Answers.Body: <p><a href="https://www.philohome.com/pf/pflights.htm" rel="nofollow noreferrer">https://www.philohome.com/pf/pflights.htm</a></p>
  <p>So the leds draw 2mA, a voltage drop of 12-9 = 3 V is required.
  So Ohms law says you need about a 1.5 kOhm resistor in series to have the leds shine at the same brightness...</p>
  <p>On the other hand, it could very well be that the only consequence of omitting the resistor is that the leds shine 30% brighter... you could test and monitor if the leds heat up. If you cut off power fast enough as soon as you feel them heating up, it's unlikely that they would be damaged.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2023-03-16T15:33:53.457Z
Answers.Text: It is very easy to burn out LEDs when you leave out the resister in my experience.  Even AA batteries can fry an LED without a resistor in seconds.
Answers.UserDisplayName: chicks
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2023-03-16T17:31:10.273Z
Answers.Text: Note that there IS already a resistor internally of 4.7 kOHm, so you are not exactly short-curcuiting the LEDs. Knowing Lego, the resistor in place is probably on the safe side. What one could do is use a potentio-meter and regulate from the 1.5 kOhm slowly to 0 Ohm and see what is appropriate...
Answers.UserDisplayName: Michael Verschaeve
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2023-03-23T09:39:49.057Z
Answers.Text: @user253751, perhaps you should expand your comment to an answer
Answers.UserDisplayName: Michael Verschaeve
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2023-03-16T07:39:41.140Z
Answers.LastActivityDate: 2023-03-16T08:46:35.890Z
Answers.LastEditDate: 2023-03-16T08:46:35.890Z
Answers.UserDisplayName: Michael Verschaeve
Body: <p>I am considering building a train that uses a 12v motor from the 1980s and Power Functions LED lighting. For this to work it would be necessary to run the Power Functions LED set from one of the 12v sockets on the motor.</p>
  <p>Obviously the physical connection is not compatible, but it should be straightforward to remove the connector brick from a Power Functions cable and fit 12v-style plugs instead. Where I'd welcome information is the electrical aspect. Since the LEDs are designed for 9v,  exposing them to 12v may damage them, and they will need a resistor in series to reduce the voltage across the LED set to 9v.</p>
  <p>Has anyone determined the correct value of the resistor? Or the current that the LED set draws at 9v, which would allow the necessary resistance to be calculated using Ohm's Law.</p>
  
Tags: <power-functions><12v>
Title: How to run Power Functions LEDs from a 12v supply?
UserDisplayName: john_e
ThreadText: # How to run Power Functions LEDs from a 12v supply?
  
  (Asked by: username_0 on 2023-03-15)
  
  I am considering building a train that uses a 12v motor from the 1980s and Power Functions LED lighting. For this to work it would be necessary to run the Power Functions LED set from one of the 12v sockets on the motor.
  
  Obviously the physical connection is not compatible, but it should be straightforward to remove the connector brick from a Power Functions cable and fit 12v-style plugs instead. Where I'd welcome information is the electrical aspect. Since the LEDs are designed for 9v, exposing them to 12v may damage them, and they will need a resistor in series to reduce the voltage across the LED set to 9v.
  
  Has anyone determined the correct value of the resistor? Or the current that the LED set draws at 9v, which would allow the necessary resistance to be calculated using Ohm's Law.
  
  **Tags:** power-functions, 12v
  
  
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  ## 1 answers
  
  ### Answer by username_1 on 2023-03-16. Score: 4
  
  <https://www.philohome.com/pf/pflights.htm>
  
  So the leds draw 2mA, a voltage drop of 12-9 = 3 V is required.
  So Ohms law says you need about a 1.5 kOhm resistor in series to have the leds shine at the same brightness...
  
  On the other hand, it could very well be that the only consequence of omitting the resistor is that the leds shine 30% brighter... you could test and monitor if the leds heat up. If you cut off power fast enough as soon as you feel them heating up, it's unlikely that they would be damaged.
  
  > **Comments:**
  > **username_2**: It is very easy to burn out LEDs when you leave out the resister in my experience.  Even AA batteries can fry an LED without a resistor in seconds.  
  > **username_1**: Note that there IS already a resistor internally of 4.7 kOHm, so you are not exactly short-curcuiting the LEDs. Knowing Lego, the resistor in place is probably on the safe side. What one could do is use a potentio-meter and regulate from the 1.5 kOhm slowly to 0 Ohm and see what is appropriate...  
  > **username_1**: @user253751, perhaps you should expand your comment to an answer  
  
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