Connect with us

Fitting an exponential curve to inductance graph

Discussion in 'General Electronics Discussion' started by mike_980, Apr 26, 2013.

Scroll to continue with content
  1. mike_980

    mike_980

    16
    0
    Mar 24, 2013
    I took the inductance reading today of a coil with my oscilloscope. My oscilloscope though is only a pocket one and cannot get the data to my PC. The current took 4.4 seconds to reach max (0.6A).

    Do I simply put 4.4 as the right hand value on the X-axis and 0.6A on the top of the Y-axis then plot an exponential curve from T=0 through to T=4.4 to get the graph I need to read my time constant from or is it more complicated than that?
     
  2. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    Well, you never actually reach the full current, so determining the point at which it reaches the full current is pretty difficult.

    The way you would determine the time constant (and thus the inductance) is to find when it reaches (1-1/e) or 62.3% of the final current.

    Bob
     
  3. mike_980

    mike_980

    16
    0
    Mar 24, 2013
    Thanks,
    I will attach a picture of the scope screen with the data stored. I can put markers in on the time axis and the voltage axis to measure them, I have measured it to the little squiggle as the line straightens and to that point the time is 4.4 seconds (I took 3 measurements reading 4.4, 4.4 and 4.6 respectively).

    So how do I go from this on my oscilloscope to having the exponential curve in matlab or excel so I can read off the time constant? I assumed just putting 4.4 seconds on the X-axis and 0.6A on my Y-axis and then plotting an exponential curve of my own as the actual curve cannot be taken from the oscilloscope, is this correct?
     
  4. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    You don't need to plot any exponential curves. All you need to know is how long it takes for the current to rise to 62.3% of the final value. That time is the time constant, you can read it directly off the scope. But I would not rely on getting better than about 10% accuracy.

    For example. If the final value is .6A as you stated, look at how long it takes for the current to go from 0 to 0.37A. Then that time is the time constant.

    I assume you are scoping the voltage across the resistor?

    Bob
     
  5. mike_980

    mike_980

    16
    0
    Mar 24, 2013
    ahh yes that makes sense thanks! yes I scoped it across a 1ohm resister so voltage and current were the same to make it easy
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-