# Max startup current for Vacuum cleaner motor

We used the vacuum cleaner motor 2200W 220v in one dental device (vacuum former) as you can see here:

So now one of our costumers said when they used this device the power cable is warmed and when the vacuum motor started the dental clinic circuit breaker (25A C type fuses) trip (could be seen here:)

and disconnect the power of the whole system, so I guess it using 10A for steady-state but I don't know how much current this motor get for startup and could this situation be the result of this trips.

Based this site:

Vpeak = IinR, where Vpeak = √2(V)

and

E = CV2/2

So we have 220v : √2*220/15=20.74 A

I found this vacuum cleaner’s current waveform from here: , which is similar, so is this the inrush current of this motor?

I need the causes, and if possible I want to know what is the proper way to solve this problem? Is it good to use soft starter?

Update:

These Vacuum cleaner motor we used in our devices are Brushes.

• I seem to recall AC motors can have a start current of up to 6x times the rated current but for a very short amount of time (Dont quote me on that!). I guess a soft starter would be appropriate. – Sorenp Jul 22 '19 at 7:26
• Vacuum cleaner motors often have brushes. Check if those are still ok. – Jeroen3 Jul 22 '19 at 7:26
• @jeroen3 you are right this vacuum cleaner motor is bushes. – Soheil Paper Jul 30 '19 at 5:22

so is this the inrush current of this motor?

YES 300% of the rated current is common for start surge of a low-efficiency motor. (high-efficiency BLDC motors use 10x to 12x rated current for start surge)

I need the causes, and if possible I want to know what is the proper way to solve this problem? Is it good to use soft starter?

You may have damaged the motor by excessive winding temp rise from lack of airflow with your design of the vacuum former. So a soft start may only extend the start duration and allow the blocked airflow temperature to rise faster.

A blocked vacuum motor uses more than rated current.

However, using a variac or relay controlled autotransformer can reduce the voltage by 50% and thus current but quadruple the startup time.

## Suggestions

• The vacuum motor never starts with the full air-flow blocked i.e. full load.
• Use a shop vac. drum to contain the vacuum to increase airflow and reduce the load on the motor.
• use a vacuum pressure sensor for speed control.
• Use a brushless DC vacuum motor with separate forced-air cooling and higher efficiency @ 1kW should be adequate.
• are this calculations correct for brushes vacuum motors? – Soheil Paper Jul 30 '19 at 5:25
• Yes. If the Surge current on start was the same as the fully rated load current it would be very inefficient. As it is with a 3x ratio is already fairly inefficient and needs the forced air cooling of the vacuum intake. Blocking the airflow not only cuts off the cooling, but it runs faster with more eddy current and brush losses and can draw more than rated current. How much I can't say for sure. – Tony Stewart EE75 Jul 30 '19 at 9:22
• So if the difference is from brushes losses, can i say it must not be too much more than almost 10% difference with BLDC motors? – Soheil Paper Aug 5 '19 at 20:27
• No the difference is the ratio of DCR to Reactive impedance at full RPM and full rated load. – Tony Stewart EE75 Aug 7 '19 at 16:26
• I asked this part of question here: electronics.stackexchange.com/questions/452445/…, so I would be grateful If you will be attending the link visit. – Soheil Paper Aug 11 '19 at 5:36

The plot shows the peak motor current as about 30 A. That would make the maximum RMS starting current 21 A. That should not trip the breaker. The 10 A peak / 7 A RMS running current should not make the power cable get too hot, but it may be possible to detect the difference in temperature compared to room temperature. It seems like the actual motor draws significantly more current than the plot shows or that there are other parts of the equipment that add to the current. There may also be other equipment connected to the same circuit breaker. This could easily be one customer with too much plugged into the same circuit.

With an inrush current that starts at 300% of normal running current and drops to normal in 1/2 second, there should be no difficulty unless the circuit breaker is quite heavily loaded. Even then, it seems like the breaker should not trip that quickly. You should check the breaker curve and determine if that type of breaker is commonly used.

You should not need a soft starter. You may need to advise customers about breaker ratings and other equipment on the same circuit.

Actual Motor vs. Plot

It appears that the actual motor is more than twice as large as the motor for which the current is plotted. Even with that, the motor should not be tripping the breaker unless there is significant additional load on the same breaker.

If you want to offer a product that can be plugged into a 25 amp circuit that has significant additional load with no risk of tripping the breaker, the best alternative would probably a brushless motor with an electronic controller that limits the current. That will also likely use less energy in normal use. However it will also likely be more expensive. An electronic speed control with current limiting would probably be a less expensive option. You may or may not be able to find a soft starter that is less expensive than an electronic speed control.

Determining Starting Current

The initial current seen at the instant that a DC motor is energized is essentially the voltage divided by the series impedance of the motor. A universal motor is a wound-field, brushed DC motor with the armature and field connected in series. Therefore the initial impedance is the series inductance and resistance of those two structures. As the speed increases, the rotor generates a proportional voltage that opposes the current. That is called the back EMF, an abbreviated form of reverse electromotive force. As the speed increases, the back emf increases proportionally. So that proportionally reduces the net voltage applied to the series impedance.

The mechanical power developed in the rotor is the back emf multiplied by the current. Note that initially, most of the power developed in the rotor is used to accelerate inertia. Once the motor reaches a steady speed the inertia no longer contributes to the load.

With a blower such as the one used to create a vacuum in this machine, the mechanical power used by the load is proportional to the rate of air flow multiplied by the pressure. If the air flow is blocked, the load is reduced because the blower is not producing any air flow. It is only stirring the air internally. Thus the motor current is reduced when the flow of air is blocked.