Thickening wire to prevent voltage drop

Question

Context:

I want to extend a wire that hooks up to the power momentary switch of a PC ( which has a built in LED). This extension will be about 10 meters long. While this distance might be that long to matter, I want to make sure that there is no voltage drop so I have read that thickening the wire will help.

A cat5 cable has 6 wires. I'd like to use this cable as it looks nicer and is easier to route.

Question:

Does soldering three of the cat5 cables to each terminal on the momentary switch effectively thicken the wire (thus preventing a drop in voltage)? Or am I misled in thinking this would work?

More information on the switch: -The switch is rated up to 3A/250VAC, but the LED is rated for 1.8-2.8V, if you want the switch and LED to work under 12V, you need to add the resistor

Answer 1

The 'power momentary switch of a PC' should take very little current at all, at least with respect to cat5 cores. It is a logic input to the mobo, so a few mA at most, probably < 1mA.

If OTOH there is significant current running through it, then I would not use cat 5, even multiple cores of it which would 'thicken the wire'.

Answer 2

If I've read you right, you're trying to put the POWER push-on/off switch from your PC on the end of about 10 m of cable.

Inside the PC, the switch connects to the motherboard and this goes into the allsorts power circuit that controls the PSU's power on/off input. This circuit is powered from the standby rail coming out of the PSU, so it's always powered while the PC has mains coming in.

There's a good chance that the power switch signal will be 5 V or 3.3 V and may pick up noise or interference on its 20 m trip up your cable and back. The thing is, you can't readily add noise filtering without knowing/experimenting what the effect will be on the PC circuit.

One solution would be something like the circuit shown below. This replaces the power switch with the transistor of an opto-isolator. That leaves you to drive the opto-isolator's IRLED and you can put plenty of noise filtering on that. Here, R1 and R2 provide the IRLED current and the ~1.6 mA drawn by R3. D2 protects the IRLED from reverse voltages, simply because IRLEDs aren't good at it rather than because they're likely to come along but you'll wish you had it there if the unforeseen happens. C1 and R3 stop high-frequency transients briefly powering the IRLED, if they ever had the current to. R3 also discharges C1 so transients have a long hill to climb. R2 stops cabling faults accidentally shorting out the PSU's standby rail. R1 protects the diodes from unexpected power supplies. It's all a lot of protection against a lot of possible hazards by a few cheap components.

The supply is shown to be taken from the PC's standby power supply but you might use other methods. In some ways, a higher supply rail going up your wires and a higher threshold at the detecting end would be nice. But in place of that, a decent load with a charging time for an RC filter is just as good.

^{simulate this circuit – Schematic created using CircuitLab}

There's other ways of handling the noise filtering, such as by introducing a time delay. This just sketches out an option.

With this, using a single RJ45 wire would be fine but its up to you if you want to do something different.

Answer 3

The answer is: it depends... It depends on how much current will be drawn by the circuit at the other end. If it is a logic level (5 volts or less for on state-such as turning on 2 computers with 1 switch), you should be OK. If, however, you are trying to turn on something that draws serious current, you would have to calculate the drop according to Ohm's law E=IR, where E is the applied voltage, I is the current in amperes to be drawn by the remote circuit, and R is the characteristic resistance of the wire in ohms. One other consideration is interaction between the two circuits. If one circuit runs at or near 5 volts and the other uses a different logic level like 2 or 3 volts, you could potentially cause a short circuit between the 2 power levels and possibly damage to the remote circuit. So you need to know EXACTLY what you are connecting. The best idea would always be to isolate the two circuits using a wired-or with 2 diodes, a gate or even somehow replace the switch with one that is double-pole/single throw to completely isolate the two circuits. Otherwise you could risk damaging either your motherboard or the remote circuit or both.

I hope this helps.