View Full Version : voltage loss over long runs
wireties
15-02-2015, 10:32
Lots of tall robots this year! Several online calculators tell me I lose 1.8V in a 30' long 10 AWG wire with a 30A load.
Question: do I really lose two times 1.8V? 1.8V in the red wire and 1.8V in the black?
TIA
RufflesRidge
15-02-2015, 10:42
Nope, 1.8V is correct if the length of the run is 30ft. (i.e. 30ft. of + and 30ft. of -).
The resistance of 10AWG is .001 Ohms per ft. * 60ft. = .06 Ohms
.06 Ohms * 30A = 1.8V
rich2202
15-02-2015, 20:15
Basically, you have +12 -> R1 -> load-> R2 -> ground.
But, R2 is sinking a lot of current, which is in essence ground, and not really resisting the current.
However, if you are saturating the wire, then "pushing" power to ground could result in voltage loss needed to push the power to ground.
Just some thoughts from a Math major, and not a EE major.
MrForbes
15-02-2015, 20:22
"ground" is not something we really have on FRC robots.
Good question, I think you lose voltage on each wire, but I don't' really know, since I'm a mechanical guy. How about measuring the voltage at both ends of the wire run, under load? or measure the voltage across each of the wires, under load.
This is something we get to deal with on underwater robots. I expect that 148 gets to deal with it on their tether, too. On the underwater robots, with a surface mounted battery, we run a higher system voltage, so a specific voltage drop is a smaller percent of the total voltage available. And with a higher voltage, there is less current for the same amount of power, so the voltage drop is less also.
But they don't let us run four 12v batteries in series in FRC....only NURC let us do that.
Basically, you have +12 -> R1 -> load-> R2 -> ground.
But, R2 is sinking a lot of current, which is in essence ground, and not really resisting the current.
R1=R2, and the current is the same through both because they are in series; so the voltage drop across R2 is the same as the drop across R1. R2 is "resisting" current just as much as R1 is.
How about measuring the voltage ... across each of the wires, under load...
You'll get then same voltage drop across each one.
Lots of tall robots this year! Several online calculators tell me I lose 1.8V in a 30' long 10 AWG wire with a 30A load.
Question: do I really lose two times 1.8V? 1.8V in the red wire and 1.8V in the black?
TIA
You shouldn't need to run 30' of wire for robot height; 78" each way is the maximum length you'll need due to height, even if you put the PDP at the floor. (Sprawl and routing will add additional length requirements, of course.) We have a robot over 6' tall, but I believe all of our motor runs are less than 3' from the PDP to the motor (and another 3' back), because all of our motors are down low on the chassis; all the electricity above the control board is for lights or limit switches. Keeping the motors low also helps reduce the likelihood of tipping over, as well.
The easy way to "see" the voltage drop is with a multimeter. Connect the leads to each end of the wire in the circuit and then operate the circuit. The multimeter will measure the drop across that section of the circuit. Alternatively you can connect the leads at the motor and see the voltage drop that occurs across the motor terminals. Yes it is not easy to do that with a motor or controller that has wires exiting it instead of terminals.
Al Skierkiewicz
16-02-2015, 07:59
Keith,
By now you have gathered that the current flows in both wires so both wires contribute to the voltage drop while running at 30 amps. If your motors stall at a greater current, then the voltage drop changes for that increased current. This was a question on the FCC First Class RadioTelephone License test. You were asked to determine the voltage drop to the tower navigation light at the top of a 1000 ft. tower. One of the answers was the drop in one wire, the correct answer included the voltage drop in both wires.
wireties
16-02-2015, 08:46
Thanks all - I appreciate the feedback. I was a little confused that the online calculator were giving an answer close to the observed results but I felt that the loss through the return was to be included. It turns out that those calculators (when you tell them it is a DC voltage) include the return distance by default.
wireties
16-02-2015, 08:48
The easy way to "see" the voltage drop is with a multimeter. Connect the leads to each end of the wire in the circuit and then operate the circuit. The multimeter will measure the drop across that section of the circuit. Alternatively you can connect the leads at the motor and see the voltage drop that occurs across the motor terminals. Yes it is not easy to do that with a motor or controller that has wires exiting it instead of terminals.
We set up a test with a TalonSRX (which will tell you the supply voltage) at the end of the wire and clearly saw the drop. Making the results jive with the online calculators was the real question I reckon. Doing the actual math solved all questions and CD opinions verified the results.
Al Skierkiewicz
17-02-2015, 07:42
Keith,
I have talked about wire losses and voltage drop for years here on CD. One of the easy to remember items is the "wire foot". Simply this is a simple calculation you can make in your head. As previously stated, #10 wire is .001 ohms per foot. Well, at 100 amps, that one foot chunk of wire will drop exactly 0.1 volts. In terms of 'wire feet' here are some other equivalents.
#6= 0.5 WF/ft.
#12=2 WF/ft.
#14=3 WF/ft.
Victor=6 WF
Talon and Jag=4 WF
Battery=11WF
0.1 volt drop/WF/100 amps
So all you have to do is just add up the losses in WF and that will get you an approximation of the loss (keeping in mind both wires need to be counted).
As an example, #6 wiring from battery to PDP is typically 1 foot from battery to APP connector and 2 feet from there through the main breaker to the PDP. So that is 6 feet of #6 at 0.5 WF=3WF or 0.3 volts at 100 amps. Add 11 WF for the battery and you get 1.4 volts. Typical FRC robots at worse case will draw 500 amps so 5 x 1.4 volts=7 volts drop at the input of the PDP. And yes, that is real, the PDP will only see 5 volts at the input with a fully charged battery. That is why the PD of old had the boost/buck power supplies that were capable of running down to 4.5 volts.
rich2202
17-02-2015, 09:13
You shouldn't need to run 30' of wire for robot height
Not tall, but go look at Team 148 (batman thread). They have a 30 foot or longer tether. Drive wheels are on the robot with the battery, but the tethered piece has motors for lift.
Not tall, but go look at Team 148 (batman thread). They have a 30 foot or longer tether. Drive wheels are on the robot with the battery, but the tethered piece has motors for lift.
This falls under the heading of sprawl.
(Sprawl and routing will add additional length requirements, of course.)
Jon Stratis
17-02-2015, 11:17
In 2008 we saw a significant voltage drop to the motors on the ends of our elevator. There are really two days around this... Go to a larger wire (see Al's post) or double up your wire (two equivalent resistors in parallel cut the voltage drop in half). At the time, we had plenty of the wire we were using, but would have had to buy thicker wire, and didn't want to sacrifice flexibility with the thicker wire (or pay a lot more for high flex wire). So, we doubled up the wire to each motor. Each Victor had 2 positive and two negative wires going from it, and they were then rejoined when they went into the motor. The inspector we got that year was a bit confused by it, and had to call Al over to check it out and make sure it was legal. As it was still on controller powering one motor, it was.
wireties
17-02-2015, 12:51
This falls under the heading of sprawl.
Or you get a number using 30' and divide by 5 or 10 - you get a better number for the shorter run that way.
Or you get a number using 30' and divide by 5 or 10 - you get a better number for the shorter run that way.
?:confused:?
wireties
17-02-2015, 13:53
?:confused:?
The relationship of the effect on a longer run scales (almost linearly) to a shorter run - not true? Impedance per unit length is a key component of the equation, correct?
Not tall, but go look at Team 148 (batman thread). They have a 30 foot or longer tether. Drive wheels are on the robot with the battery, but the tethered piece has motors for lift.
This falls under the heading of sprawl.
Or you get a number using 30' and divide by 5 or 10 - you get a better number for the shorter run that way.
?:confused:?
The relationship of the effect on a longer run scales (almost linearly) to a shorter run - not true? Impedance per unit length is a key component of the equation, correct?
This is true, and I wasn't arguing that it wasn't. I was just having difficulty parsing your earlier quote, and especially how that might not mean that the Batman and Robin tether was a case of sprawl.
wireties
17-02-2015, 14:27
This is true, and I wasn't arguing that it wasn't. I was just having difficulty parsing your earlier quote, and especially how that might not mean that the Batman and Robin tether was a case of sprawl.
No worries, several different contexts here I reckon - your OP used 3' and 6' as examples thus the 5 and 10 divisor.
John_freemen
18-04-2015, 23:59
Coming back and reading this thread I can't help but laugh at the fact that 1296 ended up having a tethered bot.
wireties
19-04-2015, 01:55
I get teased about it quite often. ;o)
With some of the long wire runs we are seeing this year, you may also want to watch out for the weight of your wiring. We have been through multiple cycles of weight reduction and I have concluded that we could have laid out our electronics better and dropped some excess weight. I used a kitchen scale and found that 18 feet of 12AWG wire weighed 0.494 pounds. That was about how much wire we would have needed to add a pair of motors up front for an active intake mechanism.
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