rocketmonkeys
Member
Frunple said:
I considered letting the house burn, but definitely doesn't have WAF. We'll figure out this power issue one way or another.
About to finish pre-wiring and need to wire for motorized blinds/shades. Which wire should I use?
- Thread starter rocketmonkeys
- Start date
I considered letting the house burn, but definitely doesn't have WAF. We'll figure out this power issue one way or another.
I would not run anything less than 16/2 for power unless you plan to have the power source within 10 ft of the motor. You'll have voltage drop issues with 22/4, double or not. Based on the windows size on your picture, you'll need sufficient torque to lift the shades. I don't know what motors you plan to install, but my Somfy motors can draw up to 1.5A at the peak.
rocketmonkeys
Member
picta said:
Frunple said:
Hmm, this is interesting. I finally got around to looking at real(er) numbers. Keep in mind - it's late, I'm sure I missed a lot, and I'm *definitely* a long way from college EE101 classes.
So 9.6v at the end w/ 22 AWG definitely seems bad. 11.4v on the high end (16AWG) is good. What about 11v? 10.8v? How do these motors work with low voltage? Do they just work slower, will they die earlier, will the RF not work?
Things look better at 1 amp:
I wonder if I could use a >12v power supply to alleviate this somewhat. I'd risk having higher voltage for closer motors than farther, but that's going to happen anyway. A 13v PS w/ 2x 22AWG would give similar results to using 16 AWG. Again, not sure how the motors respond to higher voltage (12.5v for 1A at the closest windows, 11.8v for the farthest).
We'll see once I get back in touch with the electrician. They were wiring the 22/4 already, we'll see what the best plan is forward. If I have to bite the bullet and wire an additional 16/2, so be it.
I've also thought about doing multiple power hubs. So instead of home-running the whole house to one place, I could home-run each floor to a central location, and put power supplies there. I can keep the control lines all home-run to the same location, but using per-floor power supplies might help w/ the max-distance AWG issue.
I take it the runs are 50'? That seems reasonable as 100' runs in a house would be a big house. Just noting that the 100' in your table is for total wire length for both wires which makes the actual run half that or 50 ft. Looks like you should really be using #16. May eliminate lots of issues later. It's hard to run wires later.
rocketmonkeys
Member
JimS said:
The tables above were for 100' @ 1.5A. I'm guessing I'll have some less than 100', maybe some above. The motor spec I saw said 1A, but I'm edging on the safe side by assuming 1.5A.
For the doubled-22 AWG, I just used the 1/R = 1/R + 1/R formula. But I double checked, and this random book excerpt from google search implies that I'm in the right neighborhood... basically, two identical AWG wires (ie 22) twisted together acts as a 19 AWG, so it seems like I'm in the right ballpark.
I also got a chance to see the 22/4 today... that wire is TINY! I think it's sinking into my brain now how small that stuff is, and I probably don't want to use it for anything power related (Yes, you guys are right). I'm going to figure out Monday what it'll mean to run an additional 16AWG pair to each location.
Is it possible to have all the 22/4 home-runned to a control panel in the garage, but have all the separate 16/18AWG lines run to power-supplies on each floor? That would probably make wiring cheaper/easier, but not sure if that'll cause more problems.
rocketmonkeys said:
I see the table is for 100 ft and the voltage drops look right for the different wire sizes but only for one conductor of the given length. Unless you doubled the length you have only figured drop for one conductor. If the runs are actually 100 ft end to end you need to double your voltage drops.
Try plugging the numbers into this calculator using the actual end to end length. It gives total voltage drop for both conductors:
http://www.supercircuits.com/resources/tools/voltage-drop-calculator
If you have a spot on each floor for the power supplies that seems like a good plan to reduce the length of power runs. Let the power supply float (don't ground the negative lead at the power supply) to avoid ground loops through your control wires and you should be fine with home running the control wires.
If the length end to end is 50 feet and the total length of both conductors is 100 feet then the numbers are correct for voltage drop.Frunple said:
The OP statement that the length is 100 feet isn't clear (at least to me) as to if it is end to end length or total conductor length (double end to end length). I took it to mean that the end to end length is 100 feet in which case the total conductor length is 200 feet and has twice the voltage drop.
I don't think the statement of "one pair (supply and return) at 100'" is completely clear either. Is this end to end length or total conductor length (giving an end to end length of 50 ft)?
He could have the right voltage drop for his situation. I really can't tell without some clarification on the length.
Frunple
Active Member
You're misunderstanding this.
A voltage drop is measured at the end of the wire run. So a 100' run of wire is 2 conductors at 100' each.
That's what those figures are for, and they are correct.
What you are saying is the same as saying you're measuring a voltage drop at the end of one conductor at 100'. You can't do this. No matter how long, how much power or the AWG of wire, the result will always be 0 because you need a closed loop for current to flow.
The figures are for a pair of wires at 100' each conductor.
A voltage drop is measured at the end of the wire run. So a 100' run of wire is 2 conductors at 100' each.
That's what those figures are for, and they are correct.
What you are saying is the same as saying you're measuring a voltage drop at the end of one conductor at 100'. You can't do this. No matter how long, how much power or the AWG of wire, the result will always be 0 because you need a closed loop for current to flow.
The figures are for a pair of wires at 100' each conductor.
Frunple,
I think I understand this pretty well. Just stating that you are correct several times does not make it so...
Of course a closed loop is needed for current. But you misunderstand what I am saying and are making up things you claim are the same. I could easily measure the drop of one conductor. Just double the cable back on itself so both ends are at the same place. Connect the load to one end (using both conductors so current loop is complete) and use a meter to measure the voltage from one end of one conductor to the other end of the same conductor.
You could always use a power supply and a load with a known length of known gauge wire and measure the voltage. That would show who is correct.
Or you could look at the google book link given by rocketmonkeys. On page 398 it states that for an 8 ohm speaker 500 feet from the amplifier connected with #19 wire that half the power would be lost in the wire.
On page 400 of the google book it gives the resistance of #19 wire as 8.1 ohms per 1000 ft (which is about halfway between #18 and #20 on the table given in this thread earlier so the tables agree). If half of the power is lost in the wire the resistance of the wire must equal the resistance of the load (the 8 ohm speaker). The resistance for the wire is the value for 1000 ft or twice the end to end length. The tables list resistance for a single conductor of the given length as I stated before.
I think I understand this pretty well. Just stating that you are correct several times does not make it so...
Of course a closed loop is needed for current. But you misunderstand what I am saying and are making up things you claim are the same. I could easily measure the drop of one conductor. Just double the cable back on itself so both ends are at the same place. Connect the load to one end (using both conductors so current loop is complete) and use a meter to measure the voltage from one end of one conductor to the other end of the same conductor.
You could always use a power supply and a load with a known length of known gauge wire and measure the voltage. That would show who is correct.
Or you could look at the google book link given by rocketmonkeys. On page 398 it states that for an 8 ohm speaker 500 feet from the amplifier connected with #19 wire that half the power would be lost in the wire.
On page 400 of the google book it gives the resistance of #19 wire as 8.1 ohms per 1000 ft (which is about halfway between #18 and #20 on the table given in this thread earlier so the tables agree). If half of the power is lost in the wire the resistance of the wire must equal the resistance of the load (the 8 ohm speaker). The resistance for the wire is the value for 1000 ft or twice the end to end length. The tables list resistance for a single conductor of the given length as I stated before.
Solid vs. stranded will have the same basic resistance if the cable construction and materials are the same. If you were talking about skin effect or other esoteric items, sure, but in actuality, solid is a better conductor:az1324 said:
For No. 12 AWG, Table 8 of the NEC does list two different DC resistances for stranded and solid copper. DC resistance per 1000 feet for solid is 1.93 ohms, for stranded the DC resistance is given as 1.98 ohms per 1000 feet.
Trivial differences for different conductor types.
az1324 said:
Very true. Hadn't really thought about the resistance difference between solid and multistrand wire before. Some stranded is less and some is more. Diameter varies more than I expected too. Here is a fairly comprehensive chart.
http://www.calmont.com/pdf/calmont-eng-wire-gauge.pdf
Still, for most applications the exact diameter and resistance is not quite THAT critical, IMHO.
