A motor does not have "inrush current". It has stall (or locked rotor) current. This is the current a motor draws a 0 RPM, and develops the initial torque required to overcome its own static friction and that of the mechanical system it's connected to.

Limiting the current to the motor can prevent the motor from starting at all -- and this will be load dependent.

If you must go this route, I suggest a current limiting circuit something like this (that +2V is actually +12V):

View attachment 136412

Your mileage will vary.

Dear would you mind telling me please what is each part in the picture, what transistor? and also what is in the middle? I can not recognize the elements, I am not pro, sorry.

 

Dear would you mind telling me please what is each part in the picture, what transistor? and also what is in the middle? I can not recognize the elements, I am not pro, sorry.

Yes, dear, I can. But first, I am confused. You said this:

I put a dc motor which is 12 V, 23 Ohm internal resistance, 0.545 Amps to pump the water and accordingly I used one power supply for both dc Motor and Arduino which is 12 V, 1 Amps output.
The problem is, when the dc motor has turned on by relay, the inrush current which is around 1.5 Amps or less

If the winding resistance was really 23 ohms, the max locked rotor current would be 0.521A at 12V, but you claim 0.545A (I assume running), and 1.5A locked. This is not possible (unless you have a fat cap in parallel with the motor). How did you measure ohms?

If you give me some real numbers that make sense, I can provide you details for each component required for the limiter.

 

Yes, dear, I can. But first, I am confused. You said this:



If the winding resistance was really 23 ohms, the max locked rotor current would be 0.521A at 12V, but you claim 0.545A (I assume running), and 1.5A locked. This is not possible (unless you have a fat cap in parallel with the motor). How did you measure ohms?

If you give me some real numbers that make sense, I can provide you details for each component required for the limiter.

I know I should be precise but the dc motor is not since it is not standard and a cheap motor, the last time I measured it was 23 ohm internal resistance, but when I tried again it is showing 1.5 ohm and I am confused either, however I could overcome the problem by changing the power supply by a 12 v 2 amps output and a hand made inductor (10 turns L = 6 mm, D = 23 mm), there is no interrupt and everything is working fine. I am going to put a 1 nF capacitor close to the motor to avoid motor brushes and a 560 uF close to the relay (inductor is between them) so it can help to prevent any further interruption. anyway, if you mind, please assume a number for the current and internal resistance, I am ready to learn if you wanna teach me something (please be very detailed I am not pro but I want to learn)... TNX for your kind support.

 

When you measure the DC resistance of the motor, turn the shaft slowly and you will see a variation in the measurement. The lowest measured value is what you want.

 

with this method the internal resistance is 1.5 ohm!
when the motor is running the Voltage is 11.93 and the current is 0.51 precisely.

 

Put a diode in the + supply to the arduino, then put a big capacitor across the arduino power inputs and in-between the diode and the arduino. Now even if the power drops to zero, the capaitor will sustain power to the arduino for a short time. Kind of like a UPS. The diode will prevent the capacitor power from going to the motor.

 

Put a diode in the + supply to the arduino, then put a big capacitor across the arduino power inputs and in-between the diode and the arduino. Now even if the power drops to zero, the capaitor will sustain power to the arduino for a short time. Kind of like a UPS. The diode will prevent the capacitor power from going to the motor.

Very clever Idea, it makes sense, I am going to test it. however, the off time is something around half a second which is too far from capacitance supply time.
what sort of Diode? 1N4148? 1N400x ( x from 1 to 7)?
do you have any idea for the Capacitor? how many Farad I mean.

 

First figure out the approximate current draw for the arduino, how low you're willing to allow the arduino input voltage to drop, then determine how long you want it to run on capacitor power. Do some math to figure out what size capacitor is required to maintain the required voltage for the required time given the arduino current demand, then start with that size capacitor. Or take a guess, put something big and see how it works. Look for a diode that can handle the forward current easily and with minimal voltage drop. There are probably many viable choices.

 

what sort of Diode? 1N4148? 1N400x ( x from 1 to 7)?

A 1N400x would work, but for lowest forward voltage drop, you could use a Schottky diode, such as a 1N5817.

do you have any idea for the Capacitor? how many Farad I mean.

Depends upon the Arduino current and tolerable voltage drop.
The capacitance equation is C = Q/V where Q is current times time.
So, for example, for a 50mA current draw and a 1V maximum drop, C = (50mA* 0.5s)/1 = 25mF or 25,000μF.

 

A 1N400x would work, but for lowest forward voltage drop, you could use a Schottky diode, such as a 1N5817.
Depends upon the Arduino current and tolerable voltage drop.
The capacitance equation is C = Q/V where Q is current times time.
So, for example, for a 50mA current draw and a 1V maximum drop, C = (50mA* 0.5s)/1 = 25mF or 25,000μF.

Good suggestion, but those diodes can only handle one amp of current. Would it be possible to use a p-fet configured as a diode instead?

 

Good suggestion, but those diodes can only handle one amp of current. Would it be possible to use a p-fet configured as a diode instead?

The diode only carries the Arduino current so I would think 1A would be more than sufficient in this case.

A P-FET certainly can be used as a diode in such an application for very low voltage drop.
Here's my discussion on that.

 

A 1N400x would work, but for lowest forward voltage drop, you could use a Schottky diode, such as a 1N5817.
Depends upon the Arduino current and tolerable voltage drop.
The capacitance equation is C = Q/V where Q is current times time.
So, for example, for a 50mA current draw and a 1V maximum drop, C = (50mA* 0.5s)/1 = 25mF or 25,000μF.

thanks man you helped a lot.

 

The diode only carries the Arduino current so I would think 1A would be more than sufficient in this case.

A P-FET certainly can be used as a diode in such an application for very low voltage drop.
Here's my discussion on that.

1A would be plenty to run the arduino, but when it's first turned on there would be a big (but brief) rush of current to charge up the cap. Check the data sheet to see if the diode would live long term with short bursts of a lot of current (however much your power supply can give). Worst case, you could probably put a couple of diodes in parallel... would some low value series resistors for each diode be required to keep one diode from taking all the current?

 

Worst case, you could probably put a couple of diodes in parallel.

You could also add a small resistor in series with the diode to limit the surge current, if necessary.
That should cause only a minor voltage drop from the normal operating current.