Discussion
A diode carries \( 1 \ \mathrm{mA} \) at \( 0.7 \ \mathrm{V} \). At \( 10 \ \mathrm{mA} \) its drop is about
Answer: C
The exponential law gives about sixty millivolts per decade:
\[ \begin{aligned} \Delta V &= 2.3 V_{T} \log_{10}(10) \\ &\approx 60 \ \mathrm{mV} \end{aligned} \]
so the drop rises to about 0.76 V. Assuming it stays at 0.7 V is the piecewise-linear approximation, adequate for many purposes. Seven volts would follow from treating the diode as a resistance. The sixty-millivolts-per-decade figure is worth memorising, since it makes every such estimate a matter of counting decades.
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