Ohm's Law — Question 4

Question 4

Self-heating can curve a characteristic of an otherwise ohmic material

A specimen obeys $V=R(T)I$ at any fixed temperature, with $R(T)=R_0[1+\alpha(T-T_0)]$. During slow measurements, the supplied thermal law is $T-T_0=\Theta IV$, where $\alpha,\Theta>0$.

1. Derive the steady V-I relation and its admissible current domain.

2. Find the leading small-current nonlinear term and explain why this does not disprove fixed-temperature Ohm behavior.

Question 4 - Solution

Self-heating can curve a characteristic of an otherwise ohmic material

(a) Method and calculation

$$V=R_0I[1+\alpha\Theta IV]\Rightarrow V(1-\alpha\Theta R_0I^2)=R_0I.$$$$V=\frac{R_0I}{1-\alpha\Theta R_0I^2},\quad |I|<(\alpha\Theta R_0)^{-1/2}.$$

(b) Interpretation and further calculation

$$V=R_0I+\alpha\Theta R_0^2I^3+O(I^5).$$$$\left.\frac VI\right|_{T\ \mathrm{fixed}}=R(T)=\text{constant}.$$The slow sweep changes temperature as well as current.

Physical check

The excluded pole marks failure of a finite steady solution within this linear thermal model, not a prediction of infinite physical voltage.