/diy/ - Do It Yourself

>>1411396
I figured out how to do it with inductors, by putting an inductor in series with each relay, the current through the relays will resist change. Since an inductor is defined as V = L*dI/dt, we can make the assumption that dI/dt is constant (not exponentially decreasing like in reality) to give ourself a good, if not too generous estimate of the system's behaviour. Since initially 7.5A will be going through each relay, we can imagine 6V will be dropped by the resistor from due to the current through each relay at t=0; at the time when the first relay has turned off. The current through the relay will start increasing at a rate this constant dI/dt for a time length ∆t (the duration between the first relay going and the second relay going), meaning the current increases by ∆I = ∆t*dI/dt. If you want to limit this to 10A then ∆I = 2.5A, so 2.5 = ∆t*dI/dt.
Since V = L*dI/dt, 6/L = dI/dt, 2.5 = ∆t*6/L.
Rearranging this we get L = 2.4∆t, with which you can build a circuit (not forgetting the diodes). You'd need to measure ∆t 20 or so times and go with an upper estimate, using a scope, arduino/mcu, or even sound card. If you'd end up with an inductance above 10mH or so (∆t > 4ms) you might be better off trying some other method, since 10mH is pretty tough to get cheaply/have lying about and chances are the ESR will be too high. You'll probably want an inductor ESR below 0.1Ω.

I can't think of any non-active way of doing this with capacitors, they'd just back-feed into the sole active relay and put even more current though it.