Electronics · Math shelf · Optional — for the curious · build 2026.09.23-1644

The Transistor's Arithmetic

The idea

The transistor multiplies current: collector current = β × base current — until the circuit can't supply any more. Then it's simply a closed switch.

The cast: three legs

B · base C · collector E · emitter
The NPN transistor. The heavy traffic flows collector → emitter; the base is the small control current that permits it. The arrow marks the emitter.

Two rules run everything

  1. The base door costs ≈ 0.7 V. The base-to-emitter junction behaves like a small diode: below ≈ 0.7 V nothing happens; at ≈ 0.7 V it conducts, and that drop stays ≈ 0.7 V no matter the current — a smaller sibling of the LED's 2.0 V.
  2. IC = β × IB. Every unit of base current permits β units of collector current. β (beta — datasheets write it hFE, and your multimeter's transistor socket measures it) is typically 100 to 300 for the same part number. Plan on the minimum, ≈ 100; anything more is bonus. (That design habit has its own sheet: Tolerance & Error.)

Saturation: the switch region

Rule 2 has a ceiling. The collector circuit obeys Ohm's law like everything else — the supply and the load resistor set a maximum current no β can exceed. Push the base hard enough that β × IB would overshoot that maximum, and the transistor saturates: fully on, dropping only ≈ 0.2 V collector-to-emitter. For a switch, saturation is the goal — the rule of thumb is to feed the base about ×10 the minimum, so the switch slams shut even on a weak-β part.

The design: a transistor switches the LED

+5 V rail 220 Ω ≈ 13 mA C E R_B 4.7 kΩ ≈ 1 mA control signal 5 V on · 0 V off ground rail
The whole trick in one circuit: ≈ 1 mA whispering at the base lets ≈ 13 mA drive the LED. The control signal never touches the heavy current.

Every number in that figure is computable. Watch:

  1. Collector current first. Fully on, the loop from 5 V drops: 2.0 V (red LED) + 0.2 V (saturated transistor), leaving 2.8 V across the 220 Ω. IC = 2.8 ÷ 220 ≈ 13 mA. (Barely less than the 14 mA without a transistor — the 0.2 V toll is cheap.)
  2. Minimum base current. 13 mA ÷ β = 13 ÷ 100 = 0.13 mA.
  3. Overdrive ×10. Aim for ≈ 1.3 mA into the base — saturation with margin.
  4. Size the base resistor. A 5 V control signal, minus the 0.7 V door: 4.3 V lands on RB. RB = 4.3 ÷ 1.3 mA ≈ 3.3 kΩ. From the kit: 4.7 kΩ gives 0.91 mA (7× the minimum — fine), 1 kΩ gives 4.3 mA (33× — also fine). Engineering answers are ranges, not points: anything in 1 – 4.7 kΩ switches this LED hard.

The base is a diode

Never wire a base straight to 5 V. A 0.7 V door across a 5 V supply with nothing in between is the naked-LED story again — a huge current and a cooked transistor. The base always gets its own resistor, for exactly the reason the LED always gets its 220 Ω.

Where this is headed

The Voltage Divider sheet warns that a tap can't power anything — hang a load on it and it sags. But a transistor base barely sips: ≈ 1 mA of information. So a sensor divider whispers its reading to a base, and the collector does the heavy lifting. Photoresistor divider → base → LED: that's the night-light build ahead, and now you've already done its arithmetic.

The math, for the curious — always optional

How hot does the switch run? Power is P = IV across the transistor itself: saturated, 0.2 V × 13 mA ≈ 2.6 mW — nothing. But run it half-on (2.5 V across it, ≈ 6 mA) and it burns 15 mW — several times more while doing less. Switches want to be fully on or fully off; the in-between is where the heat lives. That, at a thousand times our scale, is why amplifiers wear heat sinks.

Practice

Answers at the bottom. β = 100 unless stated.

  1. Swap in the blue LED (3.2 V drop), same 220 Ω. Saturated collector current?
  2. Minimum base current for that — and with the ×10 rule, what target?
  3. What RB hits that target from a 5 V signal, and which kit resistor do you actually use?
  4. The meter's transistor socket reads your part's β as 250. Does the design change?

Answers

1. 5 − 3.2 − 0.2 = 1.6 V across 220 Ω → ≈ 7 mA. 2. 7 ÷ 100 = 0.07 mA minimum; ×10 → 0.7 mA. 3. RB = 4.3 ÷ 0.7 mA ≈ 6.1 kΩ → use 4.7 kΩ (0.91 mA, ≈ 13× minimum — comfortably saturated). 4. No. The design was built on the minimum β; a stronger part just saturates harder. Designs that only work on lucky parts aren't designs.