🙃 The Upside-Down Transistor

What you're building

A push-button lamp, which you have built before. The difference is the switch in the middle. In the Automatic Night Light the NPN transistor sat below the LED, between the lamp and battery −. Here a PNP transistor sits above it, between battery + and the lamp. And it turns on backwards: you make the LED light by pulling the base down toward −, not up toward +.

That sounds like a party trick. It isn't. Every motor driver, every H-bridge, every chip that can push a signal both ways is built from an NPN and a PNP working as a pair. This is the other half of the transistor.

The NPN rule, flipped

Hold a 2N3906 with the flat face toward you, legs down. Left to right the legs are Emitter, Base, Collector, exactly like the 2N2222. Same package, same leg order, opposite insides.

The NPN turned on when its base sat about 0.7 V above its emitter. The PNP turns on when its base sits about 0.7 V below its emitter. Here the emitter is tied straight to battery +, which is 4.5 V. So:

• base at 4.5 V (same as the emitter): no difference, transistor off, lamp dark.
• base pulled down to about 3.8 V (0.7 V below): base current flows out of the base, the transistor turns on, and current pours from the emitter out through the collector into the lamp.

That is the whole idea. An NPN needs its base pushed up to turn on. A PNP needs its base pulled down. Think of the PNP as the NPN drawn upside down: emitter at the top, current flowing downward.

The two 10 kΩ resistors

The pull-up. One 10 kΩ runs from battery + to the base. With the button up, no current flows through it, so there is no voltage across it, so the base sits at the full 4.5 V, the same as the emitter. Transistor off. Without this resistor the base would be connected to nothing, and a transistor with a floating base picks up whatever stray voltage is around and flickers. Never leave a base floating.

The pull-down. The second 10 kΩ runs from the base, through the button, to battery −. Press the button and it tries to drag the base down toward 0 V. It can't get there: the base-emitter junction is a diode, and it clamps the base about 0.72 V below the emitter, at roughly 3.8 V. Now both resistors carry current:

pull-down: 3.8 V ÷ 10 000 Ω ≈ 0.38 mA (flowing down to −)

pull-up: (4.5 − 3.8) V ÷ 10 000 Ω ≈ 0.07 mA (flowing in from +)

The difference, about 0.31 mA, is what comes out of the base. That's the base current.

Is 0.31 mA enough?

The 2N3906's gain (hFE) is about 100 to 300; the simulation uses 200. 0.31 mA of base current could let 200 × 0.31 ≈ 62 mA flow through the collector. The lamp only wants about 11 mA (next section). So the transistor is asked for far less than it can supply, and it goes fully on, which engineers call saturated. A saturated transistor acts like a closed switch with about 0.1 V lost across it. The margin is about 5.6×, comfortably above the 5× floor that keeps a switch reliable when the gain drifts.

What the LED sees

With the transistor saturated, the collector sits only about 0.1 V below battery +, at about 4.4 V. The lamp is a 220 Ω resistor and a red LED to −, so Ohm's law with the LED's 2.0 V forward voltage:

(4.4 − 2.0) V ÷ 220 Ω ≈ 10.9 mA

Bright, and safely under the LED's 20 mA limit. The battery supplies the lamp's 10.9 mA plus the pull-down's 0.38 mA: about 11.3 mA in total.

Build it

1. Rails: a + rail along the top, a − rail along the bottom, battery between them.
2. The transistor: flat face toward you, E to the + rail, C free for now, B free for now.
3. The pull-up: 10 kΩ from the + rail to B.
4. The pull-down: 10 kΩ from B to one leg of the button; the button's other leg to the − rail.
5. The lamp: C → 220 Ω → LED long leg; LED short leg → − rail.
6. Power up. Dark. Press the button: lit. Let go: dark.

Try this before you move on: with the button up, measure (or just reason about) the base. It is at 4.5 V. Press the button and it drops to about 3.8 V. That 0.7 V step is the whole switch.

Debug like an engineer

If the lamp never lights, test the halves separately. First the lamp on its own: move the 220 Ω end from the collector straight to the + rail. Lit? Then the lamp is fine and the problem is the transistor half. Check E-B-C order next. A PNP wired with E and C swapped is weakly on backwards and gives a dim, confusing glow. Then confirm the pull-down really reaches the − rail through the button: with the button pressed the base should sit near 3.8 V, not 4.5 V.