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@Laplace:
He got the idea of using the derivative from me (https://www.electronicspoint.com/incremental-resistance-diode-t245938.html). My idea was that this might be appropriate for the homeworks section.
The derivative is not too difficult in this case. Besides, Wolfram Alpha can help.
Your way of doing the calculation is essentially the same thing: approximate the derivative by a tangent to the characteristic. For my part I'm a friend of using closed solutions instead of numerical ones as long as it is practical, This becaus it is easier to use the resulting formula in further calculations. If you have only a numerical value, you have to do the approximation each time a parameter changes.
Harald
He got the idea of using the derivative from me (https://www.electronicspoint.com/incremental-resistance-diode-t245938.html). My idea was that this might be appropriate for the homeworks section.
The derivative is not too difficult in this case. Besides, Wolfram Alpha can help.
Your way of doing the calculation is essentially the same thing: approximate the derivative by a tangent to the characteristic. For my part I'm a friend of using closed solutions instead of numerical ones as long as it is practical, This becaus it is easier to use the resulting formula in further calculations. If you have only a numerical value, you have to do the approximation each time a parameter changes.
Harald
@Harald:
Closed form solutions are truly more elegant. Yet it has been so many years since I took the derivative of anything, I would need to open my old calculus textbook. Still understand the principles behind derivatives but have forgotten the mechanics of doing it. Also, having a quick numerical check to verify correctness of the solution can be useful.
Closed form solutions are truly more elegant. Yet it has been so many years since I took the derivative of anything, I would need to open my old calculus textbook. Still understand the principles behind derivatives but have forgotten the mechanics of doing it. Also, having a quick numerical check to verify correctness of the solution can be useful.
Hi Laplace:
Thank you very much for your response. I'm sorry that I'm quite slow on the spreadsheet thing that your program can do the derivative? And, the formula you use is the dR=dV/dI on the xls? Also, how come the delta V is 4.001 and not 4.01 nor 4.0001? Sorry for many stupid questions.
Thank you in advance for your guidance.
Regards,
Bati
Thank you very much for your response. I'm sorry that I'm quite slow on the spreadsheet thing that your program can do the derivative? And, the formula you use is the dR=dV/dI on the xls? Also, how come the delta V is 4.001 and not 4.01 nor 4.0001? Sorry for many stupid questions.
Thank you in advance for your guidance.
Regards,
Bati
The problem started with an equation giving I as a function of V, I=f(V). So if you take the derivative dV/dI you get another function of V giving the incremental resistance for any value of V.
A numerical approximation needs to provide the value dV/dI for any value of V using the same function I=f(V). To calculate dV/dI at V volts the formula is, Incremental Resistance @V = (V-(V+delta V))/(f(V)-f(V+delta V))
For a numerical example choose V=4 and delta V=0.001, then dV/dI= (4.000-4.001)/(f(4.000)-f(4.001)) = 0.001/(f(4.001)-f(4.000))
Why a delta V=0.001? Just choose something reasonable. Generally the smaller the delta the better the accuracy, unless round-off errors become significant. How many digits does your calculator have? Note that one way to improve accuracy is to have the delta symmetrical around the calculation point, dV/dI=0.001/(f(4+0.0005)-f(4-0.0005)).
A numerical approximation needs to provide the value dV/dI for any value of V using the same function I=f(V). To calculate dV/dI at V volts the formula is, Incremental Resistance @V = (V-(V+delta V))/(f(V)-f(V+delta V))
For a numerical example choose V=4 and delta V=0.001, then dV/dI= (4.000-4.001)/(f(4.000)-f(4.001)) = 0.001/(f(4.001)-f(4.000))
Why a delta V=0.001? Just choose something reasonable. Generally the smaller the delta the better the accuracy, unless round-off errors become significant. How many digits does your calculator have? Note that one way to improve accuracy is to have the delta symmetrical around the calculation point, dV/dI=0.001/(f(4+0.0005)-f(4-0.0005)).
