Reducing the voltage

Discussion in 'General Electronics Chat' started by zosh, May 5, 2005.

  1. zosh

    Thread Starter New Member

    May 5, 2005
    4
    0
    I have a 24v. 1500mah. battery that I need to run one LED that requiers 5v. 20hm.
    and one phototransistor at 5v. 50mh. I would apriciate it if someone would help me figure out resister values to make this posible.

    please be nice... im reely new!
     
  2. David Bridgen

    Senior Member

    Feb 10, 2005
    278
    0
    Do you mean that the l.e.d. should be run at 20mA? What colour l.e.d. is it?

    Please explain what you want to do with the phototransistor. What is the significance of 5V. What is 50mh?
     
  3. zosh

    Thread Starter New Member

    May 5, 2005
    4
    0
    sorry, im trying to make a circuit that i found on the internet that calls for the LED and Photosenser to detect when somthing pases between the two. I have found these components on the Jamco catalog. The operating values are hard to interpret and the best I can understand the operating values are as I stated earlier.

    the LED is white.

    I need a circuit drawing program to show it better.
     
  4. David Bridgen

    Senior Member

    Feb 10, 2005
    278
    0
    Hi zosh,

    Ok, can you tell me the catalogue part numbers and I will have look. And is that Jamco or maybe Jameco?

    Oh, I use Micro$oft "Paint" to draw diagrams, and save them as gifs. Very easy.
     
  5. zosh

    Thread Starter New Member

    May 5, 2005
    4
    0
    ok i have the part #'s the LED is 320531 and the phototranststor is 120221

    you can check the data sheets at www.jameco.com

    Im going to try paint later.
     
  6. David Bridgen

    Senior Member

    Feb 10, 2005
    278
    0
    Try this zosh.
     
  7. zosh

    Thread Starter New Member

    May 5, 2005
    4
    0
    Thanks just one question... what dose 1k2 stand for? =)
     
  8. David Bridgen

    Senior Member

    Feb 10, 2005
    278
    0
    1k2 = 1.2k.

    The method of putting the multiplier where the point would go was introduced to reduce possible confusion in reading values from a poorly printed page.

    0.12 ohms would be R12

    1.2 ohms 1R2

    12 ohms 12R

    1.2M 1M2
     
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