i'm having some difficulties with calculating the value of the thermal overload fuse in this circuit.

The thermal fuse is used to protect the motor and the circuit behind it. The deal problem is that the thermal overload fuse is in the live mains circtut and not after the AC-DC PSU.

And i cant seem to understand how to calculate the value. Could someone please help me understand what the formulas are and try to explain it to me? I have talked to other people and they say that its should be able to calculate the value.

I already have the thermal fuse for this its this module(**Moeller PKZM0-0,25)**

```
Supply voltage = 230VAC
Motor nominal current 3A
Motor Voltage 12VDC
Adjustable Range of overload fuse = 160mA - 250 mA
```

## Best Answer

That motor-protection is protecting the complete circuit - transformer, rectifier, PWM modulator and motor.

Your motor is rated at 12 V, 3 A. From the equation \$ P = VI \$ we can calculate that it is a 36 W motor.

Discounting losses for a moment we can calculate the required current at 230 V AC to provide 36 W. Using the same formula and rearranging we get \$ I = \frac {P}{V} = \frac {36}{230} = 0.16~A \$.

Allowing for some losses in the transformer, wiring and PWM (in)efficiency I would probably set the trip current to 200 mA. The protection device looks like the right choice.