how to scale a device for the second derivative of Amps/Volt?

I'm testing parts for a lab at school that measure current as a function of voltage. With different sized devices I can scale the results easily lets say I want to estimate the current of a device that is twice the size of the original device, I multiply the current by 2 at that same voltage.

However, I am being asked to find the d^2I/d V^2, second derivative of the curve. I believe this results in a unit of Amps/Volt*Volt, A/VV, A/V^2 as I am looking for the how much the current changes over voltage with respect to another voltage. How would I scale a device that's twice as big for a second derivatives, A/VV^2. please let me know if i am confused and probably have confused others with this post. One thought was that I scaled it using 2^2. Looking forward to any help/understanding you guys can provide Thanks!

• Have you worked with op-amps before? This is exactly what they're for (analog computers); use two op-amps configured as differentiators, and cascade the output of first into the second. – MarkU Nov 17 '15 at 22:00
• See for example this question: electronics.stackexchange.com/questions/130578/… – MarkU Nov 17 '15 at 22:06
• @MarkU, doesn't that give derivatives wrt time? – Chu Nov 17 '15 at 22:07
• @Chu : yes, but if you keep dV/dT constant it may still be a useful technique. – Brian Drummond Nov 17 '15 at 22:21
• The first derivative of amps over volts is just conductance which of course is the reciprocal of resistance .These second and higher order derivatives that you seek only exist if there is nonlinearity .These higher order terms of course correspond to harmonics .You could do a little math and then implement by whatever means a harmonic analyser which is well documented. – Autistic Nov 18 '15 at 9:42

First of all, you said that for a device that is twice the size of the original device, then the current is twice the amount for the same voltage. That is simply a definition of your choosing. (For example, for many things, twice as big would be defined as twice the mass.)

Given that as the definition:
$i_1 = f(v)$ for device1
if device2 is twice the size of device1 then $i_2 = 2f(v)$

So the second derivative of $i_1$ with respect to $v$ is
$i_1^" = \frac{d^2 f(v)}{dv^2}$

And the second derivative of $i_2$ with respect to $v$ is
$i_2^" = 2\frac{d^2 f(v)}{dv^2}$

Therefore, the scaling is the same, because derivative is a linear operator.