Have a question about zener diode TC

Discussion in 'Homework Help' started by ///m3dave, Sep 10, 2010.

  1. ///m3dave

    Thread Starter New Member

    Dec 8, 2009
    23
    0
    In my text for example problem it asks:
    An 8.2V zener diode ( 8.2V at 25°C ) has a positive temperature coefficient of 0.05%/°C. What is the zener voltage at 60°C?
    Solution:
    The change in zener voltage is ΔVz = Vz*TC*ΔT
    ( 8.2V)(0.05%/°C)(60°C - 25°C) = ( 8.2V )(0.0005/°C )( 60°C - 25°C )= 144mV.
    My question is why does the 0.05%/°C change to 0.0005/°C?

    Thanks in advance.
     
  2. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    It doesn't change, 0.05% is defined as 0.0005, that's how percentages work.
    1% = 0.01
     
  3. ///m3dave

    Thread Starter New Member

    Dec 8, 2009
    23
    0
    ah...got it, don't know why I didn't see that. Thanks
     
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