Robotics · Electronics basics · Grade 8

The transistor — a switch

An Arduino pin gives 20 mA, but a motor needs 400 mA. How does a small current control a large one?

The situation

The motor of a toy robot car takes ≈ 400 mA from a 9 V battery. An Arduino pin gives only 20 mA (the limit is 40 mA) — connect the motor straight to the pin and the microcontroller burns out. How does a small signal control a large current? The answer is the transistor: a three-legged “switch” in a TO-92 package. Calculate the base resistor, bring the transistor into saturation, spin the motor. Then break things on purpose: a base with no resistor, 400 mA through a 100 mA BC547, and a motor switched off without a flyback diode — a 40-volt “spike” on the oscilloscope trace.

Topic

Transistor as a switch: a small current controls a large one. An Arduino pin gives 20 mA, but the motor needs 500 mA: a transistor switch, β and saturation

Key ideas

NPN transistor
β and saturation
Protection diode

Formulas

I_{C} = β \cdot I_{B}

Active region: the collector current is β times the base current (BC547 ≈ 200, 2N2222 ≈ 100) — the transistor amplifies current.

I_{B} = \frac{U_{p i n} - U_{B E}}{R_{B}}, U_{B E} \approx 0.7 V

The base current: ≈ 0.7 V of the base–emitter junction is subtracted from the pin voltage, and the rest falls across the base resistor.

I_{C, \max} = \frac{U - U_{C E (s a t)}}{R_{l o a d}}, U_{C E (s a t)} \approx 0.1 – 0.2 V

Saturation: if $β I_{B}$ is larger than the current the load can pass, the load sets the current and only 0.1–0.2 V is left across the transistor — it is like a closed switch.

R_{B} \leq \frac{β (U_{p i n} - 0.7)}{k \cdot I_{C}}, k = 2 – 3

Choosing $R_{B}$ for a switch: the base current must be $k$ times the required $I_{C} / β$ (a margin — β differs from one unit to another).

P_{Q} = U_{C E} \cdot I_{C}

The power dissipated in the transistor: small in saturation (0.2 V · 0.4 A = 80 mW), large in the active region — for a TO-92 the limit is ≈ 0.5 W.

Theory

Why a pin is not enough

An Arduino pin is a weak source: ≈ 5 V and an internal resistance of ≈ 25 Ω, an absolute limit of 40 mA, a normal value of 20 mA. A small motor takes ≈ 400 mA at 9 V, and more than 1 A at start-up. Connect the motor straight to the pin and the current is $5 / (25 + 6) \approx 160$ mA — 4 times the limit: the ATmega328P heats up and burns out, and the motor does not even start. The pin must only give a command; let another source supply the current.

The transistor: E, B, C

An NPN transistor is a three-layer semiconductor crystal with three legs: the emitter (E), the base (B) and the collector (C). If a small current $I_{B}$ enters the base, a current β times larger flows from the collector to the emitter: $I_{C} = β I_{B}$ (β is the current gain, from 50 to 500). The base–emitter junction is a diode: it needs ≈ 0.7 V, so the base is connected through a resistor. The legs of a TO-92 package with the flat face towards you: BC547 — C B E, 2N2222 (PN2222A) — E B C. If you mix them up the transistor will not work — always check the datasheet.

Three regions: cut-off, active, saturation

No base current ( $U_{B E} < 0.7$ V) — the transistor is in cut-off: the collector current is zero, the switch is open. There is a base current, but $β I_{B}$ is less than the current the load asks for — the active region: $I_{C} = β I_{B}$ , the surplus voltage $U_{C E}$ stays across the transistor and it heats up ( $P = U_{C E} I_{C}$ ). $β I_{B}$ larger than the load current — saturation: the load sets the current, $U_{C E}$ falls to 0.1–0.2 V, the transistor is like a closed switch and hardly warms up. As a switch the transistor works only in the two extreme states: cut-off or saturation.

The calculation: the base resistor

Motor 400 mA, 2N2222 (β ≈ 100): the required base current is $I_{B} \geq 400 / 100 = 4$ mA. β differs between units and with temperature, so a margin of 2–3 times is taken: $I_{B} \approx 10$ mA. Pin 5 V, 0.7 V at the base: $R_{B} = (5 - 0.7) / 0.010 = 430$ Ω → the standard 470 Ω (≈ 9 mA) or 330 Ω (≈ 12 mA). The pin current must not exceed 20 mA: $R_{B} \geq 4.3 / 0.02 = 215$ Ω. With no resistor ( $R_{B} = 0$ ) the base current is limited only by the pin’s internal resistance — ≈ 100 mA: both the pin and the base–emitter junction burn out.

Limits and the flyback diode

Every transistor has a current limit: BC547 — 100 mA, 2N2222 — 800 mA. Connect a 400 mA motor to a BC547 and it burns out in a second. Power: a TO-92 can dissipate ≈ 0.5 W — if 1 W is dissipated in the active region, it smokes within 10 seconds. A motor is an inductive load: when the current is interrupted the coil “tries” to keep it going: $u = L d i / d t$ — a 40–45 V spike appears at the collector and the transistor breaks down. The flyback diode is connected across the motor, in reverse: it is shut while the motor runs, and when switched off it gives the current a path — the voltage stops at 9.7 V. When the speed is controlled with PWM ( $𝚊𝚗𝚊𝚕𝚘𝚐𝚆𝚛𝚒𝚝𝚎$ ) the transistor switches off 490 times a second — without the diode it burns out in an instant.

About the model

The circuit is solved by the nodal method. Transistor: $U_{B E}$ = 0.7 V + 20 Ω, $I_{C} = β I_{B}$ , saturation starts from $U_{C E}$ = 0.2 V; β is constant for each type (in a real transistor it depends on current and temperature, falling to ≈ 40 at 500 mA for a 2N2222). Motor: armature 6 Ω + back-EMF $k ω$ (k = 6 mV·s/rad), 1 mH, a load torque with a propeller — at 9 V ≈ 9 300 rpm, 0.38 A. With PWM the average current is taken as $D \cdot I$ (smoothing by the inductance is ignored). Heating: the part of the current above the limit accumulates over time; the package thermal model is $τ \approx 8$ s; the spike energy is $\frac{1}{2} L I^{2}$ — after several switch-offs the transistor “breaks down” (in reality this is random, but this is the order).

Did you know?

The first transistor was made in December 1947 at Bell Laboratories in the USA by John Bardeen and Walter Brattain (a point-contact device), and in 1948 William Shockley proposed the junction transistor; all three received the Nobel Prize in Physics in 1956.
A single modern processor chip holds tens of billions of transistors: the Apple M2 Ultra (2023) has 134 billion — each a switch like the one in this experiment, only a few nanometres in size.

Inside the experiment

A real board in a hand-inked isometric drawing: turn it, place the parts, wire the circuit, write and upload the code — current and voltage are computed at once, and a wrong connection burns a part. Inside are missions, theory, a quiz and a guided tour.

More experiments in this section

LED and resistor: Ohm’s law — Grade 7
Resistors: the colour code — Grade 7
Voltage divider and potentiometer — Grade 7
Capacitor: charging and the RC time — Grade 7
Diode and rectifier — Grade 8
Measuring with a multimeter — Grade 8