Switches, LEDs, and Interface Circuits

Schematic symbols are abstractions; the physical part has to be identified and oriented on the bench. The markers that carry polarity information are worth memorising: the LED’s longer lead is the anode (+) and the flat spot on its rim marks the cathode; an electrolytic capacitor has a stripe on the negative lead; a small diode has a painted band at the cathode end. Resistors, ceramic capacitors and switches have no polarity. Transistor packages are the trap: two parts in the same three-leg package can have their pins in different orders, so the pinout diagram in the datasheet is the only reliable guide.

A switch is two pieces of metal pressed together. When they meet, they strike, spring apart, and strike again, several times, before settling. The input pin sees a burst of edges rather than one.

How long does it last? Ganssle’s measurements on 18 different switches are the classic reference. Leaving out two outliers, the average bounce was about 1.6 ms and the longest 6.2 ms; one of the outliers bounced for 157 ms when opening but under 20 µs when closing, and seven of the switches bounced much longer on closing than on opening. Manufacturers give an upper bound where they give anything: Omron specifies 5 ms maximum for its B3F tactile switches. Bounce times below 100 ns were also common, which is faster than most software can see but not faster than a hardware counter.

The practical rule that follows: any edge from a mechanical contact must be treated as the start of a possible event, confirmed only when the level has held steady for longer than the bounce. Doing that well, in hardware with an RC filter and a Schmitt-trigger input or in firmware with a timer, is a topic of its own later in the curriculum. Here the point is to recognise the burst when you see it.

A relay coil at 50 mA, a motor, a string of LEDs at 100 mA: none of these belongs on a pin. The answer is a transistor used as a switch, and the simplest reliable form is an N-channel MOSFET on the low side of the load.

Read the circuit from the pin outward:

• The gate is a capacitor. The MOSFET starts to conduct near its threshold $V_{GS(th)}$, but low-resistance switching requires a higher gate voltage supported by the $R_{DS(on)}$ specification. Ideal gate current falls to zero after charging, although the external pull-down still draws a small steady current. The small gate resistor limits the pulse of current that charges the gate capacitance so the pin is not shorted for a few nanoseconds on every edge.
• The gate pull-down (100 kΩ) holds the gate at 0 V while the pin is undriven, which is the case from power-up until firmware configures it. Without it the gate floats (lesson 3), and a floating gate can half-turn-on the MOSFET, which then heats.
• Drain to the load, source to ground. The load sits between its supply, 12 V here, and the drain. When the gate is high, the channel connects drain to source and current flows +12 V → load → drain → source → ground.
• The return path. That current returns to the 12 V supply through the ground net. The microcontroller’s ground and the load supply’s ground must be the same net, or the gate voltage is measured against nothing and the loop is open.

Which MOSFET

The gate is driven from 3.3 V, so the part must be fully on at 3.3 V. Datasheets tell you by listing $R_{DS(on)}$ at specific gate voltages, and this is where “logic level” in a title is not enough. The Nexperia 2N7002 has $V_{GS(th)}$ up to 2.5 V and its lowest characterised $R_{DS(on)}$ row is at $V_{GS}$ = 4.5 V (5.3 Ω max); at 3.3 V a worst-case unit is barely above threshold, and nothing is promised. The BSS138AKA specifies $V_{GS(th)} \le 1.5$ V and gives an $R_{DS(on)}$ row at 2.5 V (13 Ω max at 10 mA), but that 10 mA condition does not establish a 50 mA guarantee at 3.3 V. Choose a part whose on-resistance is specified at a gate voltage no higher than your guaranteed drive and at an appropriate load current; check temperature, dissipation, and the load’s minimum operating voltage. The diagram uses a generic illustrative 5 Ω switch, not a guaranteed model of either named part.

Inductive loads and the flyback diode

A relay coil or motor is an inductor, and an inductor’s current cannot stop instantly: $v = L\,di/dt$ says that forcing a fast change produces a large voltage. When the MOSFET turns off, the coil tries to keep its current flowing and drives the drain voltage upward, tens or hundreds of volts if nothing stops it, enough to break the MOSFET.

The flyback diode across the coil, cathode to the supply, anode to the drain, is reverse-biased in normal operation and does nothing. At turn-off it gives the coil current a loop: coil → diode → coil, circulating until the coil’s own resistance dissipates the stored energy. In the idealized circuit the drain is clamped near one diode drop above the supply; wiring inductance and diode dynamics can add overshoot. Select the figure’s “just switched off” state to see that loop. A 1N4148 is one possible example for a small load, not a universal relay-diode recommendation. Check reverse voltage, initial coil current, pulse duration, repetition rate, thermal limits, and the coil’s stored energy for the selected diode.