Resistor LED help (simple)

Discussion in 'Homework Help' started by davidGG, Dec 22, 2012.

  1. davidGG

    Thread Starter Member

    Dec 22, 2012
    51
    0
    Hey, let's say I have a circuit like so...
    http://imgur.com/Cecr7
    Battery Voltage=9V, Resistors are all at 180ohm. (sorry for the bad drawing).
    How do I determine the voltage the diode will be receiving?
    The diode is rated at 1.6V 22mA.
     
  2. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Why do you have THREE resistors? Only a single resistor is needed.
    Most LEDs are 1.8V for a red one or 3.5V for a white or blue one.
    Your LED has a voltage that is too low.

    An LED sets its own voltage that changes a little with current and temperature changes. The series resistor sets its current.

    Example: 9V battery, 1.8V red LED, current needed= 20mA. Then (9V - 1.8V)/20mA= 360 ohms.
     
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  3. davidGG

    Thread Starter Member

    Dec 22, 2012
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    0
    Thank you! You have no idea how long I've been puzzling with this.


    EDIT: What if I were to replace the diode with a voltmeter, what would the reading be?
     
    Last edited: Dec 22, 2012
  4. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Deleted my duplicated reply.
     
  5. Audioguru

    New Member

    Dec 20, 2007
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    It is called Ohm's Law. The voltage across a resistor is caused by the amount of current through its resistance. Things in series all have the same current.
     
  6. davidGG

    Thread Starter Member

    Dec 22, 2012
    51
    0
    What if I were to replace the diode with a voltmeter, what would the reading be?
     
    Last edited: Dec 22, 2012
  7. Mike33

    AAC Fanatic!

    Feb 4, 2005
    349
    25
    Good question! OK - take the LED, which has almost NO resistance when forward biased. Most of the "action", the voltage drop, will be in the series resistor. That 330 ohm is much larger in resistance than the little resistance of the LED, so it will 'eat' most of the current, resulting in a larger drop (~7.2V) than the one at the diode (which will be the 1.8V described above).

    Remove the LED, and put a voltmeter in series. A modern DMM may have something like 3 megohms of resistance. SO with your series resistor and the DMM, which can be thought of as a 3Meg resistor, you just have a voltage divider! 360 ohms is so little compared to the 3Meg, you'd still read almost 9V at their junction...the big drop would be across the 3Meg. Of course, a DMM voltmeter doesn't really work this way...you'd be reading current if it were set up that way!
    The high resistance of the DMM is so that when you measure small voltages, it doesn't draw much current and disturb your reading.
    Cool, huh? :eek:)
     
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