But why???

Go RIGHT → for the tutorial on using the tool.

Go DOWN ↓ to find out why you might use the tool.

If you want to know why I made this go to the last slide.

If you want to get started, go straight to the tool.

Long cables have voltage drop

When you supply power to something (a light, a computer, whatever), you do so by sending electricity along a cable.

Normally, in most low voltage situations you can more or less ignore the voltage drop over the cable. This is because:

• it is usually negligble (unless you have very high current draw); AND
• you have a power source/point per item you are powering (i.e. you are not daisy-chaining).

Power outlets are scarce underground

In tunnels or underground mines, you might want to install cameras or Wi-Fi to cover large areas underground. However, you have to run power to your devices - sometimes over a long distance.

An example underground mine

For example, an underground might might look like this:

Working through the circuit

We can convert it to a circuit diagram

We have assumed here that the long cables have a resistance of 2 ohms each with no complex components. You need to approximate this value yourself, but it is usually the length in metres multiplied by the resistance per metre.

If the cameras behave like resistive loads

Well then we could model them each as a 30 ohm resistance, and then we could analyse this like any simple circuits problem.

We would add series resistances together

We would combine parallel resistances and then perform the basic circuit analysis. Easy!

But cameras are not resistive loads

Digital devices typically need a stable DC voltage to operate. This is why your laptop, phone, security cameras etc all come with a "power pack".

The power pack does the job of cleaning and regulating the power for the device.

This is a deep and complex topic, but there are power circuits that can convert from AC to DC, from DC to DC, from higher to lower voltages and from lower to higher voltages.

Cameras are constant-power loads

You might have a camera that:

• accepts between 12-48 V DC, and
• draws 10 Watts.

The camera is a 'constant-power' load, i.e. it modulates the current it draws based on the voltage it is receiving so that it always consumes a constant power.

Assume the camera's optimal voltage is 24 V and that the regulator in the camera is 100% efficient (P = I * V).

If you supply:

24 V

The voltage is exactly what the camera needs.
So P = 10 and V = 24, then I = 0.42 amps

48 V

The regulator steps the voltage down, "gaining" current in the process.
So p = 10 and V = 48, then I = 0.21 amps

12 V

The regulator steps the voltage up, "drawing" more current to achieve this.
So p = 10 and V = 12, then I = 0.83 amps

Using the camera voltage (Vc) and current (Ic), we can see how this non-linear behaviour affects the voltage we need to supply.

The camera can be modelled in a number of ways. For example, you could consider the camera as a resistive load that changes resistance based on the available voltage.

Using the numbers from before, if you supply:

24 V

The current is 0.42 amps, and 24V drop across the camera.
So the equivalent resistance is R = V / I = 57.1 ohms.

48 V

The current is 0.21 amps, and 48V drop across the camera.
So the equivalent resistance is R = V / I = 228.6 ohms.

12 V

The current is 0.83 amps, and 12V drop across the camera.
So the equivalent resistance is R = V / I = 14.5 ohms.

But wait! That wasn't hard!

You are right, doing it for one device is not hard, and doing it for two is manageable, but doing it for 3+ take a serious amount of time.

This is because the voltages throughout the circuit affect the current / voltage and thus effective resistances throughout the circuit which continue to affect....you see the point.

I don't believe you

That's ok, just try solving this one by hand

(answer on the next slide)

Ok, so here is the same circuit with the solution.

Fortunately it is easy to check the solution once you have found one.

• The current (0.314 A) and voltage (47.82 V) in element 3 gives 15.02 Watts
• The current (1.845 A) through the first cable (2.5 ohms) gives a voltage drop of 4.6125 V
• The input voltage (54) less the voltage drop in cable 1 (4.6125) gives the voltage at element 1 of 49.39.
• The current into the node at element 1 (1.845) less the current out (0.833) give the current into element 1 as 1.012 A
• The current (1.012) and voltage (49.39) in element 1 gives 49.98 Watts
• Etc...you can check them all.

Right, so I guess I use the tool

Yep, the tool solves these problems by starting with initial conditions for the 'effective' resistances and iteratively moves them up and down to reduce the error in the answer.

Hopefully it eventually arrives at some numbers that reduce the error to a small number, and thus represent a "solution" to the circuit (within some tolerance).

However, some circuits are unsolvable!

Unsolvable circuits! Huh?

It can seem counterintuitive that a circuit like this could not have a solution.

It occurs when the current flowing down the cable would cause the voltage to drop too much to be able to supply the power required.

What next

Learn the tool →!