🪶 The Switch That Uses No Current

What you're building

A push-button lamp, again, with the transistor swapped. In the Night Light and the Trip Wire a 2N2222 sat under the LED and turned on when you pushed current into its base. Here a 2N7000 MOSFET sits in the same spot and turns on when you raise the voltage on its gate. Hold the button: the LED lights at full brightness. Measure what the gate is taking from the battery: nothing.

That difference is the whole point. A bipolar transistor is a current-operated switch and the thing driving it must supply that current. A MOSFET is a voltage-operated switch. Anything that can make a voltage, a feeble sensor, a logic chip, a finger on a touch pad, can drive it, and the MOSFET will switch 200 mA for them.

The three legs

Hold the 2N7000 with the flat face toward you, legs down. Left to right: Source, Gate, Drain. Same positions the 2N2222 uses for emitter, base and collector, and the same roles: source to −, the load between + and the drain, the gate is the control.

The gate is the unusual one. Inside, it is a metal plate separated from the rest of the transistor by a layer of glass a few nanometres thick. No current can cross glass. Voltage on the plate makes an electric field that opens a channel between drain and source, which is where the name comes from: Metal-Oxide-Semiconductor Field-Effect Transistor.

The rule

Below a threshold voltage between gate and source, about 2 V for a 2N7000 (the datasheet allows 0.8 to 3 V from part to part; the simulator uses 2.0 V), the channel is closed. Above it the channel opens, and the further above, the wider. With the drain well above the source:

I_drain ≈ ½ × 0.1 × (V_gate − 2)² amps

Setting the gate voltage

The button connects + to a 10 kΩ, and the 10 kΩ goes to the gate. From the gate a 100 kΩ runs to −. That is a voltage divider:

V_gate = 4.5 V × 100 ÷ (10 + 100) ≈ 4.09 V

The current through the pair is 4.5 ÷ 110 000 ≈ 41 µA, and all of it flows on through the 100 kΩ to −. The 10 kΩ and the 100 kΩ carry the same 41 µA because there is only one path: the gate in between takes none (apart from a tiny gulp of charge the instant its voltage changes).

Button released: nothing drives the top of the divider, the 100 kΩ bleeds the gate to 0 V, the MOSFET is off.

Is 4.09 V enough?

Plenty. 4.09 V is 2.09 V over threshold, so the formula says the MOSFET could pass ½ × 0.1 × 2.09² ≈ 218 mA. The LED lane only wants about 11 mA. Asked for a twentieth of what it can deliver, the MOSFET is fully on and behaves like a small resistor:

R_on ≈ 1 ÷ (0.1 × 2.09) ≈ 4.8 Ω, so 11 mA × 4.8 Ω ≈ 0.05 V across it

The drain sits a twentieth of a volt above ground. The LED lane sees 4.5 − 0.05 = 4.45 V, and with the red LED's 2.01 V: (4.45 − 2.01) ÷ 220 ≈ 11 mA. First Light brightness, switched by a gate that drew nothing.

Now starve the gate

On the Workbench, hold the button and tap the 10 kΩ until it reads 100 kΩ. The divider is now 100 kΩ over 100 kΩ, so the gate sits at 2.25 V, a quarter of a volt over the simulator's 2.0 V threshold. The MOSFET is not fully on any more: ½ × 0.1 × 0.25² ≈ 3 mA, a dim LED, but on. (A real 2N7000 could land anywhere from 0.8 to 3 V, so on your breadboard this one might be brighter, or fully dark.)

Why did it dim? Not because the gate got less current; it never had any. The gate voltage fell. Do the math for one more swap (the Workbench's resistors stop at 100 kΩ, so this one's on paper): make the bottom resistor 1 MΩ. Now it's 100 kΩ over 1 MΩ, the gate is back at 4.09 V, and the LED is back to full, while the divider draws only 4 µA. Try that with a 2N2222: a tenfold bigger base resistor cuts its base current tenfold, and the LED dims with it. The MOSFET does not care how much current the divider can spare. It only reads the voltage.

The pull-down is not optional

Because nothing flows into the gate, nothing flows out of it either. A gate left floating holds whatever charge it last picked up, from your finger or the air, and the MOSFET sits half on, flickering, warm. The 100 kΩ gives that charge somewhere to go. Any value from 10 kΩ to a few hundred kΩ works; 100 kΩ is the habit because it wastes only 41 µA while the button is held and nothing at all the rest of the time.

Build it

1. Rails: + along the top, − along the bottom, battery between them.
2. The MOSFET: flat face toward you. S to the − rail. G and D in free rows.
3. The pull-down: 100 kΩ from G to the − rail.
4. The button: one leg on the + rail, the other to a free row; 10 kΩ from that row to G.
5. The lamp: + rail → 220 Ω → LED long leg; LED short leg → D.
6. Power up. Dark. Hold the button: lit. Let go: dark.

Before you move on, touch the 10 kΩ leg at the gate with a damp finger while the button is up. On a dry day the LED may flicker: your body is supplying the gate's voltage. That is why the pull-down exists.