# Differential resistance vs. $R = \frac{U}{I}$

Resistance is defined as $R = \frac{U}{I}$. If Ohm's law holds, $U(I)$ is a linear function and the slope of the curve is equal to the resistance $R$.

I realized only recently that if $U(I)$ is a non-linear function, the slope of the function is no longer equal to the resistance as defined above. We then have two kinds of resistances:

• the differential resistance $r(I) = \frac{d U}{d I} (I)$ (which is equal to the slope of the curve at the point $I$)

• "ordinary" resistance $R(I) = \frac{U(I)}{I}$.

My questions:

• Is the quotient $R(I) = \frac{U(I)}{I}$ used to describe devices with non-linear current-voltage characteristics at all? (I am aware of the fact that $R$ isn't a property of the device in this case but it may still be a sensible concept)
• If both $R(I)$ and $r(I)$ are used, when do I use which one?
• What can we say about the relationship of the two quantities in general?

Example: Let's have a look at the current-voltage characteristic of a tunnel diode (see image below, taken from Wikipedia). I am not really familiar with how this device operates but it illustrates my question. For "ordinary resistance", we have $R(i_1) = \frac{v_1}{i_1} \ll R(i_2) = \frac{v_2}{i_2}$ while for the differential resistance we have $r(i_1) = r(i_2) = 0$ (/edit: As freecharly noted, this is wrong. Actually, we have $r(i_1) = r(i_2) = \infty)$

So there's some justification to say that if we apply the constant voltage $v_2$ that the resistance is bigger than the resistance if we apply the constant voltage $v_1$, and there's also some justification to say that the resistance is zero infinity in both cases. How is the termonology actually used?

/edit: I edited the phrasing of the question and included an example in order to make things more clear.

• I'm not sure what type of information you're looking for, since you've described the situation pretty clearly (i.e., that the meaning of "slope" for a nonlinear material property can have multiple possible definitions), but you might find it useful that the two versions are called the tangent and the secant definitions, respectively. The same situation arises with material stiffness (as characterized by the Young's modulus), for which you might find more information in an online search Feb 17, 2018 at 18:43
• Calling the different definitions "secant" and "tangent" doesn't sound right to me. The secant corresponds to a difference quotient and the tangent to the limit of a difference quotient, i.e. the differential quotient, i.e. the derivative. $R = \frac{U}{I}$ on the other hand doesn't involve differences, it is just a simple quotient.
– Marc
Feb 17, 2018 at 19:40
• When I learned about diodes &tc., it was called trans-resistance, hence transistors. Feb 17, 2018 at 20:24
• @PeterDiehr: My first impression is that this is a different concept because it combines an input value with an output value (see en.wikipedia.org/wiki/Transconductance). I may be wrong about this.
– Marc
Feb 17, 2018 at 20:55
• @Marc the differences are implicitly taken from (0, 0) in this standard usage. I didn’t make it up. In any case, thank you for editing your question to provide clarification. Feb 17, 2018 at 20:59

For any current-voltage characteristic you can always formally define a resistance $$R=\frac {V}{I}$$ or conductance $G=I/V=1/R$ at any point of the characteristic. The differential resistance $r$ and differential conductance $g$ are defined by $$g=\frac {1}{r}=\frac {dI}{dV}$$ In the tunnel diode characteristic, you have a region of negative differential resistance (or conductance) between the peak and valley voltages $v1$ and $v2$, which is useful for microwave frequency oscillators and other applications. The differential conductance $g=dI/dV$ is zero at the peak and the valley of the $I-V$ characteristic.Thus the differential resistance $r=1/g$ is infinite at these extrema, not zero! At $V=v1$ the resistance $R=V/I$ is smaller than the resistance at $V=v2$.

• Is this formally defined resistance $R$ used in practise? (Also thanks for catching the error regarding the value of the differential resistance. I am used to diagrams where U is on the y-axis.)
– Marc
Feb 18, 2018 at 13:25
• @Marc - To my knowledge, the resistance $R$ is not used very much as compared to the $I-V$ characteristic itself. The differential resistance or conductance are frequently used in small-signal models of the NDR. Feb 18, 2018 at 14:57

An example. Diodes have a non-linear current-voltage characteristic. In the ideal case: $$I = I_0(e^\frac{qV}{kT}-1).$$

When there is a current in the forward direction, one can neglect the unity. So $\frac{1}{r} = \frac{dI}{dV} = \frac{q}{kT}I$. The dynamical resistance $r = \frac{kT}{q}/I \approx \frac{0.025\ {\rm volt}}{I}$ or $25/I$ ohm when current is given in mA.

This is then the impedance for small ac current $i$ when superposed on a DC current $I$ through the diode. The small-signal voltage $v = i r$.

• 1) The unit of $r$ as "ohm per mA" doesn't sound right to me. $r = \frac{dU}{dI}$ needs to have "ohm" as its unit. 2) Your last paragraph sounds interesting to me. How do I determine, whether I should use $R$ or $r$? I assume that for the DC current, you would not use $r$ but $R$, correct?
– Marc
Feb 17, 2018 at 19:48
• @Marc Ok, "per" is wrong. To be completely correct one could say that the diode dynamical resistance expressed in ohm is about $25/I$ when current is given in mA. A diode does not have an $R$ in a meaningful way.
– user137289
Feb 17, 2018 at 19:58
• I have edited my question in order to clarify what I want to know. In the light of the updated question: I think you are saying that the differential resistance is used in AC situations and what I call the "ordinary resistance" isn't used at all. Is this understanding correct?
– Marc
Feb 17, 2018 at 20:53
• @Marc Differential resistance is used in small-signal situations, when one can use a Taylor expansion of a non-linear current-voltage relation around a working point set by a DC current/voltage. Tunnel diodes are a different subject.
– user137289
Feb 17, 2018 at 21:03
• @Marc Higher-order terms would cause harmonics. In a HiFi system these are distortions that one wants to avoid. But yes, the higher derivatives are a way to estimate how large the harmonic distortion is, and at what signal levels it would become a problem.
– user137289
Feb 17, 2018 at 21:47