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Easier on my electrical system, if I set my amp below max?

qdeathstar

Completely Serious
May 17, 2019
4,061
4,315
VB
Yes you are really drawing less power. The dimmer switch analogy is not correct either, as a dimmer switch increases the resistance to a light bulb, that changes the total draw of the circuit according to Ohms Law, it does not dissipate the entire amount in heat, although they do often get warm.


Dimmers switch off and on a triac, they don’t change the resistance of the light bulb. The bulb dims because it is on for only part of the sinewave and isn’t seeing the peaks of the AC voltage waveform.

Dimming your lights will save your energy..

Would be interesting to see how telsa limits the current through the onboard inverter.
 
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ChadS

Last tank of gas: March 2009
Jul 16, 2009
3,381
2,824
Redmond, WA
Years ago Tomsax measured energy delivered vs energy drawn for various L2 amperages on the Roadster, which went from 120V/12A to 240V/70A. His post is HERE.

240V/40A was the sweet spot, although up to 70A was extremely similar. 240/32A was also extremely close, but you started to lose efficiency as you got slower than that. 240V/24A was still no big deal, but you start to notice the efficiency loss by 240V/16A. 120V/12A was really pretty bad. Most likely due to overhead of running pumps and such while charging.

Other cars may have different types of overhead (and the chargers could be different too), so it's hard to say exactly how applicable this is. I suspect newer cars may not run pumps as often at low charge rates (I think the Roadster just ran it all the time). But even if the exact numbers differ, I would imagine the general trend is pretty similar.

Note that I am only discussing efficiency. Of course there are other issues, like how long it takes to finish the charge, how if affects the battery, how easy it is to stay in a TOU band, how it affects household load, etc.

FWIW I charge my cars at 240V/32A. Both of them are sharing a 60A circuit so they might get less than that, but I try to time it so they generally aren't charging at the same time. At our old house they were on separate circuits (including a 100A circuit for a car that could charge at 80A) but I still charged at 32A or 40A there too.
 
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  • Informative
Reactions: eevee-fan
Reducing your current by 50% means your line losses go down by 75%. That's the law ;)

View attachment 509779

I think I did the calcs when debating whether swapping out my 30A circuit for an unused 40A circuit would result in higher efficiency. Resistance of 10AWG copper wire for 30A circuit is 0.001 ohms/ft. At 24A, line losses are (I^2)*R = 0.576 W/ft. At 12A, line losses are a quarter = 0.144W/ft. The run from my panel to my outlet is 60 ft, so losses are 34W at 24A and 8.4W at 12A.

So yes, you save 75% of the line losses, or 25W. But you double the amount of time charging with fixed overhead of at least 200W, used by the car pumps and electronics.

What's the net takeaway? If your daily charge takes 6 minues or less, by all means charge at 12A instead of 24A to be more efficient overall. If you daily charge takes more than 6 minutes, should be more efficient to charge at highest amps allowable. But if the run from your panel to your outlet is very long, like say a quarter mile or more, then the tradeoff may differ :)
 

eevee-fan

Active Member
Dec 2, 2019
2,035
2,663
Nevada
... interesting... here's an online calculator.

Wire Gauge Size and Resistance Calculator - Inch Calculator

Voltage Drop Calculator - Inch Calculator

Have not taken EE class. Is this right?

#6 copper wire @ 5 ft
50A circuit
245V power
To charge 10 kW

@ 12A, voltage drop of 0.059v, so 0.71W loss
= 3.4 hours, so 2.41W loss on the wall.
= Add 200W loss per hour, so 682.31W loss

@ 16A, voltage drop of 0.0786v, so 1.26W loss
= 2.55 hour, so 3.21W loss
= Add 200W loss per hour, so 511.26W loss

@ 24A, voltage drop of 0.1179v, so 2.83W loss
= 1.7 hour, so 4.81W loss
= Add 200W loss per hour, so 342.83W loss

@ 32A, voltage drop of 0.1572v, so 5.03W loss
= 1.28 hour, so 6.44W loss
= Add 200W loss per hour, so 261.03W loss

Going from #6 wire to #4 wire, resistance drops from .0020 to .0012 ohm (so 40% less W loss on the wire)
 
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Tessaract

Member
Aug 12, 2019
337
338
Ottawa
Dimmers switch off and on a triac, they don’t change the resistance of the light bulb. The bulb dims because it is on for only part of the sinewave and isn’t seeing the peaks of the AC voltage waveform.

Technically, the bulb dims because the average power over the portion of AC sine wave cycle where the triac is on is reduced. Even if the peak of the AC waveform is passed, the average power can still be reduced.

Would be interesting to see how telsa limits the current through the onboard inverter.
The current is limited by the onboard charger (which is technically not an invertor because the battery requires DC, not AC) reduces the average battery charge voltage which reduces the battery charge current, which in turn reduces the AC input current to the selected value.
 

Tessaract

Member
Aug 12, 2019
337
338
Ottawa
... interesting... here's an online calculator.

Wire Gauge Size and Resistance Calculator - Inch Calculator

Voltage Drop Calculator - Inch Calculator

Have not taken EE class. Is this right?

#6 copper wire @ 5 ft
50A circuit
245V power
To charge 10 kW

@ 12A, voltage drop of 0.059v, so 0.71W loss
= 3.4 hours, so 2.41W loss on the wall.
= Add 200W loss per hour, so 682.31W loss

@ 16A, voltage drop of 0.0786v, so 1.26W loss
= 2.55 hour, so 3.21W loss
= Add 200W loss per hour, so 511.26W loss

@ 24A, voltage drop of 0.1179v, so 2.83W loss
= 1.7 hour, so 4.81W loss
= Add 200W loss per hour, so 342.83W loss

@ 32A, voltage drop of 0.1572v, so 5.03W loss
= 1.28 hour, so 6.44W loss
= Add 200W loss per hour, so 261.03W loss

Going from #6 wire to #4 wire, resistance drops from .0020 to .0012 ohm (so 40% less W loss on the wire)

Pretty close!
  • Wire resistance varies a little bit depending on the number of strands in stranded wire (which 6AWG wire used in residential wiring most certainly is) My table gives 7-strand AWG at .38 Ω/kft, which is "close enough for government work" to the .395 that your online calculator gives.
  • 5 ft is an awfully short run of cable to a HPWC or NEMA plug. Mine is more like 50ft, and voltage drops and power losses will be proportionately higher.
  • Voltage drop = I * R. Power loss is i^2 *R. For your 32A example, voltage drop would be 32* 2*.395*5/1000 = .126 V. Power loss would be 32^2 * 2*.395*5/1000 = 4.05 W.
  • upload_2020-2-11_12-34-20.png
  • Energy loss should be expressed in kWH, not kW.
  • I think the constant losses per hour are likely lower, though. I've measured 41W with Sentry off, 272W with Sentry on. I've not measured during charging, though.
 

Attachments

  • AWG sizes.pdf
    247.3 KB · Views: 4

electrongeek

Metrology Fanboy
Nov 1, 2019
69
74
Maine
... interesting... here's an online calculator.

Wire Gauge Size and Resistance Calculator - Inch Calculator

Voltage Drop Calculator - Inch Calculator

Have not taken EE class. Is this right?

#6 copper wire @ 5 ft
50A circuit
245V power
To charge 10 kW

@ 12A, voltage drop of 0.059v, so 0.71W loss
= 3.4 hours, so 2.41W loss on the wall.
= Add 200W loss per hour, so 682.31W loss

@ 16A, voltage drop of 0.0786v, so 1.26W loss
= 2.55 hour, so 3.21W loss
= Add 200W loss per hour, so 511.26W loss

@ 24A, voltage drop of 0.1179v, so 2.83W loss
= 1.7 hour, so 4.81W loss
= Add 200W loss per hour, so 342.83W loss

@ 32A, voltage drop of 0.1572v, so 5.03W loss
= 1.28 hour, so 6.44W loss
= Add 200W loss per hour, so 261.03W loss

Going from #6 wire to #4 wire, resistance drops from .0020 to .0012 ohm (so 40% less W loss on the wire)

You actually have to double those wire loss figures, because the electrons are coming in on one wire and leaving on another. Also, even if your run from load center to charger is only 5 feet, you also have to take into account losses from your high voltage transformer all the way to your charger, although much of that will be with larger gauge wiring. To know what is really happening take a look at what your car says the input voltage is at different charge rates - that should reflect all of the losses in the circuits.
 

Tessaract

Member
Aug 12, 2019
337
338
Ottawa
You actually have to double those wire loss figures, because the electrons are coming in on one wire and leaving on another. Also, even if your run from load center to charger is only 5 feet, you also have to take into account losses from your high voltage transformer all the way to your charger, although much of that will be with larger gauge wiring. To know what is really happening take a look at what your car says the input voltage is at different charge rates - that should reflect all of the losses in the circuits.
If you check my post carefully, you'll see that I did that.
 

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