Help with determining the proper resistor replacement to make an LED less bright

Thread Starter

steve167

Joined Jun 7, 2009
9
I've got an existing guitar effects pedal that has an excruciatingly bright clear blue PCB mount LED. It indicates when the pedal state is on or off. It is just waaaaaaaaaaaaaay too bright and makes it nearly impossible to look at the controls and make adjustments. There is an existing resistor that I'd like to replace with a higher resistance to decrease the brightness.

Let me share what I do understand so far as being necessary to determine the desired resistor:

1. Need the forward volts of the LED
2. Need the preferred current of the LED during operation
3. Need the volts supplied to the LED

From testing the circuit, the input volts to the 5mm blue LED is roughly 8 volts -- which makes sense since it operates using a standard 9v dry cell battery. The drop across the LED is about 2.8 volts -- that sounds about right for a 5mm blue LED too. I'm guessing that for a 5mm LED the current should probably be somewhere in the ballpark of 20 mA. The question I have is that I don't understand the use of the existing resistor based on what I think I've read as its value -- 11 ohms (see attached picture). Based on the above numbers I arrive at a much different value that should probably be used -- 260 ohms or so -- to get the current in the ballpark of 20 mA with this particular LED. I'd like to understand the existing resistor value to try and determine a better value to use as a replacement and achieve a much dimmer LED.

I would be extremely grateful if someone could double check my math above and also the reading of the existing resistor value.

Also, just FYI, looking at the image, the LED is mounted to the opposite side of the board with the positive leg attached to the left-most solder joint above the resistor. Power flows through the LED from a circuit trace on the other side, to and through the resistor, and to ground.

resistor.jpg

Thanks so much!

Steve
 

Thread Starter

steve167

Joined Jun 7, 2009
9
To make this even more confusing... I checked the value of the resistor directly and it comes in at 3.3K on my multimeter.
 

jungleb0y0124

Joined Apr 18, 2010
11
Hey Steve, that looks like a 1/4 watt 100ohm 5% resistor, (brown, brown, black, gold, gold). if I got the colors wrong you can check it here http://resistor.cherryjourney.pt/.

My advice to you would be to buy a pack of 1/4 watt resistors off some place like amazon or ebay this would be an example http://goo.gl/wJDh9s, and then take that one out and increase the resistance till you find one that you like for the brightness as it's going to be a personal preference type of thing for you.
 

WBahn

Joined Mar 31, 2012
25,757
No, it's a 1% 3.3 kΩ resistor. The colors are Orange,Orange,Black,Brown,Brown

If the tolerance band were gold, then there would only be two value bands and one multiplier band (four bands total).
 

WBahn

Joined Mar 31, 2012
25,757
With 8 V total and 2.8 V across the LED, you have 5.2 V across the resistor. With a 3.3 kΩ resistor, that would be about 1.6 mA. There are quite a few LEDs that can operate at this current level.

Where did the 11 Ω value you mentioned come from? If it was a resistance measurement taken with the resistor in the circuit, you can't rely on that. What you are often measuring predominantly are all of the various parallel paths and not the resistor itself.

Remove the resistor and replace it with something considerably larger -- the human eye is pretty insensitive to brightness changes. I'd start with a 10 kΩ resistor or something in the vicinity. Then adjust up or down until you find the brightness you want. You could also take a small potentiometer, say 33 kΩ or so, and short the wiper to one of the end legs turning it into a rheostat (variable resistor). Then you can easily adjust the brightness to what you want it to be and change it as conditions change.
 

jungleb0y0124

Joined Apr 18, 2010
11
Okay WBahn can say it's something else but i'm sticking to my guns lol.. HowverI don't know why I said 100ohm when brown brown black gold gold is 11ohm 5%.

You'll have to tell us what the colors are because it looks gold to me and orange to him, just please don't be color blind Steve.
 

Thread Starter

steve167

Joined Jun 7, 2009
9
LOL. Color blind... not really! So yeah the bands on the resistor are where I thought I was reading correctly the 11 ohms from -- brown, brown, black, gold, gold. Reading from the left I was thinking it was 1 1 0 multiplied by .01 and then 5% for the tolerance. The multimeter was telling me 3.3K however once I removed the resistor from the board completely. So I suspected that I was misreading the color bands. Admittedly, I'm a bit new at this so I knew there was a possibility of error.

Yeah I've got a multi pack of 1/4 watt resistors in various values on the way. I'm going to assume that the reading from my multimeter at 3.3K is accurate and try some values greater than that and see what I get.

I was really interested in hearing some others' interpretation of the color bands on the resistor though. So it sounds like I have one vote for 3.3K(which it apparently should be) and another for mine(probably incorrect) -- 11 ohms. At least I don't feel alone!

Thanks for all the advice!
 
Last edited:

Thread Starter

steve167

Joined Jun 7, 2009
9
I understand the color bands on this resistor now as well and how WBahn arrived at the measured 3.3K. I was simply too focused on the tolerance band HAVING to be gold(from either just gold or silver as possibilities) and then forcing the color to the left since it was the same -- as viewed from the picture -- to be gold as well. The other tolerances weren't as well documented and missing from more than a few of the guides I used as reference.

Thanks a lot to WBahn!

bands.jpg
 

Veracohr

Joined Jan 3, 2011
715
BTW, a lot of LEDs have a max current around 25mA, so assuming a current of 20mA is giving it max brightness. I did some investigation of LED current vs visibility, on random LEDs I had on hand, and found I could see as low as 20uA or something around that when I used my hands to block out the surrounding light. If you plan on using the pedal in dark environments like the typical stage, you might be able to go down to 100uA or so.
 

Thread Starter

steve167

Joined Jun 7, 2009
9
BTW, a lot of LEDs have a max current around 25mA, so assuming a current of 20mA is giving it max brightness. I did some investigation of LED current vs visibility, on random LEDs I had on hand, and found I could see as low as 20uA or something around that when I used my hands to block out the surrounding light. If you plan on using the pedal in dark environments like the typical stage, you might be able to go down to 100uA or so.
Thanks! Yeah, I've done some more reading too and it actually looks like I'm going to be pretty good at an *indicator lamp* level indoors(and some outdoors) with only about 50uA to 100uA driving it.

Thanks again!
 

WBahn

Joined Mar 31, 2012
25,757
LOL. Color blind... not really! So yeah the bands on the resistor are where I thought I was reading correctly the 11 ohms from -- brown, brown, black, gold, gold. Reading from the left I was thinking it was 1 1 0 multiplied by .01 and then 5% for the tolerance. The multimeter was telling me 3.3K however once I removed the resistor from the board completely. So I suspected that I was misreading the color bands. Admittedly, I'm a bit new at this so I knew there was a possibility of error.

Yeah I've got a multi pack of 1/4 watt resistors in various values on the way. I'm going to assume that the reading from my multimeter at 3.3K is accurate and try some values greater than that and see what I get.

I was really interested in hearing some others' interpretation of the color bands on the resistor though. So it sounds like I have one vote for 3.3K(which it apparently should be) and another for mine(probably incorrect) -- 11 ohms. At least I don't feel alone!

Thanks for all the advice!
Three value bands for a 5% resistor make no sense. Let's say that you were correct. The marking would say that the value is 11.0 Ω ± 0.55 Ω. That means that the actual value could correspond to ANY value of that final value band, making it completely meaningless.

Another thing worth pointing out, you may run into resistors that really are ambiguous, because sometimes value and precision are not the only things of interest. Sometimes a band represents something like temperature coefficient. And this gets taken to a whole new level once you start talking about markings on capacitors!
 

Phil-S

Joined Dec 4, 2015
142
As far as I know, the human eye is more sensitive at the blue end so blue LEDs appear a lot brighter than the usual reds and yellows.
Yes, they can be horribly bright and dangerous in my opinion.
For a resistor, you could go for a higher value than normally calculated or simply try it out with a 1 or 2k pot to get an acceptable output, then substitute the nearest fixed value.
 

rfagen

Joined Jan 19, 2013
2
You people and your math and standards and documentation. My first impulse was to stick a pot on there and then twiddle the knob until it was the right level of brightness for your eyes. As it turns out @KeepItSimpleStupid had the right idea with day/night, and the pot would let you do that automagically.

Don't expect me to solve the problem of how to integrate it with the case... that's what DREMEL is for :)

Cool (and practical) question.
 
if 3.3K is OK for bright, then add a 5K or 10K pot in series and put a switch across the POT. Your POT can be internal.

OR, use a 10K pot in series with the present resistor and adjust for night time viewing and replace it with a fixed resistor, again with a switch across it, is a better choice. Just use the POT to get the value.

If you think it's too bright in the daytime, do the same thing, but this time replace the 3.3K with the closest combination of 0-10K in series with 3.3K. Do this operation first.

Harder would be to actually determine when it's dark. The switch could be activated with a paperclip for that matter.
 

MrChips

Joined Oct 2, 2009
21,112
Better yet, put a light dependent resistor (LDR) in series with the current limiting resistor. Now the brightness of the LED will adjust automatically with the amount of ambient lighting. As the ambient lighting increases, the resistance of the LDR will decrease making the LED brighter.

Now all you have to do is to find the right combination of the LDR resistance sensitivity and the value of the series resistor.
 

Parkera

Joined May 3, 2016
77
I would wire a pot in series with the existing resistor. That way, you can't burn out the LED regardless of where or how far you turn the knob on the pot.

Resistor color codes get real confusing because of changes that have occurred over the last 90 years or so. For that reason, when in doubt, use an ohmmeter to put you in the ball park. That will also help you determine about what value of pot to use to dial in the brightness.

Note that the 1st band from the left in the photo is WIDER than the others. That signifies the FIRST band. On my monitor it looks like brown, brown, black, gold(?), gold(?). That translates to 11 ohm, 5% which is a standard EIA value. The last color band could represent either the temperature coefficient or the reliability. (REF http://www.resistorguide.com/resistor-color-code/). You would have to know the manufacturer of the resistor to know for sure.
 

ian field

Joined Oct 27, 2012
6,539
I've got an existing guitar effects pedal that has an excruciatingly bright clear blue PCB mount LED. It indicates when the pedal state is on or off. It is just waaaaaaaaaaaaaay too bright and makes it nearly impossible to look at the controls and make adjustments. There is an existing resistor that I'd like to replace with a higher resistance to decrease the brightness.

Let me share what I do understand so far as being necessary to determine the desired resistor:

1. Need the forward volts of the LED
2. Need the preferred current of the LED during operation
3. Need the volts supplied to the LED

From testing the circuit, the input volts to the 5mm blue LED is roughly 8 volts -- which makes sense since it operates using a standard 9v dry cell battery. The drop across the LED is about 2.8 volts -- that sounds about right for a 5mm blue LED too. I'm guessing that for a 5mm LED the current should probably be somewhere in the ballpark of 20 mA. The question I have is that I don't understand the use of the existing resistor based on what I think I've read as its value -- 11 ohms (see attached picture). Based on the above numbers I arrive at a much different value that should probably be used -- 260 ohms or so -- to get the current in the ballpark of 20 mA with this particular LED. I'd like to understand the existing resistor value to try and determine a better value to use as a replacement and achieve a much dimmer LED.

I would be extremely grateful if someone could double check my math above and also the reading of the existing resistor value.

Also, just FYI, looking at the image, the LED is mounted to the opposite side of the board with the positive leg attached to the left-most solder joint above the resistor. Power flows through the LED from a circuit trace on the other side, to and through the resistor, and to ground.

View attachment 134156

Thanks so much!

Steve
Its a 5 band resistor, very likely 1% tolerance. The black band in the middle could add another digit to the value, but in this case its the number zero. 2 brown bands = 11, the black band is a zero and the brown band is a multiplier of one zero.

Not too sure about the different shades of brown at each end though.

The way I'd go is experimenting with adding a series resistor - you can't end up with lower resistance than you stated with and fry the LED. When you get it right; just measure the total series resistance and fit the nearest standard value.

Someone suggested a light/dark switch - upmarket CRT TVs had an LDR sensing ambient light and auto adjusted the picture brightness. Doing the job for a LED shouldn't take more than a couple of transistors.
 

WBahn

Joined Mar 31, 2012
25,757
Its a 5 band resistor, very likely 1% tolerance. The black band in the middle could add another digit to the value, but in this case its the number zero. 2 brown bands = 11, the black band is a zero and the brown band is a multiplier of one zero.
So what's the fifth band?

And how do you explain the fact that the measured resistance of that resistor is no where near whatever value you seem to be zeroing in on, while if you read it as a 5-band resistor starting from the other end the value is 3.3 kΩ, 1%, which matches what the TS read with an ohmmeter?
 
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