Electronic – Basic question about diode voltage drop and resistor position

Assuming ideal components, when the diode is "on", the equivalent circuit is:

^{simulate this circuit – Schematic created using CircuitLab}

The charging current clockwise through C1 produces a voltage across the parallel combination of resistors.

When the diode is "off", the equivalent circuit is:

^{simulate this circuit}

Now, you can see that R1 is provide a path for a counter-clockwise current to discharge the capacitor.

Also, see that R2 is required to pull the output voltage to zero while the diode is off.

Why is the path that R1 provides needed? if it's not there, then there is an open-circuit, no current flows. Theoretically the end result should be the same: the negative part of the signal does not appear at the output.

If there is no path for a counter-clockwise current through the capacitor, the voltage across the capacitor will only increase and this voltage will oppose the positive signal voltage. Eventually, the capacitor voltage will be large enough that the diode will not turn on and the output voltage will be zero.

Below is a simulation I ran without R1. The top trace is the voltage across the capacitor; the bottom trace is the output voltage. The input source is 1kHz sine wave.

Note that the voltage across the capacitor increases while the diode is on but doesn't decrease while the diode is off. Note the effect of this on the output voltage.

It just looks like magic doesn't it?

If you haven't taken a course in semiconductor device physics, I suppose it does look like magic.

PN junctions are not resistors. Whereas the voltage across a resistor is proportional to the current through

\$v_R =R \cdot i_R \$

the current through a diode is approximately:

\$i_D = I_S(e^{\frac{qv_D}{kT}}-1) \$

In words, the diode current goes up exponentially with the diode voltage.

The derivation of the diode equation isn't trivial but it isn't mysterious either; it's based on well understood device physics.

For zener and avalanche diodes, the physical explanation is a bit easier to grasp. From Wiki:

Under a high reverse-bias voltage, the p-n junction's depletion region expands, leading to a high strength electric field across the junction. A sufficiently strong electric field enables tunneling of electrons from the valence to the conduction band of a semiconductor leading to a large number of free charge carriers. This sudden generation of carriers rapidly increases the reverse current and gives rise to the high slope resistance of the Zener diode.

The Zener effect is distinct from avalanche breakdown which involves minority carrier electrons in the transition region which are accelerated by the electric field to energies sufficient to free electron-hole pairs via collisions with bound electrons. Either the Zener or the avalanche effect may occur independently, or both may occur simultaneously. In general, diode junctions which break down below 5 V are caused by the Zener effect, while junctions which experience breakdown above 5 V are caused by the avalanche effect.