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Old 11/13/2017, 05:34 AM   #1
salty joe
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wall wart question

I have a 12V peristaltic pump on the way and draws something like 1.4A.

I have a 12V 2A wall wart but if I were to run it at 9V, would a 9V 2A wall wart do the trick or should I get a higher amp model?


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Old 11/13/2017, 06:13 AM   #2
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If it behaves nicely that should work. Its still capable of delivering more watts than the motor should use.


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Old 11/13/2017, 06:25 AM   #3
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12 x 1.4 = 16.8watts
9 x 2 = 18 watts..
18W>16.8W = good..

I typically like to have more extra margin there and would want something capable of 20W or more just so its not running at close to 100% (I typically size a power supply at 125% or more of my continuous max load) but since its a pump with a short cycle time thats not on all the time the 9V @2A is likely just fine..


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Old 11/13/2017, 06:51 AM   #4
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I would expect the draw on the pump to drop as it's also on lower voltage, or am I missing something?

Tim


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Old 11/13/2017, 02:24 PM   #5
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You can't generalize that for motors because it really depends on the load. It will run slower at a lower voltage but the load (power) doesn't always scale linearly. In the case of a peristaltic pump you're basically correct though.

That said I try to keep wall warts at or below 50% of their rating. I don't like to trust cheap commodity electronics without a 100% safety margin. A robust switching power supply with a rated duty cycle and a datasheet that gives detailed operational parameters, sure - run at 80% or 90% as long as the data sheet supports that. But a $3 generic wall wart? No thanks.


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Old 11/13/2017, 07:12 PM   #6
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Thanks guys. I'll give myself some extra headroom.


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Old 11/14/2017, 06:53 AM   #7
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Quote:
Originally Posted by der_wille_zur_macht View Post
You can't generalize that for motors because it really depends on the load. It will run slower at a lower voltage but the load (power) doesn't always scale linearly. In the case of a peristaltic pump you're basically correct though.

That said I try to keep wall warts at or below 50% of their rating. I don't like to trust cheap commodity electronics without a 100% safety margin. A robust switching power supply with a rated duty cycle and a datasheet that gives detailed operational parameters, sure - run at 80% or 90% as long as the data sheet supports that. But a $3 generic wall wart? No thanks.
Yeah, when the device is more than just a motor (as a peristaltic could be if it has integrated flow rate control, etc) then all bets are off. And as you say, the change is not always linear. But generally would expect simple devices to draw less on lower voltage.

Even when buying wall warts I (personally) try to stick to a decent make (mean well usually)...

Tim


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Old 11/14/2017, 07:04 AM   #8
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I'm confused...

if its a 12V pump rated at 1.4amps, and a wall plug rated at 12V 2amps, why would you turn the wall plug down to 9V? Even at 12V its still has plenty of amps to supply the pump. I would also think that the decrease in voltage may shorten the pumps life span, or not make it work at all.

Using mcgyver's method above:

12v @ 1.4 amps = 16.8 watts

12V @ 2 amps = 24 watts


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Old 11/14/2017, 08:24 AM   #9
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With these types of pumps..

Less voltage = slower speed = slower dosing ml/min

and yes with many motors less voltage = more current = more heat = shorter life..


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Old 11/14/2017, 12:44 PM   #10
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Quote:
Originally Posted by mcgyvr View Post
and yes with many motors less voltage = more current = more heat = shorter life..
This is the one that confuses me. I've seen it said before and I'm not suggesting that it isn't true, but why would lower voltage lead to more current? On the motors I have played with (which is a lot less than you, undoubtedly!) dropping the voltage drops the current draw also. Not always proportionally, but I've never come across one where it increases the draw. Under what circumstances does that happen?

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Old 11/14/2017, 01:05 PM   #11
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food for thought:
Quote:
The existing load is a major factor in determining how much of a decrease in supply voltage a motor can handle (see sidebar, below). For example, let's look at a motor that carries a light load. If the voltage decreases, the current will increase in roughly the same proportion that the voltage decreases. For example, a 10% voltage decrease would cause a 10% amperage increase. This would not damage the motor, if the current stays below the nameplate value.
Quote:
On lightly loaded motors with easy-to-start loads, reducing the voltage will not have any appreciable effect, except that it might help reduce the light load losses and improve the efficiency under this condition. This is the principle behind some add-on equipment whose purpose is to improve efficiency.
http://www.ecmweb.com/design/highs-a...-motor-voltage


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Old 11/14/2017, 07:02 PM   #12
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I plan on using this pump for my calcium reactor and thought if I could slow it down the tubing would last longer. Not much of a bargain if it's at the expense of the motor.

If it ran cool at 9V, that would pretty much say everything is OK-is that right?


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Old 11/14/2017, 07:22 PM   #13
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It's going to be fine at 9v. This discussion is academic.

The bigger question, will your tubing last longer? By the calendar, yes. Measured by the volume pumped, no.


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Old 11/15/2017, 05:05 AM   #14
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I'm fairly certain at 12V, the volume will be too high. Slowing the pump down with lower voltage seems better than choking the pump's tubing at 12V.

So, if I experiment with even lower voltage, will pump temp tell me everything I need to know?


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Old 11/15/2017, 05:11 AM   #15
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I guess I'm just slightly confused...

If the pump is designed to be run at 12V, why would you want to slow it down, as opposed to just stopping the pump on a timer?


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Old 11/15/2017, 05:23 AM   #16
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I guess I could run the calcium reactor on a timer but prefer constant flow. Plus, a timer is another piece of equipment to fail although it might be my best fallback plan if the motor won't tolerate lower voltage. Thanks.


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Old 11/15/2017, 05:30 AM   #17
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Ahhh I got you now. I was thinking 2 part dosing pump.


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Old 11/15/2017, 05:35 AM   #18
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Choking the pump will just create a pressurized stream and still flow just as much as with an open tube.

This is a positive displacement type pump, which means it will move the same amount of fluid each turn 'no matter what'. If it can't move that, as in if the output gets plugged, the pump will create pressure in the output side of the tube until either the plug goes away, the pump stalls, or the tube ruptures.

Now realistically it may not do any of those if it leaks back past the rollers enough in the event of a complete plug.


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Old 11/15/2017, 06:05 AM   #19
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Well then, wow-choking is out of the question. Thanks for the heads up!


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Old 11/15/2017, 07:26 AM   #20
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Quote:
Originally Posted by oreo57 View Post
But that's talking about AC motors isn't it? I couldn't find explicit clarification in the link (and the chart it refers to I couldn't find either ) as to whether it was AC or DC but the voltages they were talking about suggests AC to me? I was thinking about DC motors.

Don't get me wrong, I'm not trying to argue - I'm trying to clarify my confusion over something I have seen suggested a few times (not just in this thread).

As der_wille_zur_macht pointed out, this just academic...

Tim


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Old 11/15/2017, 09:15 AM   #21
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Quote:
Originally Posted by perkint View Post
But that's talking about AC motors isn't it? I couldn't find explicit clarification in the link (and the chart it refers to I couldn't find either ) as to whether it was AC or DC but the voltages they were talking about suggests AC to me? I was thinking about DC motors.

Don't get me wrong, I'm not trying to argue - I'm trying to clarify my confusion over something I have seen suggested a few times (not just in this thread).

As der_wille_zur_macht pointed out, this just academic...

Tim
Yea.. may apply mostly to AC motors..
Fun w/ academics:
Quote:
I did not write about universal dc motors but rather a parallel connected dc motor.Doesn't matter whether it's series or shunt wound, the model's the same.I'm not so sure that you are right about about a parallel connected dc motor. At no load if you reduce the voltage in such a motor you also reduce the field strength which results in less back emf and the armature current increases temporarily until the speed increases to bring it back down to that current sufficient to overcome losses. However you also reduce the torque so that under load the motor will slow down and the current increase. It does seem to me that if you reduce the voltage in such a motor the current could increase enough to damage it.It's not that difficult conceptually.

A d.c. motor will have a locked rotor torque:

T0 = K Vs / R, where

K = a constant for a particular motor,
Vs = supply voltage,
R = locked rotor resistance.

It will have a no-load speed of:

w0 = Vs / K

and in between locked rotor speed (0) and no-load speed, the torque will be:

T = T0 (1 - w / w0)

It will draw a current of:

I = T / K

Let's say we have a 12 V motor, which draws 2 A with a locked rotor, and has a no-load speed of 3,000 rads-1. Then:

R = 12 / 2 = 6 ohms,
K = 12 / 3000 = 0.004 NmA-1,
T0 = 0.004 x 12 / 6 = 0.008 Nm.

What will happen if we reduce the supply voltage to 8 V? No-load speed will drop to 8 / 0.004 = 2,000 rads-1. This doesn't tell us much about current, though, because in this model, no-load current is zero (i.e., frictional losses are treated as a load).

So let's have the thing drive a fan. We'll assume that with a 12 V supply, it'll drive the fan at 2,700 rads-1. We'll further assume that the torque required to drive the fan is of the form:

Tf = a w

where a is a constant for the fan.

So for our fan,

a x 2,700 = 0.008 (1 - 2,700 / 3,000)
a = 2.963 x 10-7,

and the motor will draw a current of:

0.008 (1 - 2,700 / 3,000) / 0.004 = 0.2 A.

What happens now when we reduce the supply voltage to 8 V?

2.963 x 10-7 x w = 0.004 x 8 / 6 (1 - w / 2,000)
w = 1,800 rads-1
I = 0.004 x 8 / 6 (1 - 1,800 / 2000) / 0.004 = 0.133 A

So speed dropped and current dropped as the supply voltage dropped.

The above result will hold for any monotonically increasing load torque function, i.e., provided it always takes more torque to make the thing go faster. This will generally hold for any type of load, once the initial static friction has been overcome.

That's the end of the story for d.c. motors.DC motor speed control is either with armature current control or field current control but the armature and field are not in parallel but are separately excited. In that case you can either vary the field current to change the stationary magnetic field or the armature current so as to vary the rotating magnetic field.We're not talking about negative resistance devices, here, so reducing current through the thing necessarily implies reducing the voltage across it.
http://boards.straightdope.com/sdmb/.../t-313597.html

Best fit.. ???


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Old 11/15/2017, 09:16 AM   #22
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perkint...
The (lower voltage/more current) is typical with AC induction motors.
Its not applicable here..


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Old 11/15/2017, 12:57 PM   #23
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Thanks guys - interesting and I am a little more enlightened

Tim


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Old 11/16/2017, 02:40 PM   #24
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AC motors are essentially speed regulated by the AC frequency. At a given frequency and load they will always try to be constant power devices and will pull more current at lower voltages.

DC motor speeds are regulated by voltage. They balance current against load at a given speed instead of the other way around. If you lower the voltage it just slows down, does less work, and (usually) draws less current.


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Old 11/16/2017, 04:14 PM   #25
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That was what I thought for DC: I didn't understand the implications on AC. Do a little better now

Tim


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