Diode logic gate basics

WHY does a diode do logic at all?

WHAT is a diode doing here? A diode conducts (turns ON) only when its anode is about $0.7\text{ V}$ (silicon) more positive than its cathode. Otherwise it blocks. When ON, it behaves roughly like a battery: it clamps the voltage across itself to $\approx V_D \approx 0.7\text{ V}$.

WHY this gives logic: Logic is just "is this node HIGH or LOW?" A conducting diode can pull a shared node toward one of the inputs. If we arrange several diodes so they all try to pull the same node, whichever diode conducts hardest dominates the node voltage. Choosing the diode orientation decides whether HIGH-inputs win (OR) or LOW-inputs win (AND).

Deriving the DIODE OR gate from scratch

Setup (derive, don't memorise): Put two inputs $A$ and $B$. Connect a diode from each input to a common node $Y$, anode at the input, cathode at $Y$ (arrow points toward $Y$). Connect a resistor $R$ from $Y$ down to ground.

Why this arrangement? Anode-at-input means a diode can only push current into $Y$ from a HIGH input. So any HIGH input can lift $Y$ up.

Let's reason case by case ($V_H = 5\text{ V}$, $V_D = 0.7\text{ V}$):

• $A=0,\;B=0$: Neither input can forward-bias its diode (both anodes at 0 V, node would have to be below 0 V for current). No diode conducts $\Rightarrow$ $R$ pulls $Y$ to ground $\Rightarrow Y = 0$. ✔ (OR of 0,0 = 0)
• $A=5,\;B=0$: Diode from $A$ conducts. Current flows $A \to$ diode $\to Y \to R \to$ gnd. $Y$ rises until diode is ON, clamping: $Y = V_A - V_D = 5 - 0.7 = 4.3\text{ V (HIGH)}.$ ✔
• $A=0,\;B=5$: Symmetric, $Y = 4.3\text{ V}$. ✔
• $A=5,\;B=5$: Both conduct, $Y = 4.3\text{ V}$. ✔

Deriving the DIODE AND gate from scratch

Flip everything. Now put diodes cathode at input, anode at node $Y$ (arrows point away from $Y$, toward the inputs). Connect the resistor $R$ from $Y$ up to $+V_{CC}$ (the supply).

Why flip? Cathode-at-input means a LOW input can pull current out of $Y$, dragging $Y$ down. The pull-up resistor tries to hold $Y$ high; any grounded input defeats it.

Case analysis ($V_{CC}=5\text{ V}$):

• $A=0,\;B=0$: Both diodes conduct, pulling $Y$ down. $Y = V_A + V_D = 0 + 0.7 = 0.7\text{ V (LOW)}$. ✔
• $A=5,\;B=0$: Diode to $B$ conducts (path: $V_{CC}\to R \to Y \to$ diode $\to B$=0). $Y = 0 + 0.7 = 0.7\text{ V (LOW)}$. The diode to $A$ is reverse biased (its anode $Y=0.7$ V < cathode $A=5$ V) → OFF. ✔
• $A=0,\;B=5$: Symmetric, $Y=0.7\text{ V}$. ✔
• $A=5,\;B=5$: No diode can conduct (both cathodes at 5 V, node can't get above them enough). $R$ pulls $Y$ up to $V_{CC}=5\text{ V (HIGH)}$. ✔

Worked Examples

Common Mistakes (Steel-man + fix)

Active-Recall Flashcards

Connections

• PN Junction Diode — the forward-drop $V_D \approx 0.7$ V that sets all our offsets.
• Diode as a Switch — ON/OFF model underlying the voting behaviour.
• DTL Diode-Transistor Logic — adds a transistor to invert + restore levels.
• Logic Levels and Noise Margin — why level degradation is dangerous.
• Boolean Algebra basics — $\max$=OR, $\min$=AND correspondence.
• Pull-up and Pull-down Resistors — the resistor that sets the default node state.

Concept Map

Hinglish (regional understanding)