# incremental resistance

Discussion in 'Electronics Homework Help' started by kbcheong, Mar 24, 2012.

1. ### kbcheong

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Mar 24, 2012
I have the questions as attached. I got the answer of 250 ohm for 4V and 158.22 ohm for 10V. Is it correct?

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2. ### Laplace

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Apr 4, 2010
Not the answer I got, but I did not try to get fancy by finding the derivative. Just used numerical approximation instead with the formula I=0.002 x V^(3/2). Then found the current at 4 volts and at 4.001 volts so delta:V=0.001 and delta:I is as calculated. Same process at 10 volts.

3. ### Harald KappModeratorModerator

12,334
2,930
Nov 17, 2011
@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

4. ### Laplace

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Apr 4, 2010
@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.

5. ### bati38

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Apr 1, 2012
Hi Laplace:

I'm no good at calculus, and I would like to know how did you find 4.001 V?
I could not figure it out when Vp = 4^3/2. Then, it's about 8

6. ### Laplace

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185
Apr 4, 2010
Initially I used my HP-15C calculator. But then I just entered the formula into the Open Office Calc spreadsheet for a sensitivity analysis of how much the final answer changed using 4.01, 4.001, 4.0001 volts.

7. ### bati38

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Apr 1, 2012
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.

Regards,
Bati

8. ### Laplace

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185
Apr 4, 2010
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)).

9. ### bati38

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Apr 1, 2012
Thank you for explaining clearly.

10. ### bati38

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Apr 1, 2012
Oh, I forgot to mention that my calculator has 12 digits, and kinda fancy. But, I don't know how to use all that fancy stuffs  