I need to provide 12V DC to a DC brushless motors that need 12V 30A at distance of 30m from the source.

The cables I can us are 20 Gauge, I know that the cables are loosing power, how can I calculate the power that I need to provide from the source to the motors cable in V DC and current that I need to provide at distance of 30m from the motors?

  • 1
    \$\begingroup\$ All of the answers so far are saying its impossible. You're saying that you can use 20 gauge cables. Can you use dozens of 20 gauge cables in parallel? That makes this more plausible, but still impractical vs. sending a high voltage down the cables and stepping it down to 12v at the end. \$\endgroup\$ – Bryan Boettcher Jan 21 '13 at 15:50

20 gauge wire has a resistance of about \$ 33.31m\Omega / m \$. For a 20 meter cable, it's effectively a resistor \$33.31\frac{\Omega}{m} \cdot 20m \approx 1 \Omega \$. Your circuit is this:


You want 30A through the motor, and since this is a series circuit, that means 30A through R1 and R2, also. The voltage drop over a resistor is given by Ohm's law: voltage equals current times resistance:

\$ V_{R1} = 30A \cdot 1\Omega = 30V \$

So, you lose 30V in R1, and another 30V in R2, so you will need 60V plus whatever voltage is required at the motor to get 30A. Worse, power equals the product of voltage and current. We know the voltage across R1 is 30V, and current is 30A, so power lost in the wire is:

\$ P_{R1} = 30A \cdot 30 V = 900 W \$

For comparison, a typical electric heater is around 1000W. You have two wires, so your total losses are 1800W. It's quite likely you will trip a circuit breaker if you can magically prevent the wires from bursting into flames, and we haven't even powered the motor yet.

There are two obvious solutions:

  • make the wires shorter
  • make the wires fatter

If you can't do either of those, there's a less obvious solution:

  • keep the power the same by raising the voltage and reducing the current

This is the solution the electric company uses to avoid huge losses in power transmission. If you combine the two previous equations, you can see that the power through a fixed resistance can be calculated through the current alone:

\$ P = I^2 R \$

12V at 30A is 360W. 360V at 1A is also 360W, and in theory, capable of producing the same mechanical power at your motor. But, the losses in your wires will be much less, by minimizing the \$I^2\$ term above.

To do this, look for a motor that operates at a higher voltage, or put a mechanism for converting the voltage near the motor.


This is not possible. The problem is that there will be a voltage drop over the wires. You'll have to use a power supply with a higher output voltage. This calculation shows how much volt you'll need:

You'll have to calculate the resistance of the cables. When you know how much resistance 1m cable has, you can calculate the total resistance by multiplying with 2*30. You can find here that 20Gauge has a resistance of 33.31Ohm per kilometer, so 0.03331 per meter. Thus:

$$R_{cables} = 0.03331 \cdot 60 = 1.9986\Omega$$

After that:

$$R_{motor} = \frac{U}{I} = \frac{12}{30} = 0.4\Omega$$

$$R_{total} = R_{motor} + R_{cables} = 0.4 + 1.9986 = 2.3986\Omega$$

$$U = I \cdot R_{total} = 30 \cdot 2.3986 = 71.958V$$

You see this is pretty much. I'd recommend you to place the power supply near the motor or to use bigger wires.

  • 3
    \$\begingroup\$ While not part of the original question, even if the voltage were increased the cables would consume roughly 80% of the total power required. About 4 times more than the motor itself, or about 1,400 watts! While it is super hard to calculate how hot the cable would get, I think it is safe to say that it would get very hot and could easily be a fire risk. Getting the power supply close to the motor, and using thick cables, is the correct solution. \$\endgroup\$ – user3624 Jan 19 '13 at 16:18
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    \$\begingroup\$ +1 for "use bigger wires". According to the table you cited, 20 gauge isn't rated for 30 amps anyway. Switch to 10 gauge wire (or thicker), and the result looks a lot more sensible. (Ps. When I first saw this answer, I tried to figure out where the arithmetic error was, since the result just looked too absurd to be true. Of course, there isn't any, it's just that the parameters given by the OP are absurd to begin with.) \$\endgroup\$ – Ilmari Karonen Jan 19 '13 at 16:20
  • \$\begingroup\$ Both very true! Minor note for clarity: switch to at least 10 gauge, means not more than 10. \$\endgroup\$ – user17592 Jan 19 '13 at 16:22
  • \$\begingroup\$ Thanks! I actually use a special silicon cable so it wont burn. i have other constraints that requires me to use a 20 gauge cable. But what about the current ? \$\endgroup\$ – Danpe Jan 19 '13 at 16:31
  • 1
    \$\begingroup\$ @Danpe: 30 amps is 30 amps, if your motor needs that much then that's what you need to supply. But seriously, no matter what kind of magic insulation you have, I'd strongly suggest finding some way to use a lower-resistance wire, even if only for part of the distance. At 30 amps both ways, your 33 mΩ/m wires are emitting about 2 watts of heat per meter. That's not a power cable, that's a heating element. \$\endgroup\$ – Ilmari Karonen Jan 19 '13 at 16:38

All answers so far are forgetting the fact that the motor uses 30A (probably) nominal. But with varying mechanical load, the voltage at motor end with these thin wires will vary greatly. Both answers so far are in the same range 60-70V for the power supply, so for the sake of argument let's say both calculations are correct. This means that when mechanical load (and with that current) decreases, the voltage on the motor can rise from its rated 12V up to 70V worst case. This will dramatically impact expected lifetime of the motor.

  • 1
    \$\begingroup\$ I don't think either answer is suggesting a 70V power supply is the solution. The lifetime of the motor under such conditions is somewhat of a moot point as the power cables to it would have burst into flames before it could even get up to speed. \$\endgroup\$ – Phil Frost Jan 19 '13 at 20:51
  • \$\begingroup\$ @PhilFrost You are correct, and I totally agree that it isn't a viable solution from their point of view either. I just wanted to point out that if the high voltage solution would be implemented, an entirely different problem would occur with load variations. \$\endgroup\$ – jippie Jan 19 '13 at 20:54

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