⚖️ The 3.3-Volt Rule

What you're building

A tiny power supply. A 9 V battery feeds one resistor, and below that resistor sits a node that stays at about 3.3 V whatever you hang on it, within limits. One LED lane is always connected; a second lane waits behind a switch. Close the switch and the second LED lights almost as bright as the first, and the first one barely dims. Something is holding the voltage still. That something is a zener diode, and this is the first circuit on the site whose job is not to light or buzz but to regulate.

Why 9 V is a problem

Every LED project so far ran on 4.5 V and used Ohm's law once: pick a resistor, get a current. That works because the battery voltage is a known number. Try it with a 9 V battery and the LED's 220 Ω resistor gives (9 − 2.0) ÷ 220 ≈ 32 mA: over the LED's 20 mA rating and past the point where the Workbench pops it. You could pick a bigger resistor, but a 9 V battery also sags toward 8 V as it drains, and a fresh one reads 9.4 V. Everything downstream drifts with it.

The fix is to stop depending on the battery voltage at all. Make a node that sits at 3.3 V, and design every lane for 3.3 V.

The zener: a diode that breaks on purpose

A normal diode passes current one way and blocks it the other. Push hard enough backwards and every diode eventually gives in; for a 1N4001 that happens past 50 V and may wreck it. A zener is a diode built to give in at a precise, low voltage and survive doing it. The 1N4728 gives in at 3.3 V.

So you wire it backwards: the striped end (the cathode) toward +, the plain end to −. Below 3.3 V it does nothing. The instant the node tries to rise above 3.3 V, the zener conducts, as much as it has to, and the node cannot rise any further. It is a voltage clamp.

The series resistor does the real work

The zener can only hold the voltage down if something limits how much current the battery can push. That is the 220 Ω between the battery and the node. With the node at about 3.36 V (the Workbench's exact figure), the resistor sees 9 − 3.36 = 5.64 V across it, so:

I_series = 5.64 V ÷ 220 Ω ≈ 25.6 mA

That 25.6 mA arrives at the node and has two places to go: into the LED lanes, or through the zener to −. The rule that runs this whole circuit is

I_zener = I_series − I_loads

The zener takes whatever the loads leave over. Add a load and the zener gives some up. Remove one and the zener absorbs it. As long as the zener keeps some current flowing, the node stays pinned. (Leave it a milliamp or more if you can; below that it starts to let go, as you'll see.)

The numbers, switch open

One lane: 100 Ω and a red LED. The LED keeps 2.02 V, so the resistor sees 3.36 − 2.02 = 1.34 V:

I_LED1 = 1.34 ÷ 100 ≈ 13.4 mA (bright, under the 20 mA rating)

I_zener = 25.6 − 13.4 ≈ 12 mA

The zener is quietly eating 12 mA. Notice the resistor is 100 Ω, not 220: on 3.3 V there is only 1.3 V to burn off, so a 220 Ω would give a dim 6 mA.

The numbers, switch closed

Now two identical lanes share the node. The node sags a hair, to about 3.28 V, because the zener's current falls off its knee. Each lane:

I_LED = (3.28 − 2.02) ÷ 100 ≈ 12.7 mA, so 25.4 mA for both

I_zener = 25.9 − 25.4 ≈ 0.5 mA

Two LEDs, both bright, and the first one barely noticed the second arrive: 13.4 mA down to 12.7 mA. The zener went from 12 mA to half a milliamp to make it happen. Half a milliamp is cutting it close, and that's why the node slipped from 3.36 V to 3.28 V. That is regulation: the thing that changed was the zener's current, so the thing you care about, the voltage, did not.

The limit

Add a third lane and the loads want 38 mA, more than the 25.6 mA the resistor can supply. The zener switches off entirely and the node falls until the three lanes settle for what the resistor can give: about 9 mA each, noticeably dimmer, and no longer regulated. A zener regulator only works while the loads take less than the series resistor delivers. Design rule: add up what the loads need, add a milliamp for the zener, and pick the resistor to supply that.

Build it

1. Rails: the 9 V clip's red wire to a top rail, black wire to a bottom rail. Check this twice; 9 V the wrong way round is unforgiving.
2. The series resistor: 220 Ω from the top rail to an empty row. That row is your 3.3 V node. Everything below hangs from it.
3. The zener: stripe end in the node row, plain end to the bottom rail. Backwards on purpose.
4. Lane 1: node → 100 Ω → LED long leg; LED short leg → bottom rail.
5. Lane 2: node → switch → 100 Ω → LED → bottom rail.
6. Power up. One LED. Close the switch: two LEDs, nearly the same brightness.

Touch the 220 Ω after a minute. Warm. It is burning 0.0256² × 220 ≈ 0.14 W all the time, whether the LEDs use the current or not. That is the price of this kind of regulator, and the reason real power supplies use a transistor instead of a resistor as the series part.

Debug like an engineer

Measure the node. Near 0.7 V with both LEDs dark: the zener is in forwards (stripe toward −) and acting as an ordinary diode. Near 4.2 V with the switch open and LED 1 extra bright (about 21 mA, over its rating): the zener is not in the circuit at all, a wrong row or a bent leg. (Check this one with the switch open. With both lanes on, the two LEDs happen to drag the node to about 3.3 V on their own and hide the fault.) Near 3.3 V with one LED dark: that lane's LED is backwards. Three readings, three different faults.