Not sure what HBChris was referring to. AFAIK none of the motors originally had relays.
The two wire motors require reversing the current direction to make the motor run the other direction. The three wire motors simply require energizing one wire relative to ground for one direction, and the other wire relative to ground for the other direction.
Both types can benefit from relays.
The primary rationale behind the use of relays is:
- Reduces circuit losses. Since the motors draw a lot of current, and the original wires must go from the fuses, to the console switches and then out to the window motors, the combination of wire length, wire size, and small switches means there are resistance losses that reduce the current in the motors, slowing them down*.
- Reduces the current controlled by the switches, causing them to last longer and operate more reliably (and less loss/resistance)
- Allows for a larger gauge wire carrying the drive current, reducing resistance (lower wire resistance, shorter wire).
You make a good point about the flyback spike damaging any microelectronics. The spike is ultimately on the 12 volt bus, not just the window switch or relay contacts, so technically it may affect other components in the car. I am thinking ECUs, radios, etc., all of which have some level of microelectronics (although the D-Jet, and L-Jet ECUs are pretty simple). So the contact diode is probably always a good idea. This also argues against using transistors instead of relays... I agree that the relay coil, and the current level in the relay coil are low enough that the coil diode is probably unnecessary. It can't hurt, and if it is already present in the relay, then obviously leave it, but I would not add them if they are not already present.
* For those who may not understand the mechanism: As current flows across a resistance, the voltage across the resistance drops. Here isa simple circuit called a voltage divider. I have drawn this as a battery, a switch, a resistance and a motor. Consider the resistance to represent all of the resistances in the chain from the battery to the motor (switch losses, wire losses, connector losses, fuse losses, etc). Since all of these are in series, we can just bundle them all together, and then assume a perfect battery and a perfect switch.
View attachment 213596
Now, if we assume the motor, if attached to the battery directly draws 10 Amps. By Ohm's Law we can determine that the "resistance" of the motor is 1.2 ohms (I put resistance of the motor in quotes because it is technically not a "resistor"...). If the circuit resistance R is zero then the voltage at the motor (Vmotor) will be 12 volts. But watch what happens to the motor voltage and the motor current as the resistance goes up.
View attachment 213597
So, you can see that even small amounts of resistance in the circuit will cause the voltage at the motor to drop significantly, and this causes the current through the motor to drop, reducing the power and torque of the motor. If we assume that torque is proportional ti current, then 0.5 ohms of resistance in the wires, fuse, and switch will cause the current (and this the motor torque) to drop by 30%. If the resistance is 1 Ohm then the current and torque will drop by nearly 50%...