[SOLVED] [moved] Dc motors, AC/DC adapters, wattage voltage and air pump.

Hello and thank you for helping me out with my doubts.

I've been trying to convert a small air pump from working with a car lighter to working with an AC/DC adapter, for home use.

The pump is identical to this one but is rated for 300 psi instead of 250: **broken link removed**

The pump has no amperage information, it only states that it's 12V DC.
I've searched around and some of this pumps say that the car lighter should be of 15 amps or higher(??).

I've tried wiring the motor with an adapter that's rated for 12V, but only 500mA.When i turned it on, the motor ran really slow and hardly any air came out of the pump.

My questions here are:

• 1 Will another adapter with higher or lower voltage work with this pump? I have other adapters. None of them is rated for 12V, but they have bigger current capacity. For that purpose I would like to understand how different voltage and amperage values affect the working of the motor.
  
• 2 How does the motor draw more or less current? Does it have a certain speed at which it runs, and if it runs slower than that speed, it pulls out more amps?
  
• 3 This is kind of a different question. The air pump is a bit loud, I deduce that most of it's noise is a result of vibration, so would rubber padding at strategic places make the pump quieter?
  
• 4 The pump will be used only to fill up inflatables and basketballs, as opposed to car tires. Will that diminish the current needed by the pump?

Once again, thank you for all your help.

jamalAbadin said:

[*]2 How does the motor draw more or less current? Does it have a certain speed at which it runs, and if it runs slower than that speed, it pulls out more amps?

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It is typical for a motor to draw more current as a greater load slows it down.

[*]3 This is kind of a different question. The air pump is a bit loud, I deduce that most of it's noise is a result of vibration, so would rubber padding at strategic places make the pump quieter?

Click to expand...

I have a similar inflator which is also loud. I'm not sure anything can quiet it except to place it inside an enclosure which is lined with foam padding. However then it would be liable to overheat.

[*]4 The pump will be used only to fill up inflatables and basketballs, as opposed to car tires. Will that diminish the current needed by the pump?

Click to expand...

Yes.

BradtheRad said:

It is typical for a motor to draw more current as a greater load slows it down.

Click to expand...

True and False, both at the same time.

A common motor under load slows down. As it slows down, the back emf produced is reduced and it draws greater current. The larger current partly offsets the larger load. Hence true.

A motor fed from a high impedance (as in the present case) source cannot produce high speed- because high speed needs high out of phase current. As the load increases, the phase comes towards the voltage and the load power increases. If the source cannot supply high current, the motor cannot attain full speed- load or no load. Full speed means larger back emf and hence needs larger input voltage. False in that case.

Common motors are speed regulated with a simple voltage regulator.

In the present case, the adapter is unable to provide sufficient current- the motor perhaps needs 25-50W of power and hence a 12V supply with about 2-3A rating will be needed.

That is my problem, I don't have a 12V supply with 2-3A.

Can I use a higher voltage supply?

If so, how much amps does it have to provide and why?

Power in watts (W) is equal to voltage (V) times current (A)

e.g. for a motor that needs 50W will require 50/12=4.2A minimum (this would assume no losses, unrealistic)

edit - the pump is rated to use 12 V you can't go with a higher voltage without the risk of damaging something. Of course you can go as high as 14V as that is normal for the electrical system of a car when the engine is running.

jamalAbadin said:

1. Will another adapter with higher or lower voltage work with this pump? I have other adapters. None of them is rated for 12V, but they have bigger current capacity. For that purpose I would like to understand how different voltage and amperage values affect the working of the motor.

2. How does the motor draw more or less current? Does it have a certain speed at which it runs, and if it runs slower than that speed, it pulls out more amps?

Click to expand...

DC Motors have a very important characteristic which is easy speed regulation.
For every kind of DC motor (we do not care if it has the inductor in shunt, series, large or small compound), it's speed follows this easy equation:

V=voltage applied (input voltage)
Ii=current through the induced, which means, the current that pases through the loop that rotates and hence creates the torque.
kE=a constant that depends on the physical construction of the motor
phi=flux provided by the magnets or electromagnets.

So to regulate the speed, you are focusing on V and Ii, higher V => higher speed. Higher Ii => lower speed.

Every electric machine has some nominal values, that means, the machine feels great working on those values. If you give it more, you will damage it (damaging its protections etc), and if feed lower than its nominal values, will not work as expected.

---Updated---

Do not confuse you if you see that if you want more speed, then just make lower the Ii. You need power to drive what is needed to be driven. And also Ii is direct proportional to the Torque, more Ii results in more Torque.

For a larger life of the air pump, use voltage manufacturer said.. and if we had known the power required then would have been easier to find the DC source.

Thank you all very much for your help, I think I'm going to try to find an old PSU to use as the power source, which provides me the appropriate voltage and that is able to give enough current.

jamalAbadin said:

That is my problem, I don't have a 12V supply with 2-3A.

Can I use a higher voltage supply?

If so, how much amps does it have to provide and why?

Click to expand...

Motors are basically current devices- it is the current in the windings that produce the magnetic field and the attendant torque.

It should be obvious that the current is directly proportional to the torque developed.

However, as the motor starts rotating, it produces a back emf and that opposes the external applied voltage (this is the voltage that is causing the current in the first place)

The back emf is proportional to the speed of the motor. This too should be clear.

Torque and speed together determines the power. But you cannot control both voltage and current independently - that is fixed by the motor design and construction. Motors are not really resistive loads (except for very small motors).

You can safely try a higher voltage, say 20% higher. But it must be able to provide higher current.