Ideas to modify a 12v air pump to pump high pressures

[eccerr0r]

Re: Ideas to modify a 12v air pump to pump high pressures

Allright, for completeness and mistake in calculations so ignore my previous post numbers. I definitely am holding at least 10-15lbs force (likely more...) against the output hose, and it's sufficient to stop air from coming out. At 10lbs on that inflator latch is about 200PSI. At 15lbs force, this is 300PSI - air should stop as this is beyond the design limit. As I was able to stop it, either this thing is not providing enough pressure, or I really need to accurately measure how much force my thumb is putting against the inflator latch.

Now it seems there must be a pressure relief valve inside the compressor so that it doesn't stop the motor and/or stop things from exploding...

[caphair]

Re: Ideas to modify a 12v air pump to pump high pressures

How would I lubrciate the piston assembly? Anyone done it before?

[eccerr0r]

Re: Ideas to modify a 12v air pump to pump high pressures

Well, how smoothly does the piston assembly rotate right now? The crankshaft should at least be lubricated a bit with regular mineral oil (it's metal, right?), the worry is what kind of seal the piston has to determine what kinds of safe lubricant can be used.

Speaking of my compressor, it seems that the "pressure relief" is actually due to the latch is extremely hard for me to hold together and seal and actually leaking air between the metal and plastic without me being able to feel the leakage. So apparently my small compressor is most definitely capable of hitting 250PSI even if it'll take forever because it's duty cycle is rated for 20 minutes on, 20 minutes off...

[momaka]

Re: Ideas to modify a 12v air pump to pump high pressures

Not sure if it also takes into account that older PM motors are likely using alnico or some other iron based magnets, and newer ones may be using neodymium magnets, which will reduce size.

Neodymium magnets? In these cheap pumps? -HA!
I've never opened one, but I highly doubt it. Most likely they have a 540, 550, or 560 size motor with regular iron magnets and geared down for more torque on the piston.

If the motor is stalling, just install a bigger motor...that is, if you can even fit a bigger one or take the gear off from the old one and install it on the new one (I wouldn't be surprised to see brass heat-pressed pinion on the original motor). If the motor is indeed a 540/550/560 size, you might be able to fit one of those modified "short-turn" racing RC motors in there. But just be aware - those things will eat a ton more current. And they cost an arm and a leg, because hobby stuff always does.

Aside from that, I don't think there is an easy way to mod that cheap pump to put out high pressure. No cap, no oil, and no magic will help it. It's just built cheap and not for high pressure, regardless what the manual tells you.

[eccerr0r]

Re: Ideas to modify a 12v air pump to pump high pressures

Well, there are some that can attain the pressure, at least I think mine can...

[caphair]

Re: Ideas to modify a 12v air pump to pump high pressures

Update:

Went to try cleaning and regreasing and as I was taking it apart the positive cable broke off the motor terminal. It was rusted/corroded. So I cleaned and resoldered and it works great now. Was able to go to 115psi and didn’t slow down or stall.

However I noticed if there’s 60psi or more already in the tire the pump won’t even start. If I let air out to around 30psi it’ll work off momentum and keep pumping no problem

[eccerr0r]

Re: Ideas to modify a 12v air pump to pump high pressures

Now I need an explanation why a better connection there stops blowing the car's fuse...

[goontron]

Re: Ideas to modify a 12v air pump to pump high pressures

^ Low voltage = high current with motors.

[caphair]

Re: Ideas to modify a 12v air pump to pump high pressures

Now I need an explanation why a better connection there stops blowing the car's fuse...

I don't use the car's 12v plug so can't say if it would blow the fuse again. I'm running it on an old pc psu.

But a bad connection creates resistance, which drops the voltage so it tries to pull as many amps as it can on a lower voltage and pop

[eccerr0r]

Re: Ideas to modify a 12v air pump to pump high pressures

errr... with higher resistance joint it means lower voltage across the intended *load* - higher voltage across the *joint* - in any case this would mean less current flowing through, making it LESS likely to pop the fuse, making the conclusion that either
- the fuse will still blow and blow faster due to now more current flowing (since this fixed situation wasn't tested)
or
- fuse blowing was due to shorting against the case or other ground and not due to current flow through the motor

and

- premise of fuse blowing was a red herring.

[caphair]

Re: Ideas to modify a 12v air pump to pump high pressures

errr... with higher resistance joint it means lower voltage across the intended *load* - higher voltage across the *joint* - in any case this would mean less current flowing through, making it LESS likely to pop the fuse, making the conclusion that either
- the fuse will still blow and blow faster due to now more current flowing (since this fixed situation wasn't tested)
or
- fuse blowing was due to shorting against the case or other ground and not due to current flow through the motor

and

- premise of fuse blowing was a red herring.

I don’t know it makes sense to me that a poor connection caused a voltage drop so current consumption increases to get the same amount of work done (motor spinning).

The motor was very hot to touch when it would stall before fixing the connection. To me that heat was generated by all the current it was trying to use to spin the motor on a lower voltage

[eccerr0r]

Re: Ideas to modify a 12v air pump to pump high pressures

The problem is that the resistance across the whole circuit went up because of the link, so the maximum amount of current necessarily goes down. Yes, if you stall the motor, the current will go up (NOT because of the voltage goes down though the motor is more likely to stall if your voltage goes down). However, the maximum current that it can draw is limited by the bad link.

[momaka]

Re: Ideas to modify a 12v air pump to pump high pressures

errr... with higher resistance joint it means lower voltage across the intended *load* - higher voltage across the *joint* - in any case this would mean less current flowing through

Exactly!
If the connection is solid, but with lots of resistance, you get a resistive divider network, so the current and power draw of each element in the network will be lower overall.

However, an intermittent connection *can* produce higher current draw and is more likely to blow a fuse. This is because the motor draws a high starting current before it starts spinning. But with an intermittent connection, the motor may fail to start up at all, thus always pulling a high starting current when the connection is made for a short enough period of time.

^ Low voltage = high current with motors.

Wrong.

Especially when it comes to DC (i.e. linear) motors and other linear devices.

With DC motors, low voltage means low current draw - that is, in the unloaded state. But once a motor is loaded, you have to look at the efficiency curves to see how it would behave.

My uncle, who is a mechanical engineer by trade, always made that incorrect assumption when working with water heaters and stoves/ovens. He used to think that if the voltage across a heating element became lower, the heating element would draw more current to compensate for that (as his thinking was that since power = voltage x current, current must go up if voltage goes down to keep the power the same). I had to explain to him that the power rating is not a fixed value. The power rating is as stated only when the line voltage is at the specified rated voltage on the heating element. When the voltage gets lower, the so does the power rating. And that has to do with the fact that a heating element is pretty much like a simple resistor. So essentially you have Ohm's law. When the voltage across a resistor falls, then so does the current through it. With both lower voltage and current through a heating element, the power rating decreases.

Otherwise, if his assumption was correct, then when you turn off the power (disconnect the line) going to a heating element, the current would go to infinity to make up for the complete drop in line voltage (i.e. power switch going Off). But we all know that does NOT happen!

Yes, I know DC motors are not quite 100% linear devices, but they are overall linear as long as you don't stall them. Once stalled, you'll be looking at a different (but still linear) power draw curves - for those, the current will be purely dependent on the coil resistances in the motor.

[goontron]

Re: Ideas to modify a 12v air pump to pump high pressures

Wrong.

Especially when it comes to DC (i.e. linear) motors and other linear devices.

With DC motors, low voltage means low current draw - that is, in the unloaded state. But once a motor is loaded, you have to look at the efficiency curves to see how it would behave.

My uncle, who is a mechanical engineer by trade, always made that incorrect assumption when working with water heaters and stoves/ovens. He used to think that if the voltage across a heating element became lower, the heating element would draw more current to compensate for that (as his thinking was that since power = voltage x current, current must go up if voltage goes down to keep the power the same). I had to explain to him that the power rating is not a fixed value. The power rating is as stated only when the line voltage is at the specified rated voltage on the heating element. When the voltage gets lower, the so does the power rating. And that has to do with the fact that a heating element is pretty much like a simple resistor. So essentially you have Ohm's law. When the voltage across a resistor falls, then so does the current through it. With both lower voltage and current through a heating element, the power rating decreases.

Otherwise, if his assumption was correct, then when you turn off the power (disconnect the line) going to a heating element, the current would go to infinity to make up for the complete drop in line voltage (i.e. power switch going Off). But we all know that does NOT happen!

Yes, I know DC motors are not quite 100% linear devices, but they are overall linear as long as you don't stall them. Once stalled, you'll be looking at a different (but still linear) power draw curves - for those, the current will be purely dependent on the coil resistances in the motor.

Well, no. This was explained to me by an EE. As an electric motor runs up, it will create an electromagnetic field that acts counter of the torque created by the applied voltage, until the back emf generated causes net torque to be zero.

If you reduce the voltage, the motor does not have enough torque to get up to equilibrium speed. With the BackEMF being less than the EMF, It runs slower. Hence why it couldn't get up to 115psi even when it was running, Because it was slow. And with it being slow, it would be pulling more current to try and reach a net zero torque and an equilibrium RPM, which, under a heavy load, it couldn't.
Even though the EMF is lower with undervoltage, the back EMF is even less.

Same thing applies to overvoltage. Less current needed to generated net zero torque. Until the core saturates anyway. Then current goes up and so does losses.

The resistance in the line probably exacerbated the issue due to slow speed under very high load.

Now, this is AC motors, DC, as you say, may be different. But i doubt it.

[caphair]

Re: Ideas to modify a 12v air pump to pump high pressures

Well, no. This was explained to me by an EE. As an electric motor runs up, it will create an electromagnetic field that acts counter of the torque created by the applied voltage, until the back emf generated causes net torque to be zero.

If you reduce the voltage, the motor does not have enough torque to get up to equilibrium speed. With the BackEMF being less than the EMF, It runs slower. Hence why it couldn't get up to 115psi even when it was running, Because it was slow. And with it being slow, it would be pulling more current to try and reach a net zero torque and an equilibrium RPM, which, under a heavy load, it couldn't.
Even though the EMF is lower with undervoltage, the back EMF is even less.

Same thing applies to overvoltage. Less current needed to generated net zero torque. Until the core saturates anyway. Then current goes up and so does losses.

The resistance in the line probably exacerbated the issue due to slow speed under very high load.

Now, this is AC motors, DC, as you say, may be different. But i doubt it.

I agree with this. The faster the motor spins the less current it draws. When the load increases the motor slows down. This reduces the counter EMF. This allows a higher current to flow.

The partial connection created resistance as well as the load of the air pressure inside the tube the motor was trying to overcome thus slowing the motor down and increasing the current through it as it tries to overcome it.

With the bad connection the pump stopped very early on even when starting with a lower psi in the tire. Now with the connection fixed if the pump is started with little to no air pressure in the tube the rotating momentum keeps it pumping no problem.

Btw I’m no motor expert, just applying concepts I know to what I experienced.

[desert-rat]

Re: Ideas to modify a 12v air pump to pump high pressures

I don't think a single stage compressor will give you that higher pressure. You will need a 2 stage compressor.

[eccerr0r]

Re: Ideas to modify a 12v air pump to pump high pressures

The motor will always generate a back EMF when running and *no load* regardless of the voltage. The back EMF will always goes back as long as work isn't done.

When work is done (compressing air, overcoming friction, etc.) the back EMF is reduced. All of it is gone when in one situation : when it's stalled.

So whether you are sending in 1 volt and have it stalled versus sending 12V and have it stalled, it's generating no back EMF and will take full current, resistance of the windings. However at 1 volt, V=IR, current is lower!

In case anyone is still wondering, here is data I collected with my bench supply and a little CDROM motor tray motor:

1.2V - 11.59mA, stalled 93mA
3V - 14.22 mA, stalled 270mA
4V - 16.30 mA, stalled 360mA

If in doubt, set up an experiment! Unfortunately my bench PSU does not go below 1.2V but the stall current is linear, which is worst case current draw.

[momaka]

Re: Ideas to modify a 12v air pump to pump high pressures

Now, this is AC motors, DC, as you say, may be different.

And it is.

I studied EET, and we had an entire class on motors and machinery. Indeed what the EE guy told you is correct - but this is mostly to do with AC motors. DC and "universal" motors are different animals - they are linear in nature. Both the current and the rotational speed (RPM) are directly proportional to the output torque (that is, current increases for increasing torque and the RPM decreases).

As an example, see the attached PDF below, which is of the Mabuchi RE-260RA - a small permanent magnet motor that is commonly used in all kinds of gadgets and toys. In particular, look at the graphs of the current, RPM, and efficiency. Note that there are two graphs, and they appear to be quite different. This is because the RE-260RA-2670 and RE-260RA-18130 have different number of turns and wire thickness for the rotor windings.

So yes, current increases as you load the motor more (well, makes sense, right?), and that's because you are asking the motor to output a higher torque. But that's at a constant voltage. As you can see from the datasheet I attached, the stall current is the highest current the motor will draw, and right before the motor cuts off and stop spinning, that's when you can pull the highest torque. So with the example that eccerr0r provided above, if the stall current is a function of the input voltage, then we know that lower input voltage = lower stall current. And lower stall current means lower maximum output torque.

Also note that maximum efficiency for most small DC motors usually occurs when the motor is ran between 1/2 to 1/5 of it's maximum output torque. (I have examples of other similar DC motors, including the popular RS-550, which is often used in older Ni-Cd drills). If anyone is interested, I'll see if I can dig out its datasheet and post it here.

I don't think a single stage compressor will give you that higher pressure. You will need a 2 stage compressor.

I just thought about that today as well.
I guess this is similar to vacuum pumps, where you just can't go past a certain pressure with a single-stage.

[goontron]

Re: Ideas to modify a 12v air pump to pump high pressures

And it is.

I studied EET, and we had an entire class on motors and machinery. Indeed what the EE guy told you is correct - but this is mostly to do with AC motors. DC and "universal" motors are different animals - they are linear in nature. Both the current and the rotational speed (RPM) are directly proportional to the output torque (that is, current increases for increasing torque and the RPM decreases).

As an example, see the attached PDF below, which is of the Mabuchi RE-260RA - a small permanent magnet motor that is commonly used in all kinds of gadgets and toys. In particular, look at the graphs of the current, RPM, and efficiency. Note that there are two graphs, and they appear to be quite different. This is because the RE-260RA-2670 and RE-260RA-18130 have different number of turns and wire thickness for the rotor windings.

So yes, current increases as you load the motor more (well, makes sense, right?), and that's because you are asking the motor to output a higher torque. But that's at a constant voltage. As you can see from the datasheet I attached, the stall current is the highest current the motor will draw, and right before the motor cuts off and stop spinning, that's when you can pull the highest torque. So with the example that eccerr0r provided above, if the stall current is a function of the input voltage, then we know that lower input voltage = lower stall current. And lower stall current means lower maximum output torque.

Also note that maximum efficiency for most small DC motors usually occurs when the motor is ran between 1/2 to 1/5 of it's maximum output torque. (I have examples of other similar DC motors, including the popular RS-550, which is often used in older Ni-Cd drills).

I see. Weird. Good to know nonetheless... Well, nevermind my theory about excess current then.

[caphair]

Re: Ideas to modify a 12v air pump to pump high pressures

And it is.

I studied EET, and we had an entire class on motors and machinery. Indeed what the EE guy told you is correct - but this is mostly to do with AC motors. DC and "universal" motors are different animals - they are linear in nature. Both the current and the rotational speed (RPM) are directly proportional to the output torque (that is, current increases for increasing torque and the RPM decreases).

As an example, see the attached PDF below, which is of the Mabuchi RE-260RA - a small permanent magnet motor that is commonly used in all kinds of gadgets and toys. In particular, look at the graphs of the current, RPM, and efficiency. Note that there are two graphs, and they appear to be quite different. This is because the RE-260RA-2670 and RE-260RA-18130 have different number of turns and wire thickness for the rotor windings.

So yes, current increases as you load the motor more (well, makes sense, right?), and that's because you are asking the motor to output a higher torque. But that's at a constant voltage. As you can see from the datasheet I attached, the stall current is the highest current the motor will draw, and right before the motor cuts off and stop spinning, that's when you can pull the highest torque. So with the example that eccerr0r provided above, if the stall current is a function of the input voltage, then we know that lower input voltage = lower stall current. And lower stall current means lower maximum output torque.

Also note that maximum efficiency for most small DC motors usually occurs when the motor is ran between 1/2 to 1/5 of it's maximum output torque. (I have examples of other similar DC motors, including the popular RS-550, which is often used in older Ni-Cd drills). If anyone is interested, I'll see if I can dig out its datasheet and post it here.


I just thought about that today as well.
I guess this is similar to vacuum pumps, where you just can't go past a certain pressure with a single-stage.

Thanks for this. So the poor connection on the motor, did it cause a voltage drop and thus a lower stall current? Or was the voltage 12v regardless but the max current that the motor was trying to pull couldn't be accomplished?