Can two switches make a decision?
Your car beeps if you are seated AND the belt is unfastened, OR the door is open AND the key is in.
Somebody had to build that decision out of electricity. What is the cheapest part that can do it?
Minecraft players build working calculators from redstone "gates". Same idea: a gate is anything that turns input levels into an output level by a fixed rule. Today's gate-builder is the cheapest decision-maker in electronics: the diode.
Before we start: which part of this beep rule do you think a handful of diodes cannot build?
Argue it in pairs for two minutes. There may be more than one defensible answer.
Inside the diode: a hill you can flatten
The diode, four ways
Every diode analysis starts by choosing a model. The real curve is Shockley's equation. For logic we almost always use the constant-drop model.
0.7 V is a model, not a law. Silicon turns on gradually around 0.6–0.7 V; germanium near 0.3 V; Schottky 0.2–0.4 V; LEDs 1.8–3.3 V depending on colour.
Circuit: VS → 470 Ω → diode → ground.
Agreeing on what a “1” is
A wire carries a voltage, not a bit. Logic families agree on bands. We use TTL-compatible thresholds all semester.
This is positive logic: higher voltage = 1. Negative logic flips the assignment, and the same circuit then performs the dual function.
Why not make VIL = VIH = 1.4 V and get rid of the forbidden zone altogether?
Argue it in pairs for two minutes. There may be more than one defensible answer.
Diode OR: the highest input wins
Anodes at the inputs, cathodes joined, a resistor to ground.
Any 5 V input forward-biases its diode and lifts the output to
Diodes on lower inputs are reverse-biased and drop out.
Find the input combination where no current flows at all.
Diode AND: the lowest input wins
Flip the diodes; move R to the supply. R pulls up, any 0 V input drags the node down to
Memory trick: the conducting diode is the one that argues with the resistor. R pulls down → highest input fights back (OR). R pulls up → lowest input fights back (AND).
Worked example: a diode AND gate
Given
Two-input diode AND, VCC = 5 V, R = 4.7 kΩ, silicon diodes (Vγ = 0.7 V). Inputs A = 0 V, B = 5 V.
Find
Vo, each diode's state, the current in R, and the logic level.
Method: assume–solve–verify, as in Boylestad Ch. 2 and Millman & Taub's diode-logic sections
Solution
Predict the answer first. Then reveal one step at a time.
Worked example: three inputs, messy voltages
Given
Three-input diode OR, R = 10 kΩ, Vγ = 0.7 V. A = 5 V, B = 3 V, C = 0 V.
Find
Vo, the diode states and IR.
Solution
Predict the answer first. Then reveal one step at a time.
Why nobody builds a CPU from diodes
Three fatal flaws: no inversion (not functionally complete), no power gain (every stage weakens the signal), and a Vγ tax at every gate.
Still valid after 3 gates. Add more stages.
Unloaded gates assumed. Real cascades are worse: each following gate's resistor also loads the node.
Your turn · 1: a “zero” that isn’t
A three-input diode AND gate: VCC = 5 V, R = 10 kΩ, Vγ = 0.7 V.
Inputs: A = 5 V, B = 5 V, C = 0.3 V.
Find Vo, which diodes conduct, IR, and the logic level.
Work it out first.
Press Solve when the class has an answer.
Your turn · 2: the loading trap
A diode OR gate (R1 = 10 kΩ to ground) has both inputs at 0 V. Its output X feeds one input of a diode AND gate (R2 = 10 kΩ to +5 V); the AND's other input is 5 V.
Find VX. Is the OR gate's output still a 0?
Work it out first.
Press Solve when the class has an answer.
Critical thinking · design a two-level diode circuit
Using only diodes and 10 kΩ resistors (Vγ = 0.7 V, VCC = 5 V), build
Find Y for (A, B, C) = (5, 5, 0), (0, 0, 5) and (5, 0, 0) V. Does every output land in a valid band? How would you improve it?
Work it out first.
Press Solve when the class has an answer.
Critical thinking · three short arguments
Input A comes from a sensor with 1 kΩ internal resistance (open-circuit 5 V) into a diode OR with R = 10 kΩ. Is the output still 4.3 V?
Argue it in pairs for two minutes. There may be more than one defensible answer.
A classmate says: “Use germanium diodes (Vγ = 0.3 V) and diode logic becomes cascadable.” Agree or disagree?
Argue it in pairs for two minutes. There may be more than one defensible answer.
Diode OR still ships everywhere: power ORing
Nobody builds logic from diodes today, but the diode OR lives inside every device that must never lose power. Feed a mains adapter and a battery to the load through two diodes: whichever voltage is higher supplies the load, and the other diode blocks backflow. No software, no switch, no delay.
At home: the IPS that keeps your fan running during load-shedding makes the same “whichever source is available” decision with relays and electronics.
A bedside monitor or infusion pump cannot reboot when someone trips over the plug. Hospital equipment uses exactly this diode-OR, often with Schottky diodes or “ideal diode” MOSFET controllers to cut the 0.7 V loss. The same trick joins several bed call-buttons into one nurse-station lamp.
Pop quiz
Question 1 / 5
A two-input diode OR gate (Vγ = 0.7 V) has inputs 0 V and 5 V. The output is:
Three things to remember
Diode OR: highest input wins, Vmax − Vγ. Diode AND: lowest input wins, Vmin + Vγ.
Diode logic cannot invert, has no gain, loses 0.7 V per stage and loads its neighbours. It does not cascade.
The fix is a transistor that restores levels and inverts. Next class: RTL, DTL, and a moon computer built from one gate.
Diode OR, R = 5 kΩ, Vγ = 0.6 V, inputs A = 4 V, B = 4.5 V. Find Vo, each diode's state and current. (Careful with DA!)
Work it out first.
Press Solve when the class has an answer.