PF and PU lights: LED specs?

[CoNSpiracy]

Hi, please forgive me if this is not the right part of the forum for this topic.

I am considering to mod a set of PF lights, by replacing the two "bright white" LEDs with LEDs in a different colour.

I am also considering to mod a set of PU lights, by removing the resistors (or replacing them with smaller ones) and adding additional LEDs. This would allow me to have more than the two LEDs from a standard PU lights cable powered from a PU hub together with a motor.

I have searched this forum and elsewhere and have not been able to find specifications for the LEDs (voltage and current) and resistors that LEGO uses for the PF and PU lights respectively. Can anyone help, please?

When taking apart the black middle brick in a set of PF lights, I see a print board where there appears to be three resistors. Next to both R1 and R2 it says 472, so I assume 472 ohm for each of these, and next to R3 it says 180, so I assume 180 ohm for that one. There is also a black box (transistor?) on the print board, but I am not sure why that is. I am wondering if anyone has made a drawing of the circuit, to better understand how it works?

[Mikdun]

1 hour ago, CoNSpiracy said:

Next to both R1 and R2 it says 472, so I assume 472 ohm for each of these, and next to R3 it says 180, so I assume 180 ohm for that one

IIFRC those resistors are 47*10^2 = 4700 or 4.7k  and 18*10^0 = 18 Ohm.

1 hour ago, CoNSpiracy said:

There is also a black box (transistor?) on the print board, but I am not sure why that is

This will be bridge rectifier.

 

The schematic is on Philo site: PFlights

Edited January 31, 2023 by Mikdun

[pow]

The first thing I have to admit is that I am only half competent. But what I do know about LEDs is that they pick up the amount of current based on the voltage applied.
So if you want to run other LEDs you would need to know how much voltage they need. And if and how you have to adjust the circuit if necessary. It is also important that you do not use too many LEDs on one connector instead of two, so that too much current does not destroy the electronics. hope that is helpful.

Best wishes

(translated with deepl.com)

[kolbjha]

6 hours ago, Mikdun said:

6 hours ago, CoNSpiracy said:

Next to both R1 and R2 it says 472, so I assume 472 ohm for each of these, and next to R3 it says 180, so I assume 180 ohm for that one

IIFRC those resistors are 47*10^2 = 4700 or 4.7k  and 18*10^0 = 18 Ohm.

According to Philo site, R3 is labeled 18D, not 180. R3 is 150 kOhm.

[dr_spock]

There are online resistor calculators for LEDs if you don't want to work out the V=IR equations.  You should know the voltage and current requirements of your replacement LEDs.

 

[CoNSpiracy]

Thanks, All.

What I understand from Philo's site is that the two original Lego LEDs in the PF lights are each approx. 3V and 1 mA. Would you agree? I am trying to find out what the original LED specs are so that I know what LEDs (and resistors) to replace them with.

Is there a similar site like Philo's site for PU elements? I am wondering if the PU lights are exactly the same as the PF lights (except for the connector).

Edited January 31, 2023 by CoNSpiracy

[Jeroen vW]

I've been tinkering with leds and lego for a while now. I would advise not to replace the leds of the PF. Instead take a 5 mm or 3 mm led in any color you want and put them in your project. The 5mm led almost fits in a standard lego technic hole. And when you buy them you have the exact voltage and current. And you can select the brightness. Leds are often too bright for Lego imho. So i almost always limit the light they give. There are a lot of examples on the internet.

Or buy a premade kit.

[Toastie]

2 hours ago, Jeroen vW said:

So i almost always limit the light they give.

I fully agree to your assessment. Current low power LEDs shine nicely on 1kOhm resistors at 5V DC. The PF voltage is 9V (PWM modulated). Just try it out. Begin with 1kOhm and see what you get - at full PWM = DC ("power 7") or dimmed down (power 1, 2 ...). And then decide. 

Best,
Thorsten

[Mikdun]

On 1/31/2023 at 6:13 PM, kolbjha said:

According to Philo site, R3 is labeled 18D, not 180. R3 is 150 kOhm.

I know, but didn't want to confuse CoNSpiracy even further.

Anyway R3 can be left alone, it's not in the light path.

 

Edited February 2, 2023 by Mikdun
Typo

[Davidz90]

21 hours ago, pow said:

But what I do know about LEDs is that they pick up the amount of current based on the voltage applied.

More specifically, they pass zero current until some minimum voltage is applied, and then the current starts increasing exponentially (as opposed to linear relation for reistors). Also, the operating voltage can vary slightly from led to led, even in the same batch so it is very hard to just set the voltage for proper current. That is why current limiting resistor is always needed with LEDs. If you don't want to bother with calculations, but have a decent selection of resistors, you can just start from the highest resistance one and keep decreasing it until LED lights up.

However, the formula is simple:

(Supply voltage) - IR = (LED Voltage)

where I is the target current and R is the necessary resistor.

 

Here's a good tutorial:

https://www.electronics-tutorials.ws/diode/diode_8.html

Edited February 1, 2023 by Davidz90

[CoNSpiracy]

Thanks for all your input. I appreciate it.

1. Am I interpreting Philo’s circuit diagram correctly if I read it to say that the voltage running through each of the two LEDs is a little less than 3V (2.81V and 2.66V)? I have two new warm white LEDs each of which have a forward voltage of 2.8 (min), 3.0 (typ.) and 3.2 (max), and a current of 1 mA, so I should be able to just replace the original Lego LEDs with these new warm white LEDs 1:1 without changing the resistors if I understand Philo’s diagram correctly.

2. Alternatively, since the voltage from the PF battery box is around 9V, it should also work to have three 3V LEDs connected in a series (without any resistors) via a PF connector to the battery, right? I could cut a PF extension cable and use the one end with a connector for that, I reckon.

3. Does any of you know if the PU lights have the same specs (same LEDs and resistors etc) as the PF lights?

 

 

 

[Davidz90]

40 minutes ago, CoNSpiracy said:

Alternatively, since the voltage from the PF battery box is around 9V, it should also work to have three 3V LEDs connected in a series (without any resistors) via a PF connector to the battery, right?

That is likely to burn them. LEDs have a very little resistance, the current will be much more than 1mA. You always want to have a resistor in series.

[dr_spock]

I think the supply voltage has to be greater than the voltage drop of the LEDs in series. 

[Davidz90]

6 hours ago, dr_spock said:

I think the supply voltage has to be greater than the voltage drop of the LEDs in series.

Yes, and the remaining voltage is the voltage drop in the resistor. So for example, you can have two LEDs in series (6 volts) and a resistor (remaining 3 volts); to get 1 mA current, you need 3 kOhm resistor. 

LED calculator says the same:

https://ledcalculator.net/#p=9&v=3&c=1&n=2&o=w

[pow]

23 hours ago, Davidz90 said:

Also, the operating voltage can vary slightly from led to led, even in the same batch so it is very hard to just set the voltage for proper current.

Didn't know that.

A question that popped up after following this thread is how complicated it would be to cut the voltage. What i mean is that the LED should get only let's say 5 Volts even if the akkubox outputs more on that channel. So the LEDs can be used parallel to a motor.

[Davidz90]

7 minutes ago, pow said:

A question that popped up after following this thread is how complicated it would be to cut the voltage.

Again, just use a proper resistor in series. It does just that - cuts the voltage (in addition to limiting current).

Learned about LED variation hard way when I was building a toy lightsaber. Wired 100 LEDs in parallel and powered them with 3 volts. In theory, with so many LEDs, internal resistance of batteries should be sufficient. What I didn't anticipate was that some LEDs reached full brightness at 3.1 volts and some at 2.9 volts, so my lightsaber turned out uneven. The 2.9 volt ones burned out few weeks later.

[pow]

5 minutes ago, Davidz90 said:

The 2.9 volt ones burned out few weeks later.

Damn

What i mean is to nail the voltage to a specific value independent of the input. If the input is higher, of course. I wouldn't expect only a resistor to be enough. Didn't such an application need some "intelligence"?

[Davidz90]

Ah, ok. Yes, if the input voltage is varying, some "intelligence" is needed to keep voltage constant. However, if the voltage variations are not great, it is sufficient to ensure that LED always gets more than minimum needed voltage. 

 

For example, let's use the above mentioned system with 2 LEDs, 3V each and 1kOhm resistor. If the supply voltage is increased from 9 volts to 11 volts, then:

-LED voltage is still very close to 3V, because current increasing exponentially with voltage also mean that small changes of current make almost no difference in voltage.

-Resistor voltage is now 11V-6V = 5V

-Current is now 5V/3kOhm = 1.67 mA, which is perfectly acceptable in most cases.

Edited February 2, 2023 by Davidz90

[CoNSpiracy]

Thanks Davidz90, dr_spock and pow for addressing my question #2.

I still don't understand why it wouldn't work to connect the three 3V LEDs in a series to the 9V power source. Ohm's law says that the resistance (R) in this case would be 0, right?

Anyone who can help answering my questions #1 and #3?

[Davidz90]

27 minutes ago, CoNSpiracy said:

Ohm's law says that the resistance (R) in this case would be 0, right?

Yes, Ohm's law indicates that 0 resistance is needed. However, by the same Ohm's law the current I=U/R, where U=9V and R would be resistance of 3 leds. Since that is very small, the current would be enormous (like, several amps).

Sorry, I have no knowledge of Lego part specs so I cannot fully answer questions 1 and 3. Regarding 1 I'd say that yes, 1:1 replacement without changes should be fine.

Edited February 2, 2023 by Davidz90

[CoNSpiracy]

9 minutes ago, Davidz90 said:

Yes, Ohm's law indicates that 0 resistance is needed. However, by the same Ohm's law the current I=U/R, where U=9V and R would be resistance of 3 leds. Since that is very small, the current would be enormous (like, several amps).

For Christmas decoration I bought a simple 20 warm white LED fairy lights string. This is powered by 2 AA 1.5V batteries (= 3V power source) and the 20 LEDs are connected in parallel to the cable string. I don't see any resistors anywhere in that setup?

3 hours ago, Davidz90 said:

Yes, and the remaining voltage is the voltage drop in the resistor. So for example, you can have two LEDs in series (6 volts) and a resistor (remaining 3 volts); to get 1 mA current, you need 3 kOhm resistor. 

LED calculator says the same:

https://ledcalculator.net/#p=9&v=3&c=1&n=2&o=w

So for the set-up described in my question #2 I would need a 1 ohm resistor?! https://ledcalculator.net/#p=9&v=3&c=1&n=3&o=w

This other LED resistor calculator says 0 ohm, as per Ohm's law: https://ohmslawcalculator.com/led-resistor-calculator

[Davidz90]

10 minutes ago, CoNSpiracy said:

For Christmas decoration I bought a simple 20 warm white LED fairy lights string. This is powered by 2 AA 1.5V batteries (= 3V power source) and the 20 LEDs are connected in parallel to the cable string. I don't see any resistors anywhere in that setup?

The key here is that they are connected in parallel, so the total current is quite high. That high current passes through AA batteries, which have quite big internal resistance. 

Regarding question 2, the supply voltage must be greater than total led voltage. In principle, the 1 ohm resistor could sort of work, although that setup works a bit differently than the one with two leds. Instead of reducing the current, we rely on the fact that the voltage drop over the resistor reduces the voltage of leds so they don't exceed their specs. But as I said, trying to control leds by voltage only is a bad idea.

Edited February 2, 2023 by Davidz90

[Toastie]

2 hours ago, CoNSpiracy said:

So for the set-up described in my question #2 I would need a 1 ohm resistor?! https://ledcalculator.net/#p=9&v=3&c=1&n=3&o=w

This other LED resistor calculator says 0 ohm, as per Ohm's law: https://ohmslawcalculator.com/led-resistor-calculator

Just my 2 cents:

The first link is more suitable than the second, I believe, as you can choose the LED type (color = semiconductor material mix). An LED is not an ohmic resistor, as many have already pointed out, so in essence Ohm's law is not applicable :D I bet you have been here, just in case: https://en.wikipedia.org/wiki/Electrical_resistance_and_conductance.

You "make" Ohm's Law applicable to your problem by defining boundary "conditions" not known to Ohm's Law, e.g., the forward voltage drop across the LED, which depends on its color (= material used) and manufacturer's tolerances when brewing the material. Thus, you first need to select the LED type/make, and look at the forward voltage drop as per data sheet. You don't have to do that when using ohmic material alone; two ohmic resistors in series simply add up, and then you can use Ohm's Law directly (yes I know, we know all that, just summarizing).

Furthermore, you also need to define your desired current going through the diode(s). As the current is identical in the entire circuit, if you select 1 mA in the above referenced calculator(s), 1 mA goes through all LEDs. And then the resistor value is calculated based on Ohm's Law - the calculator simply "assumes" that if there is a given voltage drop at a given current over the LED, then its ohmic resistance is assumed to be R=U/I. 

Having a resistor in series with any of the (n) LEDs is always good - it does not really matter whether it is 1 or 5 Ohms (your example). Temperature affects the characteristics of any diode, including LEDs. With no series resistor in the circuit and with a powerful power supply (alkaline batteries can provide several amps for shorter times), things may very swiftly turn ugly, when the LEDs are slipping away from their nominal characteristics. More current = temp goes further up etc. Or: The beauty of exponential functions. With a resistor in series, the voltage drop across the resistor increases with current as well (Ohm's Law ;) and thus "regulates" the current in the circuit - to a certain extent. In order to make this regulation more effective, I'd always go with at least a 100 Ohm series resistor and the lowest tolerable (as in enough light) current in your setup. I'd rather leave out an LED in series and increase the resistor value at constant supply voltage than the other way around, just have LEDs in the circuit. May work, but may also become ballistic.

As LED lifetime is also affected by the current flowing through it, again, I'd go with the lowest tolerable current. When I run 3/5 mm LEDs for test purposes in any of my projects, I'll take a 1kOhm series resistor (for 1 LED). They still "light" up. For powerful light generation, this of course won't work.

Good luck with your project!!!

All the best,
Thorsten