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)

Comments

Post a Comment

Popular posts from this blog

Improving and calibrating the capacitive water sensor

Image
In a previous article, I explained a type of water level sensor that works capactitively and therefore passes no direct current through the water. When I wrote that article, I had already made two of these sensors and installed one in each of the two sump pump wells in my basement. These sensors have been working quite well for a long time, but they have a few issues. In this article, I will describe the improvements I have made to the sensor design and how I calibrated the sensors to allow me to determine the water flow rate and the volume of water the sump pump pumps each time it comes on. Introduction The two water level sensors I installed in the sump pump wells were equipped with nRF24L01 wireless modules. They transmitted measurements every 30 seconds to a Raspberry Pi, which recorded them to a database. This allowed me to visualize the water level in the sump-pumps in real time and see roughly how often the sump pumps came on. However, the water level followed this interest
Read more

Turn a buck converter module into a buck-boost converter with only two components

Image
In this article I will show how you can turn a buck converter module (like the one shown below), which can only output a voltage lower than the input voltage, into a buck-boost converter module, which can output a voltage that is either lower or higher than the input voltage. Buck converter theory The schematic of a buck converter is shown below: Link to simulation The components in the box, in addition to the feedback loop (not shown), are usually integrated into a single chip. A regular buck-converter works by switching current into the output capacitor through an inductor. When the transistor in the box is on, current can flow from the supply to the output via the inductor. The voltage across the inductor is \(V_{in}-V_{out}\) and the rate of change of the current is \(\frac{V_{in}-V_{out}}{L}\). When the transistor in the box switches off, the current in the inductor continues to flow, but now through the diode. The voltage across the inductor is \(V_{out}\) and the rate of
Read more

Self-starting inverter

Image
In this article, I will explain a circuit that can be used to automatically start an inverter when a load is connected to it and turn it off once the load is disconnected. This circuit works by biasing the output of the inverter with a low-current signal from the battery and waiting for a change in that signal. The circuit then uses a current transformer to monitor the current through the load. If there is no current in the load for a few seconds, the inverter is switched off. The inverter The inverter I am using is a modified square wave inverter. The voltage between the live and neutral terminals at the output looks like this: However, this diagram misses out on an important fact: the neutral of the output is not connected to the negative battery terminal. In fact, the modified square wave is generated by alternately switching the live and neutral terminals of the output to an internal high-voltage rail. As a result, both live and neutral are always positive relative to the nega
Read more