About current n power

Discussion in 'General Electronics Chat' started by allcircuit, Apr 5, 2009.

  1. allcircuit

    Thread Starter Active Member

    Jan 7, 2009
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    Providing with only 9V battery, a load such as LED is used... will the current goes as high as 50A++ with power up to 1300W++? Possible? Or i have sumting wrong with it? I am testing my watt meter... :confused:
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Unfortunately something is wrong with your wattmeter or the way in which you are taking the measurement. Is the meter something you built yourself? Post a circuit or schematic of your measurement setup.
     
  3. allcircuit

    Thread Starter Active Member

    Jan 7, 2009
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    Oh no.... I am using the schematic provided by bhabott:
    [the link is here...]

    I am using a 9V battery and a 1 ohm shunt. However if i use a smaller shunt like 0.1Ohm..it short circuited. I wonder what had happened...
     
  4. Wendy

    Moderator

    Mar 24, 2008
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    9V batteries aren't very robust, figure around 200ma max, and then their lifespan is severly limited. My personal experience is they max out at 1A with a dead short (no voltage), and get hot.

    Is this circuit you show a wattmeter?

    The load you describe, an LED, what limited the current in it? No current limit and the LED is toast.
     
  5. allcircuit

    Thread Starter Active Member

    Jan 7, 2009
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    so you mean 9V may have up tp a max of 200ma? yup..that is a watt meter...i juz connected the LED across the load + and -.
     
  6. Wendy

    Moderator

    Mar 24, 2008
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    200 ma may be pushing it for a 9V battery, less is better.

    You're LED is a dead short without any current limiting. As I said, it is toast. Did you share the 9V between the wattmeter and the load?
     
  7. allcircuit

    Thread Starter Active Member

    Jan 7, 2009
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    currently yes....
    even if i changed to use LED having external 9V battery also same result..
     
  8. Audioguru

    New Member

    Dec 20, 2007
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    Duracell says that the internal resistance of a new 9V alkaline battery is 1.7 ohms. Then with a current of 2.65A the output voltage is 4.5V. Its current is 5.3A into a dead short.

    Energizer shows a graph of the capacity of a 9V alkaline battery (when its voltage has dropped to only 4.8V). It is 625mAh when the current is 25mA and it is 280mAh when the current is 500mA.

    The shunt on your wattmeter should be much less than 1 ohm.
    Your LED will immediately burn out without a series current-limiting resistor.
     
  9. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    You need to use a current limiting resistor in series when powering LEDs. Otherwise, the LEDs will burn up in a hurry.

    The general formula for determining the resistance needed is:
    Rlimit >= (Vsupply - VfLED) / DesiredCurrent
    where:
    Vsupply is the voltage being supplied to the LED
    VfLED is the manufacturer's typical forward voltage at the desired current.
    DesiredCurrent is usually 20mA for a standard LED.

    For example: you're using a 9v battery, and a typical red LED may be rated for a Vf of 2.1v @ 20mA.
    Rlimit >= (9v - 2.1v) / 20mA
    Rlimit >= 6.9V / 0.02 A
    Rlimit >= 345 Ohms. The closest E24 standard value is 360. Chart: http://www.logwell.com/tech/components/resistor_values.html
    Then calculate how much current you'll actually get through the resistor (and hence the LED):
    Current = Voltage / Resistance = 6.9v/360 Ohms = 19.17mA
     
  10. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Try testing your wattmeter with a 1kΩ Resistor with a 5VDC supply. With NO LED.

    It should give a value around 0.025mW or 25uW

    Then try only a 220Ω Resistor With NO LED., it should give a display of 113 mW

    If you tested the LED without a current limiting resistor, it was trying to estimate the value of a short circuit, and the LED is probably dead.
     
  11. allcircuit

    Thread Starter Active Member

    Jan 7, 2009
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    thanks for all the guidance provided....
    which means i had done something wrong during my measurement?
    I am afraid if my wattmter goes wrong....
     
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