Solenoid Metronome

This article is about a solenoid metronome. I used a 15 V power supply. However, some solenoids might drive the rod at lower voltages.

The gravity is causing the rod with a yellow-blue tag ball to fall down with the supplied current to the solenoid is zero. There are solenoid drivers with a spring available online.

The videos show slow and fast beats.

The square wave generator can be implemented with 555 timers. I advise you not to use USB outputs unless you are an expert in electronic and electrical engineering. Poor wiring might short the USB output and cause permanent damage. However, this circuit draws very little current from USB output. (5 V - 0.7 V) /10,000 ohms = 430 uA.

The circuit shown is not assuming a solenoid coil resistance. All coils have resistances, although in most coils those resistances are small.

You will need:

- push-type linear solenoid electromagnet driver,

- heat sink,

- heat transfer paste.

- power transistor or Darlington pair transistor,

- 10 ohm power resistor,

- 10 kohm resistor,

- soldering iron,

- solder,

- wire stripper (optional),

- matrix board,

- 2 mm metal wire,

- 1 mm metal wire,

- pliers,

- plastic base or piece of cardboard or wood,

- blue tag or plasticine,

- square wave generator.

I implemented the first circuit with a Darlington pair transistor to reduce the number of resistors in the circuit. However, it is my personal opinion that the Darlington pair is not the proper method to drive relays or motors because the transistors are not saturating. Using more resistors is not a problem. If the transistors fail because they are not saturated then this would cost a lot more money than a few resistors.

I used a TIP122 Darlington pair power transistor. You can calculate the maximum collector current. Icmax = (Vs - Vsat) / Rc = (15 V - 0.9 V) / 10 ohms = 1.4 Amps (approximately). Power Dissipation During Saturation = Vsat * Icmax = 0.9 V * 1.4 Amps = 1.4 Watts (approximately). Maximum power dissipation occurs when the collector-emitter voltage is half the power supply. (VRc / Rc) * Vce = (7.5 V / 10 ohms) * 7.5 V = 750 mA * 7.5 V = 5.625 Watts. This occurs during the transition from ON to OFF state. When selecting the power transistor and heat sink what you need to consider is the maximum values in the specifications. Average allowed power, average allowed voltage or average allowed current is irrelevant and thus are not usually specified in the datasheets because the component could be damaged in just a few milliseconds of exceeding the maximum characteristics.

In the transistor Darlington pair circuit, the minimum collector-emitter voltage cannot fall to a typical 0.2 V saturation voltage of a single power transistor. During the saturation of the Darlington pair, the first transistor (connected to Rb) is saturated at the collector-emitter voltage of 0.2 V. The second transistor (connected to the ground via its emitter terminal) needs a base-emitter voltage of about 0.7 V to remain on. However, the base of the second transistor is connected to the fully saturated first transistor. Thus the collector-emitter voltage of the second transistor and the entire Darlington pair via series voltage summation will be Vbe2 + Vce1 = 0.2 V + 0.7 V = 0.9 V. If the collector-emitter voltage of the Darlington pair falls below 0.9 V, the base of the second transistor will fall below 0.7 V and the second transistor will turn OFF because the first transistor cannot remain on without the second transistor because it needs emitter current from the first transistor. This is why the Darlington pair saturation voltage never falls to below 0.9 V.

You can try using two transistors instead of the Darlington pair as shown in the second circuit. Then the second transistor will saturate to 0.2 V collector-emitter voltage. 2N2222 is not a power transistor. You will need to use the power transistor.

You can try making the circuit with just a typical single-power transistor. Not the Darlington pair that I used. Then you will have to use a 1 kohm Rb resistor instead of 10 kohm Rb resistor. However, there is a risk that the transistor might not saturate if the coil resistance is low. The coil might not turn ON and lack of transistor saturation might burn the transistor because the power dissipation across the transistor is equal to the collector current multiplied by collector-emitter voltage as stated earlier. If the transistor is not saturated the collector-emitter voltage will not be close to zero. Reducing Rb to below 1 kohm will increase the chances of transistor saturating but might cause the base to exceed the maximum allowed value. Most BJT (Bipolar Junction) transistors have a low maximum base current.

If you are implementing this circuit with MOSFET then do not forget to check the maximum gate-source voltage and do not apply a negative voltage to the gate. MOSFET has a high input resistance and does not need a base resistor Rb.

I attached the heat sink to the matrix board with a thin 1 mm metal wire. I did not use any heat transfer paste. However, it is a good idea to use it.

I used the 2 mm metal wire to attach the circuit to a plastic base. You can use an old cardboard box or plastic base.