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Help Understanding Power in Circuits

  1. Nov 24, 2009 #1
    In University we almost always deal with voltage sources that can supply whatever current needed. What determines the amount of power "drawn", is it purely the resistance? How do we design for low-power?
     
  2. jcsd
  3. Nov 24, 2009 #2

    vk6kro

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    The current drawn from a power source depends on the internal resistance of the power source and the impedance of the load.

    So, you don't have to design for low power. If you put a 1000 ohm resistor across a 12 V car battery, it will deliver 12 mA even though the battery can deliver hundereds of amps.
    (This is Ohm's Law..... I = V / R = 12 /1000 = 12 mA.)

    The internal resistance of the power source decides the maximum current the supply can deliver. If this is not important, then it is convenient to regard the supply as perfect and assume the voltage will remain stable with varying load currents.
     
  4. Nov 24, 2009 #3
    So... the key to low power is to have a very high load impedance? I don't think I'm comprehending this correctly.
     
  5. Nov 24, 2009 #4

    berkeman

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    Staff: Mentor

    Yes, per Ohm's law.

    And for active circuits, you try to minimize the bias currents, in order to minimize overall current consumption. That brings on a host of issues, however.

    Here's an intro article with lots of links to other references:

    http://en.wikipedia.org/wiki/Low-power_electronics

    .
     
  6. Nov 24, 2009 #5

    vk6kro

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    So... the key to low power is to have a very high load impedance?

    Yes, exactly. The power actually drawn from a power source doesn't have to be as much as the source is capable of.
    A car battery is quite capable of delivering only 1 mA if the load is 12000 ohms. This is only 12 milliwatts!
    Or, it could deliver 50 amps if the load was 0.24 ohms. This would be 600 watts. (12 volts * 50 amps = 600 watts).
     
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