Electrical – terminal block voltage rating

Steps to diagnose / eliminate the problem:

• Add a 10 μF, 6.3 Volts or higher capacitor in parallel with C2 from the 7805 output leg, as close as possible, to ground.
• Add a 100 uF, 10 Volts or higher capacitor to the 9 volt input, to act as a buffer / reservoir capacitor.
• Add a 100 Ω resistor on the line from the regulator to the MOSFET gate, and replace the pulldown resistor on the Gate with least 1 kOhms. This will reduce the current spike due to capacitance of the MOSFET gate - I believe this should not be the problem, but it is a simple and useful preventive step nevertheless.

If the above does not resolve the issue:

• Use a separate 9 volt supply for the motor, your existing supply is unable to cope with the starting load (stall current >> operating current, usually) of the motor.

Quoting from the great Wiki

"Like other wire, magnet wire is classified by diameter (AWG number or SWG) or area (square millimetres), temperature class, and insulation class.

Breakdown voltage depends on the thickness of the covering, which can be of 3 types: Grade 1, Grade 2 and Grade 3. Higher grades have thicker insulation and thus higher breakdown voltages.

The temperature class indicates the temperature of the wire where it has a 20,000 hour service life. At lower temperatures the service life of the wire is longer (about a factor 2 for every 10 °C lower temperature). Common temperature classes are 105° C, 130° C, 155° C, 180° C and 220° C."

Calculation of breakdown voltage (Test acc. To IEC 60851.5.4.2, cylinder)

The breakdown voltage depends mainly on the thickness of the insulation (see formula below), but also on the bare wire diameter, the application temperature of the coil and the type of enamel.

Calculation of average values Ds:

Ds = t x Vμ [Volt], with

Ds : breakdown voltage

T : increase due to insulation,

t = da – dnom, : wire diameters with and without insulation

Vμm = volts per micron insulation (dependent on type of insulation)

Example:

Test with cylindrical electrode (round wire)

dnom 0.071mm (bare wire nominal diameter)

da = 0.083mm (wire with coating)

t = da – dnom = 0.083 – 0.071 = 0.012mm = 12μm (thickness of insulation between wires)

Vμ = 205 V/μm, therefore

Ds = 12μ x 205 V/μ = 2,460 V