🔐 The Two-Key Lock

What you're building

A lock with two keys. Press button A: nothing. Press button B: nothing. Press both at once: the LED lights. You built this once already with two switches in a row, in The Vault. This time the deciding is done by a logic chip, a CD4011, and the buttons never carry the LED's current at all. They only tell the chip what they are doing. The chip decides.

That is a small circuit and a very large idea. Every decision a computer makes is a chain of exactly these: a few inputs, each either high or low, and a gate that answers with a high or a low of its own.

High, low, and the gate's question

A logic gate only knows two voltages. High means near the + rail, 4.5 V here; call it 1. Low means near 0 V; call it 0. The CD4011 draws the line at about half the supply, 2.25 V: above that an input reads as 1, below as 0.

Each of the chip's four gates is a NAND: NOT-AND. It asks "are both my inputs 1?" and then says the opposite. Its truth table, inputs A and B, output Y:

• 0, 0 → 1
• 0, 1 → 1
• 1, 0 → 1
• 1, 1 → 0

The output is low in exactly one case: both inputs high. Which is almost the lock you want, except upside down. You want the LED on for 1, 1.

Two NANDs make an AND

Here is the trick that makes NAND famous. Take gate 2 and tie its two inputs together. Now its inputs are always equal: 0, 0 gives 1 and 1, 1 gives 0. It has become a NOT gate: whatever goes in, the opposite comes out.

Feed gate 1's output into that NOT. Gate 1 says 0 only when both buttons are down; the NOT flips that to 1. Every other time gate 1 says 1 and the NOT flips it to 0. The pair together say 1 only when A and B: an AND gate, built from two NANDs. Add a NOT in front of each input and you can build OR too. Every logic function there is can be built from NAND alone, which is why engineers call it the universal gate and why a quad-NAND chip is the first logic chip worth owning.

Reading the pins

The CD4011 is a 14-pin chip. Hold it with the notch (or the dot) at the top left and count anticlockwise: pins 1 to 7 down the left side, 8 to 14 up the right.

• Gate 1: inputs pin 1 (A1) and pin 2 (B1), output pin 3 (Y1).
• Gate 2: inputs pin 5 (A2) and pin 6 (B2), output pin 4 (Y2).
• Gate 3: pins 8, 9 → 10. Gate 4: pins 12, 13 → 11. Unused here.
• Power: pin 14 (VDD) to +, pin 7 (VSS) to −.

Y1 jumpers to both A2 and B2: that is the NOT. Y2 drives the LED.

The inputs: 100 kΩ pull-downs

Each button connects + to one input of gate 1. Button up, the input is connected to nothing, and a logic input connected to nothing is neither 1 nor 0; it drifts, and the chip guesses. So each input also has a 100 kΩ to −, a pull-down. Button up: 0 V, a firm 0. Button down: the button wins and the input sits at 4.5 V, a firm 1, while the pull-down passes 4.5 ÷ 100 000 = 45 µA. That is the only current the buttons ever carry. A CMOS input itself takes a few picoamps, which is why a resistor this large is enough, and why the Switch That Uses No Current used the same value on its MOSFET gate.

Tying off the spare gates

Gates 3 and 4 do nothing, but their inputs must not float either. A CMOS input left in mid-air drifts toward half the supply, where both halves of the gate conduct at once: the chip oscillates, warms up and can upset the gates you are using. So A3 and B3 go to −, and A4 and B4 go to +. Either rail works; the rule is only that every input goes somewhere.

What the LED sees

When both buttons are down, Y2 goes high. A CD4011 output is a weak thing. The simulator models it as about 200 Ω between the pin and the rail it is driving, so the LED lane is 200 Ω inside the chip plus your 220 Ω plus the LED's 2.0 V:

(4.5 − 2.0) ÷ (200 + 220) ≈ 6 mA

Clearly lit, dimmer than First Light's 11 mA. On a real CD4011B expect dimmer still: the datasheet rates the output at about 1 mA at 5 V, and it sags under load. That is normal for the 4000 series. Logic outputs are signals, not power; when you need brightness, the output drives a MOSFET gate, which costs it nothing.

The LED resistor still matters. Without it the output would try to push (4.5 − 2.0) ÷ 200 ≈ 12 mA, which a real chip cannot deliver and would not enjoy trying.

Build it

1. Rails: + along the top, − along the bottom, battery between them.
2. The chip: across the centre gap, notch at the top. Pin 14 to +, pin 7 to −. Do this first and check it twice.
3. Spare gates: pins 8 and 9 to −; pins 12 and 13 to +.
4. Input A: button from + to pin 1; 100 kΩ from pin 1 to −.
5. Input B: button from + to pin 2; 100 kΩ from pin 2 to −.
6. The NOT: jumper pin 3 to pin 5, and pin 3 to pin 6.
7. The lamp: pin 4 → 220 Ω → LED long leg; LED short leg → −.
8. Power up. Dark. Press A: dark. Press B: dark. Press both: lit.

On the Workbench the board runs in time, because a chip has state. Hold each button with the mouse, then both; the output pads on the chip glow amber when that gate's output is high, so you can watch Y1 drop and Y2 rise the moment the second key goes down.

Debug like an engineer

Test the halves separately. Pin 3 (Y1) should sit near 4.5 V with the buttons up and drop near 0 V only with both held: that proves gate 1, the pull-downs and the buttons. Then pin 4 (Y2) should be the opposite of pin 3: that proves the NOT. If pin 3 never moves, an input is floating or a button is in the wrong rows. If pin 3 moves but pin 4 does not, one of the two jumpers from pin 3 is missing, so gate 2 has a floating input.