DC doorbell driver

In this article, I will describe the need for and operation of a timer circuit when using a mechanical doorbell. This circuit switches the solenoid in the doorbell off after about a second and keeps it off even if the user continues to hold down the doorbell button. By switching off the solenoid, the power dissipation is reduced, preventing the solenoid from melting.

Background

My doorbell is not powered with an AC transformer, but from a 12 volt battery. The doorbell was recently damaged, most likely by the following prcess:
  1. The doorbell button got stuck in the closed position or was held down for too long.
  2. Since the solenoid has a resistance of around 6 ohms, the solenoid dissipated around 24 watts, causing it to heat up significantly.
  3. The plastic armature softened and compressed, causing the plunger to stick.
  4. The plunger heated up and melted the plastic tips on the plunger that would normally strike the bells.
  5. The spring that pulls the plunger back started to lose its shape because of the heat and stuck plunger.
When I disassembled the doorbell, I noticed soot marks on the casing and an odor of burnt enamel in the doorbell. This is also not the first doorbell this has happened to. To save money replacing the doorbell, only to have it fail again in the future, I built this circuit to protect it. Additionally, all of the components used in this project (except for the perforated board) were removed from old electronics.

Repaired doorbell

  1. The upper tip of the plunger was reconstructed using a thick piece of wire held covered with hot glue.
  2. The spring was streched out more than it was originally, so that it can hit the upper bell with enough force.
  3. Using a drill, I enlarged the hole in the armature so the plunger could slide through it again. Additionally, I sanded down the plunger by spinning it with a drill.
  4. (Not visible) I added hot glue to the other end of the plunger, this time without wire.
  5. A timer circuit was added to prevent the doorbell from burning out in the future.

Operation of the timer circuit

Link to simulation

Suppose the doorbell switch is pressed by the user. Since the capacitor is charged to 0 volts, the gate of the MOSFET will be at 12 volts. The MOSFET switches on, energizing the solenoid. However, the capacitor begins to charge via the right resistor, and the gate voltage decreases. Eventually, the MOSFET switches off, and the solenoid de-energizes. When the user releases the doorbell button, the capacitor discharges via both resistors, resetting the circuit so it can be activated again.

The backs of the circuit board and solenoids are shown below: (including the burnt insulation)

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