Maker Pro
Maker Pro

someone to settle a LED confusion that i have please

fersy

Jan 30, 2022
6
Joined
Jan 30, 2022
Messages
6
ok so full disclosure i'm not an electronics expert i just like tinkering with stuff and learning so hopefully you don't go too hard on me lol.
so anyway my current project is replacing the 4x 5mm red led's in my technics 1210 mk5 with blue ones and i've been watching a video of a guy on youtube showing people how to do it and in the video he's claiming that if i want to go blue i need to add a 470 ohm resistor in line.

so whats confusing me is this if both red and blue led's have the same forward current specification of 20ma and the red has a lower forward voltage requirement of 1.8v than the blue 3.8v why would i need a resistor? would the blues not actually need more current rather than less?

maybe i'm mixing this all up but it's been bugging me for a while and i'd appreciate it if someone could set me straight please.

thanks in advance
 

Bluejets

Oct 5, 2014
6,059
Joined
Oct 5, 2014
Messages
6,059
I cannot see the need for an "extra" 470 ohm resistor but the existing one may need replacing with a different value.
The calculation for the resistor value changes due to the different voltage drop across the LEDs as you have already realised.
The calculation is just basic ohms law with the LED (and resistor) current remaining at 20mA.
 
Last edited:

Audioguru

Sep 24, 2016
3,650
Joined
Sep 24, 2016
Messages
3,650
The existing red LEDs already have one resistor or a resistor in series with each LED.
Remove the red LEDs and replace them with the blue LEDs (the anode and cathode must not be reversed) then either there is not enough voltage for the blue LEDs or they will be dim.
 

fersy

Jan 30, 2022
6
Joined
Jan 30, 2022
Messages
6
Cheers guys thats what i thought. It didnt make sense to me what he was saying but then i started thinking maybe im misunderstanding
 

Martaine2005

May 12, 2015
4,382
Joined
May 12, 2015
Messages
4,382
Also,
What voltage is supplying the 4 red LEDs?. Are they individually wired?, series connected? Or parallel connected?. All of which require different resistor counts and values.
I have seen lots of terrible videos where the uploader gives instructions and clearly hasn’t got a clue what they’re talking about.

Martin
 

fersy

Jan 30, 2022
6
Joined
Jan 30, 2022
Messages
6
Also,
What voltage is supplying the 4 red LEDs?. Are they individually wired?, series connected? Or parallel connected?. All of which require different resistor counts and values.
I have seen lots of terrible videos where the uploader gives instructions and clearly hasn’t got a clue what they’re talking about.

Martin
theres the part of the schematic pertaining to that circuit. like i said i'm not an expert but i'd say thats in series. with regards the voltage supplying them im not exactly sure as im not great at reading those diagrams but on a quick meter measurement from the first led in the series to the last there appears to be about 7.2v across the 4 leds which id assume is right seeing they have a forward voltage of 1.8v each
 

Attachments

  • Screen Shot 2022-01-31 at 02.50.50.png
    Screen Shot 2022-01-31 at 02.50.50.png
    176.5 KB · Views: 9

fersy

Jan 30, 2022
6
Joined
Jan 30, 2022
Messages
6
i had removed the image m8 and attached it as a file to click on just below the post. are you not seeing it on your end? it all looks good on mine
 

Harald Kapp

Moderator
Moderator
Nov 17, 2011
12,879
Joined
Nov 17, 2011
Messages
12,879
So we have 4 LEDs in series (D203...D206) plus a 150 Ω current limiting resistor (R212). Supply voltage is 21 V.
The forward voltage of a single LED is typ. 1.8 V (datasheet). That makes for VLED = 4 × 1.8 V = 7.2 V forward voltage across the assembly of 4 LEDs. Current is the ILED = (21 V - 7.2 V)/150 Ω = 92 mA. Note that this is a pulsed current to achieve the strobe effect, therefore the seemingly high value 0f 92 mA is averaged to a much lower value, maybe ~10 mA with a duty cycle of 1:10.
Blue LEDs have a typical forward voltage of 3 V (depending on the type this can vary from 2.5 V ... 3.7 V). Instead of 4 red LEDs you could use
  • 2 blue LEDs which will result in a bit higher current and thus more brightness which will compensate for the lack of the other 2 LEDs.
  • 3 blue LEDs which will result in a bit low current. On the other hand moderns LEDs are so much more efficient that 3 modern blue ones may be even brighter than 4 old red ones even at lower current.
  • 3 or 4 blue LEDs and a changed resistor R212 to increase the current to 92 mA. I doubt that this will be necessary.
You can try the 2 LED or 3 LED setup rather easily: lift one of the pins of the existing LED assembly (either point A or B in the schematic). No need to lift both pins. Then connect the string of blue LEDs between points 11 and 12 (observe the polarity!) and compare the 2 LED vs 3 LED versions. Decide which one you like more.
You may have to patch the void in the housing where formerly the 4th LED was placed with whatever material you want to use (e.g. epoxy, sugru, hot glue ...).
 

fersy

Jan 30, 2022
6
Joined
Jan 30, 2022
Messages
6
So we have 4 LEDs in series (D203...D206) plus a 150 Ω current limiting resistor (R212). Supply voltage is 21 V.
The forward voltage of a single LED is typ. 1.8 V (datasheet). That makes for VLED = 4 × 1.8 V = 7.2 V forward voltage across the assembly of 4 LEDs. Current is the ILED = (21 V - 7.2 V)/150 Ω = 92 mA. Note that this is a pulsed current to achieve the strobe effect, therefore the seemingly high value 0f 92 mA is averaged to a much lower value, maybe ~10 mA with a duty cycle of 1:10.
Blue LEDs have a typical forward voltage of 3 V (depending on the type this can vary from 2.5 V ... 3.7 V). Instead of 4 red LEDs you could use
  • 2 blue LEDs which will result in a bit higher current and thus more brightness which will compensate for the lack of the other 2 LEDs.
  • 3 blue LEDs which will result in a bit low current. On the other hand moderns LEDs are so much more efficient that 3 modern blue ones may be even brighter than 4 old red ones even at lower current.
  • 3 or 4 blue LEDs and a changed resistor R212 to increase the current to 92 mA. I doubt that this will be necessary.
You can try the 2 LED or 3 LED setup rather easily: lift one of the pins of the existing LED assembly (either point A or B in the schematic). No need to lift both pins. Then connect the string of blue LEDs between points 11 and 12 (observe the polarity!) and compare the 2 LED vs 3 LED versions. Decide which one you like more.
You may have to patch the void in the housing where formerly the 4th LED was placed with whatever material you want to use (e.g. epoxy, sugru, hot glue ...).

thanks thats very helpful. thats actually how i was reading the diagram surprisingly lol 21v supply with 150 ohm resistor (212)
i had thought about maybe changing that to increase the current but i wasnt sure how to work out what id need to change it to so i could achieve the correct amount for the 4 blue leds.

last night before you responded i figured id just try the 4 blue leds and see what happened as i thought worst that could happen is they would be dim or not light at all due to being underpowered but they light and work well actually.

would you say thats something i need to be concerned about in the long run or if they are working will they be ok as is, because i have located resistor r212 and could change it np? thing is im not sure that 21v isnt being split between this and the speed selector circuit when you look at the full schematic (attached) so dunno if using the led/ resistor calculator online would be accurate.

also thanks a lot for the info
 

Attachments

  • Screen Shot 2022-01-31 at 11.41.38.png
    Screen Shot 2022-01-31 at 11.41.38.png
    474.3 KB · Views: 5

Harald Kapp

Moderator
Moderator
Nov 17, 2011
12,879
Joined
Nov 17, 2011
Messages
12,879
If the LEDs are working well there is no need to fiddle with the resistor. Current will be lower than before, but as I said that may be compensated by the higher efficiency of the modern LEDs.
 

Audioguru

Sep 24, 2016
3,650
Joined
Sep 24, 2016
Messages
3,650
The 21V is not "split". Instead it is fed as 21V to R212 and R213.

The calculation of original red LED current in post #9 has a decimal point error. The current is (21 V - 7.2 V)/150 Ω = 9.2mA.
Too much current burns out or shortens the life of LEDs. The current in your four series blue LEDs fed through 150 ohms is only about 6mA so they are fine and will last for a long time.

EDIT: My currents calculations are very wrong. The peak current in the four series red LEDs is a whopping 92mA and in four blue LEDs is 60mA.
Those are very high currents for the LEDs and the resistors.
 
Last edited:
Top