Ohm's Law — Question 10

Question 10

Negative differential resistance can coexist with positive absorbed power

Over the specified range $0<V<2V_*$, a device has $I(V)=I_*[2(V/V_*)-(V/V_*)^2]$, where $I_*,V_*>0$.

1. Find the current maximum and the voltage range with negative differential conductance.

2. At $V=3V_*/2$, find static and differential resistance and determine the sign of VI.

Question 10 - Solution

Negative differential resistance can coexist with positive absorbed power

(a) Method and calculation

$$dI/dV=(2I_*/V_*)(1-V/V_*).$$$$I_{\max}=I_*\text{ at }V=V_*,\quad dI/dV<0\text{ for }V_*<V<2V_*.$$

(b) Interpretation and further calculation

$$I(3V_*/2)=3I_*/4,\quad R_s=2V_*/I_*,\quad r_d=-V_*/I_*.$$$$VI=(3V_*/2)(3I_*/4)=9V_*I_*/8>0.$$

Physical check

A negative tangent slope concerns perturbations; it does not mean the device supplies net power at that operating point.