If I was powering a motor from two different power sources where would the power come from to run the motor?

Ah a picture is worth a thousand words:

And of course this leads to a thousand other questions.

First what would determine the "source" of power?

While the 120 volts ac will match the RMS range probable would not, does that matter?

During the day with good sun I'd like to use power from the solar array (I know I need a diode :-).

So what is the best way to do this.

The drawing is just for thought development so please add your own, with your thoughts.

Ralph

I would think you would need some sort of auto-switch like they use for natural gas generators. Something that would auto-disconnect the mains power and connect your solar setup when the solar power was adequate. Otherwise when it was not, it would connect the power company up.

Rick

P.S. Ralph, you mentioned something about Paul (Noter) not being around in another thread. I've noticed his absence as well for some time now. Do you know what happended to him? Is he just taking a break? I miss his informative posts. We didn't always agree on the way to get something done, but I always enjoyed reading and learning from his programs.

My brother forgot to disconnect his generator before switching back on to the grid once, it wasn't pretty. I assume the coils on the generator shorted.

The switch I believe is called an isolation switch and is just as important for the safety of the utility company's linesmen as it is for your inverter.

The isolation switch is to protect linemen working on downed wires.

A generator can run with live utility power, or at least I have never seen a problem. The problem might arrise from miss-phase but I believe that is auto correcting, again the strongest RMS prevailing.

They (the linemen) isolate the section of wire they are working on from the power grid but if a home owners generator is powering the lines they have no control over that so utilities require an isolation switch.

I'll draw a picture if necessary, but thanks for the feedback so far.

So what would be the power source of the motor, as far as I "know" the circuit I illustrated would be safe.

I can picture doing a isolated circuit with switches etc. but if everything was wired in line what would prevail?

I am thinking that the "strongest" RMS (Root Means Square) would be the power source.

Also I would like suggestions about isolated circuits, what might you picture?

I just think it is an interesting question about what would be the power source with everything inline.

The solar array would be generating no matter if it was directly consumed but where would it go if not consumed in the motor operations?

What would happen if I did this with batteries? Essentially they would be in series but which would power the motor or would they share the load 50/50?

These are the things I think of to keep me humble.

A simple question "what would be the power source in my illustrated circuit"?

One can picture two equal batteries as the "potential" power source.

Then possible picture a stronger and a weaker battery situation what happens then?

And of course all power sources are "potential" meaning that they "might" be the power source.

My head is swimming.

Ralph

Honestly Ralph, I don't know what would happen. I would guess that the higher voltage would attempt to equalize with the lower voltage backfeeding into it. I would think the safest (for your hardware) would be to have sensing circuitry on your solar side that would switch it in or out depending on power availability. This is all a guess though, I really don't know.

BTW, did I miss something with Paul??

Rick

So what might a isolation circuit look like?

I am looking at a (cheap) solar walkway light.

It has a Solar detector cell (damm I forgot what they are called but it is an analog solar detector) and they use a transistor plus of course the solar array.

I was thinking about possible switching the AC power using a triac so if the solar detector only detected so much (limited) solar radiation it would allow the AC source to power the motor other wise with a strong solar source detected the AC would be powered off and the solar array would be used.

Does that sound right?

Ralph

Will the motor be running at the time of switching? Or is it something like a sump pump that would intermittently run? I was thinking of some relays to toggle between your two voltage circuits. You could check the voltage at the inverter input and if it is adequate to power the pump, switch the power to solar, if not, switch to main power.

You probably could use triacs to turn the two on/off instead of the relay if that was what you chose.

BTW, I don't know if you check it or not, but could you please check your email for your website.

Well... I wouldn't connect a motor in this manner. The utility lines have a major advantage over the inverter that just happens to be wired in parallel to the utility lines. Since it is probably not a synchronized inverter, there will be a frequency difference and/or a phase difference. The utility lines will fry the inverter on short order.

If the motor is to run constantly, a load switching system would be required. It would be running on whichever is the primary source. Should the primary source fail, the circuit would be switched to the backup source.

The point is never to tie multiple AC sources together.

Grid-tie inverters are wired in parallel to the utility lines but their power is generally much lower than the ability of the utility lines. Typically, these inverters are connected on the consumer side of the KWH meter and can effectively stop and even reverse the direction of the meter. At this point, the motor wouldn't care where the power actually comes from. It would be powered by both sources pretty evenly. Only a true sine wave inverter can be used in this manner. The pseudo sine wave inverters actually put out a voltage in a modified square wave. Because of the signal level changes as the utility poser crosses certain voltage planes, unnecessary loading can occur.

I noticed the cheap solar circuit mention... I took one apart a while back and found it pretty interesting. No real switching between two sources because the solar cell is just to recharge the battery. In good sunlight, the power is diverted into the battery and a transistor to the LED is reverse biased. When the sun sets, the voltage level at the base of the transistor changes with the drop in cell output. at that point, the battery is essentially discharged to the LED.

Thanks, mongo now that is getting somewhere.

If I were to do a full blown solar array I would hook it into/upto the local utility and "Net Metering" would be utilized.

At times the meter would actually turn backwards as you said. So if I had a "true sine wave inverter" the utility power and solar array could be live.

I believe the key here is consumption if the solar array is putting out more power then is being consumed then that excess power is put out onto the utility grid, hence the meter turns backwards.

Yes the motor does not/will not care where the power comes from, but I do.

How would I conceptualize the source of the power in a parallel circuit?

So far I keep going around in circles.

The "potential" of the utility grid is certainly greater than the potential of the solar array yet the solar array can be the dominant source if the consumption is lower than the potential (solar array) i.e. the meter turns backwards.

This is so intriguing, I actually have a real situation where I'd like to use solar power and I could just do an isolated circuit but that would be to simple.

Ralph

If I were doing this, I would hook the solar cell up to a large battery. Measuring the voltage at the battery, you could tell if the solar cell is producing power, as the battery will discharge when it stops.

When the voltage drops below a certain point, just switch off the inverter and switch on the ac source. You could use 2 triacs, or you could use a relay with 2 contacts.

It is worth mentioning, that a 3 amp ac motor is 360 watts, and a 45 watt solar array could not power it continuously. It would have to charge the battery, run the motor, then charge the battery, run the motor... The motor will also require considerably higher current when starting, which could be too much for a small inverter.

Hello,

I have some experience with a type of commercial inverter called a "grid-tie" inverter. They are made to covert DC power from solar array (PV array) to AC and "sell" to the grid ("sell" means send power in the direction of the grid). They also will sync to the grid (their waveform will exactly match the grids before they try to start selling). They will also AUTOMATICALLY disconnect from the grid in a grid outage (called "anti-islanding"). They are safe. They are UL certified (to UL1741). They also have another very interesting feature called "Maximum Power Point Tracking (MPPT) beyond the scope here. Google it if interested.

The smallest ones I know about are ~200 Watts, the so-called "micro inverters" (for instance see Enphase M210). These are made to work with one commercial panel (around 200 Watts, 30 Volts).

Your AC motor requires 360 Watts (3 A @ 120 V), I should think you would want more than 45 Watts of solar. In any case, let's say you had 200 Watts worth of PV (solar panel) with a commercial grid-tie inverter. If your motor is running at 360 Watts, and the sun is full out and the PV is putting out a full 200 W, you will be "buying" 160 Watts from the grid (160W comes from the grid and 200W comes from the PV-inverter). As I said, the inverter sends power "toward" the grid. It is totally in phase with the grid. In actuality, it doesn't matter to the inverter whether the power in consumed locally or actually sending power into the grid (toward the tower in your drawing).

Let's summarize 2 different cases:

the ac motor is on and consuming 360W

• if sun is out, PV panel generating 200W --> you are buying net 160 W from the grid
  
• if sun is out but it is hazy, PV gen 100W --> you are buying net 260 W from the grid

ac motor is turned off and therefore not consuming any watts

• if the sun if full out, PV generating 200W --> you are SELLING 200W to the grid
  
• sun is partially out, PV gen 100W --> you are selling 100W to the grid

*** selling to the grid means your meter is spinning backward and subtracting from your net energy consumption. An inverter operating in this manner must be installed by a professional and inspected. The inverter must be UL certified to UL1741 (there are other regulations as well).

DRBOB thank you, it's getting closer to my actually understanding the answer to my question:

If I was powering a motor from two different power sources where would the power come from to run the motor? I am still a bit puzzled on the "why"!

But I like your explanation so far.

A problem I have with "most" "grid-tie inverters" is that there has to be AC power on the grid in order for them to function.

If there is a power loss they break the connection so that from my poor illustration the motor would not be energized even with full sun.

From a safety situation I understand why BUT it's not applicable to (my) real life usage.

I would still want to use the AC motor even if there was a power failure.

Of course having a either/or isolated circuit would alleviate this problem but then I would not be using a grid tie inverter but just a regular kinda one with a separate isolation switch at the utilities meter.

In a practical sense I can see why net metering works but I am still lost as to why.

Say we had:

if sun is out, PV panel generating 500W -->...

I would be selling 140W but why? Just because it's there?

Like I said I understand the practical application I just do not understand the WHY!

Thanks everyone for your input it really helps,

Ralph

Ralph
You ask really good questions. The "grid-tie" inverter sends power one way: outward bound. If it were just tied to the grid (no motor) it would send all this power to the grid. How can it do this? You might ask. One way to think of it is that the inverter generates a SLIGHTLY greater voltage than the grid thereby "pushing" power into the grid. Before you laugh and say "that's crazy, the little old inverter can't overpower the grid!", let me explain. The term "grid voltage" is a little bit misleading. Let's say it's 120V (RMS). Where is that measured? Let's say it's measured at the power pole. Between the power pole and your house is a circuit element(s).... the inductance and resistance of the intervening wires. If Ohms law is to be obeyed, and power is going into the grid, the voltage at the house MUST be greater than the voltage at the power pole. Since the impedances are small (small inductive reactance and small resistance) the voltages differences don't have to be very large to push a lot of power (P=V^2/R). Now, let's say you connect the 360W motor at the house. That motor will consume some of the available current output from the inverter. Now less power will be pushed into the grid (let's say the inverter is putting out 500W) because some is being consumed locally. Now the voltage drop is still positive from the house to the power pole but the difference is less. The voltage drop is V=IZ (Z = impedance) but now there is less I. It all adds up. Am I making any sense?
Bob
PS There ARE inverters that do exactly what you want. They can power local loads when the grid goes out. They also can sell to the grid when the grid is not out. However they DO need batteries, for reasons too complicated to explain in this PS. I will post more if anyone is interested. Check out Outbackpowersystems.com and look under "grid-interactive" to see these inverters.

DRBOB:

One way to think of it is that the inverter generates a SLIGHTLY greater voltage than the grid thereby "pushing" power into the grid.

That is the only way that I could picture for pushing power to the grid but assumed I must be wrong, that was why I was expressing my confusion.

This is the first time I have ever heard that the solar array/inverter would output a greater voltage than the grid, it makes sense but wow!!

I downloaded the documentation on the grid-interactive so I need to read up on these inverters.

Thank you so much, this really helps.

Please I'am interested (don't know about anybody else) so please post more more.

Please expand this thread to any relative length desired.

I would love to see some discussion on solar array pricing, I "hear" that prices have dropped but I do not "see" prices that much lower I must be looking in the wrong places where should I be looking?

I would so much love to go off the grid or at the least use net metering.

I can run my home off a 5000 watt generator, that's two refrigerators a pool pump an oil burner plus two tv's and house lighting.

Oh yeah a couple of computers.

I would love to go to a ground source heat pump in place of the oil burner.

Ralph

Hi Ralph

From previous discussion:
DRBOB:
One way to think of it is that the inverter generates a SLIGHTLY greater voltage than the grid thereby "pushing" power into the grid.
Ralph
That is the only way that I could picture for pushing power to the grid but assumed I must be wrong, that was why I was expressing my confusion.

Be careful here. One CAN'T "push the grid around". (most people think) It has very low impedance (e.g. close to an ideal AC voltage source). The ONLY reason you can generate a voltage "slightly greater" than the grid voltage is because there is an intervening INDUCTANCE (and a little resistance). Let's say there is 10 uH of inductance. At 60 Hz, that translates to only about 4 mOhm (inductive reactance). So you'd only get a 4 mV drop going from house to "grid"..... hard to even measure.

Bob
PS Since this is a "basic electronics" blog, I probably shouldn't post more "renewable energy" stuff here. There are already many renewable energy blogs out there. So far, though, the discussion was/is pretty much "basic electronics"

To connect to the grid, a synchronous inverter is required. It feeds excess power back into the grid, which in turn runs the meter backward or simply stops it. These are more commonly called "grid tie" systems and usually don't have a battery bank. They feed excess back during the daylight hours but you would still be using grid power at night.

Add the battery bank and you can charge them in the day, when you need very little power for things like lighting. A big enough system and you can run all of your appliances too, (just don't go for a record on how many at one time). At night, when all is dark and sleepy time arrives, you can still be selling power back to the utility company, at least until your battery reserve drops below a certain point. Then you go right back to using grid power. As the Sun comes up, the batteries begin charging. The loads increase as the day wears on but the batteries continue charging. Power is going back to the grid and all is right with the world.

Anyway, the motor will draw power from the highest voltage source. That is a trick I use a lot when using multiple power sources and diodes. The higher potential will be the source but id both sources are equal, they simply share the load. The load itself does not know which is which and it just happily runs along.

Oh, by the way, the 120 volt 3 amp motor load is more than the 45 watt PV supply can handle. It would need to be at least six times that capacity to compete with grid power. The load is essentially 300 watts. Now, we are getting big in size. My solar panels are capable of merely 250 watts at 12 volts and they are approximately 5 feet square, or 25 sq/ft in area.

Is the purpose of the synchronous inverter to sync the phase of the sinusoid wave to that of the grid?

That's exactly the reason, yes. A synchronous inverter uses the incoming frequency and matches it. This is to prevent mismatching or frequency shifts, which can destroy the output drivers of the inverter. Just imagine having the inverter on the negative swing of the sine wave while the line is on the positive swing. BANG goes the box and all the pieces.