A transistor as a switch

You need the button circuit working, one S8050 transistor, one 4.7 kΩ and one 10 kΩ resistor, a 220 Ω resistor and a red LED. The button, the LED and the 220 Ω are the ones you have already used.

Why a transistor

In the last two circuits the button carried the whole current that lit the LED. That is fine at 13 mA, and fine for a button. It is not fine when the thing being switched is a motor, or when what presses the button is the output pin of a board, which can only supply about 20 mA. A transistor is the part that lets a small current decide whether a larger one flows.

Find the legs

The S8050 is a small black plastic body with a flat face and three legs. With the flat face towards you and the legs pointing down, the order in the onsemi SS8050 datasheet is emitter, base, collector from left to right. Makers have printed other orders on other parts in the same case, so check yours with the meter before you plug it in.

In diode mode, put the red lead on a leg and the black lead on another. The base is the one leg that reads about 0.65 V to both of the others with the red lead on it. Nothing reads between the other two legs, and nothing reads with the leads the other way round. That finds the base for certain. Telling the emitter from the collector with a meter alone is not reliable, so rely on the datasheet order, and on the circuit working.

Three legs, two jobs

The base is the control. The collector and emitter are the path: current goes in at the collector, out at the emitter, and flows only if the base is fed. The base and emitter are a diode, so a base needs a resistor in series, exactly as an LED does, or it takes more current than it can stand.

Here the button feeds the base through a 4.7 kΩ resistor. A 10 kΩ from the base to the minus rail makes sure the base sits at 0 V when the button is open, for the same reason the last article gave: a base connected to nothing floats, and a floating base turns the transistor on and off at random.

Build it

The LED and its 220 Ω now sit on the collector side, from the rail to the transistor, and the emitter goes straight to the minus rail.

The button is wired from opposite corners as before. The base resistor runs from the button's output strip to the base's strip. The LED's short leg goes to the collector's strip, and the 220 Ω runs from the LED's long leg to the plus rail.

Why a few kilohms

The current into the base is set by the base resistor: (4.8 V less the 0.65 V the base junction takes) over 4.7 kΩ, less a little that the pull-down steals, is about 0.82 mA. The transistor's gain, hFE, is the factor between base current and collector current, and on the S8050 the datasheet gives 85 to 300, measured at 100 mA. So the gain would ask for 70 to 245 mA through the LED.

The LED branch cannot carry that. Its own 220 Ω limits it to about 12 mA. When the gain asks for more than the load can take, the transistor stops being an amplifier and becomes a closed switch: it is saturated, and the load sets the current. Only 0.82 mA goes in and 12 mA comes out, about fifteen times as much. The voltage left across the transistor is about 0.1 V.

Try the 1 MΩ base resistor in the plot, with the pull-down and without it. With it, the pull-down and the 1 MΩ form a divider that holds the base far below 0.65 V, and nothing turns on. Without it, the transistor works as an amplifier, and the LED's brightness follows the gain, which is different for every transistor. A switch should never depend on that.

In order

The button carries only the base current, a little under 1 mA. It is the transistor that carries the LED's current. This is the whole idea: a small current decides, and a large one obeys.

Measure

Set the meter to DC volts and press the button.

• Base to emitter: about 0.65 V, between 0.6 and 0.7. This is the junction drop, and it barely moves however hard you drive the base.
• Collector to emitter, pressed: about 0.1 V. Anything under 0.3 V is a switch that is closed. A volt or more means it is not saturated.
• Collector to emitter, released: about the rail, 4.8 V, because no current flows and the transistor takes all the voltage.
• Across the 220 Ω, pressed: about 2.7 V, which is the rail less the LED's 2 V and the transistor's 0.1 V. Divide by 220 Ω for about 12 mA.
• Across the 4.7 kΩ, pressed: about 4.1 V. Divide by 4.7 kΩ and the base current is a little under 0.9 mA, of which about 0.065 mA goes down the pull-down and the rest into the base.

Where the limit is

The S8050 is rated for about 20 to 25 V between collector and emitter, and for hundreds of milliamps through them. Makers rate it differently: onsemi's datasheet says 25 V and 1.5 A, others say 20 V and 500 to 800 mA, and the kit's maker is not known, so assume the lowest of each. A single LED at 12 mA is a long way inside it. The kit also has a second kind of switch, the 2N7000, a MOSFET, and the next experiment in this handbook is about it.