Hi,
I am new in electronics and I could not resolve the problem.
I have a hair clipper which is powered by a battery. Now I want to remove the battery and dc motor in order to power it with a normal power supply. My power supply has an output of 19.5V and 2A. Here is the datasheet of my dc motor which I want to use:

It's the JFK-390SA-21155. So as you see my inrush current will be too large for my power supply. Now I want to decrease the inrush current but I don't know how. Please consider that the shell of the clipper doesn't have too much space.
Thank you in advance!

 

Have you tried it to see what happens? Maybe the power supply has current limiting.

 

I think you may be misunderstanding the meaning - the 5.6A starting current isn't inrush in the sense of a transformer etc. Its the maximum torque it can overcome from stationary and needs 5.6A to do so. If your inertial load is much smaller it will require less starting current. Obviously if your power supply can't deliver the required starting current and goes into current limit then the motor will not start... you cannot limit the starting current except by not loading the motor, eg by using a mechanical clutch to apply load after the motor is running...

Do you know the inertial load and the current limit of the power supply?

Or, as @Ian0 said, try it and see.

[edit] Another option is to put a large capacitor across the power supply to provide additional bulk current - the capacitor has to charge fully before the motor is connected so this wont work if the motor is permanently wired directly to the supply. Also some power supplies don't like extra capacitance and will refuse to start if there is too much attached.

 

I have a hair clipper which is powered by a battery. Now I want to remove the battery and dc motor in order to power it with a normal power supply.

My question is, why are you replacing the motor. If the clippers were working but the battery was not, wouldn't it be easier to pick a power supply of the same or slightly higher voltage than the original battery? Changing a motor to match a power supply you already have seems like more work or trouble than is worth. You will need to make the motors shaft move the clippers and that may be more involved than you think.

 

My question is, why are you replacing the motor. If the clippers were working but the battery was not, wouldn't it be easier to pick a power supply of the same or slightly higher voltage than the original battery? Changing a motor to match a power supply you already have seems like more work or trouble than is worth. You will need to make the motors shaft move the clippers and that may be more involved than you think.

The old dc motor doesn't have enough power.

 

I think you may be misunderstanding the meaning - the 5.6A starting current isn't inrush in the sense of a transformer etc. Its the maximum torque it can overcome from stationary and needs 5.6A to do so. If your inertial load is much smaller it will require less starting current. Obviously if your power supply can't deliver the required starting current and goes into current limit then the motor will not start... you cannot limit the starting current except by not loading the motor, eg by using a mechanical clutch to apply load after the motor is running...

Do you know the inertial load and the current limit of the power supply?

Or, as @Ian0 said, try it and see.

[edit] Another option is to put a large capacitor across the power supply to provide additional bulk current - the capacitor has to charge fully before the motor is connected so this wont work if the motor is permanently wired directly to the supply. Also some power supplies don't like extra capacitance and will refuse to start if there is too much attached.

Aren't they equal? https://electronics.stackexchange.c...all-current-generally-equal-to-inrush-current

 

I think that the power supply has a current limitation, I had the same and I had to buy a larger power supply and now it is fine

 

The old dc motor doesn't have enough power

Is the new motor the same physical size as the old one? If so, it is likely that the two motors have similar output powers.

 

Is the new motor the same physical size as the old one? If so, it is likely that the two motors have similar output powers.

No it's not. The new dc motor is bigger.

 

Aren't they equal? https://electronics.stackexchange.c...all-current-generally-equal-to-inrush-current

In my experience, rarely, and its not that simple. For larger, fractional horsepower, motors it often is similar, but for smaller motors no. The inertial load of the rotor of a tiny motor as illustrated is very small as shown by the no-load revs and operating current. There's no way that motor needs 5.8A to start off load. I have something very similar here, about a 3ohm armature and it takes 3A on 9v at stall, but will happily start and run off load at <300mA.

Compare that with the 500W (0.75HP) wheelchair motors I use. At stall they'll try and take 240A off a 24v supply (<0.1ohm winding), though will happily start and accelerate on <50A even when loaded with a 200kg wheelchair & user. Off-load they'll start on <10A mainly due to the chunky right-angle gear box they are fitted with.

 

In my experience, rarely, and its not that simple. For larger, fractional horsepower, motors it often is similar, but for smaller motors no. The inertial load of the rotor of a tiny motor as illustrated is very small as shown by the no-load revs and operating current. There's no way that motor needs 5.8A to start off load. I have something very similar here, about a 3ohm armature and it takes 3A on 9v at stall, but will happily start and run off load at <300mA.

Compare that with the 500W (0.75HP) wheelchair motors I use. At stall they'll try and take 240A off a 24v supply (<0.1ohm winding), though will happily start and accelerate on <50A even when loaded with a 200kg wheelchair & user. Off-load they'll start on <10A mainly due to the chunky right-angle gear box they are fitted with.

Is there a way to calculate the starting current?

 

Varistors intended for inrush current limiting are available from various suppliers.

 

Varistors intended for inrush current limiting are available from various suppliers.

Yes, but in many circumstances reducing the current draw just results in a hot stationary motor - without the current it can't generate the necessary torque to start under load. You don't need a varistor, you need a decent brushed motor controller to manage the current, but ultimately you need enough current from the supply to meet the needs of the motor. Or start it off-load and apply the load gradually through mechanical means.

 

Yes, but in many circumstances reducing the current draw just results in a hot stationary motor - without the current it can't generate the necessary torque to start under load. You don't need a varistor, you need a decent brushed motor controller to manage the current, but ultimately you need enough current from the supply to meet the needs of the motor. Or start it off-load and apply the load gradually through mechanical means.

All right. THANK YOU SIR!

 

The motor specs say 20 to 30V, nominal 24V. Powering it with 19.5V is already out of spec before you even look at the current. You should get a 24V supply.

 

The motor specs say 20 to 30V, nominal 24V. Powering it with 19.5V is already out of spec before you even look at the current. You should get a 24V supply.

While true that only limits top speed. Volts = revs, amps = torque. At startup, Back-EMF = 0 so voltage is of no consequence to getting it turning as long as there are enough volts to push the necessary amps through...

 

Is there a way to calculate the starting current?

Not easily unless you have a full spec of motor and the starting load...

 

For the application as a hair clipper, the exact speed control really does not mater. And the 19 volt 2 amp computer supply can deliver 38 watts, which is far more than the current application by quite a lot. And the no load current is given as 0.11A (110 milliamps) Hair clippers are not started at full load, and so there will not be any problems.
But I also ask, why replace the motor just to be able to use an old computer power supply? What voltage is the present battery??

OR, is the TS a professional barber or hair cutter and the battery will not last all day.

Consider that there will certainly be a voltage drop in the power wires from the supply to the motor.

 

For the application as a hair clipper, the exact speed control really does not mater. And the 19 volt 2 amp computer supply can deliver 38 watts, which is far more than the current application by quite a lot. And the no load current is given as 0.11A (110 milliamps) Hair clippers are not started at full load, and so there will not be any problems.
But I also ask, why replace the motor just to be able to use an old computer power supply? What voltage is the present battery??

OR, is the TS a professional barber or hair cutter and the battery will not last all day.

Consider that there will certainly be a voltage drop in the power wires from the supply to the motor.

I replaced it because the old motor has less power and I don't like clippers with batteries.

 

Given the old motor ran off batteries, probably AA cells which are pretty poor at high current loads, my view is you are over thinking the problem.

Put the motor in the clipper and try it. If it's truly a more powerful motor, chances are it will work ok. A bigger problem might be if it's revs are much higher so cutting torque might be actually lower despite being more powerful. Try it and see.