Changing LED colour in old hifi equipment

[looxuser]

Hello forum members, I want to change the colour of LEDS in the front panels of my old hifi equipment, they are basically green & red led's (look very 80's) and I'd like them to be blue or white. What things despite the obvious of physical size should I be looking for please? What I know so far...

1. nm gives the colour
2. forward voltage - assume I could measure this across the pins of the existing LED and either get one exact or slightly over.
3. Intensity - Well would just have to guess what they are now
4. Current capacity - On the ones i've looked at the current capacity is not given in any specs so can I assume they draw very little and just get the voltage right?

Cheers all, Andy

 

[BobK]

The problem is, red and green LEDs have a forward voltage of about 2V whereas blue and white have about 3.3V, so you probably cannot simply replace the LED.

LEDs are not voltage driven, they are current driven. For indicator LEDs like these, you typically use a higher voltage and a resistor that drops the additional voltage at the desired current.

For example, for a red LED with 2.0V forward voltage, 20mA current, running from a 5V supply:

You need to drop 3V at 20mA. From Ohms law:

V = I * R
R = V / I = 3 / 0.020 =

You would use a 150Ω resistor.

Now, to run a blue or white LED with a 3.3V forward voltage, you would do the same calculation, with V = (5-3.33) = 1.66V

And you get:

R = 1.66 / 0.020 = 83Ω

If you kept the original 150Ω resistor, you would get a current of

I = V / R = 1.66 / 150 = 11mA

This might actually be totally acceptable, since newer LEDs tend to be quite a bit brighter.

Note that this is just an example. You should look into what voltage and what resistor are used in your receiver before doing your own calculations.

Bob

 

[looxuser]

Totally agree CaritaTDewall, many thanks BobK, here is an extract from the circuit diagram.


So let me try to put intio practice what you have explained there BobK. Finger crossed, here goes....

So what can I determine from the diagram...

So the forward voltage is derived through the transistors from an +8 volt supply. These transistors being used to switch on the LED's, of course there is some amplification from the transistor so we have 8.2v. So as we know the resistors R17, R18 & R19 to each set of LED's is 220ohm we can estimate to determine the voltage supplied drop or the current drawn?

So here goes V=IR, so assuming these are 2.5v Green LED's the current would be given by 8.2-2.5/220 = 5.7/220 = 0.045A = 45ma (Feels high) so ets adjust that voltage assumption...

Assume they are 2.0v using the same equation =6.6/220 = 28ma more like it as some LED ive looked at seem to have a 30ma max.

So lets say then they are 2v and we want to supply at 28ma then using BobK original calc 6.6/0.028 = 220 ohm resistor.

Bob, am I correct so far?

So.

I know we have 8.2v available so lets test what would need to be done for some Blue 3.3v - 30ma max LEDS to replace the green.

Ok so thats a volt drop of 4.9v at 28ma and R=V/I so R= 4.9/28 = 175ohm.

So nearest 1/4W resistor is 180ohm, if I then replace R17, 18 & 19 with them i'll get 4.9/180 = 27ma.

Is this ok logically and mathematically or do I need to learn some more please? Things I mighty not have taken into account are LEDs in series, the power calculation for right resistor power rating. I assume also the transistors are resistor types which limit current before the R17 etc limit it further.

Oh just noticed these LEDS are given part numbers SLZ381C but I cannot find a datasheet which would have been handy.

 

[looxuser]

... So just looking at some replacement LED's and they are saying 20mA max which seems to be the standard actually (not 30mA I stated earlier). So assuming the ones fitted are rated to 20mA max that would give a voltage drop of 4.4v which gives a supply voltage of 3.8 volts to the LED's. This again seems way higher than the 3mm LED's i've been looking up so I guess i've assumed something wrong here? VF seems to be about 2.2v to 2.5v max and IF 20mA max but I can't make them fit what is shown on the diagram with the 220R fitted?

 

[looxuser]

Ok so I measured 2 volts across the LED, which if I read the circuit diagram correctly thats a 6.6v drop and theres a 220R in series, therefore I=V/R = 6.6/220 = 30 mA. So am I accurate in saying the LED's are 2v units with a 30mA capacity and that the Forward voltage of 8.2volts is dropped 6.6V by the 220R resistor please?

 

[BobK]

Have not had a chance to check your work yet because we had a death in the family today. But I will take a look when I get a chance.

One thing I can say, The voltage is actually 8, not 8.2. I don't know what the 8.2 means there, but there is not V after it like there is after the 8, and the 8V is directly connected to the LEDs.

Bob

 

[looxuser]

Oh so sorry to hear your news Bob, hope you and family ok. It not uncommon for those circuit diagrams to have wrong characters in there which may explain the 8.2.

Ultimately what i need to know is, given theres a good voltage available, what size resistor I should use and why? Cheers, Andy

 

[Gryd3]

Green LEDs often have a forward voltage of 2.0 to 2.3 Volts. Based on that assumption and the 220Ω resistor, we can calculate current to be between 15mA to 18mA . ( 8V - [4.6 to 4.0] / 220Ω ) . Remember there are two LEDs in series, so you must make sure you subtract both of their forward voltage drops from the supply voltage before you continue with your calculations.
That lines up about right for the value required for a 20mA LED.

 

[looxuser]

Hi Gryd3 thought the series LED was a factor so thanks for confirming. So the objective here, correct me if i'm wrong, is to get the best fit possible i.e the selection of resistor (within those produced) will effect volt drop and limit current. (Note my knowledge tells me that an LED would consume as much current as it likes until it is destroyed so we are talking current & voltage "limiting" properties of a resistor here)

So the LED's I've now got are 2.8v to 4v units 20mA max so.

Lets say they are typical 3.3v so with two in series thats 6.6volts needed so that 8v-6.6v/0.02 = 70R but the nearest resistors made are 68R or 100R...

So lets take the 100R and play that back in to see what current that would allow so its 1.4/100 = 14mA. Seems low so maybe not that bright..3.3v x 0.014 = 46mWatts

Lets take 68R and that 1.4/68= 20.5mA so thats over, only just but over all the same, so

bearing in mind that the LED can take a lower voltage...

lets say 6v (3v each) so thats a 2volt drop 2/0.02 = 100R.

So am I right in saying if I change the 220R to 100R i'll drop the voltage to 6V from 8v thereby supplying each LED with 3 volts and limiting the current to them to 20mA please?

One further thing and this goes to my learning here, the LED brightness comes from the power through it does it not? Thats Volts x Amps so we could vary within the design limits either of these parameters so for instance the brightness associated with say 0.06watts or 60mW could be obtained from supplying to the LED 3.3v at 18mA or 4v at 15mA or 3v at 20ma. The Maximum LED brightness would come from 4v at 20mA giving a power consumption of 80mW.?

Sorry if i'm labouring LED's somewhat here but it could be a useful learning for novices such as myself.

Cheers

 

[shrtrnd]

I don't think you're belaboring LED talk at all. There are so many types out there it's hard to keep up with them.
I built an LED alarm clock a while back to replace my old unreliable one. Thought it would be cool to use Blue LED's, as opposed to the usual red ones.
Damned thing lights-up my whole bedroom. Wish I'd spent a little more time researching what I was doing.
Went out and bought a $9 red LED one so I could get some sleep.

 

[Gryd3]

Hi Gryd3 thought the series LED was a factor so thanks for confirming. So the objective here, correct me if i'm wrong, is to get the best fit possible i.e the selection of resistor (within those produced) will effect volt drop and limit current. (Note my knowledge tells me that an LED would consume as much current as it likes until it is destroyed so we are talking current & voltage "limiting" properties of a resistor here)

So the LED's I've now got are 2.8v to 4v units 20mA max so.

Lets say they are typical 3.3v so with two in series thats 6.6volts needed so that 8v-6.6v/0.02 = 70R but the nearest resistors made are 68R or 100R...

So lets take the 100R and play that back in to see what current that would allow so its 1.4/100 = 14mA. Seems low so maybe not that bright..3.3v x 0.014 = 46mWatts

Lets take 68R and that 1.4/68= 20.5mA so thats over, only just but over all the same, so

bearing in mind that the LED can take a lower voltage...

lets say 6v (3v each) so thats a 2volt drop 2/0.02 = 100R.

So am I right in saying if I change the 220R to 100R i'll drop the voltage to 6V from 8v thereby supplying each LED with 3 volts and limiting the current to them to 20mA please?

One further thing and this goes to my learning here, the LED brightness comes from the power through it does it not? Thats Volts x Amps so we could vary within the design limits either of these parameters so for instance the brightness associated with say 0.06watts or 60mW could be obtained from supplying to the LED 3.3v at 18mA or 4v at 15mA or 3v at 20ma. The Maximum LED brightness would come from 4v at 20mA giving a power consumption of 80mW.?

Sorry if i'm labouring LED's somewhat here but it could be a useful learning for novices such as myself.

Cheers

No worries. LEDs are one of those things that are *simple* but have a lot of *gotcha!*s for newer people.
The LED is a current driven device. It's the current that is important here... the voltage drop across it is merely a property of what normally drop when operated at the rated current.
In this case, you are not using the resistor to bring the voltage down for the LEDs, you are using the resistor to absorb or drop the 'left-over' voltage from the LED forward drop(s).
Your math is spot on from where I'm concerned. However, when planning out values it's best to determine what you think is acceptable. 15mA sounds good to me, it's below the limit and considering the original LEDs operate between 15-18mA it should not significantly dim the LED and I doubt you will notice unless you have the same LEDs nearby operating at a higher current.
The specs for the ones you want to use have a very wide voltage drop range... 2.8V - 4.0V
What colour are they?

Anyway, it's a good practice to plan for worst case scenarios, which in this case would be the LEDs only dropping 2.8V. In this case you would want (8V - 5.6V) / 0.02A = 120Ω resistor.
This would provide you with 20mA at a 2.8V drop, and would decrease to pretty damn tiny level if they happened to operate at 4.0V... That's why I'm asking for the color, this range could cause some really dim LEDs unless you planned to test the LEDs first to measure their voltage drop prior to doing calculations.

 

[Gryd3]

I don't think you're belaboring LED talk at all. There are so many types out there it's hard to keep up with them.
I built an LED alarm clock a while back to replace my old unreliable one. Thought it would be cool to use Blue LED's, as opposed to the usual red ones.
Damned thing lights-up my whole bedroom. Wish I'd spent a little more time researching what I was doing.
Went out and bought a $9 red LED one so I could get some sleep.

I did the same thing with blue LED strip light inside my computer case... who would have thought so much blue light can get out of the fan grills and perforated panel...

 

[BobK]

I think our eyes are particularly sensitive to blue light because there is not that much of it in nature. I.e. a blue LED at the same power level will seem brighter than a green or red one.

I had one on the my monitor at work, and it was so bright I had to tape over it.

Bob

 

[Gryd3]

I think our eyes are particularly sensitive to blue light because there is not that much of it in nature. I.e. a blue LED at the same power level will seem brighter than a green or red one.

I had one on the my monitor at work, and it was so bright I had to tape over it.

Bob

I thought it was related to wave-length... but that makes sense because when you begin to drift away from blue into violet and UV, it does not seem to affect the eyes quite so much.

*One of the other reasons kids with Blue HID lights on their car bother me...

 

[shrtrnd]

I was wondering what those blue headlight lamps were. Annoying when you encounter them night-driving. Sure hope they never catch-on in general usage.

 

[duke37]

A minor point, you used 8.2V in the earlier posts. In fact, the transistor drops about 0.2V when turned hard on so you have 7.8V to play with.