PV System Design Question

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Calif.C10

Member
Location
N. California
Is there anything that states you can't design a PV Module system that will produce enough KW that will exceed the capacity of the Inverter?

Specifically I have been asked to design a one-line and calcs in that a PV field will produce 77 KW-DC and feed into an Inverter rated for a maximum out put of 50 KW-AC. I know there will be about a 15% loss after the inverting process but it will still exceed the maximum. This doesn't feel right to me but I am just the guy doing the drawings for the guy paying.
 

gndrod

Senior Member
Location
Ca and Wa
Is there anything that states you can't design a PV Module system that will produce enough KW that will exceed the capacity of the Inverter?

Specifically I have been asked to design a one-line and calcs in that a PV field will produce 77 KW-DC and feed into an Inverter rated for a maximum out put of 50 KW-AC. I know there will be about a 15% loss after the inverting process but it will still exceed the maximum. This doesn't feel right to me but I am just the guy doing the drawings for the guy paying.
Trust your instincts by consulting the manufacturer's requirements. 8j
 
The inverter should be labeled with the max DC operating current.

My text shows an example of a DC max of 710A and a max AC output (cont.) of 625A. DC max voltage is 600 and AC max voltage is 230 volts.

That means we can put up to 426 kW in with a max output of up to 143.75 kW.

So just looking at kW, there is a lot of cushion for the DC max.

Without seeing the label, it seems you could very well be within the specs of the inverter with quite a bit of room to spare.
 

broadgage

Senior Member
Location
London, England
As others post, subject to compliance with the manufacturers instructions, this is probably accecptable.
The inverter wont be harmed by a PV array that can produce in excess of the inverter nominal rating, they are designed to automaticly limit output in such circumstances.
Not however IMHO a good design choice. PV modules are relatively expensive and it would normally make economic sense to provide an inverter that can utilise all or very nearly all of the PV production.
A PV array of slightly more than the inverter rating could make sense. A 50KW inverter might have losses of 2.5KW and could therefore fully utilise a 52.5KW array. An array of say 55KW would in practice only produce more than 52.5KW for a few hours a year and could make sense.
An array of significantly more than the inverter rating, implies the waste of quite a bit of potentialy valuable energy.
 

gar

Senior Member
110523-0943 EDT

I see it somewhat differently.

The maximum current, maximum power, capability of the array has nothing to do with the inverter rating other than optimization. Because an array has a capability of 77 KW does not mean it produces 77 KW. The load on the array determines the power supplied by the array up to the maximum capability of the array.

Consider a battery and various load resistors. The power from the battery is determine by the load resistance connected to the battery. Infinite load resistance and the power supplied by the battery is zero. A 1 ohm load on a 12 V battery is a load of 144 W.

Suppose you desire to get as much power from the PV system as is practical, but at no time need more than 50 KW. Then put as large an array at the input to the inverter as the budget will allow. This will extend the amount of time that 50 KW is produced from the inverter. The reason is that under lower than peak light conditions more energy can be harvested because of the larger surface area.

.
 

ron

Senior Member
110523-0943 EDT

I see it somewhat differently.

The maximum current, maximum power, capability of the array has nothing to do with the inverter rating other than optimization. Because an array has a capability of 77 KW does not mean it produces 77 KW. The load on the array determines the power supplied by the array up to the maximum capability of the array.

Consider a battery and various load resistors. The power from the battery is determine by the load resistance connected to the battery. Infinite load resistance and the power supplied by the battery is zero. A 1 ohm load on a 12 V battery is a load of 144 W.

Suppose you desire to get as much power from the PV system as is practical, but at no time need more than 50 KW. Then put as large an array at the input to the inverter as the budget will allow. This will extend the amount of time that 50 KW is produced from the inverter. The reason is that under lower than peak light conditions more energy can be harvested because of the larger surface area.

.
I agree. The inverter will imit the output to its kW rating by controlling the PV string.
 

gndrod

Senior Member
Location
Ca and Wa
Integrated design

Integrated design

Wouldn't an integrated inverter/controller for each PV module circumvent shading, angle and be optimum for grid interface efficiencies?
 

tallgirl

Senior Member
Wouldn't an integrated inverter/controller for each PV module circumvent shading, angle and be optimum for grid interface efficiencies?
Depends. It all comes down to dollars -- micro-inverters are best for difficult environments (shading, multiple array facets), but cost more. Large string inverters are best for simple environments where very large amounts of power are needed with system voltages other than what the micro-inverter folks provide.

Another issue with the OP's system is the overall wisdom of running an inverter at the nameplate rating for prolonged periods of time. I have clients who run large arrays (9KW+ DC) into inverters with less capacity (~7KW AC) and I spend a lot of time worrying about cooling and how the installers can make sure they aren't whacked with service calls or warranty claims when some poor inverter decides it didn't like running at Pmax for hours and hours on end.

My vote -- don't do it. Stick with the manufacturer's maximum recommended array size, which will be some small percentage more than the maximum AC power output.
 

philly

Senior Member
110523-0943 EDT

I see it somewhat differently.

The maximum current, maximum power, capability of the array has nothing to do with the inverter rating other than optimization. Because an array has a capability of 77 KW does not mean it produces 77 KW. The load on the array determines the power supplied by the array up to the maximum capability of the array.

Consider a battery and various load resistors. The power from the battery is determine by the load resistance connected to the battery. Infinite load resistance and the power supplied by the battery is zero. A 1 ohm load on a 12 V battery is a load of 144 W.

Suppose you desire to get as much power from the PV system as is practical, but at no time need more than 50 KW. Then put as large an array at the input to the inverter as the budget will allow. This will extend the amount of time that 50 KW is produced from the inverter. The reason is that under lower than peak light conditions more energy can be harvested because of the larger surface area.

.
In a standard system the amount of power extracted from the source depends on the overall load impedance and as the load impedance changes the power required from the source changes.

With the PV system being the source in this case does the input impedance of the inverter as seen by the PV array change in order to vary the amount of power required by the PV array? If so how does the inverter change its impedance since it is not a variable resistance but rather (6) IGBT's. How would these IGBT's change the source imput impedance or system impedance of the inverter? Does this have to do with their pulse width modulation?

Also with most systems and with the battery example you referenced the souces are a constant voltage source where the load power is determined by the source voltage squared divided by the system impedance -( Vsource^2/Rsys). Since the PV array is a constant current source does the power extracted from the array get determined by (P=Iarray^2 * Rinv)
 

tallgirl

Senior Member
A PV array isn't a constant current source. That's why.

The power from a PV array is varied by changing the input voltage of the device. Between Voc and Vmpp current changes fairly rapidly. Between Vmpp and 0, the current changes more slowly from Impp to Isc.

So, if an inverter or charge controller is close to it's maximum output, it will raise the input voltage from the array until the current or watts (or whatever -- depends on the device) is back within the permitted range.

For example, if an inverter with a maximum output of 5KW AC is connected to a 7KW DC nameplate rated array, the input voltage from the array will be raised above Vmpp as soon as the inverter is producing 5KW. As the sun heads back down, the input voltage will be lowered back towards Vmpp.

It's magic, but once you understand the magic it all makes sense.
 
A PV array isn't a constant current source. That's why.

The power from a PV array is varied by changing the input voltage of the device. Between Voc and Vmpp current changes fairly rapidly. Between Vmpp and 0, the current changes more slowly from Impp to Isc.

So, if an inverter or charge controller is close to it's maximum output, it will raise the input voltage from the array until the current or watts (or whatever -- depends on the device) is back within the permitted range.

For example, if an inverter with a maximum output of 5KW AC is connected to a 7KW DC nameplate rated array, the input voltage from the array will be raised above Vmpp as soon as the inverter is producing 5KW. As the sun heads back down, the input voltage will be lowered back towards Vmpp.

It's magic, but once you understand the magic it all makes sense.
OK, so I had to jump ahead 3 chapters to get to Vmpp. Now, my gurus have taught me that maximum power is delivered when the impedance of the load matches the impedance of the source. The impedance of a battery is called 'internal resistance'. Also, in radio the same applies. A radio with a 50 ohm internal impedance will provide maximum transfer into a load that represents 50 ohms of impedance.

So, why don't we use the same logic with PV systems? Wouldn't that simply things a bit?
 

philly

Senior Member
A PV array isn't a constant current source. That's why.

The power from a PV array is varied by changing the input voltage of the device. Between Voc and Vmpp current changes fairly rapidly. Between Vmpp and 0, the current changes more slowly from Impp to Isc.

So, if an inverter or charge controller is close to it's maximum output, it will raise the input voltage from the array until the current or watts (or whatever -- depends on the device) is back within the permitted range.

For example, if an inverter with a maximum output of 5KW AC is connected to a 7KW DC nameplate rated array, the input voltage from the array will be raised above Vmpp as soon as the inverter is producing 5KW. As the sun heads back down, the input voltage will be lowered back towards Vmpp.

It's magic, but once you understand the magic it all makes sense.
O.k. but the inverter cannot just simply ask the array to change its voltage in order to raise or lower its input voltage. The inverter or inverter/load combination must somehow change its characteristic as seen by the array. I believe this characteristic would be impedance and that by dynamically changing its impedance it causes a voltage change in the modules and arrays.

This sounds like a perfect opportunity for one of Gar's excellent examples that usually drive the concept home for me. :)
 
O.k. but the inverter cannot just simply ask the array to change its voltage in order to raise or lower its input voltage. The inverter or inverter/load combination must somehow change its characteristic as seen by the array. I believe this characteristic would be impedance and that by dynamically changing its impedance it causes a voltage change in the modules and arrays.

This sounds like a perfect opportunity for one of Gar's excellent examples that usually drive the concept home for me. :)
No, but the source impedance can change. In radio, when we have a 50 ohm transmitter and a 75 ohm antenna, a reactive matching device (not so correctly called a tuner) is put between the radio and the antenna so we can match the impedance for maximum power transfer. Tube type radios were able to do this internally as the tuning of the radio changed the internal impedance to match the load (antenna).

Internal resistance and source impedance is a difficult concept to grasp. It's almost like the values were created (not measured) to satisfy a math formula. The internal resistance of a battery is very dynamic and changes drastically depending on how much charge the battery has.
 

rgomes26

Member
I've talked to multiple inverter manufacturers about this issue. None of them have a problem with an array that can produce more dc amperage than the max input....

BUT, ALL of them strongly agree that you MUST not exceed the max voltage from the array to the inverter. An array that supplies voltage in excess of the max voltage rating of the inverter may cause catostrophic and dangerous failures.
 
I've talked to multiple inverter manufacturers about this issue. None of them have a problem with an array that can produce more dc amperage than the max input....

BUT, ALL of them strongly agree that you MUST not exceed the max voltage from the array to the inverter. An array that supplies voltage in excess of the max voltage rating of the inverter may cause catostrophic and dangerous failures.
Good point and since the cells are connected in series, it would be easy to do such a thing.
 

gar

Senior Member
110525-1644 EDT

The following comments are to try to answer several questions.

A PV cell is not a linear device. It can not be characterized as a simple voltage source and constant internal impedance. See
http://www.icpress.co.uk/etextbook/p276/p276_chap1.pdf Fig 1.4 .
http://zone.ni.com/devzone/cda/tut/p/id/7230 Multiple good sections.
http://en.wikipedia.org/wiki/Solar_cell General good information.

Maximum power transfer occurs in the knee of the V-I curve. But this knee varies as the light intensity and temperature change.

You have a multitude of characteristic curves for the PV cell. Called a family of curves. Select one for you to see what happens when an inverter is the load. If the output of the inverter can put out all the power it can extract from the PV array to somewhere, then the PV array becomes the limiting factor in the available power.

The inverter will search for the maximum power point of the array, and operate at this point. Basically the inverter adjusts its apparent input impedance to load the PV array to this maximum power point. Consider the inverter as a variable resistance in conjunction with a wattmeter. The knob on the resistor is adjusted to keep the power meter at a maximum value. This might involve dithering the knob about this point. Since the maximum power point and apparent internal impedance can be changing as the light intensity varies it is constantly necessary to search for this point.

By pulse width modulation or some other scheme the inverter changes its apparent load on the PV array.

If the inverter does not need all of the power that the PV array could produce, then the inverter effectively raises its input impedance. This lowers the load on the PV array and the PV voltage increases as a result of the lower load. The PV array voltage does not increase because the inverter forces the voltage higher, but rather because of the characteristics of the PV cell with less load.

You could also lower the PV array power output by lowering the resistance load on the PV array. But there is no object unless you want to increase power dissipation within the solar array. Would it be useful for snow melting? I don't know. In other words if you want to melt snow, then do you short the array output terminals?

,
 

BillK-AZ

Senior Member
Location
Mesa Arizona
How most utility interactive inverters work

How most utility interactive inverters work

A little background on utility interactive inverters should help in understanding the situation of this thread. This applies to most of the inverters, but some may differ. An understanding of power factor and vars is assumed in this forum.

The inverters are controlled by a micro processor (or several microprocessors) that coordinates the processing of the DC input and the formation of the AC output.

The input from the PV array is processed by a boost or buck switch mode conversion section that has at least three functions. If the design is a boost mode, it boosts the input voltage to a constant higher voltage that is used by the AC output section. If it is a buck mode, it reduces the input to the desired voltage. In either case, the conversion is controlled by feeding a variable width pulse to the switch mode device that produces the desired output voltage. The second function is to adjust this pulse width as the PV array I-V characteristic changes with temperature and irradiance so that the array is operated at the voltage that produces the most power (maximum power tracking). This is done by dithering the pulse width slowly and calculating the power produced. If the power increased over the last setting, an additional adjustment is made in the same direction until the power starts to fall off. The process is continued as long as all the power can be used by the AC inversion stage. If the maximum AC power is reached, then the third function starts and the pulse width is changed to limit the power by increasing the operating voltage of the array above the maximum power point as much as is needed to keep the power in limits. If the array is larger than needed, the operating voltage could in theory reach open circuit voltage of the array at which point there is zero power.

This same procedure is used during inverter start-up to slowly increase the power.

The AC output is sort of the reverse of this with switch mode power devices in a bridge producing the intended AC sine wave output by careful and fast operation of the switch mode power devices.

This process is intended to produce unity power factor by timing the injection of current on the micro-second scale to match the voltage waveform. It could be designed to do otherwise, but most inverters are designed for unity power factor.
The actual situation is not quite so simple, there are filters (inductive and capacitive) on the AC output to reduce the high frequency noise produced by the switch mode conversions, these filters can have an effect on power factor, especially at low power levels.

I have observed some Fronius inverters on 480V/3ph drawing 1.5 amps per phase with the DC off. These are leading vars and generally cancel the more common lagging vars on the utility. This does not happen with the Fronius 11.4 KW inverters on 208/3ph that must have a different filter design.

It is important that the maximum voltage at the DC input not exceed the ratings of the switching devices and filters. It takes only microseconds of high voltage to do major damage to inverters. Most of the newer inverters measure this voltage and record the time/date if limits are exceeded. Even if the power stage of the inverter is burnt, this data is preserved for warranty analysis.
 

philly

Senior Member
110525-1644 EDT

The following comments are to try to answer several questions.

A PV cell is not a linear device. It can not be characterized as a simple voltage source and constant internal impedance. See
http://www.icpress.co.uk/etextbook/p276/p276_chap1.pdf Fig 1.4 .
http://zone.ni.com/devzone/cda/tut/p/id/7230 Multiple good sections.
http://en.wikipedia.org/wiki/Solar_cell General good information.

Maximum power transfer occurs in the knee of the V-I curve. But this knee varies as the light intensity and temperature change.

You have a multitude of characteristic curves for the PV cell. Called a family of curves. Select one for you to see what happens when an inverter is the load. If the output of the inverter can put out all the power it can extract from the PV array to somewhere, then the PV array becomes the limiting factor in the available power.

The inverter will search for the maximum power point of the array, and operate at this point. Basically the inverter adjusts its apparent input impedance to load the PV array to this maximum power point. Consider the inverter as a variable resistance in conjunction with a wattmeter. The knob on the resistor is adjusted to keep the power meter at a maximum value. This might involve dithering the knob about this point. Since the maximum power point and apparent internal impedance can be changing as the light intensity varies it is constantly necessary to search for this point.

By pulse width modulation or some other scheme the inverter changes its apparent load on the PV array.

If the inverter does not need all of the power that the PV array could produce, then the inverter effectively raises its input impedance. This lowers the load on the PV array and the PV voltage increases as a result of the lower load. The PV array voltage does not increase because the inverter forces the voltage higher, but rather because of the characteristics of the PV cell with less load.

You could also lower the PV array power output by lowering the resistance load on the PV array. But there is no object unless you want to increase power dissipation within the solar array. Would it be useful for snow melting? I don't know. In other words if you want to melt snow, then do you short the array output terminals?

,
O.k. so for example lets say that the inverter starts out presenting a zero impedance to the array by somehow adjusting its pulse width to present a zero impedance. With the inverter presenting an input impedance of 0 then the modules in the array will be at thier Isc short circuit current and the voltage will be at zero and therefore the corrosponding power will be at zero.

As the inverter input starts to raise its impedance the current begins to drop slightly as the voltage increases. This occurs until the peak power point it reached at which point the impedance adjustment "dithers" around this point to keep the maximum power.

Why does the current only decrease slightly as the impedance is raised until it reaches the knee of the VI curve where it starts to drop rapidly.

Also why does the current decrease at all I thought a constant current source was able to supply a constant current? Or is this due to some of the shunt resistance decreasing as the voltage is increased.

Why is the voltage increaseing as the current is increaseing? Isn't the voltage across a current souce a function of the current and the load impedance?
 

gar

Senior Member
110602-2018 EDT

philly:

In the ideal world a constant voltage source would output exactly a constant voltage independent of the load, except if the load is another ideal constant voltage source of a different voltage, or an ideal zero resistance load.

In an ideal world a constant current source would output exactly a constant current independent of the load, except if the load is another ideal constant current source of a different current, or an ideal infinite resistance load.

In the real world these ideal devices do not exist. Some devices may approximate the ideal over a limited range, and others will be less than ideal. But in a loose fashion we often refer to something as a constant voltage or current source if the device approximate an ideal one. It might be a very loose association.

A pentode vacuum tube or a bi-polar junction transistor approximates a constant current source. This has certain ramifications in an audio amplifier with a transformer output. Some batteries are a useful constant voltage source.

If you look at the curve in figure 4 for a PV cell you will see a region that approximate a constant voltage source, and another that approximates a constant current at http://zone.ni.com/devzone/cda/tut/p/id/7230.

O.k. so for example lets say that the inverter starts out presenting a zero impedance to the array by somehow adjusting its pulse width to present a zero impedance. With the inverter presenting an input impedance of 0 then the modules in the array will be at thier Isc short circuit current and the voltage will be at zero and therefore the corrosponding power will be at zero.

As the inverter input starts to raise its impedance the current begins to drop slightly as the voltage increases. This occurs until the peak power point it reached at which point the impedance adjustment "dithers" around this point to keep the maximum power.
Yes.

Why does the current only decrease slightly as the impedance is raised until it reaches the knee of the VI curve where it starts to drop rapidly.
Because that is the characteristic of the PV cell.

Also why does the current decrease at all I thought a constant current source was able to supply a constant current? Or is this due to some of the shunt resistance decreasing as the voltage is increased.
See my above discussion on real world current sources.

Why is the voltage increaseing as the current is increaseing?
Because it should based on ohm's law.

Isn't the voltage across a current souce a function of the current and the load impedance?
Yes. And that is the answer to the first part of your one paragraph question that I just split into two.

.
 

philly

Senior Member
110602-2018 EDT

philly:

In the ideal world a constant voltage source would output exactly a constant voltage independent of the load, except if the load is another ideal constant voltage source of a different voltage, or an ideal zero resistance load.

In an ideal world a constant current source would output exactly a constant current independent of the load, except if the load is another ideal constant current source of a different current, or an ideal infinite resistance load.

In the real world these ideal devices do not exist. Some devices may approximate the ideal over a limited range, and others will be less than ideal. But in a loose fashion we often refer to something as a constant voltage or current source if the device approximate an ideal one. It might be a very loose association.

A pentode vacuum tube or a bi-polar junction transistor approximates a constant current source. This has certain ramifications in an audio amplifier with a transformer output. Some batteries are a useful constant voltage source.

If you look at the curve in figure 4 for a PV cell you will see a region that approximate a constant voltage source, and another that approximates a constant current at http://zone.ni.com/devzone/cda/tut/p/id/7230.

Yes.

Because that is the characteristic of the PV cell.

See my above discussion on real world current sources.

Because it should based on ohm's law.

Yes. And that is the answer to the first part of your one paragraph question that I just split into two.

.
O.k. so it looks like a PV module is both a constant current source and constant voltage source over its range of operation.

For low load impedances the PV modules is esentially a constant current souce with the voltage across the module being dependent on the load resistanc and the current passing through it.

Near the max power point of the module the module switches from a constant current source to a constant voltage source. As the load impedance increaes further the constant voltage source current rapidly drops off until we it esentially reaches its Voc voltage.

What is it that causes the module to change from a constant current source to a constant voltage source around the peak power point? Is this due to the internal shunt resistance inside the module?
 
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