Kevin Zhou — Advanced E&M III Problem 4 · Kevin Zhou #E7.4
Physics · electromagnetism
Problem
Problem 3. NBPhO 2016, problem 4. A rather complicated problem involving several exotic
circuit elements.
Next,weconsidersomequalitativelynewbehaviorthatcanemergefromlessfamiliarcircuitelements,
such as amplification, hysteresis, and instability.
Idea 1
Tunnel diodes are a variant of diodes, whose I(V) rises, falls, and rises again. That is, they
have a region with negative differential resistance, dI/dV < 0. This allows them to amplify
signals, as we’ll see below, and also can make them unstable.
Example 1: EFPhO 2003
The circuit below, containing a tunnel diode, acts as a simple amplifier.
Here, R = 10Ω and E = 0.25V. If a small signal voltage V (t) is applied across the input,
in
then an amplified and shifted version of the signal appears across the output. Find the
amplification factor.
Solution
When a constant emf E is applied, Kirchoff’s laws give
E = I R+V(I )
0 0
where V(I) is the voltage characteristic of the diode.
2
Kevin Zhou Physics Olympiad Handouts
By plotting E −IR on the graph above, we find I at the intersection. (Notice the x-axis
0
label: it is common to write a decimal point as a comma in Eastern Europe.) Now consider
the effect of applying the signal voltage, which changes the current by ∆I,
E +V = (I +∆I)R+V(I +∆I).
in 0 0
Since the signal voltage is small, we can Taylor expand the voltage characteristic, giving
V = ∆I(R+V′(I )).
in 0
This in turn tells us that
R
V = (I +∆I)R = V0 + V .
out 0 out R+V′(I ) in
0
In other words, the change in V is just V , times the amplification factor
out in
R 10 5
= = −
R+V′(I ) 10−16 3
0
where we read V′(I ) off the graph by drawing a tangent. The intuition here is that the
0
circuit is like a voltage divider, but the tunnel diode acts like a negative resistance. If we
had V′(I ) close to −R, for example, the amplification factor would have been huge. Since
0
V′(I ) is more negative than −R, the signal ends up flipped.
0