Maximum DC power available over specific distances from source

Question

I'm working on a small LVDC project using 60 VDC to distribute power over varying distances to varying loads such as network equipment. I decided to work out the maths and see exactly how much power I can get (maximum) over specific distances.

For this specific case I am assuming AWG14 (\$2.5\rm\, mm^2\$) cable is being used.

The voltage drop over the cable will be:

$$V_{drop} = I_{wire} \cdot R_{wire}.$$

The voltage available at the load will be:

$$V_{load} = V_{supply} - V_{drop},$$

which is equivalent to:

$$V_{load} = V_{supply} - \left( I_{wire} \cdot R_{wire} \right).$$

Resistance of a wire is

$$R_{wire} = {2 \cdot L_{[\rm m]} R_{wire[\rm\Omega/km]} \over 1000\,\rm m}. $$

Re-writing:

$$V_{load} = V_{supply} - I_{wire} {2 \cdot L_{[\rm m]} \cdot R_{wire[\rm\Omega/km]} \over 1000\,\rm m}.$$

Power will be given by voltage at the load times current in the wire. Voltage supplied in this case is 60 V so the resulting expression for power at the load can be written:

$$\begin{aligned} P_{load} &= \left( 60{\,\rm V} - I_{wire} {2 \cdot L_{[\rm m]} R_{wire[\rm\Omega/km]} \over 1000\,\rm m} \right) \cdot I_{wire} \\ &= 60{\,\rm V} \cdot I_{wire} - I_{wire}^2 {2 \cdot L_{[\rm m]} \cdot R_{wire[\rm\Omega/km]} \over 1000\,\rm m}. \\ \end{aligned}$$

I am looking to maximize the power available at the load with respect to the current in the wire - so using partial differentiation:

$$\begin{aligned} {\partial \over \partial I_{wire}} P_{load} &= 60{\,\rm V} - 2\ I_{wire} \cdot {2 \cdot L_{[\rm m]} \cdot R_{wire[\rm\Omega/km]} \over 1000\,\rm m} \\ &= 60{\,\rm V} - 4\ I_{wire} {L_{[\rm m]} \cdot R_{wire[\rm\Omega/km]} \over 1000\,\rm m}. \\ \end{aligned}$$

To maximize, set the derivative to zero and solve - yielding:

$$ I_{wire} = {15000 \over L_{[\rm m]} \cdot R_{wire[\rm\Omega/km]} }. $$

So for example, at 50 m with a wire resistance of 14 AWG as \$ 8.286 \, \Omega/{\rm km} \$:

$$ I_{wire} = {15000 \over 50 \cdot 8.286 } = 36.3 {\,\rm A}. $$

At this current, the maximum power is achieved. The voltage drop with this current can be calculated to be approx. 30 V, so the power delivered would be just over 1 kW at 1086 W.

Obviously 36.26 A is a lot of current especially for 14AWG so I would cap this at a much lower level or use 3 cables to divide the current by 3.

30 V at the end of the cable can then be stepped-up or down depending on device. So taking around 10-15 % losses in efficiency for the DC/DC converter - around 900 W can be transmitted at 60 VDC over 50 m.

I want to expand this to longer distances such as 200 m, 300 m and maybe further. Just want to check my thinking/maths is correct as it may be useful for someone else who may come across the same question.

Answer 1

I'm working on a small LVDC project using 60VDC to distribute power over varying distances

OK, 60 volts is the input voltage.

The voltage drop with this current can be calculated to be approx. 30V

For maximum power transfer, the volt drop of the cable series resistance will always be 50% of the applied voltage hence it will be exactly 30 volts. Maximum power available at the load will be: -

$$\dfrac{V_{SUPPLY}^2}{4\cdot R_{CABLE}}$$

Just want to check my thinking/maths is correct

Well, I think you went the long way round and proved the well-known maximum power transfer theorem so good work.

Answer 2

As your math indicates, for a given resistance of cable maximum power is transferred when volt drop in the cable is equal to half the supply voltage.

However given your goal is to supply network equipment, which presumably will be supplied by switch-mode converters I would caution you that normal switch-mode converters do not mix well with high impedance power supplies.

Generally within it's operating range, a switch-mode converter with a load approximates a constant power load.

If you plot the characteristics of your supply and the characteristics of a constant power load you will normally find they intersect in two places. The higher-voltage of the two intersections represents the desired operating point. The lower voltage represents the voltage below which the load will try to draw more current than is available.

Real world systems don't jump instantly from "power off" to normal operation, every system has capaciance so when power is applied to a discharged system starts the voltage rises gradually. If you had a true constant power load the system would never be able to start.

Fortunately switch-mode converters are not constant power loads over the full range of possible input voltages. In practice they have a minimum voltage below which they do not operate normally. A buck converter cannot produce an output voltage higher than it's input voltage. A boost converter can, but nearly all boost converters will be designed with an under-voltage lockout.

To reliably supply a switch-mode converter you need to be able to not just deliver enough power to supply the load under good operating conditions, you have to be able to deliver enough power as soon as the voltage becomes sufficiently high. Worse since the converter likely has output capacitance to charge you may need to be able to supply more power during startup than during normal operation.

Unfortunately the precise figures for what voltage a converter will start up and how much current it will draw from the supply during the start-up phase are often not readily available, so it may be a case of suck it and see to find out how much resistance in the supply you can get away with before your converter fails to start.