatly exceeding that value in order to get the brightness down since I already tested a series of resistors equal to 15k ohm on the led already. I'm thinking a 20k resistor would dim the bulb adequately while absorbing the excess voltage. Sound reasonable?

 12th June 2015, 07:44 PM #10 DF96   diyAudio Member   Join Date: May 2007 A very bright LED fed far too much current will be very very bright. Two options: - continue to use the very bright LED, but send it very little current by adding a lot of resistance - use a cheaper less bright LED and so use less resistance

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