# Voltage drop calc

#### electro7

##### Senior Member
Hi,

Voltage drop calculations are based on the max amperage a conductor will carry, and not on the max amperage plus the 125% multiplier for continously load, correct?

Voltage drop is also based on the temperature rating of the conductors being used and not on the termination temperature rating, correct? In other words, I use a calculator that asks for the conductor temperature rating. When using thwn-2 I enter 90° in that input even if I am limited by the termination rating of 75°, correct?

This is for a solar PV installation where the solar is considered a continuous load and the max output current of the inverter needs to be multiplied by 125% for conductor ampacity calculations.

Sent from my SM-G998U using Tapatalk

#### electrofelon

##### Senior Member
Hi,

Voltage drop calculations are based on the max amperage a conductor will carry, and not on the max amperage plus the 125% multiplier for continously load, correct?

Voltage drop is also based on the temperature rating of the conductors being used and not on the termination temperature rating, correct? In other words, I use a calculator that asks for the conductor temperature rating. When using thwn-2 I enter 90° in that input even if I am limited by the termination rating of 75°, correct?

This is for a solar PV installation where the solar is considered a continuous load and the max output current of the inverter needs to be multiplied by 125% for conductor ampacity calculations.

Sent from my SM-G998U using Tapatalk
I would say it's not even based on the max amperage a conductor can carry, it should be based on the actual load.

#### don_resqcapt19

##### Moderator
Staff member
Voltage drop is based on the actual current the conductor will be carrying and not on the ampacity of the conductor. The 125% adder is ficticious current and is not used in the voltage drop calculations.

The temperature used in the voltage drop calculation is the actual ambient temperature the conductor will be in. It is not based on the conductor or termination temperature rating.

If you have a 90°C conductor in an actual 90°C ambient, that conductor has an ampacity of ZERO.

Note that any votlage drop on your solar conductors reduces the kWH that you are actually getting out of the solar system.

#### electro7

##### Senior Member
Voltage drop is based on the actual current the conductor will be carrying and not on the ampacity of the conductor. The 125% adder is ficticious current and is not used in the voltage drop calculations.

The temperature used in the voltage drop calculation is the actual ambient temperature the conductor will be in. It is not based on the conductor or termination temperature rating.

If you have a 90°C conductor in an actual 90°C ambient, that conductor has an ampacity of ZERO.

Note that any votlage drop on your solar conductors reduces the kWH that you are actually getting out of the solar system.
Copy.

Thank you.

Sent from my SM-G998U using Tapatalk

#### wwhitney

##### Senior Member
The temperature used in the voltage drop calculation is the actual ambient temperature the conductor will be in. It is not based on the conductor or termination temperature rating.
But a combined calculator might also want to check the temperature corrected ampacity of the conductor to be sure it is sufficient for the load, in which case it would ask both for the (peak) ambient temperature and the conductor insulation rating.

Cheers, Wayne

#### electro7

##### Senior Member
But a combined calculator might also want to check the temperature corrected ampacity of the conductor to be sure it is sufficient for the load, in which case it would ask both for the (peak) ambient temperature and the conductor insulation rating.

Cheers, Wayne
I use this calculator on the "mobile electrician" app. See the attached screenshot.

Sent from my SM-G998U using Tapatalk

#### don_resqcapt19

##### Moderator
Staff member
65°F sounds on the low side for the ambient temperature of conductors for a solar system.

#### wwhitney

##### Senior Member
I use this calculator on the "mobile electrician" app. See the attached screenshot.
Ah, that just has one temperature input, so it wants the operating temperature of the conductor, so that it can use a temperature corrected resistance/impedance to give a more accurate answer.

I'm not aware of best practices to use in estimating the operating temperature. NEC Table 8 gives DC resistance based on 75C, so 75C would probably be quite conservative. Although in theory if you are loading a 90C insulated conductor up to its full temperature corrected ampacity, and it is embedded in thermal insulation or otherwise is very impaired at shedding heat, it could approach 90C operating temperature.

Cheers, Wayne

#### electro7

##### Senior Member
65°F sounds on the low side for the ambient temperature of conductors for a solar system.
This was for the underground feeder to the solar ac combiner.

Sent from my SM-G998U using Tapatalk

#### wwhitney

##### Senior Member
This was for the underground feeder to the solar ac combiner.
But it should be the temperature of the itself copper under load, not the ambient temperature. You can confirm this by just tweaking that temperature input a little and seeing the calculated voltage drop change accordingly.

Cheers, Wayne

#### electro7

##### Senior Member
But it should be the temperature of the itself copper under load, not the ambient temperature. You can confirm this by just tweaking that temperature input a little and seeing the calculated voltage drop change accordingly.

Cheers, Wayne
90° bumps it up very minimally

Sent from my SM-G998U using Tapatalk

#### wwhitney

##### Senior Member
90° bumps it up very minimally
As expected, but that's 90F, not 90C. Since the operating temperature effect is minimal, you could go ahead and put in something like 75C to be very conservative. And if you're happy with that, just ignore the rest of this post, it probably makes little practical difference.

Here's a reasonable way to get a high confidence if conservative upper bound on operating temperature when you just have 3 current carrying conductors or less in your feeder: Table 310.15 tells you that at 30C ambient, a 90C insulation #8 copper has an ampacity of 55A. That means that under worst case thermal dissipation conditions 55A flowing through #8 Cu could produce a temperature rise of at most 60C (90 - 30). Temperature rise varies as the square of current (that's the I2R power dissipation). So with 38.4 / 55 = 70% of the ampacity flowing through the conductor, the worst case temperature rise should be at most 70% squared = 49% of the allowable 60C rise, or 29C.

So if you know the soil temperature is at most, say, 21C, then you could pretty confidently say the conductor operating temperature is at most 50C. This assumes that the underground feeder configuration is within the worst case heat dissipation conditions contemplated in Table 310.15. That should hold true for a single underground conduit holding 3 current carrying conductors in it. If you have multiple underground conduits in proximity, I understand that may not be true, but the details of calculating temperature rise and thus ampacity for that situation are not something I'm familiar with.

Cheers, Wayne

#### electro7

##### Senior Member
As expected, but that's 90F, not 90C. Since the operating temperature effect is minimal, you could go ahead and put in something like 75C to be very conservative. And if you're happy with that, just ignore the rest of this post, it probably makes little practical difference.

Here's a reasonable way to get a high confidence if conservative upper bound on operating temperature when you just have 3 current carrying conductors or less in your feeder: Table 310.15 tells you that at 30C ambient, a 90C insulation #8 copper has an ampacity of 55A. That means that under worst case thermal dissipation conditions 55A flowing through #8 Cu could produce a temperature rise of at most 60C (90 - 30). Temperature rise varies as the square of current (that's the I2R power dissipation). So with 38.4 / 55 = 70% of the ampacity flowing through the conductor, the worst case temperature rise should be at most 70% squared = 49% of the allowable 60C rise, or 29C.

So if you know the soil temperature is at most, say, 21C, then you could pretty confidently say the conductor operating temperature is at most 50C. This assumes that the underground feeder configuration is within the worst case heat dissipation conditions contemplated in Table 310.15. That should hold true for a single underground conduit holding 3 current carrying conductors in it. If you have multiple underground conduits in proximity, I understand that may not be true, but the details of calculating temperature rise and thus ampacity for that situation are not something I'm familiar with.

Cheers, Wayne
Very in-depth. Thank you for that.

I think I will stick with 75°C to be conservative since there is minimal effect.

Sent from my SM-G998U using Tapatalk

#### ggunn

##### PE (Electrical), NABCEP certified
Hi,

Voltage drop calculations are based on the max amperage a conductor will carry, and not on the max amperage plus the 125% multiplier for continously load, correct?

Voltage drop is also based on the temperature rating of the conductors being used and not on the termination temperature rating, correct? In other words, I use a calculator that asks for the conductor temperature rating. When using thwn-2 I enter 90° in that input even if I am limited by the termination rating of 75°, correct?

This is for a solar PV installation where the solar is considered a continuous load and the max output current of the inverter needs to be multiplied by 125% for conductor ampacity calculations.

Sent from my SM-G998U using Tapatalk
None of that has anything to do with voltage drop. Vd is Ohm's Law, pure and simple. PV pushes current (I) through a wire of known resistance (R) from Table 8 for DC or Table 9 for AC, adjusted for length.

Vd = IR.

#### Solar Guy

##### Member
OK, there are several questions here.
(1) If you compare NEC 2020 table 310.16 with table 310.17, you can see that three #8 wires in a conduit on a 3-phase circuit are limited to 40A. But run a single #8 wire in open air, and you can push 60A through it. The difference is that, in the conduit, there is mutual heating of the conductors, plus no air circulation in the conduit, so the wires heat each other up and reduce how much heat they can dissipate. But run one conductor by itself in air, and it can dissipate much more heat, and thus carry more current before hitting the max temperature permitted by the insulation type.
(2) The limits in either table are used against the actual continuous current the circuit will carry, not on the maximum current permitted by the table.
(3) The 125% factor added to solar output circuits is based on the fact that some inverters can supply more than their rated capacity for a short period. Many solar systems are designed for an array wattage slightly larger than the inverter wattage, which means the system will produce its max power for a longer period and not just at the noon hour. In the past, this was a bad idea for two reasons: (a) the PV arrays cost more than the inverter, so would be a wasted expense, and (b) inverters running over their max output would clip the waveform and not deliver a good sine wave approximation. Both of these factors no longer strictly apply, because solar array costs are coming down and inverters have DC conversion controls and higher frequency square wave generators that look much more like a true sine wave. Then the normal line impedances and the utility transformer round off the edges of the square waves so the sine wave approximation is way better.
(4) The only way to get a real sine wave output is to use a rotary inverter, which is a DC motor driving an AC generator.
(5) It is really important to calculate the voltage drop in the solar circuit between the inverter and the service entrance, because that affects the inverter. If, for instance, the utility can actually deliver 240V at the service entrance, and the inverter is line-tapped right there, then the inverter will output ~240.5V so it can become the service's preferred source of power. If, however, the inverter is a long way off such that its output current causes a voltage drop of 2V in its own feeder, then it will generate ~242.5V to once again become the preferred provider. But eventually the inverter will hit some internal limit and shut down, because it can no longer push its power. Then the solar power is not harvested.
Hope this helps -

#### electrofelon

##### Senior Member
OK, there are several questions here.
(1) If you compare NEC 2020 table 310.16 with table 310.17, you can see that three #8 wires in a conduit on a 3-phase circuit are limited to 40A. But run a single #8 wire in open air, and you can push 60A through it. The difference is that, in the conduit, there is mutual heating of the conductors, plus no air circulation in the conduit, so the wires heat each other up and reduce how much heat they can dissipate. But run one conductor by itself in air, and it can dissipate much more heat, and thus carry more current before hitting the max temperature permitted by the insulation type.
(2) The limits in either table are used against the actual continuous current the circuit will carry, not on the maximum current permitted by the table.
(3) The 125% factor added to solar output circuits is based on the fact that some inverters can supply more than their rated capacity for a short period. Many solar systems are designed for an array wattage slightly larger than the inverter wattage, which means the system will produce its max power for a longer period and not just at the noon hour. In the past, this was a bad idea for two reasons: (a) the PV arrays cost more than the inverter, so would be a wasted expense, and (b) inverters running over their max output would clip the waveform and not deliver a good sine wave approximation. Both of these factors no longer strictly apply, because solar array costs are coming down and inverters have DC conversion controls and higher frequency square wave generators that look much more like a true sine wave. Then the normal line impedances and the utility transformer round off the edges of the square waves so the sine wave approximation is way better.
(4) The only way to get a real sine wave output is to use a rotary inverter, which is a DC motor driving an AC generator.
(5) It is really important to calculate the voltage drop in the solar circuit between the inverter and the service entrance, because that affects the inverter. If, for instance, the utility can actually deliver 240V at the service entrance, and the inverter is line-tapped right there, then the inverter will output ~240.5V so it can become the service's preferred source of power. If, however, the inverter is a long way off such that its output current causes a voltage drop of 2V in its own feeder, then it will generate ~242.5V to once again become the preferred provider. But eventually the inverter will hit some internal limit and shut down, because it can no longer push its power. Then the solar power is not harvested.
Hope this helps -
1. Not sure if there was a question here, but yes conductors in free air have a higher ampacity.
2. I don't understand what you are saying or asking here.
3. There are two 125% factors in PV source conductors, one for continuous duty, and the other due to possible output above standard test conditions. Inverter output circuits have just one 125% factor for continuous loading. I'm not aware of any inverters that will sometimes output more current than their name plate Max. An inverter clips the DC side not the AC side so there is no distorted sine wave from clipping.
5. IMO, the voltage rise issue seen by the inverter is rarely an issue, assuming you are dealing with reasonable efficiency when sizing your conductors. Maybe if you have a very very long run it would be most cost-effective to deal with a high voltage drop, in other words the losses will be less than the increased wire cost..... And then maybe you get to the point of worrying about the maximum voltage capabilities of the inverter. For some reason we do a lot of ground mounts that are 700 or 800 ft away, and I usually shoot for three or four volts rise. We have had inverters shut down to the high voltage on several systems, but it has always been a utility problem (bad voltage regulator) that just no one noticed. We could have spent twice as much on wire and gotten it down to two volts, but they probably still would have tripped off with that bad regulator.

#### ramsy

##### Roger Ruhle dba NoFixNoPay

Regarding the "mobile electrician" VD calculator app:
1.15% VD is way off for 38 Amps on #8cu, unless ambient is -464C below Zero.

VD calculators are getting between 9-10% VD, with Pf=1 at 18C Ambient in underground raceway.
I use this calculator on the "mobile electrician" app. See the attached screenshot.

#### Solar Guy

##### Member
Electrofelon, my post's point (2) was a worse attempt to recap what you said in your "Tuesday at 11:20am" post. You are right, and said it better.
The other point (3) hasn't changed in several iterations of the NEC, and back then inverters were all made overseas and some were cheap, both in price and performance. Today's inverters are substantially better, and deliver a good sine wave approximation with an absolute max power output limiter. So I'm not convinced the 125% over-rating rule needs to apply as much, at least not strictly. Perhaps the modern inverters that can track power factor might actually put out more V-A than Watts if the PF is way off unity, but in both residential and commercial jobs our building loads have been small resistive loads, a HVAC unit, and lots of electronic power supplies, such as with LEDs and computers and phones. I can't think of a job we have sold that went on a building with a ton of motors running at 80% PF that we had to track. But the NEC says do it, so we do.
As for (5), we too have done a lot of ground mounts quite some distance away from the building, and each time I check to see whether running the higher voltage DC on expensive conductors back from the array is better than locating the inverters at the array and running the lower voltage AC back on cheaper THWN2. It's installation cost versus keeping what you harvested as much as you can.

#### electrofelon

##### Senior Member
Electrofelon, my post's point (2) was a worse attempt to recap what you said in your "Tuesday at 11:20am" post. You are right, and said it better.
The other point (3) hasn't changed in several iterations of the NEC, and back then inverters were all made overseas and some were cheap, both in price and performance. Today's inverters are substantially better, and deliver a good sine wave approximation with an absolute max power output limiter. So I'm not convinced the 125% over-rating rule needs to apply as much, at least not strictly. Perhaps the modern inverters that can track power factor might actually put out more V-A than Watts if the PF is way off unity, but in both residential and commercial jobs our building loads have been small resistive loads, a HVAC unit, and lots of electronic power supplies, such as with LEDs and computers and phones. I can't think of a job we have sold that went on a building with a ton of motors running at 80% PF that we had to track. But the NEC says do it, so we do.
As for (5), we too have done a lot of ground mounts quite some distance away from the building, and each time I check to see whether running the higher voltage DC on expensive conductors back from the array is better than locating the inverters at the array and running the lower voltage AC back on cheaper THWN2. It's installation cost versus keeping what you harvested as much as you can.
Solar guy,

I still don't agree on the inverter output circuit. That is a continuous load and that is what the 125% is there for, it says it right in 690. On all inverters I have come across, the output current is a hard stop, although the max wattage can be a little higher, such as if the voltage is higher. I can't speak for the early early days of inverters, but I see nothing in the NEC providing a fudge Factor for higher than nameplate inverter output.

Regarding sending AC or DC back from a ground mount, there's a recent thread about this topic. My position and experience is that running DC back, although may seem like a better deal, does not pan out most of the time. Certainly it is plausible that someone may find DC to be better with different goals or different sets of conditions.

#### Solar Guy

##### Member
Electrofelon, I know what the book says, but the words "continuous load" aren't as obvious. 690.8 seems to apply to the DC side, for instance.
What I am trying to say is that even if the book says solar is a continuous load, it really isn't. Take a look at this solar performance screen shot on three cold February days:

The first day was clear blue sky, and the system produced the classic half sine wave. But its maximum power peaked for only part of one hour.
The second day, some clouds reduced the peak and cut a couple of notches in the sine wave.
The third day, we had snow and that 110kW system barely harvested enough power to brew my coffee.
That is simply NOT a "continuous load."
As I said, we follow the rule but I don't like it and don't think it accurately reflects what really happens. The notorious intermittence of solar and wind harvesters is what drives the 24/7/365 power folks nuts. They need to have a lot of sine curves like day 1, OR zero-days like day 3.. There is no way to throttle up their massive steam turbines to react to the cloud blips in day 2, and line impedances, automatic tap changers, and the like may not even allow them to see those blips.