Early on in the book "The Art of Electronics" (p.15, 2nd edition) a voltage divider circuit is described: an input voltage \$V_{in}\$ over a tunnel diode \$D\$ and a resistor \$R\$, and an output voltage \$V_{out}\$ the voltage across the resistor.
It is then written that a change in \$V_{in}\$, denoted \$v_{sig}\$, results in an (amplified) change \$v_{out}\$ in the output:
\$v_{out} = \frac{R}{R + r_{t}} \cdot v_{in}\$
where \$r_{t}\$ is (negative) dynamic resistance of the diode \$D\$.
Unfortunately no derivation is given.
Now I have been trying for the last two hours to derive this form but keep getting stuck. Unfortunately there seem to be no leads on the internet. Could someone derive the formula (in a clean manner)?
Best Answer
There will be a small signal AC current through the two devices
$$i = \frac{v_{in}}{R+r}$$
Therefore the output voltage (the voltage across the resistor) will be given by
$$v_{out} = iR = \frac{R}{R+r}v_{in}$$