Find Non-linear Current Across a Device

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To find the current across a non-linear device in a 'blackbox' circuit with known Thevenin voltage and resistance, one must apply Kirchhoff's Voltage Law (KVL) around the loop, leading to the equation V_th - R_th I - f(I) = 0. The Thevenin equivalent is valid only if the internal circuit is linear; however, the external circuit can be non-linear. It's important to note that not all blackboxes can be accurately represented by a Thevenin equivalent, especially if they are not approximately linear. The use of Thevenin's theorem may still be applied in practice, but additional specifications like Total Harmonic Distortion (THD) and Intermodulation Distortion (IMD) may be necessary to account for inaccuracies. Understanding these limitations is crucial for accurate circuit analysis.
ian2012
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I was just thinking about this the other day...

Given any 'blackbox' circuit, for which the values of the thevenin voltage and thevenin resistance are known. And given that a non-linear device is placed across the terminals of the circuit. What relation would one use to find the current across the device. Since it is non-linear, i am guessing V/R does not apply.
 
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If the nonlinear device had a simple static V/I characteristic in the form V=f(I) then you would solve the equation,

V_thv - R_thv I - f(I) = 0
 
Thank you for that. Is there an intuitive explanation?
 
ian2012 said:
Thank you for that. Is there an intuitive explanation?

Yes it's just a simple application of KVL (Kirchoff Voltage Law) around the loop.
 
ian2012 said:
Given any 'blackbox' circuit, ... (boldface added for emphasis)
However, are you also including nonlinear source circuits in your consideration?
 
turin said:
However, are you also including nonlinear source circuits in your consideration?

There's actually no problem with that turin. The "black box" itself must be linear in order to have a simple Thevenins equivalent, but there's no restriction on the nature (linear/nonlinear) of the external circuit when using a Thev equiv.
 
uart said:
The "black box" itself must be linear in order to have a simple Thevenins equivalent, ...
That's precisely the problem. I just didn't want the OP to think that they could make a few measurements to derive a Thev. eq. circuit for any blackbox. There are some blackboxes that are not even approximately linear.

EDIT: I reread the OP carefully, and now realize that the OP probably meant what you said. Sorry.
 
ian2012 said:
Thank you for that. Is there an intuitive explanation?

The part you had the Thevenin of didn't stop being linear so it still looks the same electrically.

BTW Thevenins are only strictly valid if the circuit they are representing is either linear or nearly linear (small signal assumption). Otherwise the Thevenin itself would be a problematic representation - though often enough it's still done and then you throw in new specs like THD and IMD to account for the variance.
 

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