Discussion in 'The Projects Forum' started by Russell_AH, Apr 6, 2015.

1. ### Russell_AH Thread Starter New Member

Mar 29, 2013
9
0
Hi all,

I'm building an ATX based power supply (a box with an ATX socket on the back with terminal posts on the front with breadboards on the top). Rather than reconnecting wires from each of the posts to the the four breadboard power rails to change the voltage, I was thinking of using a rotary switch (break before make) to change the voltage as shown in the left side of the attached image. If I use a 2-pole 5 position switch then I could have an LED indicate which voltage is selected.

My main question is what would be a reasonable current rating to supply a breadboard rail? I was thinking somewhere in the 0.5A - 1A range should be plenty. For 0.5A is was thinking of a switch like this: http://www.rapidonline.com/electronic-components/miniature-wafer-switch-2-x-5-50-1101. There seems to be quite a price increase for a 1A rating (although if anyone knows of any for a couple £ each then please let me know).

If I were to be doing something with motors, then I could power them from the posts rather than the rails. Can anyone think of a situation where I'd be likely to draw more than 0.5A from a single rail? There will be four rails for two 830 point boards and a 400 point board.

An alternative would be to use relays as shown in the right side of the image, but this was only if I needed currents higher than 1A, which I am starting to think is a little excessive.

Any thoughts?

File size:
19 KB
Views:
28

Jan 18, 2008
5,699
912
3. ### ScottWang Moderator

Aug 23, 2012
5,481
864
If you want then you just need a relay to using 12V and 5V to drive it, but the 12V needs to in series with a resistor to limiting the current, let's say that the current of relay is 70 mA, then you need a 100Ω resistor, you can calculate the resistor as following:
R = (12V - V_relay ) / I_relay
R = (12V - 5V)/70mA
R = 7V/70mA
R = 100Ω.

How to calculate the current of relay:
Assume that the resistance of relay = 76Ω

I_relay = 5V/R_relay
I_relay = 5V/76Ω
I_relay = 65.8 mA
So count the current as 70 mA.

4. ### Russell_AH Thread Starter New Member

Mar 29, 2013
9
0
Cheers, I think I've found the switch I'll use, it's capable of carrying an amp so should be fine, no need to worry about relays. I want a break-before-make switch (not a make-before-break) to avoid shorting the voltage rails together when I change the voltage.