**Plumbers Theory**

If you are really new to this and don't understand electricity at all I always like to compare it to the water in a hose. The water flows though the hose because the water pressure from the main forces it through. The pressure is like the voltage and the flow of water is the current. If we increase the pressure then we increase the flow just as if we increase the voltage we increase the current. The tap is like a variable resistor the more you turn it on the less resistance it has to the flow of water and so the higher the flow or current.

**Ohms Law**

?Between any two points on a conductor the current passing between them is directly proportional to the voltage applied across them when the temperature remains constant?.

**Volts (V), Amps (I) and Ohms (R)**

Ohms law may sound complex but all it means is that for a given resistance R with a voltage V across it a certain current I will flow through it.

So if we know any two of these values it is easy to determine the third.

V = I * R.............I = V / R..............R = V / I

So if the resistor is 2 Ohms and the voltage is 10 Volts the current I = 10 / 2 or 5 Amps.

**Watts**

The power consumed by a circuit in Watts is the voltage V multiplied by the current I.

W = V * I

So in the above example that is 10 * 5 = 50 W

If we double the voltage V = 20 V then we also double the current I = 20/ 2 or 10 A.

So instead of just doubling the power we now have 20 * 10 = 200 W or 4 times the power.

**Motors**

Having shown this simple relation between volts, amps and ohms in robotics the most common thing we are required to power are electric motors. The problem with these is that they do not look like a simple resistance. When a motor is stationary the windings looks like a resistor and that would be the lowest value it can be and so the highest current that can be drawn. This is why the stall current is quoted as only when the motor is stalled is it stationary when power is connected.

When the motor starts to turn it generates a back emf (electro motive force) or reverse voltage which acts to reduce the current drawn by the motor. As more mechanical load is applied to the motor output the more the motor slows down so the less backward voltage is generated and the more current the motor draws. So the motor current is dependent on the voltage applied to it, the resistance of the winding and the speed of / load on the motor. So if you have a motor that draws 0.5 A at 8 V it doesn't matter if you have a 1A or 20A controller the motor will still draw the same basic current. The current will go up as loads are applied so you need to ensure that the controller can supply in the worst case the stall current of the motor although in practice it is very difficult to stall the drive motor on an antweight.

**Regulators**

Because we need to drive the motors at a higher voltage that many receivers can handle we need a way to generate a lower voltage supply to power them. This is normally done using what is known as a linear voltage regulator. This IC takes in the full battery voltage and outputs a fixed voltage supply to power the receivers and normally the servos. Regulators normally also require decoupling capacitors. One between the input and ground and another between the output and ground.

When selecting a regulator you need to look at several factors.

The maximum input voltage

......Normally regulators are rated in excess of 20v well above the normal maximum of 12.6 V.

The minimum input voltage

......Regulators need a minimum voltage across them to work, the drop out voltage, which is normally about about 2v..

......So the battery voltage must exceed the output voltage plus the regulator drop out voltage.

......There are now a range of LDO regulators (low drop out) with about 0.2v drop out

The regulated output voltage

......This must be correct for the receiver normally 5 volts.

The regulator output current

......This must be sufficient to supply the maximum peak currents required which can exceed 1.5A on a strong servo.

The regulator power

......If a regulator is dropping 12.6V down to 5V the regulator has 12.6 - 5 = 7.6V across it.

......If we are drawing 1A that is 7.6 * 1 = 7.6 watts to be dissipated.