PV Student

Thanks for a really good prep class, Sean and Heatspring! I also got my certificate on the first try.

A:

That is excellent that you passed the NABCEP PVIP!! Glad to help!

Thanks for the feedback,

Sean White

Dr. White, I received my certificate today! Thanks so much for the great course, I will definitely recommend it to others who are prepping for the exam.

A:

That's great! Congratulations!!

Hello Sean,

I passed the exam. I got the certification yesterday. I am so happy I passed it first time trying. Your class was a great help.

Thank you so much!

A:

Congratulations!!

Post exam comment:

Hi Sean,

I did it. I think I can pass. Thanks for your help. I am glad I got this training. It helped me a lot. I think it would have been impossible if i haven't taken this course.

Yesterday, during the exam, I don't think there was any thing related to rapid shutdown, or feeders and taps.

There was one question about sub panel connection.

One of the questions is on the Sample of Nabcep Exam from NABCEP study guide.

Thanks again for your help!

Response:

That is good news. I do my best to prepare you for the exam and it is rewarding to get feedback like this from you and from all of the others.

Let me know when you get that NABCEP PV Installation Professional certificate in the mail!

Thanks,

Sean White

Q:

I am having trouble grasping an outside sourced practice question:

A 225 kW(DC) commercial system with a single 225 kW(AC) central inverter is being installed. The module specifications are:

Maximum power (Pmax): 200 W

Voltage at maximum power (Vmp): 28.9 V

Current at maximum power (Imp): 6.93 A

Open-circuit voltage (Voc): 36.2 V

Short-circuit current (Isc): 7.68 A

Maximum system voltage: 600 V

Maximum series fuse rating: 15 A

The grounded PV array is broken into subarrays, and each subarray is combined in a separate fused string combiner box that feeds a separate pair of PV output circuit conductors to the DC input bus of the inverter. Each ungrounded PV output circuit conductor connects through one pole of a three-pole DC safety switch rated at 100 A per pole to a 100 A fuse on the inverter's DC input bus. The string combiner boxes may hold up to 14 fuses.

What is the MAXIMUM number of module strings that may be connected to each fused string combiner???

I think the answer would be 8... but there is a lot going on in this question so I am not sure. Please help!

A:

This question is reality and too much information, so it is an exercise at sifting out the erroneous information, such as the rating of the inverter, voltages, etc.

What it comes down to is what would be the maximum number of strings that you can put on a 100A fuse.

The fuse is sized from 156% of Isc

Isc = 7.68A

7.68A x 1.56 = 12A

100A fuse / 12A = 8.33 strings

The answer is 8, so you have it right!

You are ready to pass the NABCEP PVIP Exam Saturday. I will bet pizza on it.

Thanks,

Sean White

Q:

Hello Sean,

About Question 33, if the building has less than 1000amp service, can we do the load side connection on subpanel?

Thanks!

A:

The answer is no, because the subpanel is on the supply side of the of the 1000A ground fault breaker.

In the NEC it says:

705.32 Ground Fault Protection: Where ground fault protection is used, the output of an interactive system shall be connected to the supply side of the ground fault protection.

One of the reasons is that if there were a ground fault on the supply side of the main breaker, the fault current could go to the inverter rather than tripping the ground fault breaker.

Thanks,

Sean White

Q:

on question 52, looking at the derating part for the number of conductors in the conduit on table 310.15b3a we have two source circuits which is 4 conductors plus an equipment grounding conductor. The table shows 4 to 6 which at the end of the day the derating would be the same for 4 or 5 but you only consider 4. Could you explain why you don't count the 5th equipment ground conductor?? The footnote states the count shall not include conductors that that are connected to electrical components but cannot be simultaneously energized.?? Thanks

A:

For table 310.15(B)(3)(a) Adjustment Factors for More Than Three Current Carrying Conductors, we only carry current carrying conductors, which means that we do not count equipment grounding conductors (EGC).

This is because current carrying conductors, when carrying current generate heat that needs to be dissipated.

In one way, you can look at a non-current carrying conductor (EGC) as a heat sink. To a small degree, an equipment grounding conductor could transmit heat out of the system as a heat sink (this is in theory).

The 310.15(B)(3)(a) that is not a table and states:

"More Than Three Current-Carrying Conductors.

Where the number of current-carrying conductors in a raceway or cable exceeds three, or where single conductors or multiconductor cables are installed without maintaining spacing for a continuous length longer than 24" and are not installed in raceways, the allowable ampacity of each conductor shall be reduced as shown in table 310.15(B)(3)(a)."

310.15(B)(3)(a) says:

"A neutral conductor that carries only the unbalanced current from other conductors of the same circuit shall not be required to be counted when applying the provisions of 310.15(B)(3)(a)"

This means that when we have a neutral that is not part of a 2 wire circuit, that we do not need to count the neutral. This would apply mostly to ac circuits for PV systems or perhaps a bipolar circuit where all of the conductors of both monopole subarrays are in the same raceway.

The footnote under table 310.15(B)(3)(a) that refers to circuits that are not simultaneously energized, would apply to circuits where either one circuit is energized or the other. If we have a backup generator and a double throw switch with 2 sets of conductors in the same raceway, then as with a double throw switch, either one set or another set of conductors will be energized, but not both at once. We should not have to count both sets of conductors as generating heat, of only one at a time would be energized.

Thanks,

Sean White

9 days until exam date!!

Now is the time to dive into the exam questions that we have here and to not procrastinate. The NABCEP PVIP Exam is famously difficult, which is why NABCEP Certificates are known as being the top of the field. Pass this exam and your salary will triple in 3 years, guaranteed! (It happened to me).

I recommend not getting overly concerned with anything obscure and especially not searching for answers on facebook.

Seriously, the best practice now is going over the practice questions here and the practice questions that go with the book and the detailed video answers to the questions.

This is not the exam that you will cram for the weekend before and pass, unless you were already ready before that weekend.

Onward!!

Sean White

Q:

Sean,

On Question 51 can you explain why we use the high cell temperature over the high ambient temperature?

Thanks

A:

When we are determining voltage, what is important is how hot or cold the solar cell is, not how hot or cold it is outside. That is why airflow leads to better production. BIPV has less airflow than a rooftop, which has less airflow than a ground mount, which has less airflow than a pole mount. The more airflow, the better the cooling, the better the voltage, power, energy and return on investment!

On of the misleading things about PV is that it is sold by its rating at Standard Test Conditions, which is 25C cell temperature, 1000W/square meter and 1.5 Air Mass spectrum of light. For STC, it does not matter what the wind speed would be, because the test condition calls for a 25C cell temperature. This test is misleading, because when there is 1000W/square meter irradiance, it is probable that the cell temperature will be considerably hotter than 25C, even if the ambient temperature was 20C. This is why we have other testing conditions, such as Performance Test Conditions (PTC), which is 20C ambient temperature, 1 meter per second wind speed, 1000W per square meter irradiance and 1.5 Air Mass spectrum of light. Also, California Energy Commission (CEC) test conditions is the same conditions as PTC, however the conditions are determined differently, so the results are slightly different.

PTC is always less than STC, because at 20C ambient, the solar cells will always heat up more than 25C cell temperature. Often the increase in temperature is about 30C for a quick estimate, so expect your solar cells to be about 55C at PTC with the ambient temperature at 20C.

To convert to F from C take

25C x 9 = 225

225 / 5 = 45

45 + 32 = 77F

In other words...

9/5 x C + 32 = F

To go from F to C

68F - 32 = 36

36 x 5 = 180

180/9 = 20C

In other words...

(F - 32) x 5/9 = C

In other words 1.8 change in F is 1 change in C and 32F = 0C

The reason we use ambient for the cold temperature instead of cell temperature is to be on the safe side and that when it is super cold, often it is windy without direct sunlight. We are just expecting ambient temperature to possibly be cell temperature when it is super darn cold.

Why do they use 25C for STC if it is misleading and makes our PV look better than it really is over time? Because that is probably the temperature of an average solar warehouse and it makes the PV sound better. If they originally made STC 55C, then they would have to heat up the testing area to sauna temperatures and the people at the solar factory would quit their jobs.

Thanks for the good question!!

Sean White

Q:

Hello
regarding question 6 explanation. I understand how you arrived at the 20 A breaker for the 100 A busbar. I am not familiar why you then took the 20 A breaker divided by 1.2 to arrive at 16 A.

A:

A breaker has to be sized based on 125% of the current of the device (and then round up). This comes from 690.9(B) of the NEC, which states:

"Overcurrent device ratings shall be not less than 125 percent of the maximum currents calculated in 690.8(A)"

This is usual throughout the NEC for other equipment that is not solar.

This means that if we are starting with the maximum breaker size, we can divide the breaker by 1.25 to get the maximum inverter size for the breaker (simple algebra).

Another shortcut is instead of dividing by 1.25, we can multiply by the inverse of 1.25, which is 0.8

1/1.25 = 0.8

20A x 0.8 = 16A

(this you can do in your head, since 8 x 2 is 16)

The reason the NEC wants us to not put a 20A inverter in a 20A breaker is to prevent false trips of the breaker, because in theory a 20A breaker should trip at 20A.

If you have a 20A inverter and wanted to determine the breaker size required, then:

20A x 1.25 = 25A

If you have a 25A inverter and you wanted to size the breaker:

25A x 1.25 = 31.25A

They don't make a 31.25A breaker, so you round up to a 35A breaker.

If you are over 800A you do not round up, but under 800A you do.

Thanks,

Sean White