The diode and the light
A motor is a coil of wire, and a coil keeps its current flowing for a moment after the switch opens. The SS14 diode across the motor gives that current a loop to die away in, so the MOSFET sees a few tenths of a volt above 3.3 V instead of a spike past its 30 V rating. The LED beside it lights whenever the MOSFET conducts.
A coil does not stop at once
Current through a coil builds a magnetic field, and the field holds energy. When the current is cut, the field collapses and pushes the current on for a moment in the same direction. Cut it with nowhere to go and the voltage at the open end rises until something lets the current through.
On this board the open end is the MOSFET's drain, the motor's − lead.
Play it with the diode fitted, then without. With the SS14 in place, the current the motor was carrying goes up through the diode to the 3.3 V rail and back down through the motor, the purple loop, and dies away in the motor's own resistance. The drain rises only as far as the rail plus the diode's forward voltage, a few tenths of a volt. The trace is illustrative; the scale is not.
Without the diode the drain has to climb until the MOSFET breaks down and conducts anyway, past its 30 V rating. It happens every time the motor is switched off, and hundreds of times a second under PWM. The diode is why the switch lasts.
Why a Schottky
The SS14 is rated 40 V and 1 A, far more than this motor's 54 mA needs. It is a Schottky diode, which turns on at a lower voltage and switches faster than an ordinary silicon one, so it takes over the current the instant the MOSFET lets go. Under PWM it carries the motor's current in every off part of every cycle, which is how the motor sees a smooth average rather than a string of hard stops.
The light
The LED and its 1 kΩ run from the 3.3 V rail to the MOSFET's drain, in parallel with the motor. So they light on the same condition: the MOSFET closed. The current is about (3.3 − 2.0) V ÷ 1 kΩ, a little over a milliamp, taking about 2.0 V for a red LED. That is worked out, not measured.
The LED does not go through the motor. A lit LED with a still motor means the switch closed and the motor did not turn: the leads, or the motor. A dark LED with SIGNAL HIGH means no rail or no gate, which is almost always a wire.
When it does not work
No. The board already has both: an SS14 Schottky diode across the motor for the switch-off kick, and a 100 nF capacitor for the motor's electrical noise. Nothing extra is needed between the block and your board.
Not through SIGNAL: the pin only touches the MOSFET's gate, which is insulated from the drain where the spike would appear. Without the diode it is the MOSFET that would suffer. With the diode fitted, the drain stays within a few tenths of a volt of the 3.3 V rail.
It follows the MOSFET, so under PWM it is on for the same share of each cycle as the motor is driven. At a low duty it looks dim. That is correct, and it makes the LED a rough strength gauge.
The LED only proves the MOSFET closed: its current does not go through the motor. So power and SIGNAL are fine, and the fault is the motor or its thin red and blue leads. Look at where they are soldered to the two pads on the right.
Three wires and a sketch that turns it on and off twice a second.
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