Solid-State Diodes and Diode Circuits

calendar_today August 22, 2025 article 349 words access_time 2 min replay Modified September 30, 2025

Table of Contents

Diode

Definition 03.1 (The Diode Equation).

$$i_{D} = I_{S}\left[ \exp \left(\frac{q v_{S}}{nkT} \right) - 1 \right] = I_{S} \left[ \exp \left( \frac{v_{S}}{n V_{T}} \right) \right]$$

where:

Reverse bias
- $V_D < 0$
- $i_{D} \approx -I_S$
Zero bias
- $V_D = 0$
- $i_{D} \approx 0$
Forward bias
- $V_D = 0$
- $i_{D} \approx I_S \exp \left( \frac{v_{D}}{nV_{T}} \right)$

$i_D$ sharply increases with enough negative voltage in the reverse direction
- This is called the breakdown voltage, $V_Z$

As the E-field increases, the energy of charge carriers cause ionization of atoms, which turn into charge carriers and cause a positive feedback loop
$V_Z$ = -5.6 V in normal silicon diodes

The E-field is strong enough to tunnel electrons from covalent bonds through the depletion region
Occurs at higher voltages than avalanche
Allows for voltage regulation, since $V_D$

Find the quiescent operating point $(I_D, V_D)$ for the diode
KVL equation characterizing diode behavior is known as the load line
- Many methods to find the load line

$$v_{S} = I_{D}R + V_{D}$$

Diode is a perfect switch when $v_S > 0.7$
Unrealistic because the existence of a potential barrier implies that diodes must have a potential difference
- Useful nonetheless since you can solve systems containing multiple diodes without much computational overhead (effectively becomes a 3SAT problem)

$V_D = 0.7$ when $v_S > 0.7$
More realistic because it accounts for the potential difference across the barrier

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