Need more power

[mvaughn]

I've learned an important lesson along the way in creating my cnc router. "Heed the advice of others when it comes to power requirements."

I'm trying to get my steppers running and I don't thing my current power supply has the cajones to make it happen.

My 3 steppers are all 5.2 Volts per Phase, 1.4 Amps per Phase Unipolar 6 wire motors. Does that mean they require 2.8 amps total or am I just crazy?

Anyhow, what I've found it that my 12 PC power supply will run one motor just barely. When I connect up 2 motors to it, the supply just shuts down. I don't have any specs on the PS but I'm assuming it is rated at 200 watts.

If I were to order a power supply from MPJA that is rated for 24 Volts and 4.5 Amps, would that be enough to run my steppers?

I know I will also need some power resistors near 13 Ohms and 25 Watts.

[Nono]

I learned in another post about power supplies (pc) that unless there is a constant drain off the 12 side of that power supply the power will drop off, that isn’t what is going on is it? It sounds like you do not have the amps. http://www.cnczone.com/forums/showthread.php?t=10680

[Nono]

A quote from Gecko
Doubling the voltage doubles the motor’s high speed power. In all cases the power supply voltage should be no less than 4 times or no more than 25 times the motor’s rated voltage.
The maximum power supply current required is 67% of the
motor’s rated phase current.

[gar]

050605-1014 EST USA

mvaughn:

I have no idea what you mean by "12 PC power supply". A search for "12 PC" on this web site produced no results. But that is a problem with their search algorithm because it did not find your post.

Assumptions are bad, but when nothing else is available one has to make assumptions.

I assume you are referring to an IBM compatible PC switching power supply. If this is the case, then the 12 v supply is very limited compared to 200 watts. The 12 v supply probably current limits around 200 ma, or approximately 6 watts. A 200 watt rating on a PC power supply would relate to total power.

Consider your question on current and resistor power rating.

You need to look at the steady state conditions that result in maximum power supply load current.

Any power supply in a steady state condition, whether regulated or not, may be viewed as a voltage source with a series resistor (called the internal resistance), and assuming no non-linear elements. This is call Thevenin's equivalent circuit.

In any power supply the equivalent internal resistance many not be constant because of non-linearities. But, as a first order approximation one will assume that it is constant.

If the power supply is regulated, then this internal resistance can be extremely low because internally thru feedback the output is forced to be nearly constant. However, I see no reason to use a regulated supply for motor power because you do not need the regulation and it would waste a lot of power if it were a linear regulator.

In a non-regulated supply the design criteria may be such that the the internal resistance is about 5 to 10 % (maybe more) of full rated load. This means that if you have a constant input voltage to the power supply that as you increase the load on the supply the terminal (output) voltage will drop by that percentage of the load.

For simplicty one generally assumes the internal resistance constant. On this assumption and for your purposes you can measure the internal resistance as follows:

Measure the ouput voltage with no load, and call it VNL. Next apply maximum load and measure the load current IFL, and power supply output voltage VFL. Then the internal resistance of the power supply is Rint = (VNL - VFL)/IFL.

Now continue to your example of a 24 v power supply and let us consider only one stepping motor. Your motor is probably a 2 phase motor with each phase center tapped. The nameplate rating is probably for 1/2 of one phase. The values you gave were 5.2 V and 1.4 A. This is a nominal resistance of 3.71 ohms, probably at room temperature. When the motor is hot this resistance will be slightly higher.

In the operation of your motor 1/2 of each phase will always be on. If in steady state you have 1.4 A to each phase then for each motor the load on your power supply is 2 x 1.4 A = 2.8 A as you indicated. The total load for three motors would be 3 x 2.8 A or 8.4 A. So 4.5 A is not adequate for 3 motors.

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(edit 6-12-05) I goofed, only one phase is energized at a time thus your current is half of what I stated. A higher current supply won't be a problem. So your total for 3 steppers should be 4.2 amps, but I would consider a 4.5 amp supply inadequate because you would be too close to the limit. I not going to redo the other calculations because they still illustrate the point.(end edit)
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(edit 6-13-05) My original statement was correct for normal useage.(end edit)
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Assume your power supply internal resistance is such that fully loaded the output drops 20%, then for 24 V open circuit the output will be 19.2 V. Assume 1 V drop in the switching transistor, then the voltage across the dropping resistor and 1/2 phase is 18.2 V. For 1.4 A the series resistor needs to be Rdrop = (18.2 - 5.2)/1.4 = 9.3 ohms. The power dissipated in the dropping resistor is (18.2 - 5.2) x 1.4 = 18.2 watts. Personally I would derate a resistor by about 50%. Therefore I would use a standard 10 ohm 50 watt resistor for the dropping resistor.

Ball park your power supply needs to be about 24 x 8.4 x 1.4 VA or about 300 VA. I used an arbitrary derating multipler of 1.4. Your worst case situation is when the stepping motors are not moving.

Keep in mind that everything you are working with must be derated as temperature rises. All ratings of the devices you are working with are based on some maximum internal temperature, and this is an absolute temperature. For example for a lot of electrical devices it is about 100 deg centigrade. The steady state temperature rise in a device is approximately a constant times power. So as ambient temperature rises the allowed temperature rise in the device decreases and this means the power dissipated must be lowered.

You need to pick your power supply, find its characteristics, then choose the resistors.

.

[H500]

Mvaughn, it sounds like you're using one of those pesky controllers that need resistors to limit the current. Your PS need to have: 3motor X 2coils X 1.4amps = 8.4 amps.

Your supply is probably shutting down because you are not using resistors. Don't do that too long or you will burn out the motors.

Also, with computer supplies, I believe you need to draw a bit of current from the 5v rail before the 12v will work properly.

[mvaughn]

When it comes to electronics and electricity... I know just enough to be dangerous.


Gar,

I'm still trying to digest and understand you posting. Thank you for taking the time to put in such a length response. I will read and re-read it until I understand.

H500,

Currently each of my steppers has 2 resistors, 4 Ohm 25 Watt Power resistor, one on each center tap. I have tried putting a load on the 5 volt line of the power supply and it doesn't seem to make a difference. In fact, my driver board uses 5 volts.

I can also hear a high pitch squeal coming from my power supply when I turn it on withe the steppers connected. For a while I thought it was the steppers making the noise, but I traced to the power supply.

I will get a multimeter so I can measure the outputs of my current setup as suggested by Gar.

[gar]

050605-1923 EST USA

mvaughn:

Look at your power supply and see if a nameplate lists a maxium current for the 12 v supply. There may be both a negative and positive 12 v supply. These each may have different ratings. Let us know what information is there.

Doing a calculation using your 4 ohm resistor and motor specs of 5.2 v and 1.4 amps. At 1.4 a the voltage drop on the 4 ohm resistor is 5.6 v. The sum of the motor and resistor is 10.8 v. Add 1 volt for the transistor switch and the sum is 11.8 v. Thus everything checks for a nominal 12 v supply, and the nominal values used for the calculation.

Your problem is the power supply.

If you have 2 IBM PC compatible power supplies which are not used for anything else, then connect the two 5 v supplies in series to get 10 v. These 5 v supplies will have a substantial current capability. Probably 10 or more amps depending date of manufacture. Another possibility is to use a 12 v car battery with a suitable fuse for the supply.

Get the meter first and learn how to use it. Just use the voltage range first. Start with a high voltage range and switch down until you get a useable reading.

.

[gar]

050605-1927 EST USA

mvaughn:

The power rating on your 4 ohm resistor is just fine. The power dissipation is about 5.6 x 1.4 = 7.8 watts.

.

[gar]

050606-1618 EST USA

mvaughn:

I found another PC power supply with a nameplate rating of 8 A on the +12 and a low value on the -12 V.

If you can find a supply like this try the load of one motor on the supply. If the power supply is supplying 12 v you should see about 5.6 v across your 4 ohm resistor. Effectively the measured voltage drop across the 4 ohm resistor will tell you the current flowing thru the resistor.

You should run this test with no step signals to the motor.

.

[mvaughn]

Gar,

I looked at the power supply label and it is like you said. The +12 side was rated for 8A, and the -12 was .5A.

I will check the voltage output of the supply to make sure it is putting out 12 volts and also check it behind the 4 Ohm resistor. I'll report back the results.

I'm not sure how to measure the load of a motor. I'm assuming you mean how much current it draws while connected.

[mvaughn]

Here are some of my initial results with the multimeter. Straight from the power supply I get 11.5 volts. My resistors measure 4 ohms. I may not be reading the multimeter correctly, it isn't digital and it's hard to measure anything under 50 ohms. The part I don't understand is how my voltage can be 11.4 when the resistors are inline.

[smarbaga]

resistors in line u say , i guess that means that one side of the resistor is connected to your power supply and the other side of the resistor is just hanging there.
and you have your meter on the unconnected end of the resistor and the ground or black wire,... if this is the case....
the voltage at the end of the resistor has no where to go, so u have maximum potential there,
if you were to put another resistor from the black wire to the resistor on the plus wire you would have a voltage dividing network.
but i c that everyone wants to put big voltage on these little motors.
this is ok as long as the wattage remains the same in the hold mode. ( i think )
is the holding torque the same in single step as it is in a multi-step mode, with the same wattage applied to the coils ???