Solar System Design

Read the specs on the motor or measure the locked rotor current.

thank you

Put a .047 ohm resistor between the motor return and ground

Connect a Digital Storage Oscilloscope in single shot mode across the resistor

Find the right DSO range and triggering

Switch on the motor to its proper power supply

Look at the DSO trace and there's yer answer!

Obviously we must start with the fundamentals. You do not need a DC motor to move the sun.

What in this world are you trying to move with a DC motor?

How much mass is this mysterious object?

Will the center of mass of this object be raise or lowered to add or release potential energy? Do you understand the idea of potential energy?

I'm thinking of a number between fifty and eighty. What do you think this number represents?

Could be the OP is planning to move the entire array to maximize irradiance capture, an improvement that is glaringly absent from the one solar farm I have seen around where I live.

Yes, it could be a tracking system drive. It could also be a drive to remove a protective cover, elevate and retract above the tree line, collapse panels to prevent wind damage and a myriad of other mechanical load possibilities. If the OP has a motor in their confused hands then they should RTFM from the motor manufacturer. This motor in the hands of the clueless may still be under or over sized for the job, it might not move fast enough or too fast, burn up or stall with the voltages available.

I've gotten very tired of posters that believe we can solve their pathetically explained problems telepathically. Far to often their requests just clearly explains their incompetence in dealing with a problem and absolutely nothing about the problem itself. If this was a homework cheater using the same technique they would literally just ask us how to solve question #12. There would be no paraphrasing or attributes of question 12, just something like the simple request "Can you please help me solve question 12?"

Maybe I should be more honest and just tell the clueless, NO! You have not given us a hint that you understand the concerns of your problem at all. I don't care if you forgot to tell us the few concerns you recognize of your problem. You did not present us with any engineering information to solve an engineering problem. I cannot teach you how to be a competent engineer. I only have a few choices here. I can laugh or complain.

so it' not just me...

Negative waves, so early in the morning? On a Sunday??

Why not just tell them to F*ξk Off!

That's what we're all thinking anyway.

wouldn't get anyone anywhere in this case

A battery (or battery bank) can provide plenty of starting power for your motor and help you through cloudy days as well.

Not knowable on the present lack of data.

Just about all motor spec sheets will give you the stall current - sometimes called "peak current", of the motor. That is the current the motor will draw under maximum stall conditions, ie when the armature is not rotating. This is the same condition at start up.

If this motor is relatively small, then consider an ATA maximiser as a means of achieving higher start current without the need for a solar panel of start current equivalent sizing.

If the motor is sufficiently large to be concerned about starting current, then you might consider a motor starter to control the inrush current to both protect the armature windings and the supply equipment.

Add a large capacitor to the circuit so that it charges through a diode over a long period and then discharges through the motor in a couple of seconds after the motor is switched on. A switch that triggers when the motor is rotating would be helpful. Then 'motor on' and 'not motor rotating' = 'capacitor discharging', also 'not motor on' = 'capacitor charging'

Usually one and half times the running current is needed to start a motor. Since providing this extra solar panel is going to be uneconomical, it is better a storage battery is introduced in the circuit that would take cars of the starting torque.

This is a DC motor. Armature current is limited only by armature resistance and the back EMF produced when the armature is rotating. At start up there is no rotation and therefore no back EMF, so the starting current can be ten or more times higher than the full load current.

Even very large DC motors have low armature resistance, generally less than one ohm and often half that, so take a 24v motor with 1Ω armature resistance and Ohms law tells you that start current will be close to 24 amps. Full load current for that motor would typically be around 1.5 amps.

Unless some form of control is used, the start current will always be much higher than one and a half times running current.

The Ohms law treatment does not take into account the inductance involved, for one thing. Starting current ratio as integrated from dead stop to full speed ? I don't think any hard and fast rules are useful. Depends on the motor and load.

When you add the implication that this is a photo-voltaic system without a battery or even a capacitor to stiffen the supply and that these photocells are misaligned with the sun and want this motor to move these cells...

I keep coming back to just saying NO!

I must have missed that. When you put it that way,...

I am obviously not getting all of the posts here as I see no indication from the OP that this motor is required to move the panels or that the cells are misaligned.

So was this extra information actually supplied or is it simply conjecture?

It was not suggested by the OP but by me as a possibility.

It is not pure conjecture but the OP has not explained at all what type of a mechanical load this motor will moving. The OP gives us so little information about this motor other than it will be powered by a photo-voltaic system and that this is titled "Solar System Design" I think that it is a reasonable assumption this motor will move some part of the "solar system". So I ask you, why would one wish to move a solar system if the solar panels and the sun were perfectly aligned.

I don't know that this is actually being used to move the panels at all, but I see your point re. the title and the consequent assumption that you have made.

At home, I have a solar panel with no attached battery that I use to run a DC motor.

The set up was designed to run a small water pump, but without me supplying this extra information I see no reason that you would decide that it was being used for any particular unstated purpose.

My understanding of the question was that he simply wanted guidance as to how to determine the starting current when he didn't have that information.

I have a feeling that the electrical community is against use of Direct current, in Solar applications. This sort of acrimony against DC existed, with Tesla and Edison, and evidently Tesla won the battle, and we are stuck with Alternating current, though it half as efficient as Direct current. No one can deny that Direct current has its glaring deficiency of losing voltage when made to travel distances, but I cannot understand why it cannot be advocated for individual houses, where the travel is minimal.

As a matter of fact, there is no single automobile with Alternating current circuitry, any where in the trillions that ply the roads, all because the travel distance is small. If individual houses are provided with solar panels, and possibly some storage batteries the grid is freed to that extent from drawal.

With battery technology showing interesting developments with much better recharge performance, there is no reason why solar energy could not be used more copiously.

The problem with DC versus AC is not the often touted by laymen voltage drop for long runs. AC power is much easier to change and therefore regulate a voltage level than DC. It is also much easier to get a motor to spin with AC power than DC power. It is particularly easy with three phase AC power.

My point is the efficiency. AC power is a sine curve, with maxima at positive and minimal at negative, whereas DC is straight line graph.

And how come the small motors used in computer systems use DC power only ?

Will it not be possible to design DC motors that work well with low voltage ?

Does anyone apply their R&D for really good DC motors that work for all applications?

If the knowledgeable say that Three phase AC is the only thing in the world, then it is NO to DC, and then let us design Solar panels that give straightaway AC power instead of using an inverter et al.

The fans are DC motors because the power distribution in the chassis is DC. I believe the arm (read head) motion of the hard drive is a stepping motor and often the disk motor, too. I did not say that it was impossible or even difficult to make a DC motor. Just that a 3 phase AC motor is easier.

"And how come the small motors used in computer systems use DC power only"

Whilst the brushless motors in computers etc. are supplied by DC, the DC is inverted to AC, albiet not a sinusoidal one. The inversion is done either externally or via electronics incorporated in the motor. So those DC motors are actually electronically commutated synchronous AC motors.

then let us design Solar panels that give straightaway AC power instead of using an inverter

Unfortunately we cannot because like batteries (for example) by their very nature solar cells don't work that way and only produce DC.

http://en.wikipedia.org/wiki/Solar_cell

It is becoming more common to have the inverter built on to the back of a solar panel and technology is developing in the inverter market to reduce their price and improve their life.

I was hoping that somebody here would be able to find some way to have light pumped back and forth to the sun at 50 or 60 times a second.

Right after you design me a chemical lead acid battery that outputs 50Hz AC.

I was hoping that somebody here would be able to find some way to have light pumped back and forth to the sun at 50 or 60 times a second.

It's actually possible using a unified variant of the standard theory / dark sucker theory light bulb concept, but in reverse.

https://astro.uni-bonn.de/~dfischer/dark_sucker_2.html

LOL!!!

There is a way to get AC directly off the solar panels, but it blanks off about 1/2 the area of the array. You must have rapidly opening and closing blocks to the array from the sun (or a rotating window disc). Maybe the OP wants to use the DC motor to spin the disc, but then will have AC. What a conundrum! LOL

DC motors are just fine, but the system voltage must match well the DC input voltage required or the motor will not live as long.

What about the car alternator? Thats AC till it runs through some diodes!!

You should read this weblink:-

Advantages_of_HVDC_over_AC_transmission

Interesting eh?

I disagree with many of your comments about both AC & DC.....

Your last sentence was probably the best part, though some here will disagree with it still!!

That's not entirely the case.

DC actually suffers less from distance transmission than does AC and can be cheaper to transmit. The advent of decent static inverters has seen a large increase in the use of HVDC for distance transmission. There are a number of these installations here in Australia and an increasing number world wide as it becomes a more viable option.

There are a few reasons for this,

One is "skin effect', where an AC current self induces a voltage in the wire and causes the current to only use the outer surface of the conductor. This requires the installation of much larger conductors than would be required for the same DC current level. Many transmission lines are hollow due to the skin effect. DC uses the entire cable cross sectional area.

Another is line capacitance, a huge amount of energy is wasted as the transmission lines continually charge and discharge in concert with the prevailing frequency, this is even a greater problem with closely spaced cables as in undersea cables etc.

No need to transmit 3 phases so, apart from using smaller cables, one can be omitted completely.

DC transmission causes less frequency interference and suffers less corona loss and stability problems.

Compatibility with other supply systems. By simply adapting the terminal equipment to suit, a supply from one area can be modified and synchronised with any other voltage or frequency of adjacent systems, thus greatly improving whole system reliability.

Due to the step up and down abilities of AC, power is still generated in AC form and also distributed to the consumer as AC. AC is a better form for end use for reasons such as switchability, motor type, etc.

Maybe Edison will be the victor after all, or at least they will call it a draw.

I guess you are referring to inductive reactance here, but since Xl = 2πƒL, and ƒ is frequency, and frequency in a DC circuit is 0, then Xl will also be 0.

It is generally accepted that instantaneous start current in an uncontrolled DC motor will be a product of resistance only. How long that current remains at a high level will then depend on load characteristics.

No, applying a step voltage into the series combination of an ideal inductor and the coil resistance will have zero current at t0(+). In five time constants of Τ=L/R the current will reach 99% of the locked rotor current condition of V/R if the rotor remains stationary. Frequently the rotor moment of inertia, and bearing friction combined with magnitudes of L and R will cause this maximum current to be achieved briefly even with no mechanical load attached to the rotor shaft. Once the rotor starts to turn, a back EMF will reduce this current.

You should notice that we do not know a single one of any of the attributes I just mentioned.

The 2(Pi)(f) expression is good for a pure sine f at steady state, but on application of DC voltage the inductive reactance will still come into play. As a load, the motor must be very predominantly inductive over resistive or the energy put in would result in mostly heat rather than motion.

You're correct in that the motor is highly inductive, but until rotation occurs, the resistance is the prime opposition and, if you were to prevent rotation for any period of time, you would quickly see how much heat that developed.

Yes, because no commutation ocurrs and the electromagnetic field is effectively at saturation in 5 time constants, but that's not normal motor operation.

I'm not entirely sure of your point, however it's the very fact that the core has nearly reached saturation before commutation occurs that makes starting current so high, this is especially the case with a loaded motor as the starting transient is prolonged.

While (almost) full saturation occurs in 5 electrical time constants (Te), over 60% saturation has been achieved in one Te, and 86% within only two Te, and as armature current rises much more rapidly than the armature can convert that energy to motion (the electro-mechanical time constant), so Te is considered to be zero and therefore so is armature inductance for the purposes of calculating starting current. This means that the calculation of starting current is restricted to I =E/R

The frequency is not zero. The change of current/voltage from the off to on state can be viewed as a frequency and will create inductive reactance.

Lots of posts but the real question is "what's the application"?

Trying to calculate the motor current or reading the motor datasheet is all well and good but if the motor is too small or too large for the application then this information may not be helpful. The motor could fail or the system could end up being over designed. Heck, the application may not even work properly at all without energy storage (if in fact there is none).

That, and were interested in the application.

So what is the motor being used for, and how big is it? The assumption so far is solar panel tracker, but I'm thinking solar powered DC bore water pump.

A motor is a prime mover. What machine it moves is secondary. Usually for machines with high torque, in higher horse power motors, slip rings and starters are introduced. For others, the speed does matter and there are pullies with differing diameters to achieve the same.

Why this repeated questioning about the application ?

Why? Because we understand Newtonian mechanics.

More information is necessary to answer mohgamal201365's question.

Application is a big part of it but as I also mentioned in the post further data is required, both to estimate the current and to check the motor and its application in general being proposed will actually work.

This guy is playing like god.

I assume the last comments follow sequence.I'm not trying to design a Solar System, just get in my viewpoint ( screwpoint ? ). I was not willing to let the Locked Rotor solution carry the day in the discussion and offered what I thought would be a parting shot. If I had been able to think of a smoother way of presenting it I would have used it. Maybe I harbour lingering resentment about times when proposals of mine were rejected. Stupidly, if you want to know. Swallowing arbitrary decisions may be necessary but I'm willing to pay the price for protest. Let's call arbitrary a lack of rational consistency as agreed upon by reasonable persons.Anyway, it's clear you don't like the style, how about the substance ? I think it's generally accepted that discussion benefits from a bit of the adversarial.BTW I doubt the OP could get a better answer.

[SHEILA!! WHERE DID YOU PUT THOSE DAMN LIGHTNING BOLTS?!!]

But a very important point in a design, is how to handle starting current in such a manner that it does not cause damage to other components.

The people who come here and ask questions often do not have even very basic knowledge or enough training to understand how things really work.

That can be frustrating for them. They don't like the answer given them as it does not fit in with their thoughts!!

Its a bit like the old UK (before GPSs) joke:-

Old Farmer, leaning over a gate, is asked by a passing Tourist "What direction do I need to drive in to get to London?"

The old Farmer replies "If I wanted to drive to London, I would not start from here!"

In a solar System design, particularly with motors, there needs to be a way of either supplying the starting current required, or reducing it or both.....So if you come here with such a naive question, you need to be able to take a joke.....

If you cannot take a joke, you are in the wrong place and it would be better to go back to school (its only relatively simple physics by the way, almost any high school will cover it) and learn the basics......learn to walk before trying to run.....

I spent in total, probably more than 10 years leaning, a lot while working, even before I went to the Uni (still working a full time job!).....I was 29 when I started there......

The required background and knowledge are not learnt in five minutes.....!

Having resentment when told the truth is about the most stupid reaction I have ever heard....even being told your "baby" is ugly, if it is, should be acceptable....or simply leave here and find a better way to learn for you....here is not really the place to try and replace 5 years at the Uni, its a place where you know and understand 95% of what you want to achieve and need help for a small problem or two....

The people here are just trying to do the job you gave them......

Furthermore, the way many beginners formulate their questions is genuinely laughable, and promotes problems as people may simply laugh at the presentation.....

The the questioner is upset as if there is a 2 minute fix not being given and all the (wrong) ideas will suddenly work!!!

Remember, there are DC motors, with very little metal in the rotor, that the difference between the resistance and the impedance is very low......

I have to admit that I haven't done any studies recently on such things (say like 30 years or more!), but a simple test, with a storage scope displaying current verses time, connected to a lab power supply, might give you a big "eye opener" for example.....one does need to see the "facts" with ones own eyes....it then often becomes suddenly clearer...

I was always a "hands-on" person myself, always wanting to "see" how things worked and I taught my classes in the same way, seemingly with success....my Guys, after training, could find and fix problems on the equipment I had taught them on far far better than those trained in the company school in the USA. Simply because I was also working in the field as tech support and they were professional trainers that had never ever been on a customer site.....

Also, some had learnt a machine only by "rote" and did not understand the whole system.....

And I bet there are some UK'ers who are miffed because they don't get to use that line much anymore.

I certainly agree with the value of hands-on. For me it expanded my thinking greatly and forced me to rethink what I thought I knew.

I don't know a lot about DC motors ( or AC ) but I know that there is a big range of kinds with a big range of characteristics. I tried to make the point earlier that trying for a rule-of-thumb was of little use for that reason. Another poster's estimate for starting current ( based I would guess on experience ) of 150% of running current seems a lot less than that of a rotor forced stationery. A design for a supply based on peak current would be a poor one, robust but having needless overcapacity.

For motors whose DC resistance is roughly equal to their impedance, starting current would not be an issue. Anyway, the setup I described would tell the story. ( probably improved with the right capacitor parallelled with the resistor )

And I'll thank you to not apply your flawed sense of aesthetics to my baby.

LOL!!